#set document(title: "5.3 Trigonometric Identities", 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")) == 5.3#h(0.6em)Trigonometric Identities === What is an Identity? Recall that an equation may be true or false, depending on the values of any variables involved. For example, the equation #math.equation(block: true, alt: "x squared plus 3 x equals 10")[$x^(2) + 3 x = 10$] is true only if #math.equation(block: false, alt: "x equals 2")[$x = 2$] or #math.equation(block: false, alt: "x equals 5")[$x = 5$]. An equation that is true only for certain values of the variable, and false for others, is called a #strong[conditional equation]. When you solve a conditional equation, you are finding the values of the variable that make the equation true. Some equations are true for all legitimate values of the variables. Such equations are called #strong[identities]. Here are some examples of identities. #math.equation(block: true, alt: "3 open parenthesis x plus y close parenthesis, equals 3 x plus 3 y; open parenthesis x plus 1 close parenthesis squared, equals x squared plus 2 x plus 1")[$3 ( x + y ) & = 3 x + 3 y \ ( x + 1 )^(2) & = x^(2) + 2 x + 1$] In an identity, the expressions on either side of the equal sign are #emph[equivalent expressions], because they have the same value for all values of the variable. When we simplify an algebraic expression, we are in fact creating an identity. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Identity] An #strong[identity] is an equation that is true for all legitimate values of the variables. ] #examplebox("Example 1")[][ Which of the following equations are identities? + #math.equation(block: false, alt: "3 s plus 7 s equals 10 s")[$3 s + 7 s = 10 s$] + #math.equation(block: false, alt: "5 c open parenthesis c minus 2 s close parenthesis equals 5 c squared minus 10 c s")[$5 c ( c − 2 s ) = 5 c^(2) − 10 c s$] + #math.equation(block: false, alt: "2 t minus 1 equals 3")[$2 t − 1 = 3$] #solutionbox[ Many of the algebraic operations you have already learned, such as combining like terms or applying the distributive law, produce equivalent expressions. + Equation (a) is an identity obtained by combining like terms on the left side. + Equation (b) is an identity obtained by applying the distributive law on the left side. + Equation (c) is not an identity, because the equation is true only for #math.equation(block: false, alt: "t equals 2")[$t = 2$]. ] ] Which of the following equations are identities? + #math.equation(block: false, alt: "open parenthesis c minus s close parenthesis open parenthesis c plus s close parenthesis equals c squared minus s squared")[$( c − s ) ( c + s ) = c^(2) − s^(2)$] + #math.equation(block: false, alt: "3 t squared equals 1")[$3 t^(2) = 1$] + #math.equation(block: false, alt: "open parenthesis 2 c plus 1 close parenthesis plus open parenthesis s minus 3 close parenthesis equals 2 c s minus 6 c plus s minus 3")[$( 2 c + 1 ) + ( s − 3 ) = 2 c s − 6 c + s − 3$] (a) and (c) are identities. Of course, you wouldn't be asked to #emph[solve] an identity, because all values of the variable are solutions. Instead, we use identities to replace one form of an expression by a more useful form. You do this when you solve a quadratic equation by factoring. For example, #math.equation(block: true, alt: "2 x squared minus x minus 1, equals 0, Factor the left side.; open parenthesis 2 x plus 1 close parenthesis open parenthesis x minus 1 close parenthesis, equals 0")[$2 x^(2) − x − 1 & = 0 & & "Factor the left side." \ ( 2 x + 1 ) ( x − 1 ) & = 0$] Because #math.equation(block: false, alt: "open parenthesis 2 x plus 1 close parenthesis open parenthesis x minus 1 close parenthesis")[$( 2 x + 1 ) ( x − 1 )$] is equivalent to #math.equation(block: false, alt: "2 x squared minus x minus 1")[$2 x^(2) − x − 1$], we have not changed the equation or its solutions. But now we can apply the Zero Factor principle and solve the equation. Using identities is especially useful when we are working with trigonometric ratios. === Using Trigonometric Ratios in Identities Because the identity #math.equation(block: true, alt: "2 x squared minus x minus 1 equals open parenthesis 2 x plus 1 close parenthesis open parenthesis x minus 1 close parenthesis")[$2 x^(2) − x − 1 = ( 2 x + 1 ) ( x − 1 )$] is true for any value of #math.equation(block: false, alt: "x")[$x$], it is true when #math.equation(block: false, alt: "x")[$x$] is replaced, for instance, by #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$cos ( θ )$]. This gives us a new identity #math.equation(block: true, alt: "2 cos squared ⁡ open parenthesis θ close parenthesis minus cos ⁡ open parenthesis θ close parenthesis minus 1 equals open parenthesis 2 cos ⁡ open parenthesis θ close parenthesis plus 1 close parenthesis open parenthesis cos ⁡ open parenthesis θ close parenthesis minus 1 close parenthesis")[$2 cos^(2) ( θ ) − cos ( θ ) − 1 = ( 2 cos ( θ ) + 1 ) ( cos ( θ ) − 1 )$] Expressions involving #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis , cos ⁡ open parenthesis θ close parenthesis , or tan ⁡ open parenthesis θ close parenthesis")[$sin ( θ ) , cos ( θ ) , "or" tan ( θ )$] can be manipulated by the same rules (such as the distributive law or the laws of exponents) that we use with simple variables. #examplebox("Example 2")[][ Which of the following equations are identities? + #math.equation(block: false, alt: "3 sin ⁡ open parenthesis θ close parenthesis plus 7 sin ⁡ open parenthesis θ close parenthesis equals 10 sin ⁡ open parenthesis θ close parenthesis")[$3 sin ( θ ) + 7 sin ( θ ) = 10 sin ( θ )$] + #math.equation(block: false, alt: "5 cos ⁡ open parenthesis θ close parenthesis open parenthesis cos ⁡ open parenthesis θ close parenthesis minus 2 sin ⁡ open parenthesis θ close parenthesis close parenthesis equals 5 open parenthesis cos ⁡ open parenthesis θ close parenthesis close parenthesis squared minus 10 cos ⁡ open parenthesis θ close parenthesis sin ⁡ open parenthesis θ close parenthesis")[$5 cos ( θ ) ( cos ( θ ) − 2 sin ( θ ) ) = 5 ( cos ( θ ) )^(2) − 10 cos ( θ ) sin ( θ )$] + #math.equation(block: false, alt: "2 tan ⁡ open parenthesis θ close parenthesis minus 1 equals 3")[$2 tan ( θ ) − 1 = 3$] #solutionbox[ These equations are obtained from the equations in the previous Example by replacing the variable #math.equation(block: false, alt: "s")[$s$] with #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$], the variable #math.equation(block: false, alt: "c")[$c$] with #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$cos ( θ )$], and the variable #math.equation(block: false, alt: "t")[$t$] with #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$]. Equations (a) and (b) are identities, for the same reasons as before, and equation (c) is not. It is false, for example, when #math.equation(block: false, alt: "θ equals 0")[$θ = 0$]. ] ] Which of the following equations are identities? + #math.equation(block: false, alt: "open parenthesis cos ⁡ open parenthesis θ close parenthesis minus sin ⁡ open parenthesis θ close parenthesis close parenthesis open parenthesis cos ⁡ open parenthesis θ close parenthesis plus sin ⁡ open parenthesis θ close parenthesis close parenthesis equals cos squared ⁡ open parenthesis θ close parenthesis minus sin squared ⁡ open parenthesis θ close parenthesis")[$( cos ( θ ) − sin ( θ ) ) ( cos ( θ ) + sin ( θ ) ) = cos^(2) ( θ ) − sin^(2) ( θ )$] + #math.equation(block: false, alt: "3 tan squared ⁡ open parenthesis θ close parenthesis equals 1")[$3 tan^(2) ( θ ) = 1$] + #math.equation(block: false, alt: "open parenthesis 2 cos ⁡ open parenthesis θ close parenthesis plus 1 close parenthesis plus open parenthesis sin ⁡ open parenthesis θ close parenthesis minus 3 close parenthesis equals 2 cos ⁡ open parenthesis θ close parenthesis sin ⁡ open parenthesis θ close parenthesis minus 6 cos ⁡ open parenthesis θ close parenthesis plus sin ⁡ open parenthesis θ close parenthesis minus 3")[$( 2 cos ( θ ) + 1 ) + ( sin ( θ ) − 3 ) = 2 cos ( θ ) sin ( θ ) − 6 cos ( θ ) + sin ( θ ) − 3$] (a) and (c) are identities. #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ To show that an equation is #emph[not] an identity, we need only find one value of the variable for which the equation is false. However, to show that a given equation #emph[is] an identity, it is not enough to show that the equation is true for one or even a dozen values of the variable. The equation must be true for #emph[all] legitimate values of the variable. Because we cannot check all values of the variable numerically, we must use algebraic methods to prove that an equation is an identity. ] === Checking Identities Graphically Is the equation #math.equation(block: false, alt: "the square root of x squared equals x")[$sqrt(x^(2)) = x$] an identity? The answer is no! Even though the equation is true for all #emph[positive] values of #math.equation(block: false, alt: "x")[$x$], it is false for negative values of #math.equation(block: false, alt: "x")[$x$]. For example, if #math.equation(block: false, alt: "x equals minus 3")[$x = − 3$], then #math.equation(block: true, alt: "the square root of x squared equals the square root of open parenthesis minus 3 close parenthesis squared equals the square root of 9 equals 3")[$sqrt(x^(2)) = sqrt(( − 3 )^(2)) = sqrt(9) = 3$] so #math.equation(block: false, alt: "the square root of x squared not equal to x")[$sqrt(x^(2)) ≠ x$]. Because the radical symbol #math.equation(block: false, alt: "the square root of")[$sqrt(zws)$] stands for the #emph[nonnegative] square root, the left side of the equation, #math.equation(block: false, alt: "the square root of x squared")[$sqrt(x^(2))$], is never negative. Therefore, #math.equation(block: false, alt: "the square root of x squared")[$sqrt(x^(2))$] cannot equal #math.equation(block: false, alt: "x")[$x$] when #math.equation(block: false, alt: "x")[$x$] is a negative number. The equation is false for #math.equation(block: false, alt: "x less than 0")[$x < 0$]. One way to see that #math.equation(block: false, alt: "the square root of x squared")[$sqrt(x^(2))$] and #math.equation(block: false, alt: "x")[$x$] are not equivalent is to compare the graphs of #math.equation(block: false, alt: "Y sub 1 equals the square root of x squared")[$Y_(1) = sqrt(x^(2))$] and #math.equation(block: false, alt: "Y sub 2 equals x")[$Y_(2) = x$], shown below. You can see that #math.equation(block: false, alt: "the square root of x squared")[$sqrt(x^(2))$] and #math.equation(block: false, alt: "x")[$x$] do not have the same value for #math.equation(block: false, alt: "x less than 0")[$x < 0$]. #figure(figph[two graphs], alt: "two graphs", caption: none) Thus, to check to whether an equation is an identity, we can compare graphs of #math.equation(block: false, alt: "Y sub 1 equals")[$Y_(1) =$] (left side of the equation) and #math.equation(block: false, alt: "Y sub 2 equals")[$Y_(2) =$] (right side of the equation). If the two graphs are identical, the equation is an identity. If the two graphs are not the same, the equation is not an identity. #examplebox("Example 3")[][ Which of the following equations are identities? + #math.equation(block: false, alt: "sin ⁡ open parenthesis 2 α close parenthesis equals 2 sin ⁡ open parenthesis α close parenthesis")[$sin ( 2 α ) = 2 sin ( α )$] + #math.equation(block: false, alt: "cos ⁡ open parenthesis x plus 1 close parenthesis equals cos ⁡ open parenthesis x close parenthesis")[$cos ( x + 1 ) = cos ( x )$] #solutionbox[ + Compare the graphs of #math.equation(block: false, alt: "Y sub 1 equals sin ⁡ open parenthesis 2 x close parenthesis")[$Y_(1) = sin ( 2 x )$] and #math.equation(block: false, alt: "Y sub 2 equals 2 sin ⁡ open parenthesis x close parenthesis")[$Y_(2) = 2 sin ( x )$]. Enter the two equations in the ZTrig window and press ZOOM #math.equation(block: false, alt: "7")[$7$] to see the graphs shown below.#figure(figph[graph], alt: "graph", caption: none) #linebreak() To help distinguish which graph is which, we can use the calculator's trace feature, illustrated in the figure at right. Because there are two distinct graphs, the expressions #math.equation(block: false, alt: "sin ⁡ open parenthesis 2 x close parenthesis")[$sin ( 2 x )$] and #math.equation(block: false, alt: "2 sin ⁡ open parenthesis x close parenthesis")[$2 sin ( x )$] are not equivalent, and consequently #math.equation(block: false, alt: "sin ⁡ open parenthesis 2 α close parenthesis equals 2 sin ⁡ open parenthesis α close parenthesis")[$sin ( 2 α ) = 2 sin ( α )$] is not an identity. + This time we graph #math.equation(block: false, alt: "Y sub 1 equals cos ⁡ open parenthesis x plus 1 close parenthesis")[$Y_(1) = cos ( x + 1 )$] and #math.equation(block: false, alt: "Y sub 2 equals cos ⁡ open parenthesis x close parenthesis")[$Y_(2) = cos ( x )$]. Although the graphs appear to be identical, when we use the trace feature we see that their #math.equation(block: false, alt: "y")[$y$]-values differ at #math.equation(block: false, alt: "x equals 0")[$x = 0$] (and at many other values of #math.equation(block: false, alt: "x")[$x$].) See the figures below.#figure(figph[graph], alt: "graph", caption: none) #linebreak() The graphs are so close together that calculator's resolution does not distinguish them, but tracing the graphs reveals that they are not identical. Because the two graphs differ, the equation #math.equation(block: false, alt: "cos ⁡ open parenthesis x plus 1 close parenthesis equals cos ⁡ open parenthesis x close parenthesis")[$cos ( x + 1 ) = cos ( x )$] is not an identity. ] ] #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ - The Example above illustrates the fact that graphs can be deceiving: even if two graphs look identical, it is always a good idea to check some numerical values as well. - Remember that we can use graphs to prove that an equation is #emph[not] an identity, if the two graphs are clearly different, but to prove that an equation is an identity, we will need algebraic methods. ] Use graphs to decide which of the following equations are identities. + #math.equation(block: false, alt: "cos ⁡ open parenthesis 2 θ close parenthesis equals 2 cos ⁡ open parenthesis θ close parenthesis")[$cos ( 2 θ ) = 2 cos ( θ )$] + #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")[$cos ( 2 θ ) = cos^(2) ( θ ) − sin^(2) ( θ )$] + #math.equation(block: false, alt: "cos ⁡ open parenthesis θ squared close parenthesis equals cos squared ⁡ open parenthesis θ close parenthesis")[$cos ( θ^(2) ) = cos^(2) ( θ )$] (b) is an identity; (a) and (c) are not identities. === Pythagorean Identity All of the trigonometric functions are related. This is plausible when you think about it, because the three trig ratios all involve the three sides of a right triangle. Taking advantage of these relationships will simplify many calculations involving the trig functions. We'll begin by considering the relationship between #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$] and #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$cos ( θ )$]. Complete the following table with exact values. #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$cos ( θ )$]], [#math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$]], [#math.equation(block: false, alt: "cos squared ⁡ open parenthesis θ close parenthesis")[$cos^(2) ( θ )$]], [#math.equation(block: false, alt: "sin squared ⁡ open parenthesis θ close parenthesis")[$sin^(2) ( θ )$]], [#math.equation(block: false, alt: "cos squared ⁡ open parenthesis θ close parenthesis plus sin squared ⁡ open parenthesis θ close parenthesis")[$cos^(2) ( θ ) + sin^(2) ( θ )$]]), [#math.equation(block: false, alt: "0 degrees")[$0^(∘)$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "30 degrees")[$30^(∘)$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "45 degrees")[$45^(∘)$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "60 degrees")[$60^(∘)$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "90 degrees")[$90^(∘)$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) You should find that all the entries in the last column are 1. For all of the angles in the table, #math.equation(block: true, alt: "cos squared ⁡ open parenthesis θ close parenthesis plus sin squared ⁡ open parenthesis θ close parenthesis equals 1")[$cos^(2) ( θ ) + sin^(2) ( θ ) = 1$] We can verify that this equation holds for all angles by graphing the expressions on either side of the equal sign. The graphs of these two functions in the ZTrig window are shown at right. #figure(figph[graph], alt: "graph", caption: none) We see that the graph of #math.equation(block: false, alt: "Y sub 1 equals cos squared ⁡ open parenthesis θ close parenthesis plus sin squared ⁡ open parenthesis θ close parenthesis")[$Y_(1) = cos^(2) ( θ ) + sin^(2) ( θ )$] appears identical to the horizontal line #math.equation(block: false, alt: "Y sub 2 equals 1")[$Y_(2) = 1$]. In fact, the graphs are identical, and the equation #math.equation(block: false, alt: "cos squared ⁡ open parenthesis θ close parenthesis plus sin squared ⁡ open parenthesis θ close parenthesis equals 1")[$cos^(2) ( θ ) + sin^(2) ( θ ) = 1$] is an identity. It is important enough to earn a special name. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Pythagorean Identity] For any angle #math.equation(block: false, alt: "θ")[$θ$], #math.equation(block: true, alt: "cos squared ⁡ open parenthesis θ close parenthesis plus sin squared ⁡ open parenthesis θ close parenthesis equals 1")[$bold(cos)^(2) ( bold(italic(θ)) ) + bold(sin)^(2) ( bold(italic(θ)) ) = 1$] ] As you might guess from its name, the Pythagorean identity is true because it is related to the Pythagorean theorem. We have not actually proved the identity, and a skeptical student may wonder if #math.equation(block: false, alt: "cos squared ⁡ open parenthesis θ close parenthesis plus sin squared ⁡ open parenthesis θ close parenthesis")[$" " " " cos^(2) ( θ ) + sin^(2) ( θ ) " " " "$] is only very close to #math.equation(block: false, alt: "1")[$1$], or if it equals #math.equation(block: false, alt: "1")[$1$] for only some values of #math.equation(block: false, alt: "θ")[$θ$]. Homework Problem 77 offers a proof of the Pythagorean identity. #examplebox("Example 4")[][ Evaluate #math.equation(block: false, alt: "cos squared ⁡ open parenthesis 27 degrees close parenthesis plus sin squared ⁡ open parenthesis 27 degrees close parenthesis")[$" " " " cos^(2) ( 27^(∘) ) + sin^(2) ( 27^(∘) ) " " " "$] without using a calculator. #solutionbox[ Because the identity #math.equation(block: false, alt: "cos squared ⁡ open parenthesis θ close parenthesis plus sin squared ⁡ open parenthesis θ close parenthesis equals 1")[$" " cos^(2) ( θ ) + sin^(2) ( θ ) = 1 " "$] is true for any value of #math.equation(block: false, alt: "θ")[$θ$], it is true in particular for #math.equation(block: false, alt: "θ equals 27 degrees")[$θ = 27^(∘)$]. So #math.equation(block: false, alt: "cos squared ⁡ open parenthesis 27 degrees close parenthesis plus sin squared ⁡ open parenthesis 27 degrees close parenthesis equals 1")[$" " cos^(2) ( 27^(∘) ) + sin^(2) ( 27^(∘) ) = 1$]. ] ] Evaluate #math.equation(block: false, alt: "5 open parenthesis cos squared ⁡ open parenthesis 12 degrees close parenthesis plus sin squared ⁡ open parenthesis 12 degrees close parenthesis close parenthesis")[$" " " " 5 ( cos^(2) ( 12^(∘) ) + sin^(2) ( 12^(∘) ) ) " " " "$] without using a calculator. #math.equation(block: true, alt: "5")[$5$] When we solve more complicated trigonometric equations in later chapters, we will need to simplify trigonometric expressions so that they involve only one of the trig functions. The Pythagorean identity is useful when we wish to write an equivalent expression for either #math.equation(block: false, alt: "cos squared ⁡ open parenthesis θ close parenthesis")[$cos^(2) ( θ )$] or for #math.equation(block: false, alt: "sin squared ⁡ open parenthesis θ close parenthesis")[$sin^(2) ( θ )$]. Note that we can write the identity in two alternate forms: {"expression":"sin(t)\*cos(t)^2"}sin(t) - sin(t)^3The example that follows uses the Pythagorean identity to rewrite sinθ·cos²θ as sinθ − sin³θ. At build time a computer algebra system checks that the two expressions are genuinely equivalent — an identity, not just equal at the points a graph shows — and a rebuild in which the claim fails stops the build. That is the section's own test for an identity, run by machine on every rebuild.#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Alternate Forms of the Pythagorean Identity] #math.equation(block: true, alt: "cos squared ⁡ open parenthesis θ close parenthesis, equals 1 minus sin squared ⁡ open parenthesis θ close parenthesis; sin squared ⁡ open parenthesis θ close parenthesis, equals 1 minus cos squared ⁡ open parenthesis θ close parenthesis")[$bold(cos)^(2) ( bold(italic(θ)) ) & = 1 − bold(sin)^(2) ( bold(italic(θ)) ) \ bold(sin)^(2) ( bold(italic(θ)) ) & = 1 − bold(cos)^(2) ( bold(italic(θ)) )$] ] #examplebox("Example 5")[][ Rewrite #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis cos squared ⁡ open parenthesis θ close parenthesis")[$" " sin ( θ ) cos^(2) ( θ ) " "$] as an expression involving #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$] only. #solutionbox[ Using one of the alternate forms of the Pythagorean identity, we replace #math.equation(block: false, alt: "cos squared ⁡ open parenthesis θ close parenthesis")[$cos^(2) ( θ )$] by #math.equation(block: false, alt: "1 minus sin squared ⁡ open parenthesis θ close parenthesis")[$1 − sin^(2) ( θ )$] to get #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis cos squared ⁡ open parenthesis θ close parenthesis, equals sin ⁡ open parenthesis θ close parenthesis open parenthesis 1 minus sin squared ⁡ open parenthesis θ close parenthesis close parenthesis, Apply the distributive law.; equals sin ⁡ open parenthesis θ close parenthesis minus sin cubed ⁡ open parenthesis θ close parenthesis")[$sin ( θ ) bold(cos)^(2) ( bold(italic(θ)) ) & = sin ( θ ) ( 1 − bold(sin)^(2) ( bold(italic(θ)) ) ) & & "Apply the distributive law." \ & = sin ( θ ) − sin^(3) ( θ )$] ] ] In the previous Example, we have shown that #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis cos squared ⁡ open parenthesis θ close parenthesis equals sin ⁡ open parenthesis θ close parenthesis minus sin cubed ⁡ open parenthesis θ close parenthesis")[$sin ( θ ) cos^(2) ( θ ) = sin ( θ ) − sin^(3) ( θ )$] is an identity. If we graph #math.equation(block: true, alt: "Y sub 1, equals sin ⁡ open parenthesis X close parenthesis cos squared ⁡ open parenthesis X close parenthesis; Y sub 2, equals sin ⁡ open parenthesis X close parenthesis minus sin cubed ⁡ open parenthesis X close parenthesis")[$Y_(1) & = sin ( X ) cos^(2) ( X ) \ Y_(2) & = sin ( X ) − sin^(3) ( X )$] #figure(figph[graph], alt: "graph", caption: none) for #math.equation(block: false, alt: "minus 360 degrees less than or equal to X less than or equal to 360 degrees ,")[$" " − 360^(∘) ≤ X ≤ 360^(∘) , " "$] we see that they have the same graph, as shown in the figure. + Rewrite #math.equation(block: false, alt: "sin squared ⁡ open parenthesis α close parenthesis cos squared ⁡ open parenthesis α close parenthesis")[$sin^(2) ( α ) cos^(2) ( α )$] as an expression in #math.equation(block: false, alt: "cos ⁡ open parenthesis α close parenthesis")[$cos ( α )$]. + Verify your identity by graphing. Replace #math.equation(block: false, alt: "sin squared ⁡ open parenthesis α close parenthesis")[$sin^(2) ( α )$] by #math.equation(block: false, alt: "1 minus cos squared ⁡ open parenthesis α close parenthesis")[$" " 1 − cos^(2) ( α ) " "$] to get #math.equation(block: false, alt: "open parenthesis 1 minus cos squared ⁡ open parenthesis α close parenthesis close parenthesis cos squared ⁡ open parenthesis α close parenthesis")[$" " ( 1 − cos^(2) ( α ) ) cos^(2) ( α ) " "$] and apply the distributive law to get #math.equation(block: false, alt: "cos squared ⁡ open parenthesis α close parenthesis minus cos to the power 4 ⁡ open parenthesis α close parenthesis")[$" " cos^(2) ( α ) − cos^(4) ( α )$]. === Tangent Identity There is also a relationship between the tangent ratio and the sine and cosine. Complete the following table with exact values. You should find that the entries in the last two columns are identical. #figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$" " θ$]], [#math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$cos ( θ )$]], [#math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$]], [#math.equation(block: false, alt: "the fraction sin ⁡ open parenthesis θ close parenthesis over cos ⁡ open parenthesis θ close parenthesis")[$display(frac(sin ( θ ), cos ( θ )))$]], [#math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$]]), [#math.equation(block: false, alt: "0 degrees")[$0^(∘)$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "30 degrees")[$30^(∘)$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "45 degrees")[$45^(∘)$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "60 degrees")[$60^(∘)$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) For the angles in the table, #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction sin ⁡ open parenthesis θ close parenthesis over cos ⁡ open parenthesis θ close parenthesis")[$tan ( θ ) = display(frac(sin ( θ ), cos ( θ )))$] Because #math.equation(block: false, alt: "cos ⁡ open parenthesis 90 degrees close parenthesis equals 0")[$cos ( 90^(∘) ) = 0$], the quotient #math.equation(block: false, alt: "the fraction sin ⁡ open parenthesis θ close parenthesis over cos ⁡ open parenthesis θ close parenthesis")[$display(frac(sin ( θ ), cos ( θ )))$] is undefined for #math.equation(block: false, alt: "θ equals 90 degrees")[$θ = 90^(∘)$], and #math.equation(block: false, alt: "tan ⁡ 90 degrees")[$tan 90^(∘)$] is undefined, too. The same is true for #math.equation(block: false, alt: "θ equals 270 degrees")[$θ = 270^(∘)$]. For all other angles between #math.equation(block: false, alt: "0 degrees")[$0^(∘)$] and #math.equation(block: false, alt: "360 degrees")[$360^(∘)$], #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction sin ⁡ open parenthesis θ close parenthesis over cos ⁡ open parenthesis θ close parenthesis")[$tan ( θ ) = display(frac(sin ( θ ), cos ( θ )))$]. A proof of the tangent identity is outlined in Homework Problem 78. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Tangent Identity] For any angle not coterminal with #math.equation(block: false, alt: "90 degrees")[$90^(∘)$] or #math.equation(block: false, alt: "270 degrees")[$270^(∘)$], #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction sin ⁡ open parenthesis θ close parenthesis over cos ⁡ open parenthesis θ close parenthesis")[$bold(tan) ( bold(italic(θ)) ) = display(frac(bold(sin) ( bold(italic(θ)) ), bold(cos) ( bold(italic(θ)) )))$] ] Now we can see how to use identities to simplify trigonometric expressions. One strategy for simplifying a trigonometric expression is to reduce the number of different trig ratios involved. We can use the tangent identity to replace the tangent ratio by sines and cosines. {"expression":"cos(t)\*tan(t) + sin(t)"}2\*sin(t)The strategy in action: replacing tanθ by sinθ/cosθ collapses cosθ·tanθ + sinθ to 2sinθ, the simplification the example derives. The CAS certifies the equivalence at build time (wherever tanθ is defined — the identity's own proviso), so the printed simplification carries a machine check that survives every rebuild.#examplebox("Example 6")[][ Simplify the expression #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis tan ⁡ open parenthesis θ close parenthesis plus sin ⁡ open parenthesis θ close parenthesis")[$" " " " cos ( θ ) tan ( θ ) + sin ( θ )$]. #solutionbox[ We apply the tangent identity to replace #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$" " tan ( θ ) " "$] by #math.equation(block: false, alt: "the fraction sin ⁡ open parenthesis θ close parenthesis over cos ⁡ open parenthesis θ close parenthesis")[$" " display(frac(sin ( θ ), cos ( θ ))) " "$] and obtain #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis tan ⁡ open parenthesis θ close parenthesis plus sin ⁡ open parenthesis θ close parenthesis, equals cos ⁡ open parenthesis θ close parenthesis open parenthesis the fraction sin ⁡ open parenthesis θ close parenthesis over cos ⁡ open parenthesis θ close parenthesis close parenthesis plus sin ⁡ open parenthesis θ close parenthesis; equals sin ⁡ open parenthesis θ close parenthesis plus sin ⁡ open parenthesis θ close parenthesis equals 2 sin ⁡ open parenthesis θ close parenthesis")[$cos ( θ ) bold(tan) ( bold(italic(θ)) ) + sin ( θ ) & = cos ( θ ) ( display(frac(bold(sin) ( bold(italic(θ)) ), bold(cos) ( bold(italic(θ)) ))) ) + sin ( θ ) \ & = sin ( θ ) + sin ( θ ) = 2 sin ( θ )$] ] ] Simplify the expression #math.equation(block: false, alt: "open parenthesis 1 plus tan squared ⁡ open parenthesis x close parenthesis close parenthesis open parenthesis 1 minus cos squared ⁡ open parenthesis x close parenthesis close parenthesis")[$" " " " ( 1 + tan^(2) ( x ) ) ( 1 − cos^(2) ( x ) )$]. Multiply the binomials to obtain #math.equation(block: true, alt: "1 minus cos squared ⁡ open parenthesis x close parenthesis plus tan squared ⁡ open parenthesis x close parenthesis minus tan squared ⁡ open parenthesis x close parenthesis cos squared ⁡ open parenthesis x close parenthesis")[$1 − cos^(2) ( x ) + tan^(2) ( x ) − tan^(2) ( x ) cos^(2) ( x )$] Because #math.equation(block: false, alt: "tan ⁡ open parenthesis x close parenthesis equals the fraction sin ⁡ open parenthesis x close parenthesis over cos ⁡ open parenthesis x close parenthesis")[$" " tan ( x ) = display(frac(sin ( x ), cos ( x )))$], we can replace the second #math.equation(block: false, alt: "tan squared ⁡ open parenthesis x close parenthesis")[$tan^(2) ( x )$] by #math.equation(block: false, alt: "the fraction sin squared ⁡ open parenthesis x close parenthesis over cos squared ⁡ open parenthesis x close parenthesis")[$" " display(frac(sin^(2) ( x ), cos^(2) ( x )))$], leaving us with #math.equation(block: true, alt: "1 minus cos squared ⁡ open parenthesis x close parenthesis plus tan squared ⁡ open parenthesis x close parenthesis minus sin squared ⁡ open parenthesis x close parenthesis")[$1 − cos^(2) ( x ) + tan^(2) ( x ) − sin^(2) ( x )$] Finally, because #math.equation(block: false, alt: "sin squared ⁡ open parenthesis x close parenthesis plus cos squared ⁡ open parenthesis x close parenthesis equals 1")[$" " sin^(2) ( x ) + cos^(2) ( x ) = 1$], we are left with #math.equation(block: false, alt: "tan squared ⁡ open parenthesis x close parenthesis")[$tan^(2) ( x )$]. === Trig Ratios are Related All three of the trigonometric functions of an angle are related. If we know the value of one of the three, we can calculate the other two (up to sign) by using the Pythagorean and tangent identities. We do #emph[not] need to find the angle itself in order to do this. We need only know in which quadrant the angle lies to determine the correct sign for the trig ratios. #examplebox("Example 7")[][ If #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction 7 over 25")[$" " sin ( θ ) = display(frac(7, 25)) " "$] and #math.equation(block: false, alt: "90 degrees less than θ less than 180 degrees ,")[$" " 90^(∘) < θ < 180^(∘) " " ,$] find #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$cos ( θ )$] and #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$]. #solutionbox[ We substitute #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction 7 over 25")[$" " sin ( θ ) = display(frac(7, 25)) " "$] into the Pythagorean identity #math.equation(block: false, alt: "cos squared ⁡ open parenthesis θ close parenthesis plus sin squared ⁡ open parenthesis θ close parenthesis equals 1")[$" " cos^(2) ( θ ) + sin^(2) ( θ ) = 1 " "$] to get #math.equation(block: true, alt: "cos squared ⁡ open parenthesis θ close parenthesis plus open parenthesis the fraction 7 over 25 close parenthesis squared, equals 1, Isolate cos squared ⁡ open parenthesis θ close parenthesis .; cos squared ⁡ open parenthesis θ close parenthesis, equals 1 minus open parenthesis the fraction 7 over 25 close parenthesis squared, 1 minus the fraction 49 over 625 equals the fraction 625 minus 49 over 625; cos squared ⁡ open parenthesis θ close parenthesis, equals the fraction 576 over 625, Take square roots.; cos ⁡ θ, equals plus or minus the square root of the fraction 576 over 625; equals plus or minus the fraction 24 over 25")[$cos^(2) ( θ ) + attach(( display(frac(7, 25)) ), t: 2) & = 1 & & "Isolate " bold(cos)^(2) ( bold(italic(θ)) ) . \ cos^(2) ( θ ) & = 1 − attach(( display(frac(7, 25)) ), t: 2) & & 1 − frac(49, 625) = frac(625 − 49, 625) \ cos^(2) ( θ ) & = display(frac(576, 625)) & & "Take square roots." \ cos θ & = ± sqrt(display(frac(576, 625))) \ & = ± display(frac(24, 25))$] Because #math.equation(block: false, alt: "θ")[$θ$] is between #math.equation(block: false, alt: "90 degrees")[$90^(∘)$] and #math.equation(block: false, alt: "180 degrees")[$180^(∘)$], #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$cos ( θ )$] must be negative, so #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction minus 24 over 25")[$cos ( θ ) = display(frac(− 24, 25))$]. To find #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$], we use the tangent identity. #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction sin ⁡ open parenthesis θ close parenthesis over cos ⁡ open parenthesis θ close parenthesis equals the fraction 7 / 25 over minus 24 / 25 equals the fraction minus 7 over 24")[$tan ( θ ) = display(frac(sin ( θ ), cos ( θ ))) = display(frac(7 / 25, − 24 / 25)) = display(frac(− 7, 24))$] ] ] We can also work the previous Example by sketching an appropriate triangle. We begin by drawing an obtuse angle #math.equation(block: false, alt: "θ")[$θ$] and its reference triangle. Because #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction 7 over 25")[$sin ( θ ) = display(frac(7, 25))$], we can choose a point on the terminal side to have #math.equation(block: false, alt: "y")[$y$]-coordinate 7 and #math.equation(block: false, alt: "r equals 25")[$r = 25$], as shown below. In order to calculate #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$cos ( θ )$] and #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$], we must find the #math.equation(block: false, alt: "x")[$x$]-coordinate of the point. By the Pythagorean theorem, #math.equation(block: true, alt: "x squared plus 7 squared, equals 25 squared, Isolate x squared .; x squared, equals 576, Take square roots.; x, equals minus 24")[$x^(2) + 7^(2) & = 25^(2) & & "Isolate " bold(italic(x))^(2) . \ x^(2) & = 576 & & "Take square roots." \ x & = − 24$] #figure(figph[angle], alt: "angle", caption: none) Note that #math.equation(block: false, alt: "x")[$x$] is negative because the point #math.equation(block: false, alt: "open parenthesis x , 7 close parenthesis")[$( x , 7 )$] is in the second quadrant. Then #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction x over r equals the fraction minus 24 over 25 and tan ⁡ open parenthesis θ close parenthesis equals the fraction y over x equals the fraction 7 over minus 24 equals the fraction minus 7 over 24")[$cos ( θ ) = display(frac(x, r)) = display(frac(− 24, 25)) " " " " " " " " " " " " "and" " " " " " " " " " " " " tan ( θ ) = display(frac(y, x)) = display(frac(7, − 24)) = display(frac(− 7, 24))$] + If #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction 5 over 13")[$" " cos ( θ ) = display(frac(5, 13)) " "$] and #math.equation(block: false, alt: "minus 90 degrees less than θ less than 0 degrees ,")[$" " − 90^(∘) < θ < 0^(∘) " " ,$] find #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$] and #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$]. + Find #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$] and #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$] by using an appropriate sketch. + #math.equation(block: false, alt: "y squared equals 13 squared minus 5 squared")[$y^(2) = 13^(2) − 5^(2)$], so #math.equation(block: false, alt: "y equals minus 12")[$y = − 12$]. Thus, #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction minus 12 over 13")[$" " sin ( θ ) = display(frac(− 12, 13))$], and #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction minus 12 over 5")[$" " tan ( θ ) = display(frac(− 12, 5))$] + #figure(figph[cos(theta)=5/13 in Q4], alt: "cos(theta)=5/13 in Q4", caption: none) === Solving Equations Now we'll see how identities are useful for solving trigonometric equations. So far we have only solved equations that involve a single trigonometric ratio. If the equation involves more than one trig function, we use identities to rewrite the equation in terms of a single trig function. #examplebox("Example 8")[][ Solve #math.equation(block: false, alt: "cos squared ⁡ open parenthesis A close parenthesis minus sin squared ⁡ open parenthesis A close parenthesis minus sin ⁡ open parenthesis A close parenthesis equals 0")[$" " " " cos^(2) ( A ) − sin^(2) ( A ) − sin ( A ) = 0 " " " "$] for #math.equation(block: false, alt: "0 degrees less than or equal to A less than or equal to 360 degrees")[$0^(∘) ≤ A ≤ 360^(∘)$]. #solutionbox[ The equation involves both the cosine and sine functions, and we will rewrite the left side in terms of the sine only. To eliminate the cosines, we use the Pythagorean identity #math.equation(block: false, alt: "cos squared ⁡ open parenthesis A close parenthesis equals 1 minus sin squared ⁡ open parenthesis A close parenthesis")[$cos^(2) ( A ) = 1 − sin^(2) ( A )$]. #math.equation(block: true, alt: "cos squared ⁡ open parenthesis A close parenthesis minus sin squared ⁡ open parenthesis A close parenthesis minus sin ⁡ open parenthesis A close parenthesis, equals 0, Replace cos squared ⁡ open parenthesis A close parenthesis by; 1 minus sin squared ⁡ open parenthesis A close parenthesis .; open parenthesis 1 minus sin squared ⁡ open parenthesis A close parenthesis close parenthesis minus sin squared ⁡ open parenthesis A close parenthesis minus sin ⁡ open parenthesis A close parenthesis, equals 0, Combine like terms.; 1 minus 2 sin squared ⁡ open parenthesis A close parenthesis minus sin ⁡ open parenthesis A close parenthesis, equals 0, Multiply both sides by minus 1.; 2 sin squared ⁡ open parenthesis A close parenthesis plus sin ⁡ open parenthesis A close parenthesis minus 1, equals 0, F")[$cos^(2) ( A ) − sin^(2) ( A ) − sin ( A ) & = 0 & & "Replace " bold(cos)^(2) ( bold(italic(A)) ) " by " \ 1 − bold(sin)^(2) ( bold(italic(A)) ) . \ ( 1 − sin^(2) ( A ) ) − sin^(2) ( A ) − sin ( A ) & = 0 & & "Combine like terms." \ 1 − 2 sin^(2) ( A ) − sin ( A ) & = 0 & & "Multiply both sides by " − 1 . \ 2 sin^(2) ( A ) + sin ( A ) − 1 & = 0 & & "Factor the left side." \ ( 2 sin ( A ) − 1 ) ( sin ( A ) + 1 ) & = 0$] Set each factor equal to zero to see that either #math.equation(block: false, alt: "sin ⁡ open parenthesis A close parenthesis equals the fraction 1 over 2")[$sin ( A ) = frac(1, 2)$] or #math.equation(block: false, alt: "sin ⁡ open parenthesis A close parenthesis equals minus 1")[$sin ( A ) = − 1$]. Now, #math.equation(block: false, alt: "sin ⁡ open parenthesis A close parenthesis equals the fraction 1 over 2")[$sin ( A ) = frac(1, 2)$] when #math.equation(block: false, alt: "A equals 30 degrees")[$A = 30^(∘)$] or #math.equation(block: false, alt: "A equals 150 degrees")[$A = 150^(∘)$], and #math.equation(block: false, alt: "sin ⁡ open parenthesis A close parenthesis equals minus 1")[$sin ( A ) = − 1$] when #math.equation(block: false, alt: "A equals 270 degrees")[$A = 270^(∘)$]. The solutions are #math.equation(block: false, alt: "A equals 30 degrees , A equals 150 degrees")[$A = 30^(∘) , " " A = 150^(∘)$], or #math.equation(block: false, alt: "A equals 270 degrees")[$A = 270^(∘)$]. ] ] Solve #math.equation(block: false, alt: "1 minus sin squared ⁡ open parenthesis θ close parenthesis plus cos ⁡ open parenthesis θ close parenthesis equals 0")[$" " " " 1 − sin^(2) ( θ ) + cos ( θ ) = 0 " " " "$] for #math.equation(block: false, alt: "0 degrees less than or equal to θ less than 360 degrees")[$0^(∘) ≤ θ < 360^(∘)$]. Replace #math.equation(block: false, alt: "1 minus sin squared ⁡ open parenthesis θ close parenthesis")[$" " 1 − sin^(2) ( θ ) " "$] by #math.equation(block: false, alt: "cos squared ⁡ open parenthesis θ close parenthesis")[$" " cos^(2) ( θ ) " "$] to get the equation #math.equation(block: true, alt: "cos squared ⁡ open parenthesis θ close parenthesis plus cos ⁡ open parenthesis θ close parenthesis equals 0")[$cos^(2) ( θ ) + cos ( θ ) = 0$] Solve this quadratic equation by factoring to find #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals 0")[$" " cos ( θ ) = 0 " "$] and #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals minus 1")[$" " cos ( θ ) = − 1$]. The solutions are #math.equation(block: false, alt: "90 degrees")[$90^(∘)$], #math.equation(block: false, alt: "180 degrees")[$" " 180^(∘)$], and #math.equation(block: false, alt: "270 degrees")[$" " 270^(∘)$] Like the Pythagorean identity, the tangent identity can be helpful in solving trigonometric equations. #examplebox("Example 9")[][ Solve #math.equation(block: false, alt: "cos ⁡ open parenthesis B close parenthesis plus the square root of 3 sin ⁡ open parenthesis B close parenthesis equals 0")[$" " " " cos ( B ) + sqrt(3) sin ( B ) = 0 " " " "$] for #math.equation(block: false, alt: "0 degrees less than or equal to B less than or equal to 360 degrees")[$0^(∘) ≤ B ≤ 360^(∘)$]. #solutionbox[ We first try to write the equation in terms of a single trig function. #math.equation(block: true, alt: "the square root of 3 sin ⁡ open parenthesis B close parenthesis, equals minus cos ⁡ open parenthesis B close parenthesis, Divide both sides by cos ⁡ open parenthesis B close parenthesis and the square root of 3.; the fraction sin ⁡ open parenthesis B close parenthesis over cos ⁡ open parenthesis B close parenthesis, equals the fraction minus 1 over the square root of 3, Apply the tangent identity.; tan ⁡ open parenthesis B close parenthesis, equals the fraction minus 1 over the square root of 3")[$sqrt(3) sin ( B ) & = − cos ( B ) & & "Divide both sides by " bold(cos) ( bold(italic(B)) ) " and " sqrt(3) . \ display(frac(sin ( B ), cos ( B ))) & = display(frac(− 1, sqrt(3))) & & "Apply the tangent identity." \ tan ( B ) & = display(frac(− 1, sqrt(3)))$] Because #math.equation(block: false, alt: "tan ⁡ open parenthesis 30 degrees close parenthesis equals the fraction minus 1 over the square root of 3")[$" " tan ( 30^(∘) ) = display(frac(− 1, sqrt(3))) " "$], we see that #math.equation(block: false, alt: "B")[$B$] has a reference angle of #math.equation(block: false, alt: "30 degrees")[$30^(∘)$]. And because the tangent is negative in the second and fourth quadrants, we have #math.equation(block: false, alt: "B equals 150 degrees")[$B = 150^(∘)$] or #math.equation(block: false, alt: "B equals 330 degrees")[$B = 330^(∘)$], as shown in the figure. #figure(figph[angle], alt: "angle", caption: none) ] ] Solve #math.equation(block: false, alt: "2 tan ⁡ open parenthesis β close parenthesis cos ⁡ open parenthesis β close parenthesis minus the square root of 3 equals 0")[$" " " " 2 tan ( β ) cos ( β ) − sqrt(3) = 0 " " " "$] for #math.equation(block: false, alt: "0 degrees less than or equal to β less than 360 degrees")[$0^(∘) ≤ β < 360^(∘)$]. Because #math.equation(block: false, alt: "tan ⁡ open parenthesis β close parenthesis equals the fraction sin ⁡ open parenthesis β close parenthesis over cos ⁡ open parenthesis β close parenthesis")[$" " tan ( β ) = display(frac(sin ( β ), cos ( β )))$], the equation reduces to #math.equation(block: false, alt: "sin ⁡ open parenthesis β close parenthesis equals the square root of 3")[$" " sin ( β ) = sqrt(3)$]. The solutions are #math.equation(block: false, alt: "60 degrees")[$60^(∘)$] and #math.equation(block: false, alt: "120 degrees")[$" " 120^(∘)$]. === Proving Identities When we show that one trigonometric expression is equivalent to another expression, we have proved a trigonometric identity. In a previous example we proved that the equation #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis tan ⁡ open parenthesis θ close parenthesis plus sin ⁡ open parenthesis θ close parenthesis equals 2 sin ⁡ open parenthesis θ close parenthesis")[$cos ( θ ) tan ( θ ) + sin ( θ ) = 2 sin ( θ )$] is an identity; it is true for all values of #math.equation(block: false, alt: "θ")[$θ$]. A common strategy for proving an identity is to transform one side of the equation using equivalent expressions until it is identical to the other side. To help us choose the transformations at each step of the proof, we try to match the algebraic form of the final expression. #examplebox("Example 10")[][ Prove the identity #math.equation(block: false, alt: "1 plus tan squared ⁡ open parenthesis t close parenthesis equals the fraction 1 over cos squared ⁡ open parenthesis t close parenthesis")[$" " " " 1 + tan^(2) ( t ) = display(frac(1, cos^(2) ( t )))$] #solutionbox[ By manipulating the left side of the equation, we will show that the expression #math.equation(block: false, alt: "1 plus tan squared ⁡ open parenthesis t close parenthesis")[$1 + tan^(2) ( t )$] is equivalent to #math.equation(block: false, alt: "the fraction 1 over cos squared ⁡ open parenthesis t close parenthesis")[$display(frac(1, cos^(2) ( t )))$]. First, we use the tangent identity to write the expression in terms of sines and cosines: #math.equation(block: true, alt: "1 plus tan squared ⁡ open parenthesis t close parenthesis equals 1 plus open parenthesis the fraction sin ⁡ open parenthesis t close parenthesis over cos ⁡ open parenthesis t close parenthesis close parenthesis squared equals 1 plus the fraction sin squared ⁡ open parenthesis t close parenthesis over cos squared ⁡ open parenthesis t close parenthesis")[$1 + tan^(2) ( t ) = 1 + attach(( display(frac(sin ( t ), cos ( t ))) ), t: 2) = 1 + display(frac(sin^(2) ( t ), cos^(2) ( t )))$] Next, we notice that the right side of the proposed identity has only one term, so we combine the terms on the left side. So that the fractions have the same denominator, we write 1 as #math.equation(block: false, alt: "the fraction cos squared ⁡ open parenthesis t close parenthesis over cos squared ⁡ open parenthesis t close parenthesis")[$" " display(frac(cos^(2) ( t ), cos^(2) ( t )))$]. #math.equation(block: true, alt: "1 plus the fraction sin squared ⁡ open parenthesis t close parenthesis over cos squared ⁡ open parenthesis t close parenthesis equals the fraction cos squared ⁡ open parenthesis t close parenthesis over cos squared ⁡ open parenthesis t close parenthesis plus the fraction sin squared ⁡ open parenthesis t close parenthesis over cos squared ⁡ open parenthesis t close parenthesis equals the fraction cos squared ⁡ open parenthesis t close parenthesis plus sin squared ⁡ open parenthesis t close parenthesis over cos squared ⁡ open parenthesis t close parenthesis")[$1 + display(frac(sin^(2) ( t ), cos^(2) ( t ))) = display(frac(bold(cos)^(2) ( bold(italic(t)) ), bold(cos)^(2) ( bold(italic(t)) ))) + display(frac(sin^(2) ( t ), cos^(2) ( t ))) = display(frac(cos^(2) ( t ) + sin^(2) ( t ), cos^(2) ( t )))$] Finally, we apply the Pythagorean identity to the numerator. #math.equation(block: true, alt: "the fraction cos squared ⁡ open parenthesis t close parenthesis plus sin squared ⁡ open parenthesis t close parenthesis over cos squared ⁡ open parenthesis t close parenthesis equals the fraction 1 over cos squared ⁡ open parenthesis t close parenthesis")[$display(frac(cos^(2) ( t ) + sin^(2) ( t ), cos^(2) ( t ))) = display(frac(1, cos^(2) ( t )))$] Thus, #math.equation(block: false, alt: "1 plus tan squared ⁡ open parenthesis t close parenthesis equals the fraction 1 over cos squared ⁡ open parenthesis t close parenthesis")[$" " 1 + tan^(2) ( t ) = display(frac(1, cos^(2) ( t )))$], and the identity is proved. ] ] When you write out the proof of an identity, you should transform the expression on one side of the identity into the expression on the other side, showing one step of the calculation on each line of your proof. You can provide a justification for each step to the right of the calculation. The proof of the identity in the previous example would look like this: #math.equation(block: true, alt: "1 plus tan squared ⁡ open parenthesis t close parenthesis, equals 1 plus open parenthesis the fraction sin ⁡ open parenthesis t close parenthesis over cos ⁡ open parenthesis t close parenthesis close parenthesis squared, Replace tan ⁡ open parenthesis t close parenthesis by the fraction sin ⁡ open parenthesis t close parenthesis over cos ⁡ open parenthesis t close parenthesis .; equals 1 plus the fraction sin squared ⁡ open parenthesis t close parenthesis over cos squared ⁡ open parenthesis t close parenthesis, Square the fraction.; equals the fraction cos squared ⁡ open parenthesis t close parenthesis over cos squared ⁡ open parenthesis t close parenthesis plus the fraction sin squared ⁡ open parenthesis t close parenthesis over cos squared ⁡ open parenthesis t close parenthesis, Replace")[$1 + tan^(2) ( t ) & = 1 + attach(( display(frac(sin ( t ), cos ( t ))) ), t: 2) & & "Replace " bold(tan) ( bold(italic(t)) ) " by " display(frac(bold(sin) ( bold(italic(t)) ), bold(cos) ( bold(italic(t)) ))) . \ & = 1 + display(frac(sin^(2) ( t ), cos^(2) ( t ))) & & "Square the fraction." \ & = display(frac(cos^(2) ( t ), cos^(2) ( t ))) + display(frac(sin^(2) ( t ), cos^(2) ( t ))) & & "Replace 1 by " display(frac(bold(cos)^(2) ( bold(italic(t)) ), bold(cos)^(2) ( bold(italic(t)) ))) . \ & = display(frac(cos^(2) ( t ) + sin^(2) ( t ), cos^(2) ( t ))) & & "Add fractions." \ & = display(frac(1, cos^(2) ( t ))) & & "Apply Pythagorean identity."$] Prove the identity #math.equation(block: false, alt: "2 cos squared ⁡ open parenthesis x close parenthesis minus 1 equals 1 minus 2 sin squared ⁡ open parenthesis x close parenthesis")[$" " " " 2 cos^(2) ( x ) − 1 = 1 − 2 sin^(2) ( x )$] #math.equation(block: true, alt: "2 cos squared ⁡ open parenthesis x close parenthesis minus 1, equals 2 open parenthesis 1 minus sin squared ⁡ open parenthesis x close parenthesis close parenthesis; equals 2 minus 2 sin squared ⁡ open parenthesis x close parenthesis minus 1; equals 1 minus 2 sin squared ⁡ open parenthesis x close parenthesis")[$2 cos^(2) ( x ) − 1 & = 2 ( 1 − sin^(2) ( x ) ) \ & = 2 − 2 sin^(2) ( x ) − 1 \ & = 1 − 2 sin^(2) ( x )$] Review the following skills you will need for this section. #notebox("Algebra Refresher", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Write each expression as a single fraction in simplest form. + #math.equation(block: false, alt: "the fraction 1 over x plus the fraction 1 over y")[$display(frac(1, x)) + display(frac(1, y))$] + #math.equation(block: false, alt: "1 minus the fraction 1 over x")[$1 − display(frac(1, x))$] + #math.equation(block: false, alt: "the fraction 1 over x minus 1 plus the fraction 1 over x plus 1")[$display(frac(1, x − 1)) + display(frac(1, x + 1))$] + #math.equation(block: false, alt: "x minus the fraction x over x plus 1")[$x − display(frac(x, x + 1))$] + #math.equation(block: false, alt: "the fraction the fraction y over x minus the fraction x over y over the fraction y over x plus 1")[$display(frac(display(frac(y, x)) − display(frac(x, y)), display(frac(y, x)) + 1))$] + #math.equation(block: false, alt: "the fraction 1 minus the fraction x squared over y squared over 1 plus the fraction x squared over y squared")[$display(frac(1 − display(frac(x^(2), y^(2))), 1 + display(frac(x^(2), y^(2)))))$] + #math.equation(block: false, alt: "the fraction the fraction 2 a over x over 1 minus the fraction a squared over x squared")[$display(frac(display(frac(2 a, x)), 1 − display(frac(a^(2), x^(2)))))$] + #math.equation(block: false, alt: "the fraction the fraction a over x plus the fraction b over y over 1 minus the fraction a b over x y")[$display(frac(display(frac(a, x)) + display(frac(b, y)), 1 − display(frac(a b, x y))))$] #math.equation(block: true, alt: "bar")[$underline(#h(2em) #h(2em) #h(2em) #h(2em))$] Algebra Refresher Answers + #math.equation(block: false, alt: "the fraction x plus y over x y")[$display(frac(x + y, x y))$] + #math.equation(block: false, alt: "the fraction x minus 1 over x")[$display(frac(x − 1, x))$] + #math.equation(block: false, alt: "the fraction 2 x over x squared minus 1")[$display(frac(2 x, x^(2) − 1))$] + #math.equation(block: false, alt: "the fraction x squared over x plus 1")[$display(frac(x^(2), x + 1))$] + #math.equation(block: false, alt: "the fraction y minus x over y")[$display(frac(y − x, y))$] + #math.equation(block: false, alt: "the fraction y squared minus x squared over y squared plus x squared")[$display(frac(y^(2) − x^(2), y^(2) + x^(2)))$] + #math.equation(block: false, alt: "the fraction 2 a x over x squared minus a squared")[$display(frac(2 a x, x^(2) − a^(2)))$] + #math.equation(block: false, alt: "the fraction a y plus b x over x y minus a b")[$display(frac(a y + b x, x y − a b))$] ] === Section 5.3 Summary ==== Vocabulary - Conditional equation - Identity - Equivalent expressions ==== Concepts + An equation that is true only for certain values of the variable, and false for others, is called a #strong[conditional equation]. An equation that is true for all legitimate values of the variables is called an #strong[identity]. + The expressions on either side of the equal sign in an identity are called #strong[equivalent expressions], because they have the same value for all values of the variable. + We often use identities to replace one form of an expression by a more useful form. + To check to whether an equation is an identity we can compare graphs of #math.equation(block: false, alt: "Y sub 1 equals")[$Y_(1) =$] (left side of the equation) and #math.equation(block: false, alt: "Y sub 2 equals")[$Y_(2) =$] (right side of the equation). If the two graphs agree, the equation is an identity. If the two graphs are not the same, the equation is not an identity. + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Pythagorean Identity] For any angle #math.equation(block: false, alt: "θ")[$θ$], #math.equation(block: true, alt: "cos squared ⁡ open parenthesis θ close parenthesis plus sin squared ⁡ open parenthesis θ close parenthesis equals 1")[$cos^(2) ( θ ) + sin^(2) ( θ ) = 1$]Alternate forms: #math.equation(block: true, alt: "cos squared ⁡ open parenthesis θ close parenthesis, equals 1 minus sin squared ⁡ open parenthesis θ close parenthesis; sin squared ⁡ open parenthesis θ close parenthesis, equals 1 minus cos squared ⁡ open parenthesis θ close parenthesis")[$cos^(2) ( θ ) & = 1 − sin^(2) ( θ ) \ sin^(2) ( θ ) & = 1 − cos^(2) ( θ )$] ] + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Tangent Identity] For any angle not coterminal with #math.equation(block: false, alt: "90 degrees")[$90^(∘)$] or #math.equation(block: false, alt: "270 degrees")[$270^(∘)$], #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction sin ⁡ open parenthesis θ close parenthesis over cos ⁡ open parenthesis θ close parenthesis")[$tan ( θ ) = display(frac(sin ( θ ), cos ( θ )))$] ] + To solve an equation involving more than one trig function, we use identities to rewrite the equation in terms of a single trig function. + To prove an identity, we write one side of the equation in equivalent forms until it is identical to the other side of the equation. ==== Study Questions + What is the difference between a conditional equation and an identity? Give an example of each. + What happens when you try to "solve" an identity? + Explain how to use graphs to verify an identity. + Delbert claims that since #math.equation(block: false, alt: "cos squared ⁡ open parenthesis θ close parenthesis plus sin squared ⁡ open parenthesis θ close parenthesis equals 1")[$cos^(2) ( θ ) + sin^(2) ( θ ) = 1$], we can take the square root of both sides to get the simpler form #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis plus sin ⁡ open parenthesis θ close parenthesis equals 1")[$cos ( θ ) + sin ( θ ) = 1$]. Is he correct? Why or why not? + If #math.equation(block: false, alt: "the fraction a over b equals the fraction 2 over 5")[$display(frac(a, b)) = display(frac(2, 5))$], is it necessarily true that #math.equation(block: false, alt: "a equals 2")[$a = 2$] and #math.equation(block: false, alt: "b equals 5")[$b = 5$]? Explain. If #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction 3 over 11")[$tan ( θ ) = display(frac(3, 11))$], is it true that #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals 3")[$sin ( θ ) = 3$] and #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals 11")[$cos ( θ ) = 11$]? ==== Skills + Recognize identities \#1–16, 41–46 + Verify identities \#17–26, 73–78 + Rewrite expressions using identities \#27–34, 47–50 + Use identities to evaluate expressions \#35–40 + olve trigonometric equations \#51–58, 67–72 + Given one trig ratio, find the others \#59–72 === Homework 5.3 For Problems 1–8, decide which of the following equations are identities. Explain your reasoning. #math.equation(block: true, alt: "open parenthesis the square root of a plus the square root of b close parenthesis squared equals a plus b")[$attach(( sqrt(a) + sqrt(b) ), t: 2) = a + b$] not an identity #math.equation(block: true, alt: "the square root of a squared minus b squared equals a minus b")[$sqrt(a^(2) − b^(2)) = a − b$] #math.equation(block: true, alt: "the fraction 1 over a plus b equals the fraction 1 over a plus the fraction 1 over b")[$display(frac(1, a + b)) = display(frac(1, a)) + display(frac(1, b))$] not an identity #math.equation(block: true, alt: "the fraction a plus b over a equals b")[$display(frac(a + b, a)) = b$] #math.equation(block: true, alt: "tan ⁡ open parenthesis α plus β close parenthesis equals the fraction sin ⁡ open parenthesis α plus β close parenthesis over cos ⁡ open parenthesis α plus β close parenthesis")[$tan ( α + β ) = display(frac(sin ( α + β ), cos ( α + β )))$] identity #math.equation(block: true, alt: "the fraction 1 over tan ⁡ open parenthesis θ close parenthesis equals the fraction cos ⁡ open parenthesis θ close parenthesis over sin ⁡ open parenthesis θ close parenthesis")[$display(frac(1, tan ( θ ))) = display(frac(cos ( θ ), sin ( θ )))$] #math.equation(block: true, alt: "open parenthesis 1 plus tan ⁡ open parenthesis θ close parenthesis close parenthesis squared equals 1 plus tan squared ⁡ open parenthesis θ close parenthesis")[$( 1 + tan ( θ ) )^(2) = 1 + tan^(2) ( θ )$] not an identity #math.equation(block: true, alt: "the square root of 1 minus sin squared ⁡ open parenthesis ϕ close parenthesis equals 1 minus sin ⁡ open parenthesis ϕ close parenthesis")[$sqrt(1 − sin^(2) ( ϕ )) = 1 − sin ( ϕ )$] For Problems 9–16, use graphs to decide which of the following equations are identities. #math.equation(block: true, alt: "sin ⁡ open parenthesis 2 t close parenthesis equals 2 sin ⁡ open parenthesis t close parenthesis")[$sin ( 2 t ) = 2 sin ( t )$] not an identity #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis plus sin ⁡ open parenthesis θ close parenthesis equals 1")[$cos ( θ ) + sin ( θ ) = 1$] #math.equation(block: true, alt: "sin ⁡ open parenthesis 30 degrees plus β close parenthesis equals the fraction 1 over 2 plus sin ⁡ open parenthesis β close parenthesis")[$sin ( 30^(∘) + β ) = display(frac(1, 2)) + sin ( β )$] not an identity #math.equation(block: true, alt: "cos ⁡ open parenthesis 90 degrees minus C close parenthesis equals sin ⁡ C")[$cos ( 90^(∘) − C ) = sin C$] #math.equation(block: true, alt: "tan ⁡ open parenthesis 90 degrees minus θ close parenthesis equals the fraction 1 over tan ⁡ open parenthesis θ close parenthesis")[$tan ( 90^(∘) − θ ) = display(frac(1, tan ( θ )))$] identity #math.equation(block: true, alt: "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")[$tan ( 2 θ ) = display(frac(2 tan ( θ ), 1 − tan^(2) ( θ )))$] #math.equation(block: true, alt: "the fraction tan squared ⁡ open parenthesis x close parenthesis over 1 plus tan squared ⁡ open parenthesis x close parenthesis equals sin squared ⁡ open parenthesis x close parenthesis")[$display(frac(tan^(2) ( x ), 1 + tan^(2) ( x ))) = sin^(2) ( x )$] identity #math.equation(block: true, alt: "tan ⁡ open parenthesis x close parenthesis plus the fraction 1 over tan ⁡ open parenthesis x close parenthesis equals sin ⁡ open parenthesis x close parenthesis cos ⁡ open parenthesis x close parenthesis")[$tan ( x ) + display(frac(1, tan ( x ))) = sin ( x ) cos ( x )$] For Problems 17–26, show that the equation is an identity by transforming the left side into the right side. #math.equation(block: true, alt: "open parenthesis 1 plus sin ⁡ open parenthesis w close parenthesis close parenthesis open parenthesis 1 minus sin ⁡ open parenthesis w close parenthesis close parenthesis equals cos squared ⁡ open parenthesis w close parenthesis")[$( 1 + sin ( w ) ) ( 1 − sin ( w ) ) = cos^(2) ( w )$] #math.equation(block: true, alt: "open parenthesis 1 plus sin ⁡ open parenthesis w close parenthesis close parenthesis open parenthesis 1 minus sin ⁡ open parenthesis w close parenthesis close parenthesis equals 1 minus sin squared ⁡ open parenthesis w close parenthesis equals cos squared ⁡ open parenthesis w close parenthesis")[$( 1 + sin ( w ) ) ( 1 − sin ( w ) ) = 1 − sin^(2) ( w ) = cos^(2) ( w )$] #math.equation(block: true, alt: "open parenthesis cos ⁡ open parenthesis θ close parenthesis minus 1 close parenthesis open parenthesis cos ⁡ open parenthesis θ close parenthesis plus 1 close parenthesis equals minus sin squared ⁡ open parenthesis θ close parenthesis")[$( cos ( θ ) − 1 ) ( cos ( θ ) + 1 ) = − sin^(2) ( θ )$] #math.equation(block: true, alt: "open parenthesis cos ⁡ open parenthesis θ close parenthesis minus sin ⁡ open parenthesis θ close parenthesis close parenthesis squared equals 1 minus 2 sin ⁡ open parenthesis θ close parenthesis cos ⁡ open parenthesis θ close parenthesis")[$( cos ( θ ) − sin ( θ ) )^(2) = 1 − 2 sin ( θ ) cos ( θ )$] #math.equation(block: true, alt: "open parenthesis cos ⁡ open parenthesis θ close parenthesis minus sin ⁡ open parenthesis θ close parenthesis close parenthesis squared, equals cos squared ⁡ open parenthesis θ close parenthesis minus 2 cos ⁡ open parenthesis θ close parenthesis sin ⁡ open parenthesis θ close parenthesis plus sin squared ⁡ open parenthesis θ close parenthesis; equals open parenthesis cos squared ⁡ open parenthesis θ close parenthesis plus sin squared ⁡ open parenthesis θ close parenthesis close parenthesis minus 2 sin ⁡ open parenthesis θ close parenthesis cos ⁡ open parenthesis θ close parenthesis equals 1 minus 2 sin ⁡ open parenthesis θ close parenthesis cos ⁡ open parenthesis θ close parenthesis")[$( cos ( θ ) − sin ( θ ) )^(2) & = cos^(2) ( θ ) − 2 cos ( θ ) sin ( θ ) + sin^(2) ( θ ) \ & = ( cos^(2) ( θ ) + sin^(2) ( θ ) ) − 2 sin ( θ ) cos ( θ ) = 1 − 2 sin ( θ ) cos ( θ )$] #math.equation(block: true, alt: "sin squared ⁡ open parenthesis x close parenthesis minus cos squared ⁡ open parenthesis x close parenthesis equals 1 minus 2 cos squared ⁡ open parenthesis x close parenthesis")[$sin^(2) ( x ) − cos^(2) ( x ) = 1 − 2 cos^(2) ( x )$] #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis cos ⁡ open parenthesis θ close parenthesis equals sin ⁡ open parenthesis θ close parenthesis")[$tan ( θ ) cos ( θ ) = sin ( θ )$] #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis cos ⁡ open parenthesis θ close parenthesis equals the fraction sin ⁡ open parenthesis θ close parenthesis over cos ⁡ open parenthesis θ close parenthesis times cos ⁡ open parenthesis θ close parenthesis equals sin ⁡ open parenthesis θ close parenthesis")[$tan ( θ ) cos ( θ ) = display(frac(sin ( θ ), cos ( θ ))) ⋅ cos ( θ ) = sin ( θ )$] #math.equation(block: true, alt: "the fraction sin ⁡ open parenthesis μ close parenthesis over tan ⁡ open parenthesis μ close parenthesis equals cos ⁡ open parenthesis μ close parenthesis")[$display(frac(sin ( μ ), tan ( μ ))) = cos ( μ )$] #math.equation(block: true, alt: "cos to the power 4 ⁡ open parenthesis x close parenthesis minus sin to the power 4 ⁡ open parenthesis x close parenthesis equals cos squared ⁡ open parenthesis x close parenthesis minus sin squared ⁡ open parenthesis x close parenthesis")[$cos^(4) ( x ) − sin^(4) ( x ) = cos^(2) ( x ) − sin^(2) ( x )$] #math.equation(block: true, alt: "cos to the power 4 ⁡ open parenthesis x close parenthesis minus sin to the power 4 ⁡ open parenthesis x close parenthesis, equals open parenthesis cos squared ⁡ open parenthesis x close parenthesis minus sin squared ⁡ open parenthesis x close parenthesis close parenthesis open parenthesis cos squared ⁡ open parenthesis x close parenthesis plus sin squared ⁡ open parenthesis x close parenthesis close parenthesis; equals open parenthesis cos squared ⁡ open parenthesis x close parenthesis minus sin squared ⁡ open parenthesis x close parenthesis close parenthesis open parenthesis 1 close parenthesis equals cos squared ⁡ open parenthesis x close parenthesis minus sin squared ⁡ open parenthesis x close parenthesis")[$cos^(4) ( x ) − sin^(4) ( x ) & = ( cos^(2) ( x ) − sin^(2) ( x ) ) ( cos^(2) ( x ) + sin^(2) ( x ) ) \ & = ( cos^(2) ( x ) − sin^(2) ( x ) ) ( 1 ) = cos^(2) ( x ) − sin^(2) ( x )$] #math.equation(block: true, alt: "1 minus 2 cos squared ⁡ open parenthesis v close parenthesis plus cos to the power 4 ⁡ open parenthesis v close parenthesis equals sin to the power 4 ⁡ open parenthesis v close parenthesis")[$1 − 2 cos^(2) ( v ) + cos^(4) ( v ) = sin^(4) ( v )$] #math.equation(block: true, alt: "the fraction sin ⁡ open parenthesis u close parenthesis over 1 plus cos ⁡ open parenthesis u close parenthesis equals the fraction 1 minus cos ⁡ open parenthesis u close parenthesis over sin ⁡ open parenthesis u close parenthesis")[$display(frac(sin ( u ), 1 + cos ( u ))) = display(frac(1 − cos ( u ), sin ( u )))$] Multiply numerator and denominator of the left side by #math.equation(block: false, alt: "1 minus cos ⁡ open parenthesis u close parenthesis")[$1 − cos ( u )$]. #math.equation(block: true, alt: "the fraction sin ⁡ open parenthesis u close parenthesis over 1 plus cos ⁡ open parenthesis u close parenthesis times the fraction 1 minus cos ⁡ open parenthesis u close parenthesis over 1 minus cos ⁡ open parenthesis u close parenthesis equals the fraction sin ⁡ open parenthesis u close parenthesis open parenthesis 1 minus cos ⁡ open parenthesis u close parenthesis close parenthesis over 1 minus cos squared ⁡ open parenthesis u close parenthesis equals the fraction sin ⁡ open parenthesis u close parenthesis open parenthesis 1 minus cos ⁡ open parenthesis u close parenthesis close parenthesis over sin squared ⁡ open parenthesis u close parenthesis equals the fraction 1 minus cos ⁡ open parenthesis u close parenthesis over sin ⁡ open parenthesis u close parenthesis")[$display(frac(sin ( u ), 1 + cos ( u ))) ⋅ display(frac(1 − cos ( u ), 1 − cos ( u ))) = display(frac(sin ( u ) ( 1 − cos ( u ) ), 1 − cos^(2) ( u ))) = display(frac(sin ( u ) ( 1 − cos ( u ) ), sin^(2) ( u ))) = display(frac(1 − cos ( u ), sin ( u )))$] #math.equation(block: true, alt: "the fraction sin ⁡ open parenthesis v close parenthesis over 1 minus sin ⁡ open parenthesis v close parenthesis equals the fraction tan ⁡ open parenthesis v close parenthesis open parenthesis 1 plus sin ⁡ open parenthesis v close parenthesis close parenthesis over cos ⁡ open parenthesis v close parenthesis")[$display(frac(sin ( v ), 1 − sin ( v ))) = display(frac(tan ( v ) ( 1 + sin ( v ) ), cos ( v )))$] Multiply numerator and denominator of the left side by #math.equation(block: false, alt: "1 plus sin ⁡ open parenthesis v close parenthesis")[$1 + sin ( v )$]. For Problems 27–34, simplify, using identities as necessary. #math.equation(block: true, alt: "the fraction 1 over cos squared ⁡ open parenthesis β close parenthesis minus the fraction sin squared ⁡ open parenthesis β close parenthesis over cos squared ⁡ open parenthesis β close parenthesis")[$display(frac(1, cos^(2) ( β ))) − display(frac(sin^(2) ( β ), cos^(2) ( β )))$] #math.equation(block: true, alt: "1")[$1$] #math.equation(block: true, alt: "the fraction 1 over sin squared ⁡ open parenthesis ϕ close parenthesis minus the fraction 1 over tan squared ⁡ open parenthesis ϕ close parenthesis")[$display(frac(1, sin^(2) ( ϕ ))) − display(frac(1, tan^(2) ( ϕ )))$] #math.equation(block: true, alt: "cos squared ⁡ open parenthesis α close parenthesis open parenthesis 1 plus tan squared ⁡ open parenthesis α close parenthesis close parenthesis")[$cos^(2) ( α ) ( 1 + tan^(2) ( α ) )$] #math.equation(block: true, alt: "1")[$1$] #math.equation(block: true, alt: "cos cubed ⁡ open parenthesis ϕ close parenthesis plus sin squared ⁡ open parenthesis ϕ close parenthesis cos ⁡ open parenthesis ϕ close parenthesis")[$cos^(3) ( ϕ ) + sin^(2) ( ϕ ) cos ( ϕ )$] #math.equation(block: true, alt: "tan squared ⁡ open parenthesis A close parenthesis minus tan squared ⁡ open parenthesis A close parenthesis sin squared ⁡ open parenthesis A close parenthesis")[$tan^(2) ( A ) − tan^(2) ( A ) sin^(2) ( A )$] #math.equation(block: true, alt: "sin squared ⁡ open parenthesis A close parenthesis")[$sin^(2) ( A )$] #math.equation(block: true, alt: "cos squared ⁡ open parenthesis B close parenthesis tan squared ⁡ open parenthesis B close parenthesis plus cos squared ⁡ open parenthesis B close parenthesis")[$cos^(2) ( B ) tan^(2) ( B ) + cos^(2) ( B )$] #math.equation(block: true, alt: "the fraction 1 minus cos squared ⁡ open parenthesis z close parenthesis over cos squared ⁡ open parenthesis z close parenthesis")[$display(frac(1 − cos^(2) ( z ), cos^(2) ( z )))$] #math.equation(block: true, alt: "tan squared ⁡ open parenthesis z close parenthesis")[$tan^(2) ( z )$] #math.equation(block: true, alt: "the fraction sin ⁡ open parenthesis t close parenthesis over cos ⁡ open parenthesis t close parenthesis tan ⁡ open parenthesis t close parenthesis")[$display(frac(sin ( t ), cos ( t ) tan ( t )))$] For Problems 35–40, evaluate without using a calculator. #math.equation(block: true, alt: "3 cos squared ⁡ open parenthesis 1.7 degrees close parenthesis plus 3 sin squared ⁡ open parenthesis 1.7 degrees close parenthesis")[$3 cos^(2) ( 1.7^(∘) ) + 3 sin^(2) ( 1.7^(∘) )$] #math.equation(block: true, alt: "3")[$3$] #math.equation(block: true, alt: "4 minus cos squared ⁡ open parenthesis 338 degrees close parenthesis minus sin squared ⁡ open parenthesis 338 degrees close parenthesis")[$4 − cos^(2) ( 338^(∘) ) − sin^(2) ( 338^(∘) )$] #math.equation(block: true, alt: "open parenthesis cos squared ⁡ open parenthesis 20 degrees close parenthesis plus sin squared ⁡ open parenthesis 20 degrees close parenthesis close parenthesis to the power 4")[$( cos^(2) ( 20^(∘) ) + sin^(2) ( 20^(∘) ) )^(4)$] #math.equation(block: true, alt: "1")[$1$] #math.equation(block: true, alt: "the fraction 18 over cos squared ⁡ open parenthesis 17 degrees close parenthesis plus sin squared ⁡ open parenthesis 17 degrees close parenthesis")[$display(frac(18, cos^(2) ( 17^(∘) ) + sin^(2) ( 17^(∘) )))$] #math.equation(block: true, alt: "the fraction 6 over cos squared ⁡ open parenthesis 53 degrees close parenthesis minus 6 tan squared ⁡ open parenthesis 53 degrees close parenthesis")[$display(frac(6, cos^(2) ( 53^(∘) ))) − 6 tan^(2) ( 53^(∘) )$] #math.equation(block: true, alt: "6")[$6$] #math.equation(block: true, alt: "the fraction 1 over sin squared ⁡ open parenthesis 102 degrees close parenthesis minus the fraction cos squared ⁡ open parenthesis 102 degrees close parenthesis over sin squared ⁡ open parenthesis 102 degrees close parenthesis")[$display(frac(1, sin^(2) ( 102^(∘) ))) − display(frac(cos^(2) ( 102^(∘) ), sin^(2) ( 102^(∘) )))$] For Problems 41–46, one side of an identity is given. Graph the expression, and make a conjecture about the other side of the identity. #math.equation(block: true, alt: "2 cos squared ⁡ open parenthesis θ close parenthesis minus 1 equals ?")[$2 cos^(2) ( θ ) − 1 = ?$] #math.equation(block: true, alt: "cos ⁡ open parenthesis 2 θ close parenthesis")[$cos ( 2 θ )$] #math.equation(block: true, alt: "1 minus 2 sin squared ⁡ open parenthesis the fraction θ over 2 close parenthesis equals ?")[$1 − 2 sin^(2) ( display(frac(θ, 2)) ) = ?$] #math.equation(block: true, alt: "1 minus the fraction sin squared ⁡ open parenthesis x close parenthesis over 1 plus cos ⁡ open parenthesis x close parenthesis equals ?")[$1 − display(frac(sin^(2) ( x ), 1 + cos ( x ))) = ?$] #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis")[$cos ( θ )$] #math.equation(block: true, alt: "the fraction sin ⁡ open parenthesis x close parenthesis cos ⁡ open parenthesis x close parenthesis over 1 minus sin squared ⁡ open parenthesis x close parenthesis equals ?")[$display(frac(sin ( x ) cos ( x ), 1 − sin^(2) ( x ))) = ?$] #math.equation(block: true, alt: "2 tan ⁡ open parenthesis t close parenthesis cos squared ⁡ open parenthesis t close parenthesis equals ?")[$2 tan ( t ) cos^(2) ( t ) = ?$] #math.equation(block: true, alt: "sin ⁡ open parenthesis 2 t close parenthesis")[$sin ( 2 t )$] #math.equation(block: true, alt: "the fraction 2 tan ⁡ open parenthesis t close parenthesis over 1 minus tan squared ⁡ open parenthesis t close parenthesis equals ?")[$display(frac(2 tan ( t ), 1 − tan^(2) ( t ))) = ?$] For Problems 47–50, use identities to rewrite each expression. #math.equation(block: false, alt: "2 minus cos squared ⁡ open parenthesis θ close parenthesis plus 2 sin ⁡ open parenthesis θ close parenthesis")[$2 − cos^(2) ( θ ) + 2 sin ( θ ) " " " " " "$] as an expression in #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$] only #math.equation(block: true, alt: "1 plus 2 sin ⁡ open parenthesis θ close parenthesis plus sin squared ⁡ open parenthesis θ close parenthesis")[$1 + 2 sin ( θ ) + sin^(2) ( θ )$] #math.equation(block: false, alt: "3 sin squared ⁡ open parenthesis B close parenthesis plus 2 cos ⁡ open parenthesis B close parenthesis minus 4")[$3 sin^(2) ( B ) + 2 cos ( B ) − 4 " " " " " "$] as an expression in #math.equation(block: false, alt: "cos ⁡ open parenthesis B close parenthesis")[$cos ( B )$] only #math.equation(block: false, alt: "cos squared ⁡ open parenthesis ϕ close parenthesis minus 2 sin squared ⁡ open parenthesis ϕ close parenthesis")[$cos^(2) ( ϕ ) − 2 sin^(2) ( ϕ ) " " " " " "$] as an expression in #math.equation(block: false, alt: "cos ⁡ open parenthesis ϕ close parenthesis")[$cos ( ϕ )$] only #math.equation(block: true, alt: "3 cos squared ⁡ open parenthesis ϕ close parenthesis minus 2")[$3 cos^(2) ( ϕ ) − 2$] #math.equation(block: false, alt: "cos squared ⁡ open parenthesis ϕ close parenthesis sin squared ⁡ open parenthesis ϕ close parenthesis")[$cos^(2) ( ϕ ) sin^(2) ( ϕ ) " " " " " "$] as an expression in #math.equation(block: false, alt: "sin ⁡ open parenthesis ϕ close parenthesis")[$sin ( ϕ )$] only For Problems 51–58, solve the equation for #math.equation(block: false, alt: "0 degrees less than or equal to θ less than 360 degrees")[$0^(∘) ≤ θ < 360^(∘)$]. Round angles to three decimal places if necessary. #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis minus sin squared ⁡ open parenthesis θ close parenthesis plus 1 equals 0")[$cos ( θ ) − sin^(2) ( θ ) + 1 = 0$] #math.equation(block: true, alt: "θ equals 90 degrees , θ equals 180 degrees , θ equals 270 degrees")[$θ = 90^(∘) , " " θ = 180^(∘) , " " θ = 270^(∘)$] #math.equation(block: true, alt: "4 sin ⁡ open parenthesis θ close parenthesis plus 2 cos squared ⁡ open parenthesis θ close parenthesis minus 3 equals minus 1")[$4 sin ( θ ) + 2 cos^(2) ( θ ) − 3 = − 1$] #math.equation(block: true, alt: "1 minus sin ⁡ open parenthesis θ close parenthesis minus 2 cos squared ⁡ open parenthesis θ close parenthesis equals 0")[$1 − sin ( θ ) − 2 cos^(2) ( θ ) = 0$] #math.equation(block: true, alt: "θ equals 90 degrees , θ equals 210 degrees , θ equals 330 degrees")[$θ = 90^(∘) , " " θ = 210^(∘) , " " θ = 330^(∘)$] #math.equation(block: true, alt: "3 cos squared ⁡ open parenthesis θ close parenthesis minus sin squared ⁡ open parenthesis θ close parenthesis equals 2")[$3 cos^(2) ( θ ) − sin^(2) ( θ ) = 2$] #math.equation(block: true, alt: "2 cos ⁡ open parenthesis θ close parenthesis tan ⁡ open parenthesis θ close parenthesis plus 1 equals 0")[$2 cos ( θ ) tan ( θ ) + 1 = 0$] #math.equation(block: true, alt: "θ equals 210 degrees , θ equals 330 degrees")[$θ = 210^(∘) , " " θ = 330^(∘)$] #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis minus sin ⁡ open parenthesis θ close parenthesis equals 0")[$cos ( θ ) − sin ( θ ) = 0$] #math.equation(block: true, alt: "the fraction 1 over 3 cos ⁡ open parenthesis θ close parenthesis equals sin ⁡ open parenthesis θ close parenthesis")[$display(frac(1, 3)) cos ( θ ) = sin ( θ )$] #math.equation(block: true, alt: "θ equals 18.43 degrees , θ equals 198.43 degrees")[$θ = 18.43^(∘) , " " θ = 198.43^(∘)$] #math.equation(block: true, alt: "5 sin ⁡ open parenthesis θ close parenthesis equals 2 cos ⁡ open parenthesis θ close parenthesis")[$5 sin ( θ ) = 2 cos ( θ )$] For Problems 59–62, use identities to find exact values for the other two trig ratios. #math.equation(block: false, alt: "cos ⁡ open parenthesis A close parenthesis equals the fraction 12 over 13")[$cos ( A ) = display(frac(12, 13)) " " " " " "$] and #math.equation(block: false, alt: "270 degrees less than A less than 360 degrees")[$" " 270^(∘) < " " A < 360^(∘)$] #math.equation(block: true, alt: "sin ⁡ open parenthesis A close parenthesis equals the fraction minus 5 over 13 , tan ⁡ open parenthesis A close parenthesis equals the fraction minus 5 over 12")[$sin ( A ) = display(frac(− 5, 13)) , " " tan ( A ) = display(frac(− 5, 12))$] #math.equation(block: false, alt: "sin ⁡ open parenthesis B close parenthesis equals the fraction minus 3 over 5")[$sin ( B ) = display(frac(− 3, 5)) " " " " " "$] and #math.equation(block: false, alt: "180 degrees less than B less than 270 degrees")[$" " 180^(∘) < " " B < 270^(∘)$] #math.equation(block: false, alt: "sin ⁡ open parenthesis ϕ close parenthesis equals the fraction 1 over 7")[$sin ( ϕ ) = display(frac(1, 7)) " " " " " "$] and #math.equation(block: false, alt: "90 degrees less than ϕ less than 180 degrees")[$" " 90^(∘) < " " ϕ < 180^(∘)$] #math.equation(block: true, alt: "cos ⁡ open parenthesis ϕ close parenthesis equals the fraction minus 4 the square root of 3 over 7 , tan ⁡ open parenthesis ϕ close parenthesis equals the fraction minus 1 over 4 the square root of 3")[$cos ( ϕ ) = display(frac(− 4 sqrt(3), 7)) , " " tan ( ϕ ) = display(frac(− 1, 4 sqrt(3)))$] #math.equation(block: false, alt: "cos ⁡ open parenthesis t close parenthesis equals the fraction minus 2 over 3")[$cos ( t ) = display(frac(− 2, 3)) " " " " " "$] and #math.equation(block: false, alt: "180 degrees less than t less than 270 degrees")[$" " 180^(∘) < " " t < 270^(∘)$] For Problems 63–66, use the identity below to find the sine and cosine of the angle. #math.equation(block: true, alt: "1 plus tan squared ⁡ open parenthesis θ close parenthesis equals the fraction 1 over cos squared ⁡ open parenthesis θ close parenthesis")[$1 + bold(tan)^(2) ( bold(italic(θ)) ) = display(frac(1, bold(cos)^(2) ( bold(italic(θ)) )))$] #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals minus the fraction 1 over 2")[$tan ( θ ) = − display(frac(1, 2)) " " " " " "$] and #math.equation(block: false, alt: "270 degrees less than θ less than 360 degrees")[$" " 270^(∘) < " " θ < 360^(∘)$] #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction minus 1 over the square root of 5")[$sin ( θ ) = display(frac(− 1, sqrt(5)))$], #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction 2 over the square root of 5")[$" " cos ( θ ) = display(frac(2, sqrt(5)))$] #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals 2")[$tan ( θ ) = 2 " " " " " "$] and #math.equation(block: false, alt: "180 degrees less than θ less than 270 degrees")[$" " 180^(∘) < " " θ < 270^(∘)$] #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction 3 over 4")[$tan ( θ ) = display(frac(3, 4)) " " " " " "$] and #math.equation(block: false, alt: "180 degrees less than θ less than 270 degrees")[$" " 180^(∘) < " " θ < 270^(∘)$] #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction minus 3 over 5")[$sin ( θ ) = display(frac(− 3, 5))$], #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction minus 4 over 5")[$" " cos ( θ ) = display(frac(− 4, 5))$] #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals minus 3")[$tan ( θ ) = − 3 " " " " " "$] and #math.equation(block: false, alt: "90 degrees less than θ less than 180 degrees")[$" " 90^(∘) < " " θ < 180^(∘)$] For Problems 67–72, find exact values for the sine, cosine, and tangent of the angle. #math.equation(block: false, alt: "2 cos ⁡ open parenthesis A close parenthesis plus 9 equals 8")[$2 cos ( A ) + 9 = 8 " " " " " "$] and #math.equation(block: false, alt: "90 degrees less than A less than 180 degrees")[$" " 90^(∘) < " " A < 180^(∘)$] #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction the square root of 3 over 2")[$sin ( θ ) = display(frac(sqrt(3), 2))$], #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction minus 1 over 2")[$" " cos ( θ ) = display(frac(− 1, 2))$], #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis equals the square root of 3")[$" " tan ( θ ) = sqrt(3)$] #math.equation(block: false, alt: "25 sin ⁡ open parenthesis B close parenthesis plus 8 equals minus 12")[$25 sin ( B ) + 8 = − 12 " " " " " "$] and #math.equation(block: false, alt: "180 degrees less than B less than 270 degrees")[$" " 180^(∘) < " " B < 270^(∘)$] #math.equation(block: false, alt: "8 tan ⁡ open parenthesis β close parenthesis plus 5 equals minus 11")[$8 tan ( β ) + 5 = − 11 " " " " " "$] and #math.equation(block: false, alt: "90 degrees less than β less than 180 degrees")[$" " 90^(∘) < " " β < 180^(∘)$] #math.equation(block: true, alt: "sin ⁡ open parenthesis β close parenthesis equals the fraction 2 over the square root of 5")[$sin ( β ) = display(frac(2, sqrt(5)))$], #math.equation(block: true, alt: "cos ⁡ open parenthesis β close parenthesis equals the fraction minus 1 over the square root of 5")[$" " cos ( β ) = display(frac(− 1, sqrt(5)))$], #math.equation(block: true, alt: "tan ⁡ open parenthesis β close parenthesis equals minus 2")[$" " tan ( β ) = − 2$] #math.equation(block: false, alt: "6 open parenthesis tan ⁡ open parenthesis β close parenthesis minus 4 close parenthesis equals minus 24")[$6 ( tan ( β ) − 4 ) = − 24 " " " " " "$] and #math.equation(block: false, alt: "90 degrees less than β less than 270 degrees")[$" " 90^(∘) < " " β < 270^(∘)$] #math.equation(block: false, alt: "tan squared ⁡ open parenthesis C close parenthesis minus the fraction 1 over 4 equals 0")[$tan^(2) ( C ) − display(frac(1, 4)) = 0 " " " " " "$] and #math.equation(block: false, alt: "0 degrees less than C less than 180 degrees")[$" " 0^(∘) < " " C < 180^(∘)$] #math.equation(block: true, alt: "sin ⁡ open parenthesis C close parenthesis equals the fraction 1 over the square root of 5 , cos ⁡ open parenthesis C close parenthesis equals the fraction 2 over the square root of 5 , tan ⁡ open parenthesis C close parenthesis equals the fraction 1 over 2; or, sin ⁡ open parenthesis C close parenthesis equals the fraction 1 over the square root of 5 , cos ⁡ open parenthesis C close parenthesis equals the fraction minus 2 over the square root of 5 , tan ⁡ open parenthesis C close parenthesis equals the fraction minus 1 over 2")[$& sin ( C ) = display(frac(1, sqrt(5))) , " " cos ( C ) = display(frac(2, sqrt(5))) , " " tan ( C ) = display(frac(1, 2)) \ "or" " " " " & sin ( C ) = display(frac(1, sqrt(5))) , " " cos ( C ) = display(frac(− 2, sqrt(5))) , " " tan ( C ) = display(frac(− 1, 2))$] #math.equation(block: false, alt: "4 cos squared ⁡ open parenthesis A close parenthesis minus cos ⁡ open parenthesis A close parenthesis equals 0")[$4 cos^(2) ( A ) − cos ( A ) = 0 " " " " " "$] and #math.equation(block: false, alt: "0 degrees less than A less than 180 degrees")[$" " 0^(∘) < " " A < 180^(∘)$] For Problems 73–76, prove the identity by rewriting tangents in terms of sines and cosines. (These problems involve simplifying complex fractions. See the Algebra Refresher to review this skill.) #math.equation(block: true, alt: "the fraction tan ⁡ open parenthesis α close parenthesis over 1 plus tan ⁡ open parenthesis α close parenthesis equals the fraction sin ⁡ open parenthesis α close parenthesis over sin ⁡ open parenthesis α close parenthesis plus cos ⁡ open parenthesis α close parenthesis")[$display(frac(tan ( α ), 1 + tan ( α ))) = display(frac(sin ( α ), sin ( α ) + cos ( α )))$] #math.equation(block: true, alt: "the fraction tan ⁡ open parenthesis α close parenthesis over 1 plus tan ⁡ open parenthesis α close parenthesis equals the fraction the fraction sin ⁡ open parenthesis α close parenthesis over cos ⁡ open parenthesis α close parenthesis over 1 plus the fraction sin ⁡ open parenthesis α close parenthesis over cos ⁡ open parenthesis α close parenthesis times the fraction cos ⁡ open parenthesis α close parenthesis over cos ⁡ open parenthesis α close parenthesis equals the fraction sin ⁡ open parenthesis α close parenthesis over sin ⁡ open parenthesis α close parenthesis plus cos ⁡ open parenthesis α close parenthesis")[$display(frac(tan ( α ), 1 + tan ( α ))) = display(frac(display(frac(sin ( α ), cos ( α ))), 1 + display(frac(sin ( α ), cos ( α ))))) ⋅ display(frac(cos ( α ), cos ( α ))) = display(frac(sin ( α ), sin ( α ) + cos ( α )))$] #math.equation(block: true, alt: "the fraction 1 minus tan ⁡ open parenthesis u close parenthesis over 1 plus tan ⁡ open parenthesis u close parenthesis equals the fraction cos ⁡ open parenthesis u close parenthesis minus sin ⁡ open parenthesis u close parenthesis over cos ⁡ open parenthesis u close parenthesis plus sin ⁡ open parenthesis u close parenthesis")[$display(frac(1 − tan ( u ), 1 + tan ( u ))) = display(frac(cos ( u ) − sin ( u ), cos ( u ) + sin ( u )))$] #math.equation(block: true, alt: "the fraction 1 plus tan squared ⁡ open parenthesis β close parenthesis over 1 minus tan squared ⁡ open parenthesis β close parenthesis equals the fraction 1 over cos squared ⁡ open parenthesis β close parenthesis minus sin squared ⁡ open parenthesis β close parenthesis")[$display(frac(1 + tan^(2) ( β ), 1 − tan^(2) ( β ))) = display(frac(1, cos^(2) ( β ) − sin^(2) ( β )))$] #math.equation(block: true, alt: "the fraction 1 plus tan squared ⁡ open parenthesis β close parenthesis over 1 minus tan squared ⁡ open parenthesis β close parenthesis equals the fraction the fraction 1 over cos squared ⁡ open parenthesis β close parenthesis over 1 minus the fraction sin squared ⁡ open parenthesis β close parenthesis over cos squared ⁡ open parenthesis β close parenthesis times the fraction cos squared ⁡ open parenthesis β close parenthesis over cos squared ⁡ open parenthesis β close parenthesis equals the fraction 1 over cos squared ⁡ open parenthesis β close parenthesis minus sin squared ⁡ open parenthesis β close parenthesis")[$display(frac(1 + tan^(2) ( β ), 1 − tan^(2) ( β ))) = display(frac(display(frac(1, cos^(2) ( β ))), 1 − display(frac(sin^(2) ( β ), cos^(2) ( β ))))) ⋅ display(frac(cos^(2) ( β ), cos^(2) ( β ))) = display(frac(1, cos^(2) ( β ) − sin^(2) ( β )))$] #math.equation(block: true, alt: "tan squared ⁡ open parenthesis v close parenthesis minus sin squared ⁡ open parenthesis v close parenthesis equals tan squared ⁡ open parenthesis v close parenthesis sin squared ⁡ open parenthesis v close parenthesis")[$tan^(2) ( v ) − sin^(2) ( v ) = tan^(2) ( v ) sin^(2) ( v )$] Prove the Pythagorean identity #math.equation(block: false, alt: "cos squared ⁡ open parenthesis θ close parenthesis plus sin squared ⁡ open parenthesis θ close parenthesis equals 1")[$" " cos^(2) ( θ ) + sin^(2) ( θ ) = 1 " "$] by carrying out the following steps. Sketch an angle #math.equation(block: false, alt: "θ")[$θ$] in standard position, and label a point #math.equation(block: false, alt: "open parenthesis x , y close parenthesis")[$( x , y )$] on the terminal side, at a distance #math.equation(block: false, alt: "r")[$r$] from the vertex. + Begin with the equation #math.equation(block: false, alt: "the square root of x squared plus y squared equals r")[$sqrt(x^(2) + y^(2)) = r$], and square both sides. + Divide both sides of your equation from part (a) by #math.equation(block: false, alt: "r squared")[$r^(2)$]. + Write the left side of the equation as the sum of the squares of two fractions. + Substitute the appropriate trigonometric ratio for each fraction. #figure(figph[angle], alt: "angle", caption: none) + By the distance formula, #math.equation(block: false, alt: "the square root of x squared plus y squared equals r")[$sqrt(x^(2) + y^(2)) = r$], or #math.equation(block: false, alt: "x squared plus y squared equals r squared")[$x^(2) + y^(2) = r^(2)$]. + #math.equation(block: false, alt: "the fraction x squared over r squared plus the fraction y squared over r squared equals 1")[$display(frac(x^(2), r^(2))) + display(frac(y^(2), r^(2))) = 1$] + #math.equation(block: false, alt: "open parenthesis the fraction x over r close parenthesis squared plus open parenthesis the fraction y over r close parenthesis squared equals 1")[$attach(( display(frac(x, r)) ), t: 2) + attach(( display(frac(y, r)) ), t: 2) = 1$] + #math.equation(block: false, alt: "open parenthesis cos ⁡ open parenthesis θ close parenthesis close parenthesis squared plus open parenthesis sin ⁡ open parenthesis θ close parenthesis close parenthesis squared equals 1")[$( cos ( θ ) )^(2) + ( sin ( θ ) )^(2) = 1$] Prove the tangent identity #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction sin ⁡ open parenthesis θ close parenthesis over cos ⁡ open parenthesis θ close parenthesis")[$" " tan ( θ ) = display(frac(sin ( θ ), cos ( θ ))) " "$] by carrying out the following steps. Sketch an angle #math.equation(block: false, alt: "θ")[$θ$] in standard position, and label a point #math.equation(block: false, alt: "open parenthesis x , y close parenthesis")[$( x , y )$] on the terminal side, at a distance #math.equation(block: false, alt: "r")[$r$] from the vertex. + Write #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$] in terms of #math.equation(block: false, alt: "y")[$y$] and #math.equation(block: false, alt: "r")[$r$], and solve for #math.equation(block: false, alt: "y")[$y$]. + Write #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$cos ( θ )$] in terms of #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "r")[$r$], and solve for #math.equation(block: false, alt: "x")[$x$]. + Write #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$] in terms of #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "y")[$y$], then substitute your results from parts (a) and (b). + Simplify your fraction in part (c).