#set document(title: "8.1 Sum and Difference Formulas", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 8.1#h(0.6em)Sum and Difference Formulas In Chapter 5 we studied identities that relate the three trigonometric functions sine, cosine, and tangent. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Pythagorean and Tangent Identities] #math.equation(block: true, alt: "Pythagorean identity, sin squared ⁡ open parenthesis θ close parenthesis plus cos squared ⁡ open parenthesis θ close parenthesis equals 1; Tangent identity, tan ⁡ open parenthesis θ close parenthesis equals the fraction sin ⁡ open parenthesis θ close parenthesis over cos ⁡ open parenthesis θ close parenthesis")[$& "Pythagorean identity" & & bold(sin)^(2) ( bold(italic(θ)) ) + bold(cos)^(2) ( bold(italic(θ)) ) = 1 \ & "Tangent identity" & & bold(tan) ( bold(italic(θ)) ) = display(frac(bold(sin) ( bold(italic(θ)) ), bold(cos) ( bold(italic(θ)) )))$] ] If we know one of the three trig values for an angle, we can find the other two by using these identities. Identities are useful for changing from one form to another when solving equations, and for finding exact values for trigonometric functions. Are there identities relating the trig ratios of different angles? For example, if we know the sine of #math.equation(block: false, alt: "27 degrees")[$27^(∘)$], can we find the sine of #math.equation(block: false, alt: "2 open parenthesis 27 degrees close parenthesis equals 54 degrees")[$2 ( 27^(∘) ) = 54^(∘)$] without using a calculator? Or, if we know #math.equation(block: false, alt: "cos ⁡ open parenthesis α close parenthesis")[$cos ( α )$] and #math.equation(block: false, alt: "cos ⁡ open parenthesis β close parenthesis")[$cos ( β )$], can we calculate #math.equation(block: false, alt: "cos ⁡ open parenthesis α plus β close parenthesis")[$cos ( α + β )$]? === The Sum of Angles Identities All of the identities that relate the trig ratios of different angles are derived from the sum and difference formulas. Let's see why we need these formulas. Is it true that #math.equation(block: true, alt: "cos ⁡ open parenthesis α plus β close parenthesis and cos ⁡ open parenthesis α close parenthesis plus cos ⁡ open parenthesis β close parenthesis")[$cos ( α + β ) " " " " " " " " "and" " " " " " " " " cos ( α ) + cos ( β )$] are equal for any values of #math.equation(block: false, alt: "α")[$α$] and #math.equation(block: false, alt: "β")[$β$] ? We can test this hypothesis by evaluating both expressions for some specific values of #math.equation(block: false, alt: "α")[$α$] and #math.equation(block: false, alt: "β")[$β$], say #math.equation(block: false, alt: "α equals 45 degrees")[$α = 45^(∘)$] and #math.equation(block: false, alt: "β equals 30 degrees")[$β = 30^(∘)$], as shown below. #figure(figph[3 angles on unit circles], alt: "3 angles on unit circles", caption: none) From the figure, you should be able to see that #math.equation(block: false, alt: "cos ⁡ open parenthesis 75 degrees close parenthesis")[$cos ( 75^(∘) )$] is in fact smaller than either #math.equation(block: false, alt: "cos ⁡ open parenthesis 45 degrees close parenthesis")[$cos ( 45^(∘) )$] or #math.equation(block: false, alt: "cos ⁡ open parenthesis 30 degrees close parenthesis")[$cos ( 30^(∘) )$], so it cannot be true that #math.equation(block: false, alt: "cos ⁡ open parenthesis 75 degrees close parenthesis")[$cos ( 75^(∘) )$] is equal to #math.equation(block: false, alt: "cos ⁡ open parenthesis 45 degrees close parenthesis plus cos ⁡ 3 open parenthesis 0 degrees close parenthesis")[$cos ( 45^(∘) ) + cos 3 ( 0^(∘) )$]. #examplebox("Example 1")[][ Verify that #math.equation(block: false, alt: "cos ⁡ open parenthesis 45 degrees plus 30 degrees close parenthesis")[$cos ( 45^(∘) + 30^(∘) )$] is not equal to #math.equation(block: false, alt: "cos ⁡ open parenthesis 45 degrees close parenthesis plus cos ⁡ open parenthesis 30 degrees close parenthesis")[$cos ( 45^(∘) ) + cos ( 30^(∘) )$]. #solutionbox[ Use your calculator to evaluate each expression. #math.equation(block: true, alt: "cos ⁡ open parenthesis 45 degrees plus 30 degrees close parenthesis, equals cos ⁡ open parenthesis 75 degrees close parenthesis equals 0.2588; but cos ⁡ open parenthesis 45 degrees close parenthesis plus cos ⁡ open parenthesis 30 degrees close parenthesis, equals 0.7071 plus 0.8660 equals 1.5731")[$cos ( 45^(∘) + 30^(∘) ) & = cos ( 75^(∘) ) = 0.2588 \ "but" " " " " " " " " cos ( 45^(∘) ) + cos ( 30^(∘) ) & = 0.7071 + 0.8660 = 1.5731$] The two expressions are not equal. ] ] Show that #math.equation(block: false, alt: "sin ⁡ open parenthesis 60 degrees plus 30 degrees close parenthesis")[$sin ( 60^(∘) + 30^(∘) )$] is not equal to #math.equation(block: false, alt: "sin ⁡ open parenthesis 60 degrees close parenthesis plus sin ⁡ open parenthesis 30 degrees close parenthesis")[$sin ( 60^(∘) ) + sin ( 30^(∘) )$]. #math.equation(block: false, alt: "sin ⁡ open parenthesis 60 degrees plus 30 degrees close parenthesis equals 1")[$sin ( 60^(∘) + 30^(∘) ) = 1$], but #math.equation(block: false, alt: "sin ⁡ open parenthesis 60 degrees close parenthesis plus sin ⁡ open parenthesis 30 degrees close parenthesis equals the fraction the square root of 3 over 2 plus the fraction 1 over 2")[$" " sin ( 60^(∘) ) + sin ( 30^(∘) ) = display(frac(sqrt(3), 2)) + display(frac(1, 2))$] #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Because the values of the expressions in the previous Example and Exercise are different, it is not true that #math.equation(block: false, alt: "cos ⁡ open parenthesis α plus β close parenthesis")[$cos ( α + β )$] is equal to #math.equation(block: false, alt: "cos ⁡ open parenthesis α close parenthesis plus cos ⁡ open parenthesis β close parenthesis")[$cos ( α ) + cos ( β )$] for all angles #math.equation(block: false, alt: "α")[$α$] and #math.equation(block: false, alt: "β")[$β$], or that #math.equation(block: false, alt: "sin ⁡ open parenthesis α plus β close parenthesis")[$sin ( α + β )$] is equal to #math.equation(block: false, alt: "sin ⁡ open parenthesis α close parenthesis plus sin ⁡ open parenthesis β close parenthesis")[$sin ( α ) + sin ( β )$]. It turns out that there is a relationship between the trig ratios for #math.equation(block: false, alt: "α plus β")[$α + β$] and the trig ratios of #math.equation(block: false, alt: "α")[$α$] and #math.equation(block: false, alt: "β")[$β$], but it is a little more complicated. ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Sum of Angles Identities] #math.equation(block: true, alt: "cos ⁡ open parenthesis α plus β close parenthesis equals cos ⁡ open parenthesis α close parenthesis cos ⁡ open parenthesis β close parenthesis minus sin ⁡ open parenthesis α close parenthesis sin ⁡ open parenthesis β close parenthesis")[$bold(cos) ( bold(italic(α)) + bold(italic(β)) ) = bold(cos) ( bold(italic(α)) ) bold(cos) ( bold(italic(β)) ) − bold(sin) ( bold(italic(α)) ) bold(sin) ( bold(italic(β)) )$] #math.equation(block: true, alt: "sin ⁡ open parenthesis α plus β close parenthesis equals sin ⁡ open parenthesis α close parenthesis cos ⁡ open parenthesis β close parenthesis plus cos ⁡ open parenthesis α close parenthesis sin ⁡ open parenthesis β close parenthesis")[$bold(sin) ( bold(italic(α)) + bold(italic(β)) ) = bold(sin) ( bold(italic(α)) ) bold(cos) ( bold(italic(β)) ) + bold(cos) ( bold(italic(α)) ) bold(sin) ( bold(italic(β)) )$] ] Notice that to find the sine or cosine of #math.equation(block: false, alt: "α plus β")[$α + β$] we must know (or be able to find) both trig ratios for both #math.equation(block: false, alt: "α")[$α$] and #math.equation(block: false, alt: "β")[$β$]. The sum and difference formulas can be used to find exact values for trig ratios of various angles. #examplebox("Example 2")[][ Find an exact value for #math.equation(block: false, alt: "cos ⁡ open parenthesis 105 degrees close parenthesis")[$cos ( 105^(∘) )$]. #solutionbox[ We can write #math.equation(block: false, alt: "105 degrees")[$105^(∘)$] as the sum of two special angles: #math.equation(block: false, alt: "105 degrees equals 60 degrees plus 45 degrees")[$105^(∘) = 60^(∘) + 45^(∘)$]. Now apply the sum of angles identity for cosine. #math.equation(block: true, alt: "cos ⁡ open parenthesis 60 degrees plus 45 degrees close parenthesis, equals cos ⁡ open parenthesis 60 degrees close parenthesis cos ⁡ open parenthesis 45 degrees close parenthesis minus sin ⁡ open parenthesis 60 degrees close parenthesis sin ⁡ open parenthesis 45 degrees close parenthesis; equals the fraction 1 over 2 the fraction the square root of 2 over 2 minus the fraction the square root of 3 over 2 the fraction the square root of 2 over 2 equals the fraction the square root of 2 minus the square root of 6 over 4")[$cos ( 60^(∘) + 45^(∘) ) & = cos ( 60^(∘) ) cos ( 45^(∘) ) − sin ( 60^(∘) ) sin ( 45^(∘) ) \ & = display(frac(1, 2)) display(frac(sqrt(2), 2)) − display(frac(sqrt(3), 2)) display(frac(sqrt(2), 2)) = display(frac(sqrt(2) − sqrt(6), 4))$] Thus, #math.equation(block: false, alt: "cos ⁡ open parenthesis 105 degrees close parenthesis equals the fraction the square root of 2 minus the square root of 6 over 4")[$cos ( 105^(∘) ) = display(frac(sqrt(2) − sqrt(6), 4))$]. You can check that your calculator gives the same decimal approximation of about #math.equation(block: false, alt: "minus 0.2588")[$− 0.2588$] for both #math.equation(block: false, alt: "cos ⁡ open parenthesis 105 degrees close parenthesis")[$cos ( 105^(∘) )$] and #math.equation(block: false, alt: "the fraction the square root of 2 minus the square root of 6 over 4")[$display(frac(sqrt(2) − sqrt(6), 4))$]. ] ] Find an exact value for #math.equation(block: false, alt: "sin ⁡ open parenthesis 75 degrees close parenthesis")[$sin ( 75^(∘) )$]. Use the sum of angles identity for sine with #math.equation(block: false, alt: "α equals 45 degrees")[$α = 45^(∘)$] and #math.equation(block: false, alt: "β equals 30 degrees")[$β = 30^(∘)$]. #math.equation(block: true, alt: "sin ⁡ open parenthesis 75 degrees close parenthesis, equals sin ⁡ open parenthesis 45 degrees plus 30 degrees close parenthesis; equals sin ⁡ open parenthesis 45 degrees close parenthesis cos ⁡ open parenthesis 30 degrees close parenthesis plus cos ⁡ open parenthesis 45 degrees close parenthesis sin ⁡ open parenthesis 30 degrees close parenthesis; equals the fraction the square root of 2 over 2 the fraction the square root of 3 over 2 plus the fraction the square root of 2 over 2 open parenthesis the fraction 1 over 2 close parenthesis; equals the fraction the square root of 6 over 4 plus the fraction the square root of 2 over 4 equals the fraction the square root of 6 plus the square root of 2 over 4")[$sin ( 75^(∘) ) & = sin ( 45^(∘) + 30^(∘) ) \ & = sin ( 45^(∘) ) cos ( 30^(∘) ) + cos ( 45^(∘) ) sin ( 30^(∘) ) \ & = display(frac(sqrt(2), 2)) display(frac(sqrt(3), 2)) + display(frac(sqrt(2), 2)) ( display(frac(1, 2)) ) \ & = display(frac(sqrt(6), 4)) + display(frac(sqrt(2), 4)) = display(frac(sqrt(6) + sqrt(2), 4))$] Of course, the sum formulas hold for angles in radians as well as degrees. #examplebox("Example 3")[][ Suppose that #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals 0.6")[$sin ( θ ) = 0.6$] and #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals minus 0.8")[$cos ( θ ) = − 0.8$]. Find an exact value for #math.equation(block: false, alt: "sin ⁡ open parenthesis θ plus the fraction 2 π over 3 close parenthesis")[$sin ( θ + display(frac(2 π, 3)) )$]. #solutionbox[ Recall that #math.equation(block: false, alt: "sin ⁡ open parenthesis the fraction 2 π over 3 close parenthesis equals the fraction the square root of 3 over 2")[$sin ( display(frac(2 π, 3)) ) = display(frac(sqrt(3), 2))$] and #math.equation(block: false, alt: "cos ⁡ open parenthesis the fraction 2 π over 3 close parenthesis equals the fraction minus 1 over 2")[$cos ( display(frac(2 π, 3)) ) = display(frac(− 1, 2))$]. Substituting these values into the sum formula for sine, we find #math.equation(block: true, alt: "sin ⁡ open parenthesis θ plus the fraction 2 π over 3 close parenthesis, equals sin ⁡ open parenthesis θ close parenthesis cos ⁡ open parenthesis the fraction 2 π over 3 close parenthesis plus cos ⁡ open parenthesis θ close parenthesis sin ⁡ open parenthesis the fraction 2 π over 3 close parenthesis; equals 0.6 open parenthesis the fraction minus 1 over 2 close parenthesis plus open parenthesis minus 0.8 close parenthesis open parenthesis the fraction the square root of 3 over 2 close parenthesis; equals the fraction 3 over 5 times the fraction minus 1 over 2 plus the fraction minus 4 over 5 times the fraction the square root of 3 over 2 equals the fraction minus 3 minus 4 the square root of 3 over 10")[$sin ( θ + display(frac(2 π, 3)) ) & = sin ( θ ) cos ( display(frac(2 π, 3)) ) + cos ( θ ) sin ( display(frac(2 π, 3)) ) \ & = 0.6 ( display(frac(− 1, 2)) ) + ( − 0.8 ) ( display(frac(sqrt(3), 2)) ) \ & = display(frac(3, 5)) ⋅ display(frac(− 1, 2)) + display(frac(− 4, 5)) ⋅ display(frac(sqrt(3), 2)) = display(frac(− 3 − 4 sqrt(3), 10))$] ] ] Suppose that #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction minus 5 over 13")[$sin ( θ ) = display(frac(− 5, 13))$] and #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction 12 over 13")[$cos ( θ ) = display(frac(12, 13))$]. Find an exact value for #math.equation(block: false, alt: "cos ⁡ open parenthesis the fraction π over 4 plus θ close parenthesis")[$cos ( display(frac(π, 4)) + θ )$]. #math.equation(block: true, alt: "cos ⁡ open parenthesis the fraction π over 4 plus θ close parenthesis, equals cos ⁡ open parenthesis the fraction π over 4 close parenthesis cos ⁡ open parenthesis θ close parenthesis minus sin ⁡ open parenthesis the fraction π over 4 close parenthesis sin ⁡ open parenthesis θ close parenthesis; equals the fraction the square root of 2 over 2 open parenthesis the fraction 12 over 13 close parenthesis minus the fraction the square root of 2 over 2 open parenthesis the fraction minus 5 over 13 close parenthesis; equals the fraction 17 the square root of 2 over 26")[$cos ( display(frac(π, 4)) + θ ) & = cos ( display(frac(π, 4)) ) cos ( θ ) − sin ( display(frac(π, 4)) ) sin ( θ ) \ & = display(frac(sqrt(2), 2)) ( display(frac(12, 13)) ) − display(frac(sqrt(2), 2)) ( display(frac(− 5, 13)) ) \ & = display(frac(17 sqrt(2), 26))$] There are also identities for the negative of an angle, which you will discover in the next Activity. #notebox("Activity", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Negative Angles] I Using Reference Triangles + + Suppose that #math.equation(block: false, alt: "θ")[$θ$] is a first-quadrant angle. In which quadrant would you find #math.equation(block: false, alt: "minus θ")[$− θ$]? + Sketch an example for #math.equation(block: false, alt: "θ")[$θ$], #math.equation(block: false, alt: "minus θ")[$− θ$], and the reference triangle for each. + How is #math.equation(block: false, alt: "sin ⁡ open parenthesis minus θ close parenthesis")[$sin ( − θ )$] related to #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$]? What about #math.equation(block: false, alt: "cos ⁡ open parenthesis minus θ close parenthesis")[$cos ( − θ )$] and #math.equation(block: false, alt: "tan ⁡ open parenthesis minus θ close parenthesis")[$tan ( − θ )$]? + Repeat part (1) for the case where is a second-quadrant angle. II Using Graphs + + Look at the graph of #math.equation(block: false, alt: "f open parenthesis θ close parenthesis equals sin ⁡ open parenthesis θ close parenthesis")[$f ( θ ) = sin ( θ )$], and choose three positive values of #math.equation(block: false, alt: "θ")[$θ$]. Draw dots on the graph that show their function values.#figure(figph[sine graph], alt: "sine graph", caption: none) + Find the negatives of your three values of #math.equation(block: false, alt: "θ")[$θ$] from part (a), and draw dots to show their function values. + How do the values of #math.equation(block: false, alt: "sin ⁡ open parenthesis minus θ close parenthesis")[$sin ( − θ )$] compare to the values of #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$]? + Repeat part (1) for the graph of #math.equation(block: false, alt: "f open parenthesis θ close parenthesis equals cos ⁡ open parenthesis θ close parenthesis")[$f ( θ ) = cos ( θ )$].#figure(figph[cosine graph], alt: "cosine graph", caption: none) + Repeat part (1) for the graph of #math.equation(block: false, alt: "f open parenthesis θ close parenthesis equals tan ⁡ open parenthesis θ close parenthesis")[$f ( θ ) = tan ( θ )$].#figure(figph[tangent graph], alt: "tangent graph", caption: none) III Negative Angle Identities State your conclusions from parts I and II as identities: #math.equation(block: true, alt: "sin ⁡ open parenthesis minus θ close parenthesis equals")[$sin ( − θ ) = #h(2em) #h(2em)$] #math.equation(block: true, alt: "cos ⁡ open parenthesis minus θ close parenthesis equals")[$cos ( − θ ) = #h(2em) #h(2em)$] #math.equation(block: true, alt: "tan ⁡ open parenthesis minus θ close parenthesis equals")[$tan ( − θ ) = #h(2em) #h(2em)$] ] === The Difference of Angles Identities The difference formulas for sine and cosine can be derived easily from the sum formulas, using the identities for negative angles. Note that the difference formulas are identical to the corresponding sum formulas, except for the signs. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Difference of Angles Identities] #math.equation(block: true, alt: "cos ⁡ open parenthesis α minus β close parenthesis equals cos ⁡ open parenthesis α close parenthesis cos ⁡ open parenthesis β close parenthesis plus sin ⁡ open parenthesis α close parenthesis sin ⁡ open parenthesis β close parenthesis")[$bold(cos) ( bold(italic(α)) − bold(italic(β)) ) = bold(cos) ( bold(italic(α)) ) bold(cos) ( bold(italic(β)) ) + bold(sin) ( bold(italic(α)) ) bold(sin) ( bold(italic(β)) )$] #math.equation(block: true, alt: "sin ⁡ open parenthesis α minus β close parenthesis equals sin ⁡ open parenthesis α close parenthesis cos ⁡ open parenthesis β close parenthesis minus cos ⁡ open parenthesis α close parenthesis sin ⁡ open parenthesis β close parenthesis")[$bold(sin) ( bold(italic(α)) − bold(italic(β)) ) = bold(sin) ( bold(italic(α)) ) bold(cos) ( bold(italic(β)) ) − bold(cos) ( bold(italic(α)) ) bold(sin) ( bold(italic(β)) )$] ] #examplebox("Example 4")[][ Use the fact that #math.equation(block: false, alt: "the fraction π over 12 equals the fraction π over 4 minus the fraction π over 6")[$display(frac(π, 12)) = display(frac(π, 4)) − display(frac(π, 6))$] to evaluate #math.equation(block: false, alt: "cos ⁡ open parenthesis the fraction π over 12 close parenthesis")[$cos ( display(frac(π, 12)) )$] exactly. #solutionbox[ Remember that #math.equation(block: false, alt: "cos ⁡ open parenthesis the fraction π over 4 close parenthesis equals the fraction the square root of 2 over 2 , cos ⁡ open parenthesis the fraction π over 6 close parenthesis equals the fraction the square root of 3 over 2 , sin ⁡ open parenthesis the fraction π over 4 close parenthesis equals the fraction the square root of 2 over 2")[$cos ( display(frac(π, 4)) ) = display(frac(sqrt(2), 2)) , " " " " cos ( display(frac(π, 6)) ) = display(frac(sqrt(3), 2)) , " " " " sin ( display(frac(π, 4)) ) = display(frac(sqrt(2), 2))$], and #math.equation(block: false, alt: "sin ⁡ open parenthesis the fraction π over 6 close parenthesis equals the fraction 1 over 2")[$sin ( display(frac(π, 6)) ) = display(frac(1, 2))$]. Substituting all these values into the difference identity for cosine, we obtain the following. #math.equation(block: true, alt: "cos ⁡ open parenthesis the fraction π over 12 close parenthesis, equals cos ⁡ open parenthesis the fraction π over 4 minus the fraction π over 6 close parenthesis; equals cos ⁡ open parenthesis the fraction π over 4 close parenthesis cos ⁡ open parenthesis the fraction π over 6 close parenthesis plus sin ⁡ open parenthesis the fraction π over 4 close parenthesis sin ⁡ open parenthesis the fraction π over 6 close parenthesis; equals the fraction the square root of 2 over 2 times the fraction the square root of 3 over 2 plus the fraction the square root of 2 over 2 times the fraction 1 over 2 equals the fraction the square root of 6 plus the square root of 2 over 4")[$cos ( display(frac(π, 12)) ) & = cos ( display(frac(π, 4)) − display(frac(π, 6)) ) \ & = cos ( display(frac(π, 4)) ) cos ( display(frac(π, 6)) ) + sin ( display(frac(π, 4)) ) sin ( display(frac(π, 6)) ) \ & = display(frac(sqrt(2), 2)) ⋅ display(frac(sqrt(3), 2)) + display(frac(sqrt(2), 2)) ⋅ display(frac(1, 2)) = display(frac(sqrt(6) + sqrt(2), 4))$] You can check that your calculator gives the same decimal approximation of about #math.equation(block: false, alt: "0.9659")[$0.9659$] for both #math.equation(block: false, alt: "cos ⁡ open parenthesis the fraction π over 12 close parenthesis")[$cos ( display(frac(π, 12)) )$] and #math.equation(block: false, alt: "the fraction the square root of 6 plus the square root of 2 over 4")[$display(frac(sqrt(6) + sqrt(2), 4))$]. ] ] Evaluate #math.equation(block: false, alt: "sin ⁡ open parenthesis the fraction π over 12 close parenthesis")[$sin ( display(frac(π, 12)) )$] exactly. As in the previous Example, use the fact that #math.equation(block: false, alt: "the fraction π over 12 equals the fraction π over 4 minus the fraction π over 6")[$display(frac(π, 12)) = display(frac(π, 4)) − display(frac(π, 6))$] and apply the difference identity for sine to find #math.equation(block: false, alt: "the fraction the square root of 6 minus the square root of 2 over 4")[$display(frac(sqrt(6) − sqrt(2), 4))$] === Sum and Difference Identities for Tangent There are also sum and difference formulas for the tangent. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Sum and Difference Identities for Tangent] #math.equation(block: true, alt: "tan ⁡ open parenthesis α plus β close parenthesis equals the fraction tan ⁡ open parenthesis α close parenthesis plus tan ⁡ open parenthesis β close parenthesis over 1 minus tan ⁡ open parenthesis α close parenthesis tan ⁡ open parenthesis β close parenthesis")[$bold(tan) ( bold(italic(α)) + bold(italic(β)) ) = display(frac(bold(tan) ( bold(italic(α)) ) + bold(tan) ( bold(italic(β)) ), 1 − bold(tan) ( bold(italic(α)) ) bold(tan) ( bold(italic(β)) )))$] #math.equation(block: true, alt: "tan ⁡ open parenthesis α minus β close parenthesis equals the fraction tan ⁡ open parenthesis α close parenthesis minus tan ⁡ open parenthesis β close parenthesis over 1 plus tan ⁡ open parenthesis α close parenthesis tan ⁡ open parenthesis β close parenthesis")[$bold(tan) ( bold(italic(α)) − bold(italic(β)) ) = display(frac(bold(tan) ( bold(italic(α)) ) − bold(tan) ( bold(italic(β)) ), 1 + bold(tan) ( bold(italic(α)) ) bold(tan) ( bold(italic(β)) )))$] ] #examplebox("Example 5")[][ Find an exact value for #math.equation(block: false, alt: "tan ⁡ open parenthesis 75 degrees close parenthesis")[$tan ( 75^(∘) )$]. #solutionbox[ We observe that #math.equation(block: false, alt: "75 degrees equals 45 degrees plus 30 degrees")[$75^(∘) = 45^(∘) + 30^(∘)$], so #math.equation(block: false, alt: "tan ⁡ open parenthesis 75 degrees close parenthesis equals tan ⁡ open parenthesis 45 degrees plus 30 degrees close parenthesis")[$tan ( 75^(∘) ) = tan ( 45^(∘) + 30^(∘) )$]. We can apply the sum formula for tangent. #math.equation(block: true, alt: "tan ⁡ open parenthesis 45 degrees plus 30 degrees close parenthesis, equals the fraction tan ⁡ open parenthesis 45 degrees close parenthesis plus tan ⁡ open parenthesis 30 degrees close parenthesis over 1 minus tan ⁡ open parenthesis 45 degrees close parenthesis tan ⁡ open parenthesis 30 degrees close parenthesis; equals the fraction 1 plus the fraction 1 over the square root of 3 over 1 minus 1 open parenthesis the fraction 1 over the square root of 3 close parenthesis times the fraction the square root of 3 over the square root of 3 equals the fraction the square root of 3 plus 1 over the square root of 3 minus 1")[$tan ( 45^(∘) + 30^(∘) ) & = display(frac(tan ( 45^(∘) ) + tan ( 30^(∘) ), 1 − tan ( 45^(∘) ) tan ( 30^(∘) ))) \ & = display(frac(1 + display(frac(1, sqrt(3))), 1 − 1 ( display(frac(1, sqrt(3))) ))) ⋅ display(frac(sqrt(3), sqrt(3))) = display(frac(sqrt(3) + 1, sqrt(3) − 1))$] ] ] Evaluate #math.equation(block: false, alt: "tan ⁡ open parenthesis the fraction π over 12 close parenthesis")[$tan ( display(frac(π, 12)) )$] exactly. Use the fact that #math.equation(block: false, alt: "the fraction π over 12 equals the fraction π over 4 minus the fraction π over 6")[$display(frac(π, 12)) = display(frac(π, 4)) − display(frac(π, 6))$] and apply the difference identity for tangent to find #math.equation(block: false, alt: "the fraction the square root of 3 minus 1 over the square root of 3 plus 1")[$display(frac(sqrt(3) − 1, sqrt(3) + 1))$]. === Double Angle Identities There are a number of other very useful identities that can be derived from the sum and difference formulas. In particular, if we set #math.equation(block: false, alt: "α equals β equals θ")[$α = β = θ$] in the sum of angles identities (also called #strong[addition formulas]), we obtain the #strong[double angle formulas]. These identities are used frequently, so it is helpful to know them well. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Double Angle Identities] #math.equation(block: true, alt: "cos ⁡ open parenthesis 2 θ close parenthesis, equals cos squared ⁡ open parenthesis θ close parenthesis minus sin squared ⁡ open parenthesis θ close parenthesis; sin ⁡ open parenthesis 2 θ close parenthesis, equals 2 sin ⁡ open parenthesis θ close parenthesis cos ⁡ open parenthesis θ close parenthesis; tan ⁡ open parenthesis 2 θ close parenthesis, equals the fraction 2 tan ⁡ open parenthesis θ close parenthesis over 1 minus tan squared ⁡ open parenthesis θ close parenthesis")[$bold(cos) ( 2 bold(italic(θ)) ) & = bold(cos)^(2) ( bold(italic(θ)) ) − bold(sin)^(2) ( bold(italic(θ)) ) \ bold(sin) ( 2 bold(italic(θ)) ) & = 2 bold(sin) ( bold(italic(θ)) ) bold(cos) ( bold(italic(θ)) ) \ bold(tan) ( 2 bold(italic(θ)) ) & = display(frac(2 bold(tan) ( bold(italic(θ)) ), 1 − bold(tan)^(2) ( bold(italic(θ)) )))$] ] You can also justify the identities to yourself by graphing both sides of the formula to see that the graphs are identical. #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ The first thing we can learn from the double angle formulas is that #math.equation(block: false, alt: "sin ⁡ open parenthesis 2 θ close parenthesis")[$sin ( 2 θ )$] is not equal to #math.equation(block: false, alt: "2 sin ⁡ open parenthesis θ close parenthesis")[$2 sin ( θ )$]! You can check this very easily by choosing a value for #math.equation(block: false, alt: "θ")[$θ$], say #math.equation(block: false, alt: "45 degrees")[$45^(∘)$]. Then #math.equation(block: true, alt: "sin ⁡ open parenthesis 2 θ close parenthesis, equals sin ⁡ open parenthesis 90 degrees close parenthesis equals 1; but 2 sin ⁡ open parenthesis θ close parenthesis, equals 2 sin ⁡ open parenthesis 45 degrees close parenthesis equals 2 open parenthesis the fraction the square root of 2 over 2 close parenthesis equals the square root of 2")[$sin ( 2 θ ) & = sin ( 90^(∘) ) = 1 \ "but" " " " " " " 2 sin ( θ ) & = 2 sin ( 45^(∘) ) = 2 ( display(frac(sqrt(2), 2)) ) = sqrt(2)$] and #math.equation(block: false, alt: "the square root of 2 not equal to 1")[$sqrt(2) ≠ 1$]. Similarly, #math.equation(block: false, alt: "cos ⁡ open parenthesis 2 θ close parenthesis not equal to 2 cos ⁡ open parenthesis θ close parenthesis")[$cos ( 2 θ ) ≠ 2 cos ( θ )$] and #math.equation(block: false, alt: "tan ⁡ open parenthesis 2 θ close parenthesis not equal to 2 tan ⁡ open parenthesis θ close parenthesis")[$tan ( 2 θ ) ≠ 2 tan ( θ )$] ] #examplebox("Example 6")[][ Find #math.equation(block: false, alt: "sin ⁡ open parenthesis 2 θ close parenthesis")[$sin ( 2 θ )$] for the angle #math.equation(block: false, alt: "θ")[$θ$] shown. #figure(figph[triangle], alt: "triangle", caption: none) #solutionbox[ We start by using the Pythagorean Theorem to find the hypotenuse of the triangle. #math.equation(block: true, alt: "c squared equals 2 squared plus 3 squared equals 13")[$c^(2) = 2^(2) + 3^(2) = 13$] so #math.equation(block: false, alt: "c equals the square root of 13")[$c = sqrt(13)$]. Thus, #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction 3 over the square root of 13")[$cos ( θ ) = display(frac(3, sqrt(13)))$] and #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction 2 over the square root of 13")[$sin ( θ ) = display(frac(2, sqrt(13)))$]. Now we can use these values in the double angle identity to find #math.equation(block: false, alt: "sin ⁡ open parenthesis 2 θ close parenthesis")[$sin ( 2 θ )$]. #math.equation(block: true, alt: "sin ⁡ open parenthesis 2 θ close parenthesis, equals 2 sin ⁡ open parenthesis θ close parenthesis cos ⁡ open parenthesis θ close parenthesis; equals 2 open parenthesis the fraction 2 over the square root of 13 close parenthesis open parenthesis the fraction 3 over the square root of 13 close parenthesis equals the fraction 12 over 13")[$sin ( 2 θ ) & = 2 sin ( θ ) cos ( θ ) \ & = 2 ( display(frac(2, sqrt(13))) ) ( display(frac(3, sqrt(13))) ) = display(frac(12, 13))$] ] ] Find #math.equation(block: false, alt: "cos ⁡ open parenthesis 2 θ close parenthesis")[$cos ( 2 θ )$] and #math.equation(block: false, alt: "tan ⁡ open parenthesis 2 θ close parenthesis")[$tan ( 2 θ )$] for the angle #math.equation(block: false, alt: "θ")[$θ$] shown in the previous Example. We can use the double angle formulas for cosine and tangent to find #math.equation(block: false, alt: "cos ⁡ open parenthesis 2 θ close parenthesis equals the fraction 5 over 13")[$cos ( 2 θ ) = display(frac(5, 13))$] and #math.equation(block: false, alt: "tan ⁡ open parenthesis 2 θ close parenthesis equals the fraction 12 over 5")[$tan ( 2 θ ) = display(frac(12, 5))$]. (But also note that, once we know #math.equation(block: false, alt: "sin ⁡ open parenthesis 2 θ close parenthesis")[$sin ( 2 θ )$], we can calculate its cosine and tangent directly.) We will often work with algebraic expressions instead of numerical values for the trig ratios. #examplebox("Example 7")[][ Use the figure to express #math.equation(block: false, alt: "cos ⁡ open parenthesis 2 ϕ close parenthesis")[$cos ( 2 ϕ )$] in terms of #math.equation(block: false, alt: "a")[$a$]. #figure(figph[triangle], alt: "triangle", caption: none) #solutionbox[ We use the Pythagorean Theorem to find an expression for the third side of the triangle. #math.equation(block: true, alt: "a squared plus b squared, equals 3 squared, Solve for b .; b, equals the square root of 9 minus a squared")[$a^(2) + b^(2) & = 3^(2) & & "Solve for" " " " " bold(italic(b)) . \ b & = sqrt(9 − a^(2))$] Now we can write expressions for the sine and cosine of #math.equation(block: false, alt: "ϕ")[$ϕ$]. #math.equation(block: true, alt: "cos ⁡ open parenthesis ϕ close parenthesis equals the fraction a over 3 and sin ⁡ open parenthesis ϕ close parenthesis equals the fraction the square root of 9 minus a squared over 3")[$cos ( ϕ ) = display(frac(a, 3)) " " " " " " " " "and" " " " " " " " " sin ( ϕ ) = display(frac(sqrt(9 − a^(2)), 3))$] Finally, we substitute these expressions into the double angle identity. #math.equation(block: true, alt: "cos ⁡ open parenthesis 2 ϕ close parenthesis, equals cos squared ⁡ open parenthesis ϕ close parenthesis minus sin squared ⁡ open parenthesis ϕ close parenthesis; equals open parenthesis the fraction a over 3 close parenthesis squared minus open parenthesis the fraction the square root of 9 minus a squared over 3 close parenthesis squared; equals the fraction a squared over 9 minus the fraction 9 minus a squared over 9 equals the fraction a squared minus 9 over 9")[$cos ( 2 ϕ ) & = cos^(2) ( ϕ ) − sin^(2) ( ϕ ) \ & = attach(( display(frac(a, 3)) ), t: 2) − attach(( display(frac(sqrt(9 − a^(2)), 3)) ), t: 2) \ & = display(frac(a^(2), 9)) − display(frac(9 − a^(2), 9)) = display(frac(a^(2) − 9, 9))$] ] ] For the triangle in the previous example, find expressions for #math.equation(block: false, alt: "sin ⁡ open parenthesis 2 ϕ close parenthesis")[$sin ( 2 ϕ )$] and #math.equation(block: false, alt: "tan ⁡ open parenthesis 2 ϕ close parenthesis")[$tan ( 2 ϕ )$]. #math.equation(block: true, alt: "sin ⁡ open parenthesis 2 ϕ close parenthesis equals the fraction 2 a the square root of 9 minus a squared over 9 , tan ⁡ open parenthesis 2 ϕ close parenthesis equals the fraction 2 a the square root of 9 minus a squared over 2 a squared minus 9")[$sin ( 2 ϕ ) = display(frac(2 a sqrt(9 − a^(2)), 9)) , " " " " tan ( 2 ϕ ) = display(frac(2 a sqrt(9 − a^(2)), 2 a^(2) − 9))$] By using the Pythagorean identity, we can write the double angle formula for cosine in two alternate forms. {"expression":"cos(2\*t)"}2\*cos(t)^2 - 1The derivation above turns cos 2θ = cos²θ − sin²θ into the alternate form 2cos²θ − 1 by one application of the Pythagorean identity. The CAS re-certifies the alternate form directly — cos 2θ against 2cos²θ − 1, no intermediate steps — at build time, so the identity chain this chapter builds on the sum formulas is re-verified whenever the book is rebuilt.#math.equation(block: true, alt: "cos ⁡ open parenthesis 2 θ close parenthesis, equals cos squared ⁡ open parenthesis θ close parenthesis minus sin squared ⁡ open parenthesis θ close parenthesis; equals cos squared ⁡ open parenthesis θ close parenthesis minus open parenthesis 1 minus cos squared ⁡ open parenthesis θ close parenthesis close parenthesis; equals 2 cos squared ⁡ open parenthesis θ close parenthesis minus 1")[$cos ( 2 θ ) & = cos^(2) ( θ ) − sin^(2) ( θ ) \ & = cos^(2) ( θ ) − ( 1 − cos^(2) ( θ ) ) \ & = 2 cos^(2) ( θ ) − 1$] #math.equation(block: true, alt: "cos ⁡ open parenthesis 2 θ close parenthesis, equals cos squared ⁡ open parenthesis θ close parenthesis minus sin squared ⁡ open parenthesis θ close parenthesis; equals open parenthesis 1 minus sin squared ⁡ open parenthesis θ close parenthesis close parenthesis minus sin squared ⁡ open parenthesis θ close parenthesis; equals 1 minus 2 sin squared ⁡ open parenthesis θ close parenthesis")[$cos ( 2 θ ) & = cos^(2) ( θ ) − sin^(2) ( θ ) \ & = ( 1 − sin^(2) ( θ ) ) − sin^(2) ( θ ) \ & = 1 − 2 sin^(2) ( θ )$] Thus, we have three forms for the double angle formula for cosine, and we can use whichever form is most convenient for a particular problem. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Double Angle Identities for Cosine] #math.equation(block: true, alt: "cos ⁡ open parenthesis 2 θ close parenthesis, equals cos squared ⁡ open parenthesis θ close parenthesis minus sin squared ⁡ open parenthesis θ close parenthesis; equals 2 cos squared ⁡ open parenthesis θ close parenthesis minus 1; equals 1 minus 2 sin squared ⁡ open parenthesis θ close parenthesis")[$bold(cos) ( 2 bold(italic(θ)) ) & = bold(cos)^(2) ( bold(italic(θ)) ) − bold(sin)^(2) ( bold(italic(θ)) ) \ & = 2 bold(cos)^(2) ( bold(italic(θ)) ) − 1 \ & = 1 − 2 bold(sin)^(2) ( bold(italic(θ)) )$] ] #examplebox("Example 8")[][ Find an expression for #math.equation(block: false, alt: "cos ⁡ open parenthesis 2 β close parenthesis")[$cos ( 2 β )$] if you know that #math.equation(block: false, alt: "cos ⁡ open parenthesis β close parenthesis equals the fraction x over 4")[$cos ( β ) = display(frac(x, 4))$]. #solutionbox[ We'll use the identity #math.equation(block: false, alt: "cos ⁡ open parenthesis 2 θ close parenthesis equals 2 cos squared ⁡ open parenthesis θ close parenthesis minus 1")[$cos ( 2 θ ) = 2 cos^(2) ( θ ) − 1$] #math.equation(block: true, alt: "cos ⁡ open parenthesis 2 β close parenthesis, equals 2 cos squared ⁡ open parenthesis β close parenthesis minus 1; equals 2 open parenthesis the fraction x over 4 close parenthesis squared minus 1 equals the fraction 2 x squared over 16 minus 1; equals the fraction 2 x squared minus 16 over 16 equals the fraction x squared minus 8 over 8")[$cos ( 2 β ) & = 2 cos^(2) ( β ) − 1 \ & = 2 attach(( display(frac(x, 4)) ), t: 2) − 1 = display(frac(2 x^(2), 16)) − 1 \ & = display(frac(2 x^(2) − 16, 16)) = display(frac(x^(2) − 8, 8))$] ] ] Find an expression for #math.equation(block: false, alt: "cos ⁡ open parenthesis 2 α close parenthesis")[$cos ( 2 α )$] if you know that #math.equation(block: false, alt: "sin ⁡ open parenthesis α close parenthesis equals the fraction 6 over w")[$sin ( α ) = display(frac(6, w))$]. #math.equation(block: true, alt: "the fraction w squared minus 72 over w squared")[$display(frac(w^(2) − 72, w^(2)))$] === Solving Equations If a trigonometric equation involves more than one angle, we use identities to rewrite the equation in terms of a single angle. #examplebox("Example 9")[][ Solve #math.equation(block: false, alt: "sin ⁡ open parenthesis 2 x close parenthesis minus cos ⁡ open parenthesis x close parenthesis")[$" " " " sin ( 2 x ) − cos ( x ) " " " "$] for #math.equation(block: false, alt: "0 less than or equal to x less than or equal to 2 π")[$0 ≤ x ≤ 2 π$]. #solutionbox[ We first use the double angle formula to write #math.equation(block: false, alt: "sin ⁡ open parenthesis 2 x close parenthesis")[$sin ( 2 x )$] in terms of trig functions of #math.equation(block: false, alt: "x")[$x$] alone. #math.equation(block: true, alt: "sin ⁡ open parenthesis 2 x close parenthesis minus cos ⁡ open parenthesis x close parenthesis, equals 0; 2 sin ⁡ open parenthesis x close parenthesis cos ⁡ open parenthesis x close parenthesis minus cos ⁡ open parenthesis x close parenthesis, equals 0")[$sin ( 2 x ) − cos ( x ) & = 0 \ 2 sin ( x ) cos ( x ) − cos ( x ) & = 0$] Once we have all the trig functions in terms of a single angle, we try to write the equation in terms of a single trig function. In this case, we can factor the left side to separate the trig functions. #math.equation(block: true, alt: "2 sin ⁡ open parenthesis x close parenthesis cos ⁡ open parenthesis x close parenthesis minus cos ⁡ open parenthesis x close parenthesis, equals 0; cos ⁡ open parenthesis x close parenthesis open parenthesis 2 sin ⁡ open parenthesis x close parenthesis minus 1 close parenthesis, equals 0, Set each factor equal to zero.; cos ⁡ open parenthesis x close parenthesis equals 02 sin ⁡ open parenthesis x close parenthesis minus 1, equals 0, Solve each equation.; sin ⁡ open parenthesis x close parenthesis, equals the fraction 1 over 2; x equals the fraction π over 2 , the fraction 3 π over 2 x equals the fraction π over 6, , the fraction 5 π over 6")[$2 sin ( x ) cos ( x ) − cos ( x ) & = 0 \ cos ( x ) ( 2 sin ( x ) − 1 ) & = 0 & & "Set each factor equal to zero." \ cos ( x ) = 0 #h(2em) 2 sin ( x ) − 1 & = 0 & & "Solve each equation." \ sin ( x ) & = display(frac(1, 2)) \ x = display(frac(π, 2)) , " " display(frac(3 π, 2)) #h(2em) x = display(frac(π, 6)) & , " " display(frac(5 π, 6))$] There are four solutions, #math.equation(block: false, alt: "x equals the fraction π over 2 , the fraction 3 π over 2 , the fraction π over 6 ,")[$x = display(frac(π, 2)) , " " display(frac(3 π, 2)) , " " display(frac(π, 6)) ,$] and #math.equation(block: false, alt: "the fraction 5 π over 6")[$display(frac(5 π, 6))$]. ] ] Solve #math.equation(block: false, alt: "cos ⁡ open parenthesis 2 t close parenthesis equals cos ⁡ t")[$" " " " cos ( 2 t ) = cos t " " " "$] for #math.equation(block: false, alt: "0 less than or equal to x less than or equal to 2 π")[$0 ≤ x ≤ 2 π$]. First rewrite #math.equation(block: false, alt: "cos ⁡ open parenthesis 2 t close parenthesis")[$cos ( 2 t )$] as #math.equation(block: false, alt: "2 cos squared ⁡ open parenthesis t close parenthesis minus 1")[$2 cos^(2) ( t ) − 1$] to obtain the equation #math.equation(block: true, alt: "2 cos squared ⁡ open parenthesis t close parenthesis minus cos ⁡ open parenthesis t close parenthesis minus 1 equals 0")[$2 cos^(2) ( t ) − cos ( t ) − 1 = 0$] Solving this equation by factoring gives us #math.equation(block: false, alt: "cos ⁡ open parenthesis t close parenthesis equals 1")[$cos ( t ) = 1$] or #math.equation(block: false, alt: "cos ⁡ open parenthesis t close parenthesis equals the fraction minus 1 over 2")[$cos ( t ) = display(frac(− 1, 2))$], form which we find #math.equation(block: false, alt: "t equals 0 , t equals the fraction 2 π over 3")[$t = 0 , " " t = display(frac(2 π, 3))$], and #math.equation(block: false, alt: "t equals the fraction 4 π over 3")[$t = display(frac(4 π, 3))$]. #notebox("Activity", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Identities] I Sum of Angles Identities + Is it true that #math.equation(block: false, alt: "cos ⁡ open parenthesis θ plus ϕ close parenthesis equals cos ⁡ open parenthesis θ close parenthesis plus cos ⁡ open parenthesis ϕ close parenthesis")[$cos ( θ + ϕ ) = cos ( θ ) + cos ( ϕ )$]? Try it for #math.equation(block: false, alt: "θ equals 60 degrees")[$θ = 60^(∘)$] and #math.equation(block: false, alt: "ϕ equals 45 degrees")[$ϕ = 45^(∘)$]. + + Recall the distributive law, #math.equation(block: false, alt: "a open parenthesis b plus c close parenthesis equals a b plus a c")[$a ( b + c ) = a b + a c$], where the parentheses denote multiplication. Is the same law true when the parentheses denote a function? In other words, is it true that #math.equation(block: false, alt: "f open parenthesis a plus b close parenthesis equals f open parenthesis a close parenthesis plus f open parenthesis b close parenthesis")[$f ( a + b ) = f ( a ) + f ( b )$]? + In the expression #math.equation(block: false, alt: "sin ⁡ open parenthesis θ plus ϕ close parenthesis")[$sin ( θ + ϕ )$], do the parentheses denote multiplication or the application of a function? Does the distributive law apply to #math.equation(block: false, alt: "sin ⁡ open parenthesis θ plus ϕ close parenthesis")[$sin ( θ + ϕ )$]? Do you think that #math.equation(block: false, alt: "sin ⁡ open parenthesis θ plus ϕ close parenthesis equals sin ⁡ open parenthesis θ close parenthesis plus sin ⁡ open parenthesis ϕ close parenthesis")[$sin ( θ + ϕ ) = sin ( θ ) + sin ( ϕ )$] is an identity? + + Look at the Sum of Angles Identities in this section. Make some observations that will help you memorize these formulas. + Do you think you would have to memorize these formulas if the equation #math.equation(block: false, alt: "sin ⁡ open parenthesis θ plus ϕ close parenthesis equals sin ⁡ open parenthesis θ close parenthesis plus sin ⁡ open parenthesis ϕ close parenthesis")[$sin ( θ + ϕ ) = sin ( θ ) + sin ( ϕ )$] were an identity? II Difference of Angles Identities, Tangent Identities + + Yikes! More formulas. Compare the Difference of Angles Identities with the Sum of Angles Identities. If you have memorized the Sum formulas, how can you also memorize the Difference formulas? + Comment on the sign patterns in the Sum and Difference Identities for Tangent. + + Now let's use the formulas backwards: look at the expression below: #math.equation(block: true, alt: "the fraction tan ⁡ open parenthesis 285 degrees close parenthesis minus tan ⁡ open parenthesis 75 degrees close parenthesis over 1 plus tan ⁡ open parenthesis 285 degrees close parenthesis tan ⁡ open parenthesis 75 degrees close parenthesis")[$display(frac(tan ( 285^(∘) ) − tan ( 75^(∘) ), 1 + tan ( 285^(∘) ) tan ( 75^(∘) )))$] Does it remind you of the left side of one of the six new identities? Use that identity to simplify the expression. + Do the same thing for this expression: #math.equation(block: true, alt: "sin ⁡ open parenthesis 4 t close parenthesis cos ⁡ open parenthesis 0.7 close parenthesis minus cos ⁡ open parenthesis 4 t close parenthesis sin ⁡ open parenthesis 0.7 close parenthesis")[$sin ( 4 t ) cos ( 0.7 ) − cos ( 4 t ) sin ( 0.7 )$] III Double Angle Identities + + Is it true that #math.equation(block: false, alt: "sin ⁡ open parenthesis 2 θ close parenthesis equals 2 sin ⁡ open parenthesis θ close parenthesis")[$sin ( 2 θ ) = 2 sin ( θ )$]? Choose a value for #math.equation(block: false, alt: "θ")[$θ$] and try it. Sketch #math.equation(block: false, alt: "θ")[$θ$] and #math.equation(block: false, alt: "2 θ")[$2 θ$] on a unit circle and show the sine of each. + Is it true that #math.equation(block: false, alt: "cos ⁡ open parenthesis 2 θ close parenthesis equals 2 cos ⁡ open parenthesis θ close parenthesis")[$cos ( 2 θ ) = 2 cos ( θ )$]? Repeat part (a) for cosine. + + Substitute #math.equation(block: false, alt: "α equals θ")[$α = θ$] and #math.equation(block: false, alt: "β equals θ")[$β = θ$] into the sum of angles formulas for sine, cosine, and tangent to derive the double angle formulas. + Choose one of the three forms of the Double Angle Identity for Cosine. Use the Pythagorean Identity to convert from that form to each of the other two forms. ] Review the following skills you will need for this section. #notebox("Algebra Refresher", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ + Compute #math.equation(block: false, alt: "f open parenthesis 2 close parenthesis , f open parenthesis 3 close parenthesis")[$f ( 2 ) , " " f ( 3 )$], and #math.equation(block: false, alt: "f open parenthesis 2 plus 3 close parenthesis")[$f ( 2 + 3 )$]. + For which of the following functions is it true that #math.equation(block: false, alt: "f open parenthesis a plus b close parenthesis equals f open parenthesis a close parenthesis plus f open parenthesis b close parenthesis")[$f ( a + b ) = f ( a ) + f ( b )$] whenever the function values are defined? + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 3 x plus 2")[$f ( x ) = 3 x + 2$] + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x squared")[$f ( x ) = x^(2)$] + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the square root of x")[$f ( x ) = sqrt(x)$] + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction 1 over x")[$f ( x ) = display(frac(1, x))$] + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals vertical bar x vertical bar")[$f ( x ) = | x |$] + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 to the power x")[$f ( x ) = 2^(x)$] #math.equation(block: true, alt: "bar")[$underline(#h(2em) #h(2em) #h(2em) #h(2em))$] Algebra Refresher Answers + + 8, 11, 17 + Not true + + 4, 9, 25 + Not true + + #math.equation(block: false, alt: "the square root of 2 , the square root of 3 , the square root of 5")[$sqrt(2) , " " sqrt(3) , " " sqrt(5)$] + Not true + + #math.equation(block: false, alt: "the fraction 1 over 2 , the fraction 1 over 3 , the fraction 1 over 5")[$display(frac(1, 2)) , " " display(frac(1, 3)) , " " display(frac(1, 5))$] + Not true + + 2, 3, 5 + Not true + + 4, 8, 32 + Not true ] === Section 8.1 Summary ==== Concepts + Identities are useful for changing from one form to another when solving equations, for simplifying expressions, and for finding exact values for trigonometric functions. + it is not true in general that #math.equation(block: false, alt: "cos ⁡ open parenthesis α plus β close parenthesis")[$cos ( α + β )$] is equal to #math.equation(block: false, alt: "cos ⁡ open parenthesis α close parenthesis plus cos ⁡ open parenthesis β close parenthesis")[$cos ( α ) + cos ( β )$] for all angles #math.equation(block: false, alt: "α")[$α$] and #math.equation(block: false, alt: "β")[$β$], or that #math.equation(block: false, alt: "sin ⁡ open parenthesis α plus β close parenthesis")[$sin ( α + β )$] is equal to #math.equation(block: false, alt: "sin ⁡ open parenthesis α close parenthesis plus sin ⁡ open parenthesis β close parenthesis")[$sin ( α ) + sin ( β )$]. + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Negative Angle Identities] #math.equation(block: true, alt: "cos ⁡ open parenthesis minus θ close parenthesis equals cos ⁡ open parenthesis θ close parenthesis")[$cos ( − θ ) = cos ( θ )$]#math.equation(block: true, alt: "sin ⁡ open parenthesis minus θ close parenthesis equals minus sin ⁡ open parenthesis θ close parenthesis")[$sin ( − θ ) = − sin ( θ )$]#math.equation(block: true, alt: "tan ⁡ open parenthesis minus θ close parenthesis equals minus tan ⁡ open parenthesis θ close parenthesis")[$tan ( − θ ) = − tan ( θ )$] ] + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Sum and Difference of Angles Identities] #math.equation(block: true, alt: "cos ⁡ open parenthesis α plus β close parenthesis equals cos ⁡ open parenthesis α close parenthesis cos ⁡ open parenthesis β close parenthesis minus sin ⁡ open parenthesis α close parenthesis sin ⁡ open parenthesis β close parenthesis")[$cos ( α + β ) = cos ( α ) cos ( β ) − sin ( α ) sin ( β )$]#math.equation(block: true, alt: "sin ⁡ open parenthesis α plus β close parenthesis equals sin ⁡ open parenthesis α close parenthesis cos ⁡ open parenthesis β close parenthesis plus cos ⁡ open parenthesis α close parenthesis sin ⁡ open parenthesis β close parenthesis")[$sin ( α + β ) = sin ( α ) cos ( β ) + cos ( α ) sin ( β )$]#math.equation(block: true, alt: "tan ⁡ open parenthesis α plus β close parenthesis equals the fraction tan ⁡ open parenthesis α close parenthesis plus tan ⁡ open parenthesis β close parenthesis over 1 minus tan ⁡ open parenthesis α close parenthesis tan ⁡ open parenthesis β close parenthesis")[$tan ( α + β ) = display(frac(tan ( α ) + tan ( β ), 1 − tan ( α ) tan ( β )))$]#math.equation(block: true, alt: "cos ⁡ open parenthesis α minus β close parenthesis equals cos ⁡ open parenthesis α close parenthesis cos ⁡ open parenthesis β close parenthesis plus sin ⁡ open parenthesis α close parenthesis sin ⁡ open parenthesis β close parenthesis")[$cos ( α − β ) = cos ( α ) cos ( β ) + sin ( α ) sin ( β )$]#math.equation(block: true, alt: "sin ⁡ open parenthesis α minus β close parenthesis equals sin ⁡ open parenthesis α close parenthesis cos ⁡ open parenthesis β close parenthesis minus cos ⁡ open parenthesis α close parenthesis sin ⁡ open parenthesis β close parenthesis")[$sin ( α − β ) = sin ( α ) cos ( β ) − cos ( α ) sin ( β )$]#math.equation(block: true, alt: "tan ⁡ open parenthesis α minus β close parenthesis equals the fraction tan ⁡ open parenthesis α close parenthesis minus tan ⁡ open parenthesis β close parenthesis over 1 plus tan ⁡ open parenthesis α close parenthesis tan ⁡ open parenthesis β close parenthesis")[$tan ( α − β ) = display(frac(tan ( α ) − tan ( β ), 1 + tan ( α ) tan ( β )))$] ] + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Double Angle Identities] #math.equation(block: true, alt: "cos ⁡ 2 open parenthesis θ close parenthesis, equals cos squared ⁡ open parenthesis θ close parenthesis minus sin squared ⁡ open parenthesis θ close parenthesis, sin ⁡ open parenthesis 2 θ close parenthesis equals 2 sin ⁡ open parenthesis θ close parenthesis cos ⁡ open parenthesis θ close parenthesis; equals 2 cos squared ⁡ open parenthesis θ close parenthesis minus 1, tan ⁡ open parenthesis 2 θ close parenthesis equals the fraction 2 tan ⁡ open parenthesis θ close parenthesis over 1 minus tan squared ⁡ open parenthesis θ close parenthesis; equals 1 minus 2 sin squared ⁡ open parenthesis θ close parenthesis")[$cos 2 ( θ ) & = cos^(2) ( θ ) − sin^(2) ( θ ) & & sin ( 2 θ ) = 2 sin ( θ ) cos ( θ ) \ & = 2 cos^(2) ( θ ) − 1 & & tan ( 2 θ ) = display(frac(2 tan ( θ ), 1 − tan^(2) ( θ ))) \ & = 1 − 2 sin^(2) ( θ )$] ] ==== Study Questions + Explain why #math.equation(block: false, alt: "f open parenthesis a plus b close parenthesis equals f open parenthesis a close parenthesis plus f open parenthesis b close parenthesis")[$f ( a + b ) = f ( a ) + f ( b )$] is not a valid application of the distributive law. + Delbert says that #math.equation(block: false, alt: "sin ⁡ open parenthesis θ plus the fraction π over 6 close parenthesis equals the fraction 1 over 2 plus sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ + display(frac(π, 6)) ) = display(frac(1, 2)) + sin ( θ )$]. Is he correct? Explain. + Francine says that #math.equation(block: false, alt: "tan ⁡ open parenthesis θ plus the fraction π over 4 close parenthesis equals the fraction 1 plus tan ⁡ open parenthesis θ close parenthesis over 1 minus tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ + display(frac(π, 4)) ) = display(frac(1 + tan ( θ ), 1 − tan ( θ )))$]. Is she correct? Explain. + Provide an example to show that doubling an angle does not double its sine or cosine. ==== Skills + Find trig values for the negative of an angle \#1–6 + Verify or disprove possible formulas \#7–12, 31–42, 73–76, 79–88 + Find exact values for trigonometric functions \#13–24, 55–62 + Simplify or expand expressions \#25–30, 43–54 + Solve equations \#63–72 + Prove standard identities \#77–78, 89–91 === Homework 8-1 + Sketch an angle #math.equation(block: false, alt: "α")[$α$] in standard position, with #math.equation(block: false, alt: "the fraction π over 2 less than α less than π")[$display(frac(π, 2)) < α < π$]. Also sketch the angle #math.equation(block: false, alt: "minus α")[$− α$]. + Choose a point on the terminal side of #math.equation(block: false, alt: "α")[$α$], and show that the negative angle identities hold for #math.equation(block: false, alt: "α")[$α$]. #figure(figph[angles], alt: "angles", caption: none) #math.equation(block: false, alt: "x sub 2 equals x sub 1")[$x_(2) = x_(1)$], #math.equation(block: false, alt: "y sub 2 equals minus y sub 1")[$y_(2) = − y_(1)$], and #math.equation(block: false, alt: "r sub 2 equals r sub 1")[$r_(2) = r_(1)$]. Thus, #math.equation(block: false, alt: "cos ⁡ open parenthesis minus α close parenthesis equals the fraction x sub 2 over r sub 2 equals the fraction x sub 1 over r sub 1 equals cos ⁡ open parenthesis α close parenthesis")[$cos ( − α ) = display(frac(x_(2), r_(2))) = display(frac(x_(1), r_(1))) = cos ( α )$], #math.equation(block: false, alt: "sin ⁡ open parenthesis minus α close parenthesis equals the fraction y sub 2 over r sub 2 equals the fraction minus y sub 1 over r sub 1 equals minus sin ⁡ open parenthesis α close parenthesis")[$sin ( − α ) = display(frac(y_(2), r_(2))) = display(frac(− y_(1), r_(1))) = − sin ( α )$], and #math.equation(block: false, alt: "tan ⁡ open parenthesis minus α close parenthesis equals the fraction y sub 2 over x sub 2 equals the fraction minus y sub 1 over x sub 1 equals minus tan ⁡ open parenthesis α close parenthesis")[$tan ( − α ) = display(frac(y_(2), x_(2))) = display(frac(− y_(1), x_(1))) = − tan ( α )$]. + Sketch an angle #math.equation(block: false, alt: "β")[$β$] in standard position, with #math.equation(block: false, alt: "π less than β less than the fraction 3 π over 2")[$π < β < display(frac(3 π, 2))$]. Also sketch the angle #math.equation(block: false, alt: "minus β")[$− β$]. + Choose a point on the terminal side of #math.equation(block: false, alt: "β")[$β$], and show that the negative angle identities hold for #math.equation(block: false, alt: "β")[$β$]. Given that #math.equation(block: false, alt: "sin ⁡ open parenthesis the fraction 7 π over 12 close parenthesis equals the fraction the square root of 2 plus the square root of 6 over 4")[$" " " " sin ( display(frac(7 π, 12)) ) = display(frac(sqrt(2) + sqrt(6), 4)) " " " "$], find #math.equation(block: false, alt: "sin ⁡ open parenthesis the fraction minus 7 π over 12 close parenthesis")[$sin ( display(frac(− 7 π, 12)) )$]. Sketch both angles. #math.equation(block: true, alt: "the fraction minus open parenthesis the square root of 2 plus the square root of 6 close parenthesis over 4")[$display(frac(− ( sqrt(2) + sqrt(6) ), 4))$] #figure(figph[angles], alt: "angles", caption: none) Given that #math.equation(block: false, alt: "cos ⁡ open parenthesis the fraction 7 π over 12 close parenthesis equals the fraction the square root of 2 minus the square root of 6 over 4")[$" " " " cos ( display(frac(7 π, 12)) ) = display(frac(sqrt(2) − sqrt(6), 4)) " " " "$], find #math.equation(block: false, alt: "cos ⁡ open parenthesis the fraction minus 7 π over 12 close parenthesis")[$cos ( display(frac(− 7 π, 12)) )$]. Sketch both angles. If #math.equation(block: false, alt: "cos ⁡ open parenthesis 2 x minus 0.3 close parenthesis equals 0.24")[$" " " " cos ( 2 x − 0.3 ) = 0.24 " " " "$] and #math.equation(block: false, alt: "sin ⁡ open parenthesis 2 x minus 0.3 close parenthesis less than 0")[$" " " " sin ( 2 x − 0.3 ) < 0 " " " "$], find #math.equation(block: false, alt: "cos ⁡ open parenthesis 0.3 minus 2 x close parenthesis")[$cos ( 0.3 − 2 x )$] and #math.equation(block: false, alt: "sin ⁡ open parenthesis 0.3 minus 2 x close parenthesis")[$sin ( 0.3 − 2 x )$]. #math.equation(block: true, alt: "cos ⁡ open parenthesis 0.3 minus 2 x close parenthesis equals 0.24")[$cos ( 0.3 − 2 x ) = 0.24$], #math.equation(block: true, alt: "sin ⁡ open parenthesis 0.3 minus 2 x close parenthesis equals 0.97")[$sin ( 0.3 − 2 x ) = 0.97$] If #math.equation(block: false, alt: "sin ⁡ open parenthesis 1.5 minus ϕ close parenthesis equals minus 0.28")[$" " " " sin ( 1.5 − ϕ ) = − 0.28 " " " "$] and #math.equation(block: false, alt: "cos ⁡ open parenthesis 1.5 minus ϕ close parenthesis greater than 0")[$" " " " cos ( 1.5 − ϕ ) > 0 " " " "$], find #math.equation(block: false, alt: "sin ⁡ open parenthesis ϕ minus 1.5 close parenthesis")[$sin ( ϕ − 1.5 )$] and #math.equation(block: false, alt: "cos ⁡ open parenthesis ϕ minus 1.5 close parenthesis")[$cos ( ϕ − 1.5 )$]. Show that #math.equation(block: false, alt: "cos ⁡ open parenthesis 45 degrees plus 45 degrees close parenthesis")[$cos ( 45^(∘) + 45^(∘) )$] is not equal to #math.equation(block: false, alt: "cos ⁡ open parenthesis 45 degrees close parenthesis plus cos ⁡ open parenthesis 45 degrees close parenthesis")[$cos ( 45^(∘) ) + cos ( 45^(∘) )$]. #math.equation(block: false, alt: "cos ⁡ open parenthesis 45 degrees plus 45 degrees close parenthesis equals cos ⁡ open parenthesis 90 degrees close parenthesis equals 0")[$cos ( 45^(∘) + 45^(∘) ) = cos ( 90^(∘) ) = 0$], but #math.equation(block: false, alt: "cos ⁡ open parenthesis 45 degrees close parenthesis plus cos ⁡ open parenthesis 45 degrees close parenthesis equals the fraction 1 over the square root of 2 plus the fraction 1 over the square root of 2 equals the square root of 2")[$cos ( 45^(∘) ) + cos ( 45^(∘) ) = frac(1, sqrt(2)) + frac(1, sqrt(2)) = sqrt(2)$] Show that #math.equation(block: false, alt: "tan ⁡ open parenthesis 60 degrees minus 30 degrees close parenthesis")[$tan ( 60^(∘) − 30^(∘) )$] is not equal to #math.equation(block: false, alt: "tan ⁡ open parenthesis 60 degrees close parenthesis minus tan ⁡ open parenthesis 30 degrees close parenthesis")[$tan ( 60^(∘) ) − tan ( 30^(∘) )$]. Use your calculator to verify that #math.equation(block: false, alt: "tan ⁡ open parenthesis 87 degrees minus 29 degrees close parenthesis")[$tan ( 87^(∘) − 29^(∘) )$] is not equal to #math.equation(block: false, alt: "tan ⁡ open parenthesis 87 degrees close parenthesis minus tan ⁡ open parenthesis 29 degrees close parenthesis")[$tan ( 87^(∘) ) − tan ( 29^(∘) )$]. #math.equation(block: false, alt: "tan ⁡ open parenthesis 87 degrees minus 29 degrees close parenthesis approximately equals 1.600")[$tan ( 87^(∘) − 29^(∘) ) ≈ 1.600$], but #math.equation(block: false, alt: "tan ⁡ open parenthesis 87 degrees close parenthesis minus tan ⁡ open parenthesis 29 degrees close parenthesis approximately equals 18.527")[$tan ( 87^(∘) ) − tan ( 29^(∘) ) ≈ 18.527$] Use your calculator to verify that #math.equation(block: false, alt: "cos ⁡ open parenthesis 52 degrees plus 64 degrees close parenthesis")[$cos ( 52^(∘) + 64^(∘) )$] is not equal to #math.equation(block: false, alt: "cos ⁡ open parenthesis 52 degrees close parenthesis plus cos ⁡ open parenthesis 64 degrees close parenthesis")[$cos ( 52^(∘) ) + cos ( 64^(∘) )$]. Use graphs to show that #math.equation(block: false, alt: "sin ⁡ open parenthesis x minus the fraction π over 6 close parenthesis")[$sin ( x − display(frac(π, 6)) )$] is not equivalent to #math.equation(block: false, alt: "sin ⁡ open parenthesis x close parenthesis minus sin ⁡ open parenthesis the fraction π over 6 close parenthesis")[$sin ( x ) − sin ( display(frac(π, 6)) )$]. #figure(figph[two sinusoidal graphs], alt: "two sinusoidal graphs", caption: none) The curves are different. Use graphs to show that #math.equation(block: false, alt: "tan ⁡ open parenthesis x plus the fraction π over 4 close parenthesis")[$tan ( x + display(frac(π, 4)) )$] is not equivalent to #math.equation(block: false, alt: "tan ⁡ x plus tan ⁡ the fraction π over 4")[$tan x + tan display(frac(π, 4))$]. For Problems 13–24, find exact values for the trig ratios. (Do not use a calculator!) Suppose #math.equation(block: false, alt: "cos ⁡ open parenthesis α close parenthesis equals the fraction 3 over 5 , sin ⁡ open parenthesis α close parenthesis equals the fraction 4 over 5 , cos ⁡ open parenthesis β close parenthesis equals the fraction 5 over 13")[$cos ( α ) = display(frac(3, 5)) , " " sin ( α ) = display(frac(4, 5)) , " " cos ( β ) = display(frac(5, 13))$], and #math.equation(block: false, alt: "sin ⁡ open parenthesis β close parenthesis equals the fraction minus 12 over 13")[$sin ( β ) = display(frac(− 12, 13))$]. Evaluate the following. + #math.equation(block: false, alt: "cos ⁡ open parenthesis α plus β close parenthesis")[$cos ( α + β )$] + #math.equation(block: false, alt: "sin ⁡ open parenthesis α plus β close parenthesis")[$sin ( α + β )$] + #math.equation(block: false, alt: "tan ⁡ open parenthesis α plus β close parenthesis")[$tan ( α + β )$] + #math.equation(block: false, alt: "the fraction 63 over 65")[$display(frac(63, 65))$] + #math.equation(block: false, alt: "the fraction minus 16 over 65")[$display(frac(− 16, 65))$] + #math.equation(block: false, alt: "the fraction minus 16 over 63")[$display(frac(− 16, 63))$] Suppose #math.equation(block: false, alt: "cos ⁡ open parenthesis α close parenthesis equals the fraction minus 2 over 3 , sin ⁡ open parenthesis α close parenthesis equals the fraction the square root of 5 over 3 , cos ⁡ open parenthesis β close parenthesis equals the fraction the square root of 3 over 2")[$cos ( α ) = display(frac(− 2, 3)) , " " sin ( α ) = display(frac(sqrt(5), 3)) , " " cos ( β ) = display(frac(sqrt(3), 2))$], and #math.equation(block: false, alt: "sin ⁡ open parenthesis β close parenthesis equals the fraction minus 1 over 2")[$sin ( β ) = display(frac(− 1, 2))$]. Evaluate the following. + #math.equation(block: false, alt: "cos ⁡ open parenthesis α minus β close parenthesis")[$cos ( α − β )$] + #math.equation(block: false, alt: "sin ⁡ open parenthesis α minus β close parenthesis")[$sin ( α − β )$] + #math.equation(block: false, alt: "tan ⁡ open parenthesis α minus β close parenthesis")[$tan ( α − β )$] If #math.equation(block: false, alt: "tan ⁡ open parenthesis t close parenthesis equals the fraction 3 over 4")[$tan ( t ) = display(frac(3, 4))$] and #math.equation(block: false, alt: "tan ⁡ open parenthesis s close parenthesis equals the fraction minus 7 over 24")[$tan ( s ) = display(frac(− 7, 24))$], find exact values for: + #math.equation(block: false, alt: "tan ⁡ open parenthesis s plus t close parenthesis")[$tan ( s + t )$] + #math.equation(block: false, alt: "tan ⁡ open parenthesis s minus t close parenthesis")[$tan ( s − t )$] + #math.equation(block: false, alt: "the fraction 44 over 117")[$display(frac(44, 117))$] + #math.equation(block: false, alt: "the fraction 4 over 3")[$display(frac(4, 3))$] If #math.equation(block: false, alt: "tan ⁡ open parenthesis x close parenthesis equals minus 3")[$tan ( x ) = − 3$] and #math.equation(block: false, alt: "tan ⁡ open parenthesis y close parenthesis equals minus 5")[$tan ( y ) = − 5$], find exact values for: + #math.equation(block: false, alt: "tan ⁡ open parenthesis x plus y close parenthesis")[$tan ( x + y )$] + #math.equation(block: false, alt: "tan ⁡ open parenthesis x minus y close parenthesis")[$tan ( x − y )$] Suppose #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction 15 over 17")[$cos ( θ ) = display(frac(15, 17))$] and #math.equation(block: false, alt: "sin ⁡ open parenthesis ϕ close parenthesis equals the fraction 3 over 5")[$sin ( ϕ ) = display(frac(3, 5))$], where #math.equation(block: false, alt: "θ")[$θ$] and #math.equation(block: false, alt: "ϕ")[$ϕ$] are in quadrant I. Evaluate the following. + #math.equation(block: false, alt: "cos ⁡ open parenthesis θ plus ϕ close parenthesis")[$cos ( θ + ϕ )$] + #math.equation(block: false, alt: "tan ⁡ open parenthesis θ minus ϕ close parenthesis")[$tan ( θ − ϕ )$] + #math.equation(block: false, alt: "the fraction 36 over 85")[$display(frac(36, 85))$] + #math.equation(block: false, alt: "the fraction minus 13 over 84")[$display(frac(− 13, 84))$] Suppose #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction 15 over 17")[$cos ( θ ) = display(frac(15, 17))$], where #math.equation(block: false, alt: "θ")[$θ$] is in quadrant IV, and #math.equation(block: false, alt: "sin ⁡ open parenthesis ϕ close parenthesis equals the fraction 3 over 5")[$sin ( ϕ ) = display(frac(3, 5))$], where #math.equation(block: false, alt: "ϕ")[$ϕ$] is in quadrant II. Evaluate the following. + #math.equation(block: false, alt: "sin ⁡ open parenthesis θ minus ϕ close parenthesis")[$sin ( θ − ϕ )$] + #math.equation(block: false, alt: "tan ⁡ open parenthesis θ plus ϕ close parenthesis")[$tan ( θ + ϕ )$] If #math.equation(block: false, alt: "sin ⁡ open parenthesis α close parenthesis equals the fraction 12 over 13 , the fraction π over 2 less than α less than π")[$sin ( α ) = display(frac(12, 13)) , " " display(frac(π, 2)) < α < π$], and #math.equation(block: false, alt: "cos ⁡ open parenthesis β close parenthesis equals the fraction minus 3 over 5 , π less than β less than the fraction 3 π over 2")[$cos ( β ) = display(frac(− 3, 5)) , " " π < β < display(frac(3 π, 2))$], find exact values for: + #math.equation(block: false, alt: "sin ⁡ open parenthesis α plus β close parenthesis")[$sin ( α + β )$] + #math.equation(block: false, alt: "cos ⁡ open parenthesis α plus β close parenthesis")[$cos ( α + β )$] + #math.equation(block: false, alt: "tan ⁡ open parenthesis α plus β close parenthesis")[$tan ( α + β )$] + Sketch the angles #math.equation(block: false, alt: "α , β")[$α , β$] and #math.equation(block: false, alt: "α plus β")[$α + β$]. + #math.equation(block: false, alt: "the fraction minus 16 over 65")[$display(frac(− 16, 65))$] + #math.equation(block: false, alt: "the fraction 63 over 65")[$display(frac(63, 65))$] + #math.equation(block: false, alt: "the fraction minus 16 over 63")[$display(frac(− 16, 63))$] + #figure(figph[angles], alt: "angles", caption: none) If #math.equation(block: false, alt: "cos ⁡ open parenthesis α close parenthesis equals the fraction 3 over 8 , the fraction 3 π over 2 less than α less than 2 π")[$cos ( α ) = display(frac(3, 8)) , " " display(frac(3 π, 2)) < α < 2 π$], and #math.equation(block: false, alt: "sin ⁡ open parenthesis β close parenthesis equals the fraction minus 1 over 4 , π less than β less than the fraction 3 π over 2")[$sin ( β ) = display(frac(− 1, 4)) , " " π < β < display(frac(3 π, 2))$], find exact values for: + #math.equation(block: false, alt: "sin ⁡ open parenthesis α minus β close parenthesis")[$sin ( α − β )$] + #math.equation(block: false, alt: "cos ⁡ open parenthesis α minus β close parenthesis")[$cos ( α − β )$] + #math.equation(block: false, alt: "tan ⁡ open parenthesis α minus β close parenthesis")[$tan ( α − β )$] + Sketch the angles #math.equation(block: false, alt: "α , β")[$α , β$] and #math.equation(block: false, alt: "α minus β")[$α − β$]. Find the exact values of #math.equation(block: false, alt: "cos ⁡ open parenthesis 15 degrees close parenthesis")[$cos ( 15^(∘) )$] and #math.equation(block: false, alt: "tan ⁡ open parenthesis 15 degrees close parenthesis")[$tan ( 15^(∘) )$]. #math.equation(block: true, alt: "cos ⁡ open parenthesis 15 degrees close parenthesis equals the fraction the square root of 6 plus the square root of 2 over 4")[$cos ( 15^(∘) ) = display(frac(sqrt(6) + sqrt(2), 4))$], #math.equation(block: true, alt: "tan ⁡ open parenthesis 15 degrees close parenthesis equals 2 minus the square root of 3")[$tan ( 15^(∘) ) = 2 − sqrt(3)$] Find the exact values of #math.equation(block: false, alt: "sin ⁡ open parenthesis 165 degrees close parenthesis")[$sin ( 165^(∘) )$] and #math.equation(block: false, alt: "tan ⁡ open parenthesis 165 degrees close parenthesis")[$tan ( 165^(∘) )$]. If #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals 0.2")[$sin ( θ ) = 0.2$] and #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis greater than 0")[$cos ( θ ) > 0$], find #math.equation(block: false, alt: "sin ⁡ open parenthesis θ plus the fraction π over 3 close parenthesis")[$sin ( θ + display(frac(π, 3)) )$]. #math.equation(block: true, alt: "the fraction 6 the square root of 2 plus 1 over 10")[$display(frac(6 sqrt(2) + 1, 10))$] If #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals 0.6")[$cos ( θ ) = 0.6$] and #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis less than 0")[$sin ( θ ) < 0$], find #math.equation(block: false, alt: "cos ⁡ open parenthesis θ plus the fraction 3 π over 4 close parenthesis")[$cos ( θ + display(frac(3 π, 4)) )$]. For Problems 25–30, use the sum and difference formulas to expand each expression. #math.equation(block: true, alt: "sin ⁡ open parenthesis θ minus 270 degrees close parenthesis")[$sin ( θ − 270^(∘) )$] #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis")[$cos ( θ )$] #math.equation(block: true, alt: "cos ⁡ open parenthesis 270 degrees plus θ close parenthesis")[$cos ( 270^(∘) + θ )$] #math.equation(block: true, alt: "cos ⁡ open parenthesis t plus the fraction π over 6 close parenthesis")[$cos ( t + display(frac(π, 6)) )$] #math.equation(block: true, alt: "the fraction the square root of 3 over 2 cos ⁡ open parenthesis t close parenthesis minus the fraction 1 over 2 sin ⁡ open parenthesis t close parenthesis")[$display(frac(sqrt(3), 2)) cos ( t ) − display(frac(1, 2)) sin ( t )$] #math.equation(block: true, alt: "sin ⁡ open parenthesis t minus the fraction 2 π over 3 close parenthesis")[$sin ( t − display(frac(2 π, 3)) )$] #math.equation(block: true, alt: "tan ⁡ open parenthesis β minus the fraction π over 6 close parenthesis")[$tan ( β − display(frac(π, 6)) )$] #math.equation(block: true, alt: "the fraction the square root of 3 tan ⁡ β minus 1 over the square root of 3 plus tan ⁡ β")[$display(frac(sqrt(3) tan β − 1, sqrt(3) + tan β))$] #math.equation(block: true, alt: "tan ⁡ open parenthesis ϕ plus the fraction π over 4 close parenthesis")[$tan ( ϕ + display(frac(π, 4)) )$] For Problems 31–34, use the unit circle to estimate trig values. Then verify with your calculator. #figure(figph[unit circle], alt: "unit circle", caption: none) Does #math.equation(block: false, alt: "sin ⁡ open parenthesis 2 times 80 degrees close parenthesis equals 2 sin ⁡ open parenthesis 80 degrees close parenthesis")[$sin ( 2 ⋅ 80^(∘) ) = 2 sin ( 80^(∘) )$]? No Does #math.equation(block: false, alt: "cos ⁡ open parenthesis 2 times 25 degrees close parenthesis equals 2 cos ⁡ open parenthesis 25 degrees close parenthesis")[$cos ( 2 ⋅ 25^(∘) ) = 2 cos ( 25^(∘) )$]? Does #math.equation(block: false, alt: "tan ⁡ open parenthesis 2 times 70 degrees close parenthesis equals 2 tan ⁡ open parenthesis 70 degrees close parenthesis")[$tan ( 2 ⋅ 70^(∘) ) = 2 tan ( 70^(∘) )$]? No Does #math.equation(block: false, alt: "tan ⁡ open parenthesis 2 times 100 degrees close parenthesis equals 2 tan ⁡ open parenthesis 100 degrees close parenthesis")[$tan ( 2 ⋅ 100^(∘) ) = 2 tan ( 100^(∘) )$]? For Problems 35–38, verify that each statement is true. #math.equation(block: true, alt: "sin ⁡ open parenthesis 90 degrees close parenthesis equals 2 sin ⁡ open parenthesis 45 degrees close parenthesis cos ⁡ open parenthesis 45 degrees close parenthesis")[$sin ( 90^(∘) ) = 2 sin ( 45^(∘) ) cos ( 45^(∘) )$] #math.equation(block: true, alt: "1 equals 2 open parenthesis the fraction 1 over the square root of 2 close parenthesis open parenthesis the fraction 1 over the square root of 2 close parenthesis")[$1 = 2 ( display(frac(1, sqrt(2))) ) ( display(frac(1, sqrt(2))) )$] #math.equation(block: true, alt: "sin ⁡ open parenthesis 60 degrees close parenthesis equals 2 sin ⁡ open parenthesis 30 degrees close parenthesis cos ⁡ open parenthesis 30 degrees close parenthesis")[$sin ( 60^(∘) ) = 2 sin ( 30^(∘) ) cos ( 30^(∘) )$] #math.equation(block: true, alt: "cos ⁡ open parenthesis 60 degrees close parenthesis equals cos squared ⁡ open parenthesis 30 degrees close parenthesis minus sin squared ⁡ open parenthesis 30 degrees close parenthesis")[$cos ( 60^(∘) ) = cos^(2) ( 30^(∘) ) − sin^(2) ( 30^(∘) )$] #math.equation(block: true, alt: "the fraction 1 over 2 equals open parenthesis the fraction the square root of 3 over 2 close parenthesis squared minus open parenthesis the fraction 1 over 2 close parenthesis squared")[$frac(1, 2) = attach(( display(frac(sqrt(3), 2)) ), t: 2) − attach(( display(frac(1, 2)) ), t: 2)$] #math.equation(block: true, alt: "tan ⁡ open parenthesis 60 degrees close parenthesis equals the fraction 2 tan ⁡ open parenthesis 30 degrees close parenthesis over 1 minus tan squared ⁡ open parenthesis 30 degrees close parenthesis")[$tan ( 60^(∘) ) = display(frac(2 tan ( 30^(∘) ), 1 − tan^(2) ( 30^(∘) )))$] In Problems 39–42, is the statement true or false? Explain your answer. If #math.equation(block: false, alt: "cos ⁡ open parenthesis α close parenthesis equals 0.32")[$cos ( α ) = 0.32$], then #math.equation(block: false, alt: "cos ⁡ open parenthesis 2 α close parenthesis equals 2 open parenthesis 0.32 close parenthesis equals 0.64")[$cos ( 2 α ) = 2 ( 0.32 ) = 0.64$]. False, but #math.equation(block: false, alt: "cos ⁡ open parenthesis 2 α close parenthesis equals 2 open parenthesis 0.32 close parenthesis squared minus 1")[$cos ( 2 α ) = 2 ( 0.32 )^(2) − 1$] If #math.equation(block: false, alt: "cos ⁡ open parenthesis 2 β close parenthesis equals 0.86")[$cos ( 2 β ) = 0.86$], then #math.equation(block: false, alt: "cos ⁡ open parenthesis β close parenthesis equals 0.43")[$cos ( β ) = 0.43$], so #math.equation(block: false, alt: "β equals cos to the power minus 1 ⁡ open parenthesis 0.43 close parenthesis")[$β = cos^(− 1) ( 0.43 )$]. If #math.equation(block: false, alt: "sin ⁡ open parenthesis 2 θ close parenthesis equals h")[$sin ( 2 θ ) = h$], then #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction h over 2")[$sin ( θ ) = display(frac(h, 2))$], so #math.equation(block: false, alt: "θ equals sin to the power minus 1 ⁡ open parenthesis the fraction h over 2 close parenthesis")[$θ = sin^(− 1) ( display(frac(h, 2)) )$]. False, but #math.equation(block: false, alt: "2 θ equals sin to the power minus 1 ⁡ open parenthesis h close parenthesis")[$2 θ = sin^(− 1) ( h )$] If #math.equation(block: false, alt: "cos ⁡ open parenthesis ϕ close parenthesis equals r")[$cos ( ϕ ) = r$], then #math.equation(block: false, alt: "cos ⁡ open parenthesis 2 ϕ close parenthesis equals 2 r")[$cos ( 2 ϕ ) = 2 r$]. For Problems 43–54, use the double angle identities to simplify the expression. #math.equation(block: true, alt: "2 sin ⁡ open parenthesis 34 degrees close parenthesis cos ⁡ open parenthesis 34 degrees close parenthesis")[$2 sin ( 34^(∘) ) cos ( 34^(∘) )$] #math.equation(block: true, alt: "sin ⁡ open parenthesis 68 degrees close parenthesis")[$sin ( 68^(∘) )$] #math.equation(block: true, alt: "cos squared ⁡ open parenthesis the fraction π over 10 close parenthesis minus sin squared ⁡ open parenthesis the fraction π over 10 close parenthesis")[$cos^(2) ( display(frac(π, 10)) ) − sin^(2) ( display(frac(π, 10)) )$] #math.equation(block: true, alt: "1 minus 2 sin squared ⁡ open parenthesis the fraction π over 16 close parenthesis")[$1 − 2 sin^(2) ( display(frac(π, 16)) )$] #math.equation(block: true, alt: "cos ⁡ open parenthesis the fraction π over 8 close parenthesis")[$cos ( display(frac(π, 8)) )$] #math.equation(block: true, alt: "2 cos squared ⁡ open parenthesis 18 degrees close parenthesis minus 1")[$2 cos^(2) ( 18^(∘) ) − 1$] #math.equation(block: true, alt: "cos squared ⁡ open parenthesis 3 θ close parenthesis minus sin squared ⁡ open parenthesis 3 θ close parenthesis")[$cos^(2) ( 3 θ ) − sin^(2) ( 3 θ )$] #math.equation(block: true, alt: "cos ⁡ open parenthesis 6 θ close parenthesis")[$cos ( 6 θ )$] #math.equation(block: true, alt: "2 sin ⁡ open parenthesis 2 α close parenthesis cos ⁡ open parenthesis 2 α close parenthesis")[$2 sin ( 2 α ) cos ( 2 α )$] #math.equation(block: true, alt: "2 sin ⁡ open parenthesis 5 t close parenthesis cos ⁡ open parenthesis 5 t close parenthesis")[$2 sin ( 5 t ) cos ( 5 t )$] #math.equation(block: true, alt: "sin ⁡ 10 t")[$sin 10 t$] #math.equation(block: true, alt: "cos squared ⁡ open parenthesis 4 w close parenthesis minus sin squared ⁡ open parenthesis 4 w close parenthesis")[$cos^(2) ( 4 w ) − sin^(2) ( 4 w )$] #math.equation(block: true, alt: "the fraction 2 tan ⁡ open parenthesis 64 degrees close parenthesis over 1 minus tan squared ⁡ open parenthesis 64 degrees close parenthesis")[$display(frac(2 tan ( 64^(∘) ), 1 − tan^(2) ( 64^(∘) )))$] #math.equation(block: true, alt: "tan ⁡ 128 degrees")[$tan 128^(∘)$] #math.equation(block: true, alt: "the fraction 2 tan ⁡ open parenthesis the fraction π over 3 close parenthesis over 1 minus tan squared ⁡ open parenthesis the fraction π over 3 close parenthesis")[$display(frac(2 tan ( display(frac(π, 3)) ), 1 − tan^(2) ( display(frac(π, 3)) )))$] #math.equation(block: true, alt: "2 cos squared ⁡ open parenthesis 2 β close parenthesis minus 1")[$2 cos^(2) ( 2 β ) − 1$] #math.equation(block: true, alt: "cos ⁡ open parenthesis 4 β close parenthesis")[$cos ( 4 β )$] #math.equation(block: true, alt: "1 minus 2 sin squared ⁡ open parenthesis 6 s close parenthesis")[$1 − 2 sin^(2) ( 6 s )$] For Problems 55–58, use the figures to find the trigonometric ratios. #figure(figph[triangles], alt: "triangles", caption: none) + #math.equation(block: false, alt: "sin ⁡ open parenthesis α close parenthesis")[$sin ( α )$] + #math.equation(block: false, alt: "cos ⁡ open parenthesis α close parenthesis")[$cos ( α )$] + #math.equation(block: false, alt: "tan ⁡ open parenthesis α close parenthesis")[$tan ( α )$] + #math.equation(block: false, alt: "sin ⁡ open parenthesis 2 α close parenthesis")[$sin ( 2 α )$] + #math.equation(block: false, alt: "cos ⁡ open parenthesis 2 α close parenthesis")[$cos ( 2 α )$] + #math.equation(block: false, alt: "tan ⁡ open parenthesis 2 α close parenthesis")[$tan ( 2 α )$] + #math.equation(block: false, alt: "the fraction 5 over 6")[$display(frac(5, 6))$] + #math.equation(block: false, alt: "the fraction the square root of 11 over 6")[$display(frac(sqrt(11), 6))$] + #math.equation(block: false, alt: "the fraction 5 over the square root of 11")[$display(frac(5, sqrt(11)))$] + #math.equation(block: false, alt: "the fraction 5 the square root of 11 over 18")[$display(frac(5 sqrt(11), 18))$] + #math.equation(block: false, alt: "the fraction minus 7 over 18")[$display(frac(− 7, 18))$] + #math.equation(block: false, alt: "the fraction minus 5 the square root of 11 over 7")[$display(frac(− 5 sqrt(11), 7))$] + #math.equation(block: false, alt: "sin ⁡ open parenthesis β close parenthesis")[$sin ( β )$] + #math.equation(block: false, alt: "cos ⁡ open parenthesis β close parenthesis")[$cos ( β )$] + #math.equation(block: false, alt: "tan ⁡ open parenthesis β close parenthesis")[$tan ( β )$] + #math.equation(block: false, alt: "sin ⁡ open parenthesis 2 β close parenthesis")[$sin ( 2 β )$] + #math.equation(block: false, alt: "cos ⁡ open parenthesis 2 β close parenthesis")[$cos ( 2 β )$] + #math.equation(block: false, alt: "tan ⁡ open parenthesis 2 β close parenthesis")[$tan ( 2 β )$] + #math.equation(block: false, alt: "sin ⁡ open parenthesis s close parenthesis")[$sin ( s )$] + #math.equation(block: false, alt: "cos ⁡ open parenthesis s close parenthesis")[$cos ( s )$] + #math.equation(block: false, alt: "tan ⁡ open parenthesis s close parenthesis")[$tan ( s )$] + #math.equation(block: false, alt: "sin ⁡ open parenthesis 2 s close parenthesis")[$sin ( 2 s )$] + #math.equation(block: false, alt: "cos ⁡ open parenthesis 2 s close parenthesis")[$cos ( 2 s )$] + #math.equation(block: false, alt: "tan ⁡ open parenthesis 2 s close parenthesis")[$tan ( 2 s )$] + #math.equation(block: false, alt: "the fraction 1 over the square root of w squared plus 1")[$display(frac(1, sqrt(w^(2) + 1)))$] + #math.equation(block: false, alt: "the fraction w over the square root of w squared plus 1")[$display(frac(w, sqrt(w^(2) + 1)))$] + #math.equation(block: false, alt: "the fraction 1 over w")[$display(frac(1, w))$] + #math.equation(block: false, alt: "the fraction 2 w over w squared plus 1")[$display(frac(2 w, w^(2) + 1))$] + #math.equation(block: false, alt: "the fraction w squared minus 1 over w squared plus 1")[$display(frac(w^(2) − 1, w^(2) + 1))$] + #math.equation(block: false, alt: "the fraction 2 w over w squared minus 1")[$display(frac(2 w, w^(2) − 1))$] + #math.equation(block: false, alt: "sin ⁡ open parenthesis t close parenthesis")[$sin ( t )$] + #math.equation(block: false, alt: "cos ⁡ open parenthesis t close parenthesis")[$cos ( t )$] + #math.equation(block: false, alt: "tan ⁡ open parenthesis t close parenthesis")[$tan ( t )$] + #math.equation(block: false, alt: "sin ⁡ open parenthesis 2 t close parenthesis")[$sin ( 2 t )$] + #math.equation(block: false, alt: "cos ⁡ open parenthesis 2 t close parenthesis")[$cos ( 2 t )$] + #math.equation(block: false, alt: "tan ⁡ open parenthesis 2 t close parenthesis")[$tan ( 2 t )$] Suppose #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction 12 over 13")[$cos ( θ ) = display(frac(12, 13))$] and #math.equation(block: false, alt: "the fraction 3 π over 2 less than θ less than 2 π")[$display(frac(3 π, 2)) < θ < 2 π$]. Compute exact values for: + #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$] + #math.equation(block: false, alt: "sin ⁡ open parenthesis 2 θ close parenthesis")[$sin ( 2 θ )$] + #math.equation(block: false, alt: "cos ⁡ open parenthesis 2 θ close parenthesis")[$cos ( 2 θ )$] + #math.equation(block: false, alt: "tan ⁡ open parenthesis 2 θ close parenthesis")[$tan ( 2 θ )$] + Sketch the angles #math.equation(block: false, alt: "θ")[$θ$] and #math.equation(block: false, alt: "2 θ")[$2 θ$]. + #math.equation(block: false, alt: "the fraction minus 5 over 13")[$display(frac(− 5, 13))$] + #math.equation(block: false, alt: "the fraction minus 120 over 169")[$display(frac(− 120, 169))$] + #math.equation(block: false, alt: "the fraction 119 over 169")[$display(frac(119, 169))$] + #math.equation(block: false, alt: "the fraction minus 120 over 119")[$display(frac(− 120, 119))$] + #figure(figph[angles], alt: "angles", caption: none) Suppose #math.equation(block: false, alt: "sin ⁡ open parenthesis ϕ close parenthesis equals the fraction 5 over 6")[$sin ( ϕ ) = display(frac(5, 6))$] and #math.equation(block: false, alt: "the fraction π over 2 less than ϕ less than π")[$display(frac(π, 2)) < ϕ < π$]. Compute exact values for: + #math.equation(block: false, alt: "cos ⁡ open parenthesis ϕ close parenthesis")[$cos ( ϕ )$] + #math.equation(block: false, alt: "sin ⁡ open parenthesis 2 ϕ close parenthesis")[$sin ( 2 ϕ )$] + #math.equation(block: false, alt: "cos ⁡ open parenthesis 2 ϕ close parenthesis")[$cos ( 2 ϕ )$] + #math.equation(block: false, alt: "tan ⁡ open parenthesis 2 ϕ close parenthesis")[$tan ( 2 ϕ )$] + Sketch the angles #math.equation(block: false, alt: "ϕ")[$ϕ$] and #math.equation(block: false, alt: "2 ϕ")[$2 ϕ$]. If #math.equation(block: false, alt: "tan ⁡ open parenthesis u close parenthesis equals minus 4")[$tan ( u ) = − 4$] and #math.equation(block: false, alt: "270 degrees less than u less than 360 degrees")[$270^(∘) < u < 360^(∘)$], find exact values for: + #math.equation(block: false, alt: "tan ⁡ open parenthesis 2 u close parenthesis")[$tan ( 2 u )$] + #math.equation(block: false, alt: "cos ⁡ open parenthesis 2 u close parenthesis")[$cos ( 2 u )$] + #math.equation(block: false, alt: "sin ⁡ open parenthesis 2 u close parenthesis")[$sin ( 2 u )$] + #math.equation(block: false, alt: "the fraction 8 over 15")[$display(frac(8, 15))$] + #math.equation(block: false, alt: "the fraction minus 15 over 17")[$display(frac(− 15, 17))$] + #math.equation(block: false, alt: "the fraction minus 8 over 17")[$display(frac(− 8, 17))$] If #math.equation(block: false, alt: "tan ⁡ open parenthesis v close parenthesis equals the fraction 2 over 3")[$tan ( v ) = display(frac(2, 3))$] and #math.equation(block: false, alt: "180 degrees less than v less than 270 degrees")[$180^(∘) < v < 270^(∘)$], find exact values for: + #math.equation(block: false, alt: "tan ⁡ open parenthesis 2 v close parenthesis")[$tan ( 2 v )$] + #math.equation(block: false, alt: "cos ⁡ open parenthesis 2 v close parenthesis")[$cos ( 2 v )$] + #math.equation(block: false, alt: "sin ⁡ open parenthesis 2 v close parenthesis")[$sin ( 2 v )$] For Problems 63–72, + Use identities to rewrite the equation in terms of a single angle. + Solve. Give exact solutions between #math.equation(block: false, alt: "0")[$0$] and #math.equation(block: false, alt: "2 π")[$2 π$]. #math.equation(block: true, alt: "sin ⁡ open parenthesis 2 θ close parenthesis plus the square root of 2 cos ⁡ open parenthesis θ close parenthesis equals 0")[$sin ( 2 θ ) + sqrt(2) cos ( θ ) = 0$] + #math.equation(block: false, alt: "2 sin ⁡ open parenthesis θ close parenthesis times cos ⁡ open parenthesis θ close parenthesis plus the square root of 2 cos ⁡ open parenthesis θ close parenthesis equals 0")[$2 sin ( θ ) ⋅ cos ( θ ) + sqrt(2) cos ( θ ) = 0$] + #math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$], #math.equation(block: false, alt: "the fraction 5 π over 4")[$display(frac(5 π, 4))$], #math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$], #math.equation(block: false, alt: "the fraction 7 π over 4")[$display(frac(7 π, 4))$] #math.equation(block: true, alt: "sin ⁡ open parenthesis 2 α close parenthesis sin ⁡ open parenthesis α close parenthesis equals cos ⁡ open parenthesis α close parenthesis")[$sin ( 2 α ) sin ( α ) = cos ( α )$] #math.equation(block: true, alt: "cos ⁡ open parenthesis 2 t close parenthesis minus 5 cos ⁡ open parenthesis t close parenthesis plus 3 equals 0")[$cos ( 2 t ) − 5 cos ( t ) + 3 = 0$] + #math.equation(block: false, alt: "2 cos squared ⁡ open parenthesis t close parenthesis minus 5 cos ⁡ open parenthesis t close parenthesis plus 2 equals 0")[$2 cos^(2) ( t ) − 5 cos ( t ) + 2 = 0$] + #math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$], #math.equation(block: false, alt: "the fraction 5 π over 3")[$display(frac(5 π, 3))$] #math.equation(block: true, alt: "cos ⁡ open parenthesis 2 x close parenthesis plus 3 sin ⁡ open parenthesis x close parenthesis equals 2")[$cos ( 2 x ) + 3 sin ( x ) = 2$] #math.equation(block: true, alt: "tan ⁡ open parenthesis 2 β close parenthesis plus 2 sin ⁡ open parenthesis β close parenthesis equals 0")[$tan ( 2 β ) + 2 sin ( β ) = 0$] + #math.equation(block: false, alt: "the fraction 2 tan ⁡ open parenthesis β close parenthesis over 1 minus tan squared ⁡ open parenthesis β close parenthesis plus 2 sin ⁡ open parenthesis β close parenthesis equals 0")[$frac(2 tan ( β ), 1 − tan^(2) ( β )) + 2 sin ( β ) = 0$] + #math.equation(block: false, alt: "0")[$0$], #math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$], #math.equation(block: false, alt: "π")[$π$] , #math.equation(block: false, alt: "the fraction 5 π over 3")[$display(frac(5 π, 3))$] #math.equation(block: true, alt: "tan ⁡ open parenthesis 2 z close parenthesis minus 2 cos ⁡ open parenthesis z close parenthesis equals 0")[$tan ( 2 z ) − 2 cos ( z ) = 0$] #math.equation(block: true, alt: "3 cos ⁡ open parenthesis ϕ close parenthesis minus sin ⁡ open parenthesis the fraction π over 2 minus ϕ close parenthesis equals the square root of 3")[$3 cos ( ϕ ) − sin ( display(frac(π, 2)) − ϕ ) = sqrt(3)$] + #math.equation(block: false, alt: "3 cos ⁡ open parenthesis ϕ close parenthesis minus cos ⁡ open parenthesis ϕ close parenthesis equals the square root of 3")[$3 cos ( ϕ ) − cos ( ϕ ) = sqrt(3)$] + #math.equation(block: false, alt: "the fraction π over 6")[$display(frac(π, 6))$], #math.equation(block: false, alt: "the fraction 11 π over 6")[$display(frac(11 π, 6))$] #math.equation(block: true, alt: "sin ⁡ open parenthesis w close parenthesis plus cos ⁡ open parenthesis the fraction π over 2 minus w close parenthesis equals 1")[$sin ( w ) + cos ( display(frac(π, 2)) − w ) = 1$] #math.equation(block: true, alt: "sin ⁡ open parenthesis 2 ϕ close parenthesis cos ⁡ open parenthesis ϕ close parenthesis plus cos ⁡ open parenthesis 2 ϕ close parenthesis sin ⁡ open parenthesis ϕ close parenthesis equals 1")[$sin ( 2 ϕ ) cos ( ϕ ) + cos ( 2 ϕ ) sin ( ϕ ) = 1$] + #math.equation(block: false, alt: "sin ⁡ open parenthesis 3 ϕ close parenthesis equals 1")[$sin ( 3 ϕ ) = 1$] + #math.equation(block: false, alt: "the fraction π over 6")[$display(frac(π, 6))$] , #math.equation(block: false, alt: "the fraction 5 π over 6")[$display(frac(5 π, 6))$] , #math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$] #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis cos ⁡ open parenthesis 3 θ close parenthesis plus sin ⁡ open parenthesis θ close parenthesis sin ⁡ open parenthesis 3 θ close parenthesis equals the fraction the square root of 2 over 2")[$cos ( θ ) cos ( 3 θ ) + sin ( θ ) sin ( 3 θ ) = display(frac(sqrt(2), 2))$] Use the sum of angles formulas for sine and cosine to derive a formula for each expression. Then use graphs to verify your formula. + #math.equation(block: false, alt: "cos ⁡ open parenthesis θ plus 90 degrees close parenthesis")[$cos ( θ + 90^(∘) )$] + #math.equation(block: false, alt: "sin ⁡ open parenthesis θ plus 90 degrees close parenthesis")[$sin ( θ + 90^(∘) )$] + #math.equation(block: false, alt: "cos ⁡ open parenthesis θ plus 90 degrees close parenthesis equals minus sin ⁡ θ")[$cos ( θ + 90^(∘) ) = − sin θ$] + #math.equation(block: false, alt: "sin ⁡ open parenthesis θ plus 90 degrees close parenthesis equals cos ⁡ θ")[$sin ( θ + 90^(∘) ) = cos θ$] Use the sum of angles formulas for sine and cosine to derive a formula for each expression. Then use graphs to verify your formula. + #math.equation(block: false, alt: "cos ⁡ open parenthesis θ plus π close parenthesis")[$cos ( θ + π )$] + #math.equation(block: false, alt: "sin ⁡ open parenthesis θ plus π close parenthesis")[$sin ( θ + π )$] Use the difference of angles formulas for sine and cosine to prove that: + #math.equation(block: false, alt: "cos ⁡ open parenthesis the fraction π over 2 minus θ close parenthesis equals sin ⁡ open parenthesis θ close parenthesis")[$cos ( display(frac(π, 2)) − θ ) = sin ( θ )$] + #math.equation(block: false, alt: "sin ⁡ open parenthesis the fraction π over 2 minus θ close parenthesis equals cos ⁡ open parenthesis θ close parenthesis")[$sin ( display(frac(π, 2)) − θ ) = cos ( θ )$] + #math.equation(block: false, alt: "cos ⁡ open parenthesis the fraction π over 2 minus θ close parenthesis equals cos ⁡ the fraction π over 2 cos ⁡ open parenthesis θ close parenthesis plus sin ⁡ the fraction π over 2 sin ⁡ θ equals sin ⁡ open parenthesis θ close parenthesis")[$cos ( display(frac(π, 2)) − θ ) = cos frac(π, 2) cos ( θ ) + sin frac(π, 2) sin θ = sin ( θ )$] + #math.equation(block: false, alt: "sin ⁡ open parenthesis the fraction π over 2 minus θ close parenthesis equals sin ⁡ the fraction π over 2 cos ⁡ open parenthesis θ close parenthesis minus cos ⁡ the fraction π over 2 sin ⁡ open parenthesis θ close parenthesis equals cos ⁡ open parenthesis θ close parenthesis")[$sin ( display(frac(π, 2)) − θ ) = sin frac(π, 2) cos ( θ ) − cos frac(π, 2) sin ( θ ) = cos ( θ )$] Use the difference of angles formulas for sine and cosine to derive formulas for: + #math.equation(block: false, alt: "cos ⁡ open parenthesis θ minus the fraction π over 2 close parenthesis")[$cos ( θ − display(frac(π, 2)) )$] + #math.equation(block: false, alt: "sin ⁡ open parenthesis θ minus the fraction π over 2 close parenthesis")[$sin ( θ − display(frac(π, 2)) )$] Prove the double angle identity #math.equation(block: false, alt: "sin ⁡ open parenthesis 2 θ close parenthesis equals 2 sin ⁡ open parenthesis θ close parenthesis cos ⁡ open parenthesis θ close parenthesis")[$sin ( 2 θ ) = 2 sin ( θ ) cos ( θ )$]. (Hint: Start with the sum of angles formula for sine and replace both #math.equation(block: false, alt: "α")[$α$] and #math.equation(block: false, alt: "β")[$β$] by #math.equation(block: false, alt: "θ")[$θ$].) #math.equation(block: true, alt: "sin ⁡ open parenthesis 2 θ close parenthesis, equals sin ⁡ open parenthesis θ plus θ close parenthesis; equals sin ⁡ open parenthesis θ close parenthesis cos ⁡ open parenthesis θ close parenthesis plus sin ⁡ open parenthesis θ close parenthesis cos ⁡ open parenthesis θ close parenthesis; equals 2 sin ⁡ open parenthesis θ close parenthesis cos ⁡ open parenthesis θ close parenthesis")[$sin ( 2 θ ) & = sin ( θ + θ ) \ & = sin ( θ ) cos ( θ ) + sin ( θ ) cos ( θ ) \ & = 2 sin ( θ ) cos ( θ )$] Prove the double angle identity #math.equation(block: false, alt: "cos ⁡ open parenthesis 2 θ close parenthesis equals cos squared ⁡ open parenthesis θ close parenthesis minus sin squared ⁡ open parenthesis θ close parenthesis")[$cos ( 2 θ ) = cos^(2) ( θ ) − sin^(2) ( θ )$]. (Hint: Start with the sum of angles formula for sine and replace both #math.equation(block: false, alt: "α")[$α$] and #math.equation(block: false, alt: "β")[$β$] by #math.equation(block: false, alt: "θ")[$θ$].) For Problems 79–88, + Use graphs to decide if the equation is an identity. + If the equation is not an identity, find a value of the variable that makes the equation false. #math.equation(block: true, alt: "sin ⁡ open parenthesis the fraction π over 2 plus β close parenthesis equals 1 plus sin ⁡ open parenthesis β close parenthesis")[$sin ( display(frac(π, 2)) + β ) = 1 + sin ( β )$] + Not an identity. + #math.equation(block: false, alt: "β equals π")[$β = π$] (many answers possible) #math.equation(block: true, alt: "cos ⁡ open parenthesis the fraction π over 3 minus β close parenthesis equals cos ⁡ open parenthesis β minus the fraction π over 3 close parenthesis")[$cos ( display(frac(π, 3)) − β ) = cos ( β − display(frac(π, 3)) )$] #math.equation(block: true, alt: "sin ⁡ open parenthesis A plus 180 degrees close parenthesis equals minus sin ⁡ open parenthesis A close parenthesis")[$sin ( A + 180^(∘) ) = − sin ( A )$] Identity #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis plus tan ⁡ open parenthesis minus θ close parenthesis equals 0")[$tan ( θ ) + tan ( − θ ) = 0$] #math.equation(block: true, alt: "cos ⁡ open parenthesis 4 θ close parenthesis equals 4 cos ⁡ open parenthesis θ close parenthesis")[$cos ( 4 θ ) = 4 cos ( θ )$] + Not an identity. + #math.equation(block: false, alt: "θ equals 0")[$θ = 0$] (many answers possible) #math.equation(block: true, alt: "cos ⁡ open parenthesis ϕ plus the fraction π over 3 close parenthesis equals the fraction 1 over 2 plus cos ⁡ open parenthesis ϕ close parenthesis")[$cos ( ϕ + display(frac(π, 3)) ) = display(frac(1, 2)) + cos ( ϕ )$] #math.equation(block: true, alt: "sin ⁡ open parenthesis x plus the fraction π over 4 close parenthesis equals the fraction the square root of 2 over 2 open parenthesis sin ⁡ open parenthesis x close parenthesis plus cos ⁡ open parenthesis x close parenthesis close parenthesis")[$sin ( x + display(frac(π, 4)) ) = display(frac(sqrt(2), 2)) ( sin ( x ) + cos ( x ) )$] Identity #math.equation(block: true, alt: "2 cos ⁡ open parenthesis x minus the fraction π over 6 close parenthesis equals sin ⁡ open parenthesis x close parenthesis plus the square root of 3 cos ⁡ open parenthesis x close parenthesis")[$2 cos ( x − display(frac(π, 6)) ) = sin ( x ) + sqrt(3) cos ( x )$] #math.equation(block: true, alt: "sin ⁡ open parenthesis x minus the fraction π over 3 close parenthesis plus cos ⁡ open parenthesis x plus the fraction π over 6 close parenthesis equals 0")[$sin ( x − display(frac(π, 3)) ) + cos ( x + display(frac(π, 6)) ) = 0$] Identity #math.equation(block: true, alt: "cos ⁡ open parenthesis 2 x close parenthesis equals open parenthesis cos ⁡ open parenthesis x close parenthesis plus sin ⁡ open parenthesis x close parenthesis close parenthesis open parenthesis cos ⁡ open parenthesis x close parenthesis minus sin ⁡ open parenthesis x close parenthesis close parenthesis")[$cos ( 2 x ) = ( cos ( x ) + sin ( x ) ) ( cos ( x ) − sin ( x ) )$] Problems 89 and 90 verify the addition and subtraction formulas for acute angles. The figure below shows a right triangle inscribed in a rectangle. #figure(figph[triangle inscribed in rectangle], alt: "triangle inscribed in rectangle", caption: none) + Label the legs #math.equation(block: false, alt: "l sub 1")[$l_(1)$] and #math.equation(block: false, alt: "l sub 2")[$l_(2)$] of the right triangle with their lengths. + Explain why #math.equation(block: false, alt: "θ sub 1 equals β")[$θ_(1) = β$] and #math.equation(block: false, alt: "θ sub 2 equals α plus β")[$θ_(2) = α + β$]. Label the diagram with these angles. + Label the legs #math.equation(block: false, alt: "s sub 1")[$s_(1)$] and #math.equation(block: false, alt: "s sub 2")[$s_(2)$] of the bottom triangle with their lengths. + Label the legs #math.equation(block: false, alt: "s sub 3")[$s_(3)$] and #math.equation(block: false, alt: "s sub 4")[$s_(4)$] of the top left triangle with their lengths. + Label the legs #math.equation(block: false, alt: "s sub 5")[$s_(5)$] and #math.equation(block: false, alt: "s sub 6")[$s_(6)$] of the top right triangle with their lengths. + Use the fact that the opposite sides of a rectangle are equal to state the addition formulas for sine and cosine. #figure(figph[triangle inscribed in rectangle], alt: "triangle inscribed in rectangle", caption: none) + #math.equation(block: false, alt: "l sub 1 equals sin ⁡ open parenthesis α close parenthesis , l sub 2 equals cos ⁡ open parenthesis α close parenthesis")[$l_(1) = sin ( α ) , #h(0.167em) l_(2) = cos ( α )$] + #math.equation(block: false, alt: "θ sub 1")[$θ_(1)$] and #math.equation(block: false, alt: "β")[$β$] are both complements of #math.equation(block: false, alt: "ϕ")[$ϕ$]; #math.equation(block: false, alt: "θ sub 2")[$θ_(2)$] and #math.equation(block: false, alt: "α plus β")[$α + β$] are alternate interior angles + #math.equation(block: false, alt: "s sub 1 equals cos ⁡ open parenthesis α plus β close parenthesis")[$s_(1) = cos ( α + β )$], #math.equation(block: false, alt: "s sub 2 equals sin ⁡ open parenthesis α plus β close parenthesis")[$s_(2) = sin ( α + β )$] + #math.equation(block: false, alt: "s sub 3 equals sin ⁡ open parenthesis α close parenthesis sin ⁡ open parenthesis β close parenthesis")[$s_(3) = sin ( α ) sin ( β )$], #math.equation(block: false, alt: "s sub 4 equals sin ⁡ open parenthesis α close parenthesis cos ⁡ open parenthesis β close parenthesis")[$s_(4) = sin ( α ) cos ( β )$] + #math.equation(block: false, alt: "s sub 5 equals cos ⁡ open parenthesis α close parenthesis cos ⁡ open parenthesis β close parenthesis")[$s_(5) = cos ( α ) cos ( β )$], #math.equation(block: false, alt: "s sub 6 equals cos ⁡ open parenthesis α close parenthesis sin ⁡ open parenthesis β close parenthesis")[$s_(6) = cos ( α ) sin ( β )$] + #math.equation(block: false, alt: "sin ⁡ open parenthesis α plus β close parenthesis equals sin ⁡ open parenthesis α close parenthesis cos ⁡ open parenthesis β close parenthesis plus cos ⁡ open parenthesis α close parenthesis sin ⁡ open parenthesis β close parenthesis")[$sin ( α + β ) = sin ( α ) cos ( β ) + cos ( α ) sin ( β )$], #math.equation(block: false, alt: "cos ⁡ open parenthesis α plus β close parenthesis equals cos ⁡ open parenthesis α close parenthesis cos ⁡ open parenthesis β close parenthesis plus sin ⁡ open parenthesis α close parenthesis sin ⁡ open parenthesis β close parenthesis")[$cos ( α + β ) = cos ( α ) cos ( β ) + sin ( α ) sin ( β )$] The figure below shows a right triangle inscribed in a rectangle. #figure(figph[triangle inscribed in rectangle], alt: "triangle inscribed in rectangle", caption: none) + Label the legs #math.equation(block: false, alt: "l sub 1")[$l_(1)$] and #math.equation(block: false, alt: "l sub 2")[$l_(2)$] of the right triangle with their lengths. + Explain why #math.equation(block: false, alt: "θ sub 1 equals β")[$θ_(1) = β$] and #math.equation(block: false, alt: "θ sub 2 equals α minus β")[$θ_(2) = α − β$]. Label the diagram with these angles. + Label the legs #math.equation(block: false, alt: "s sub 1")[$s_(1)$] and #math.equation(block: false, alt: "s sub 2")[$s_(2)$] of the bottom triangle with their lengths. + Label the legs #math.equation(block: false, alt: "s sub 3")[$s_(3)$] and #math.equation(block: false, alt: "s sub 4")[$s_(4)$] of the top left triangle with their lengths. + Label the legs #math.equation(block: false, alt: "s sub 5")[$s_(5)$] and #math.equation(block: false, alt: "s sub 6")[$s_(6)$] of the top right triangle with their lengths. + Use the fact that the opposite sides of a rectangle are equal to state the subtraction formulas for sine and cosine. Follow the steps to prove the difference of angles formula for cosine, #math.equation(block: true, alt: "cos ⁡ open parenthesis α minus β close parenthesis equals cos ⁡ open parenthesis α close parenthesis cos ⁡ open parenthesis β close parenthesis plus sin ⁡ open parenthesis α close parenthesis sin ⁡ open parenthesis β close parenthesis")[$cos ( α − β ) = cos ( α ) cos ( β ) + sin ( α ) sin ( β )$] #figure(figph[circle], alt: "circle", caption: none) + Write an expression for #math.equation(block: false, alt: "open parenthesis A B close parenthesis squared")[$( A B )^(2)$], the square of the distance between the points #math.equation(block: false, alt: "A")[$A$] and #math.equation(block: false, alt: "B")[$B$], using the Law of Cosines for #math.equation(block: false, alt: "△ A O B")[$upright(△) A O B$]. + Write another expression for #math.equation(block: false, alt: "open parenthesis A B close parenthesis squared")[$( A B )^(2)$] using the distance formula and the coordinates of #math.equation(block: false, alt: "A")[$A$] and #math.equation(block: false, alt: "B")[$B$]. + Equate the two expressions for #math.equation(block: false, alt: "open parenthesis A B close parenthesis squared")[$( A B )^(2)$] you obtained in (a) and (b). Simplify the equation to obtain #math.equation(block: false, alt: "cos ⁡ open parenthesis α minus β close parenthesis equals cos ⁡ open parenthesis α close parenthesis cos ⁡ open parenthesis β close parenthesis plus sin ⁡ open parenthesis α close parenthesis sin ⁡ open parenthesis β close parenthesis")[$cos ( α − β ) = cos ( α ) cos ( β ) + sin ( α ) sin ( β )$]. + #math.equation(block: false, alt: "open parenthesis A B close parenthesis squared equals 2 minus 2 cos ⁡ open parenthesis α minus β close parenthesis")[$( A B )^(2) = 2 − 2 cos ( α − β )$] + #math.equation(block: false, alt: "open parenthesis A B close parenthesis squared equals open parenthesis cos ⁡ open parenthesis α close parenthesis minus cos ⁡ open parenthesis β close parenthesis close parenthesis squared plus open parenthesis sin ⁡ open parenthesis α close parenthesis minus sin ⁡ open parenthesis β close parenthesis close parenthesis squared")[$( A B )^(2) = ( cos ( α ) − cos ( β ) )^(2) + ( sin ( α ) − sin ( β ) )^(2)$] + #math.equation(block: false, alt: "2 minus 2 cos ⁡ open parenthesis α minus β close parenthesis, equals open parenthesis cos ⁡ open parenthesis α close parenthesis minus cos ⁡ open parenthesis β close parenthesis close parenthesis squared plus open parenthesis sin ⁡ open parenthesis α close parenthesis minus sin ⁡ open parenthesis β close parenthesis close parenthesis squared; 2 minus 2 cos ⁡ open parenthesis α minus β close parenthesis, equals cos squared ⁡ open parenthesis α close parenthesis minus 2 cos ⁡ open parenthesis α close parenthesis cos ⁡ open parenthesis β close parenthesis plus cos squared ⁡ open parenthesis β close parenthesis plus; plus sin squared ⁡ open parenthesis α close parenthesis minus 2 sin ⁡ open parenthesis α close parenthesis sin ⁡ open parenthesis β close parenthesis plus sin squared ⁡ open paren")[$2 − 2 cos \( α − β \) & = \( cos \( α \) − cos \( β \) \)^(2) + \( sin \( α \) − sin \( β \) \)^(2) \ 2 − 2 cos \( α − β \) & = cos^(2) \( α \) − 2 cos \( α \) cos \( β \) + cos^(2) \( β \) + #h(0.167em) \ & #hide($000000000$) + sin^(2) \( α \) − 2 sin \( α \) sin \( β \) + sin^(2) \( β \) \ 2 − 2 cos \( α − β \) & = 1 + 1 − 2 \( cos \( α \) cos \( β \) − sin \( α \) sin \( β \) \) \ − 2 cos \( α − β \) & = − 2 \( cos \( α \) cos \( β \) − sin \( α \) sin \( β \) \) \ cos \( α − β \) & = cos \( α \) cos \( β \) − sin \( α \) sin \( β \) \)$]