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== 8.4#h(0.6em)Chapter Summary and Review


    
      
=== Key Concepts


      + Identities are useful for changing from one form to another when solving equations, for simplifying expressions, and for finding exact values for trigonometric functions.
+ it is not true in general that #math.equation(block: false, alt: "cos ⁡ open parenthesis α plus β close parenthesis")[$cos ( α + β )$] is equal to #math.equation(block: false, alt: "cos ⁡ open parenthesis α close parenthesis plus cos ⁡ open parenthesis β close parenthesis")[$cos ( α ) + cos ( β )$] for all angles #math.equation(block: false, alt: "α")[$α$] and #math.equation(block: false, alt: "β")[$β$], or that #math.equation(block: false, alt: "sin ⁡ open parenthesis α plus β close parenthesis")[$sin ( α + β )$] is equal to #math.equation(block: false, alt: "sin ⁡ open parenthesis α close parenthesis plus sin ⁡ open parenthesis β close parenthesis")[$sin ( α ) + sin ( β )$].
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Sum and Difference of Angles Identities]

#math.equation(block: true, alt: "cos ⁡ open parenthesis α plus β close parenthesis equals cos ⁡ open parenthesis α close parenthesis cos ⁡ open parenthesis β close parenthesis minus sin ⁡ open parenthesis α close parenthesis sin ⁡ open parenthesis β close parenthesis")[$cos ( α + β ) = cos ( α ) cos ( β ) − sin ( α ) sin ( β )$]#math.equation(block: true, alt: "sin ⁡ open parenthesis α plus β close parenthesis equals sin ⁡ open parenthesis α close parenthesis cos ⁡ open parenthesis β close parenthesis plus cos ⁡ open parenthesis α close parenthesis sin ⁡ open parenthesis β close parenthesis")[$sin ( α + β ) = sin ( α ) cos ( β ) + cos ( α ) sin ( β )$]#math.equation(block: true, alt: "tan ⁡ open parenthesis α plus β close parenthesis equals the fraction tan ⁡ open parenthesis α close parenthesis plus tan ⁡ open parenthesis β close parenthesis over 1 minus tan ⁡ open parenthesis α close parenthesis tan ⁡ open parenthesis β close parenthesis")[$tan ( α + β ) = display(frac(tan ( α ) + tan ( β ), 1 − tan ( α ) tan ( β )))$]#math.equation(block: true, alt: "cos ⁡ open parenthesis α minus β close parenthesis equals cos ⁡ open parenthesis α close parenthesis cos ⁡ open parenthesis β close parenthesis plus sin ⁡ open parenthesis α close parenthesis sin ⁡ open parenthesis β close parenthesis")[$cos ( α − β ) = cos ( α ) cos ( β ) + sin ( α ) sin ( β )$]#math.equation(block: true, alt: "sin ⁡ open parenthesis α minus β close parenthesis equals sin ⁡ open parenthesis α close parenthesis cos ⁡ open parenthesis β close parenthesis minus cos ⁡ open parenthesis α close parenthesis sin ⁡ open parenthesis β close parenthesis")[$sin ( α − β ) = sin ( α ) cos ( β ) − cos ( α ) sin ( β )$]#math.equation(block: true, alt: "tan ⁡ open parenthesis α minus β close parenthesis equals the fraction tan ⁡ open parenthesis α close parenthesis minus tan ⁡ open parenthesis β close parenthesis over 1 plus tan ⁡ open parenthesis α close parenthesis tan ⁡ open parenthesis β close parenthesis")[$tan ( α − β ) = display(frac(tan ( α ) − tan ( β ), 1 + tan ( α ) tan ( β )))$]
]
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Double Angle Identities]

#math.equation(block: true, alt: "cos ⁡ open parenthesis 2 θ close parenthesis, equals cos squared ⁡ open parenthesis θ close parenthesis minus sin squared ⁡ open parenthesis θ close parenthesis, sin ⁡ open parenthesis 2 θ close parenthesis equals 2 sin ⁡ open parenthesis θ close parenthesis cos ⁡ open parenthesis θ close parenthesis; equals 2 cos squared ⁡ open parenthesis θ close parenthesis minus 1, tan ⁡ open parenthesis 2 θ close parenthesis equals the fraction 2 tan ⁡ open parenthesis θ close parenthesis over 1 minus tan squared ⁡ open parenthesis θ close parenthesis; equals 1 minus 2 sin squared ⁡ open parenthesis θ close parenthesis")[$cos ( 2 θ ) & = cos^(2) ( θ ) − sin^(2) ( θ ) &  & sin ( 2 θ ) = 2 sin ( θ ) cos ( θ ) \  & = 2 cos^(2) ( θ ) − 1 &  & tan ( 2 θ ) = display(frac(2 tan ( θ ), 1 − tan^(2) ( θ ))) \  & = 1 − 2 sin^(2) ( θ )$]
]
+ Using one of the calculator keys #math.equation(block: false, alt: "S I N to the power minus 1 , C O S to the power minus 1")[$S I N^(− 1) , " " C O S^(− 1)$], or #math.equation(block: false, alt: "T A N to the power minus 1")[$T A N^(− 1)$] performs the inverse operation for computing a sine, cosine or tangent.
+ Two functions are called inverse functions if each "undoes" the results of the other function.
+ If #math.equation(block: false, alt: "y equals f open parenthesis x close parenthesis")[$y = f ( x )$] is a function, we can often find a formula for the inverse function by interchanging #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "y")[$y$] in the formula for the function, and then solving for #math.equation(block: false, alt: "y")[$y$].
+ The graphs of #math.equation(block: false, alt: "y equals f open parenthesis x close parenthesis")[$y = f ( x )$] and #math.equation(block: false, alt: "y equals f to the power minus 1 open parenthesis x close parenthesis")[$y = f^(− 1) ( x )$] are symmetric about the line #math.equation(block: false, alt: "y equals x")[$y = x$].
+ The domain of #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] is the same as the range of #math.equation(block: false, alt: "f")[$f$], and the range of #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] is the same as the domain of #math.equation(block: false, alt: "f")[$f$].
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Horizontal Line Test]

A function passes the #strong[Horizontal Line Test] if every horizontal line intersects the graph at most once.  In that case, there is only one #math.equation(block: false, alt: "x")[$x$]-value for each #math.equation(block: false, alt: "y")[$y$]-value, and the function is called #strong[one-to-one].
]
+ A function #math.equation(block: false, alt: "f")[$f$] has an inverse function if and only if #math.equation(block: false, alt: "f")[$f$] is one-to-one.
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Definitions of the inverse trig functions]

#math.equation(block: true, alt: "sin to the power minus 1 ⁡ open parenthesis x close parenthesis equals θ if and only if sin ⁡ open parenthesis θ close parenthesis equals x and the fraction minus π over 2 less than or equal to θ less than or equal to the fraction π over 2")[$sin^(− 1) ( x ) = θ " " " " " " "if and only if" " " " " " " " " sin ( θ ) = x " " " " " " "and" " " " " " " display(frac(− π, 2)) ≤ θ ≤ display(frac(π, 2))$]#math.equation(block: true, alt: "cos to the power minus 1 ⁡ open parenthesis x close parenthesis equals θ if and only if cos ⁡ open parenthesis θ close parenthesis equals x and 0 less than or equal to θ less than or equal to π")[$cos^(− 1) ( x ) = θ " " " " " " "if and only if" " " " " " " " " cos ( θ ) = x " " " " " " "and" " " " " " " 0 ≤ θ ≤ π$]#math.equation(block: true, alt: "tan to the power minus 1 ⁡ open parenthesis x close parenthesis equals θ if and only if tan ⁡ open parenthesis θ close parenthesis equals x and the fraction minus π over 2 less than θ less than the fraction π over 2")[$tan^(− 1) ( x ) = θ " " " " " " "if and only if" " " " " " " " " tan ( θ ) = x " " " " " " "and" " " " " " " display(frac(− π, 2)) < θ < display(frac(π, 2))$]
]
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Inverse Trigonometric Functions]

#math.equation(block: true, alt: "sin ⁡ open parenthesis sin to the power minus 1 ⁡ open parenthesis x close parenthesis close parenthesis equals x for minus 1 less than or equal to x less than or equal to 1")[$sin ( sin^(− 1) ( x ) ) = x " " " " " " " " "for" " " − 1 ≤ x ≤ 1$]#math.equation(block: true, alt: "cos ⁡ open parenthesis cos to the power minus 1 ⁡ open parenthesis x close parenthesis close parenthesis equals x for minus 1 less than or equal to x less than or equal to 1")[$cos ( cos^(− 1) ( x ) ) = x " " " " " " " " "for" " " − 1 ≤ x ≤ 1$]#math.equation(block: true, alt: "tan ⁡ open parenthesis tan to the power minus 1 ⁡ open parenthesis x close parenthesis close parenthesis equals x for all x")[$tan ( tan^(− 1) ( x ) ) = x " " " " " " " " "for all" " " x$]However,

#math.equation(block: true, alt: "sin to the power minus 1 ⁡ open parenthesis sin ⁡ open parenthesis x close parenthesis close parenthesis may not be equal to x")[$sin^(− 1) ( sin ( x ) ) " " " " "may not be equal to" " " x$]#math.equation(block: true, alt: "cos to the power minus 1 ⁡ open parenthesis cos ⁡ open parenthesis x close parenthesis close parenthesis may not be equal to x")[$cos^(− 1) ( cos ( x ) ) " " " " "may not be equal to" " " x$]#math.equation(block: true, alt: "tan to the power minus 1 ⁡ open parenthesis tan ⁡ open parenthesis x close parenthesis close parenthesis may not be equal to x")[$tan^(− 1) ( tan ( x ) ) " " " " "may not be equal to" " " x$]
]
+ The inverse sine function is also called the arcsine function and denoted by #math.equation(block: false, alt: "arcsin ⁡ open parenthesis x close parenthesis")[$arcsin ( x )$].  Similarly, the inverse cosine function is sometimes denoted by #math.equation(block: false, alt: "arccos ⁡ open parenthesis x close parenthesis")[$arccos ( x )$], and the inverse tangent function by #math.equation(block: false, alt: "arctan ⁡ open parenthesis x close parenthesis")[$arctan ( x )$].
+ When simplifying expressions involving inverse trigonometric functions, it can often clarify the computations if we assign a name such as #math.equation(block: false, alt: "θ")[$θ$] or #math.equation(block: false, alt: "ϕ")[$ϕ$] to the inverse trig value.
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Three More Functions]

If #math.equation(block: false, alt: "θ")[$θ$] is an angle in standard position, and #math.equation(block: false, alt: "P open parenthesis x , y close parenthesis")[$P ( x , y )$] is a point on the terminal side, then we define the following functions.

#math.equation(block: true, alt: "The secant : sec ⁡ open parenthesis θ close parenthesis equals the fraction r over x")[$"The" " " " " "secant" : " " " " " " " " " " " " sec ( θ ) = display(frac(r, x))$]#math.equation(block: true, alt: "The cosecant : csc ⁡ open parenthesis θ close parenthesis equals the fraction r over y")[$"The" " " " " "cosecant" : " " " " " " " " csc ( θ ) = display(frac(r, y))$]#math.equation(block: true, alt: "The cotangent : cot ⁡ open parenthesis θ close parenthesis equals the fraction x over y")[$"The" " " " " "cotangent" : " " " " " " cot ( θ ) = display(frac(x, y))$]#figure(figph[angles], alt: "angles", caption: none)
]
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Reciprocal Trigonometric Ratios]

If #math.equation(block: false, alt: "θ")[$θ$] is one of the acute angles in a right triangle,

#math.equation(block: true, alt: "sec ⁡ open parenthesis θ close parenthesis, equals the fraction hypotenuse over adjacent; csc ⁡ open parenthesis θ close parenthesis, equals the fraction hypotenuse over opposite; cot ⁡ open parenthesis θ close parenthesis, equals the fraction adjacent over opposite")[$sec ( θ ) & = display(frac("hypotenuse", "adjacent")) \ csc ( θ ) & = display(frac("hypotenuse", "opposite")) \ cot ( θ ) & = display(frac("adjacent", "opposite"))$]#figure(figph[triangle], alt: "triangle", caption: none)
]
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Reciprocal Trigonometric Functions]

#math.equation(block: true, alt: "The secant function : sec ⁡ open parenthesis θ close parenthesis equals the fraction 1 over cos ⁡ open parenthesis θ close parenthesis")[$"The" " " " " "secant" " " "function" : " " " " " " " " " " " " sec ( θ ) = display(frac(1, cos ( θ )))$]#math.equation(block: true, alt: "The cosecant function : csc ⁡ open parenthesis θ close parenthesis equals the fraction 1 over sin ⁡ open parenthesis θ close parenthesis")[$"The" " " " " "cosecant" " " "function" : " " " " " " " " csc ( θ ) = display(frac(1, sin ( θ )))$]#math.equation(block: true, alt: "The cotangent function : cot ⁡ open parenthesis θ close parenthesis equals the fraction 1 over tan ⁡ open parenthesis θ close parenthesis")[$"The" " " " " "cotangent" " " "function" : " " " " " " cot ( θ ) = display(frac(1, tan ( θ )))$]
]
+ We can obtain graphs of the secant, cosecant, and cotangent functions as the reciprocals of the three basic functions.
+ We can solve equations of the form #math.equation(block: false, alt: "sec ⁡ open parenthesis θ close parenthesis equals k")[$sec ( θ ) = k$], #math.equation(block: false, alt: "csc ⁡ open parenthesis θ close parenthesis equals k")[$csc ( θ ) = k$], and #math.equation(block: false, alt: "cot ⁡ open parenthesis θ close parenthesis equals k")[$cot ( θ ) = k$] by taking the reciprocal of both sides.
+ If we know one of the trigonometric ratios for an angle, we can use identities to find any of the others.
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Cotangent Identity]

#math.equation(block: true, alt: "cot ⁡ open parenthesis θ close parenthesis equals 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 , sin ⁡ open parenthesis θ close parenthesis not equal to 0")[$cot ( θ ) = display(frac(1, tan ( θ ))) = display(frac(cos ( θ ), sin ( θ ))) , " " " " " " " " sin ( θ ) ≠ 0$]
]
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Two More Pythagorean Identities]

#math.equation(block: true, alt: "1 plus tan squared ⁡ open parenthesis θ close parenthesis equals sec squared ⁡ open parenthesis θ close parenthesis 1 plus cot squared ⁡ open parenthesis θ close parenthesis equals csc squared ⁡ open parenthesis θ close parenthesis")[$1 + tan^(2) ( θ ) = sec^(2) ( θ ) " " " " " " " " " " " " " " " " " " " " 1 + cot^(2) ( θ ) = csc^(2) ( θ )$]
]
+ We can often simplify trigonometric expressions by first converting all the trig ratios to sines and cosines.


    
    
      
=== Review Problems


      For Problems 1–8, answer true or false.


      
        #math.equation(block: true, alt: "sin ⁡ open parenthesis β plus the fraction π over 4 close parenthesis equals sin ⁡ open parenthesis β close parenthesis plus the fraction 1 over the square root of 2")[$sin ( β + display(frac(π, 4)) ) = sin ( β ) + display(frac(1, sqrt(2)))$]


        
          False


        
      
      
        #math.equation(block: true, alt: "cos ⁡ open parenthesis the fraction π over 3 minus t close parenthesis equals the fraction 1 over 2 minus cos ⁡ open parenthesis t close parenthesis")[$cos ( display(frac(π, 3)) − t ) = display(frac(1, 2)) − cos ( t )$]


      
      
        #math.equation(block: true, alt: "tan ⁡ open parenthesis z minus w close parenthesis equals the fraction sin ⁡ open parenthesis z minus w close parenthesis over cos ⁡ open parenthesis z minus w close parenthesis")[$tan ( z − w ) = display(frac(sin ( z − w ), cos ( z − w )))$]


        
          True


        
      
      
        #math.equation(block: true, alt: "sin ⁡ open parenthesis 2 ϕ close parenthesis equals 1 minus cos ⁡ open parenthesis 2 ϕ close parenthesis")[$sin ( 2 ϕ ) = 1 − cos ( 2 ϕ )$]


      
      
        #math.equation(block: true, alt: "sin ⁡ open parenthesis the fraction π over 2 minus x close parenthesis equals 1 minus sin ⁡ open parenthesis x close parenthesis")[$sin ( display(frac(π, 2)) − x ) = 1 − sin ( x )$]


        
          False


        
      
      
        #math.equation(block: true, alt: "sin ⁡ open parenthesis π minus x close parenthesis equals sin ⁡ open parenthesis x close parenthesis")[$sin ( π − x ) = sin ( x )$]


      
      
        #math.equation(block: true, alt: "cos squared ⁡ open parenthesis α close parenthesis minus sin squared ⁡ open parenthesis α close parenthesis equals minus 1")[$cos^(2) ( α ) − sin^(2) ( α ) = − 1$]


        
          False


        
      
      
        #math.equation(block: true, alt: "tan to the power minus 1 ⁡ open parenthesis s close parenthesis equals the fraction 1 over tan ⁡ open parenthesis s close parenthesis")[$tan^(− 1) ( s ) = display(frac(1, tan ( s )))$]


      
      
        If #math.equation(block: false, alt: "sin ⁡ open parenthesis x close parenthesis equals minus 0.4")[$" " sin ( x ) = − 0.4$] and #math.equation(block: false, alt: "cos ⁡ open parenthesis x close parenthesis greater than 0 ,")[$cos ( x ) > 0 ,$] find an exact value for #math.equation(block: false, alt: "cos ⁡ open parenthesis x plus the fraction 3 π over 4 close parenthesis")[$cos ( x + display(frac(3 π, 4)) )$].


        
          #math.equation(block: true, alt: "the fraction 2 minus the square root of 21 over 5 the square root of 2")[$display(frac(2 − sqrt(21), 5 sqrt(2)))$]


        
      
      
        If #math.equation(block: false, alt: "cos ⁡ open parenthesis x close parenthesis equals minus 0.75")[$" " cos ( x ) = − 0.75$] and #math.equation(block: false, alt: "sin ⁡ open parenthesis x close parenthesis less than 0 ,")[$sin ( x ) < 0 ,$] find an exact value for #math.equation(block: false, alt: "cos ⁡ open parenthesis x minus the fraction 4 π over 3 close parenthesis")[$cos ( x − display(frac(4 π, 3)) )$].


      
      
        If #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction minus 3 over 8 , π less than θ less than the fraction 3 π over 2 ,")[$" " cos ( θ ) = display(frac(− 3, 8)) , " " π < θ < display(frac(3 π, 2)) ,$] and #math.equation(block: false, alt: "sin ⁡ open parenthesis ϕ close parenthesis equals the fraction 1 over 4 , the fraction π over 2 less than ϕ less than π ,")[$sin ( ϕ ) = display(frac(1, 4)) , " " display(frac(π, 2)) < ϕ < π ,$] find exact values for


        + #math.equation(block: false, alt: "sin ⁡ open parenthesis θ plus ϕ close parenthesis")[$sin ( θ + ϕ )$]
+ #math.equation(block: false, alt: "tan ⁡ open parenthesis θ plus ϕ close parenthesis")[$tan ( θ + ϕ )$]


        
          + #math.equation(block: false, alt: "the fraction 5 the square root of 33 minus 3 over 32")[$display(frac(5 sqrt(33) − 3, 32))$]
+ #math.equation(block: false, alt: "the fraction 5 the square root of 33 minus 3 over the square root of 5 open parenthesis 3 the square root of 3 plus the square root of 11 close parenthesis")[$display(frac(5 sqrt(33) − 3, sqrt(5) ( 3 sqrt(3) + sqrt(11) )))$]


        
      
      
        If #math.equation(block: false, alt: "sin ⁡ open parenthesis ρ close parenthesis equals the fraction 5 over 6 , the fraction π over 2 less than ρ less than π ,")[$" " sin ( ρ ) = display(frac(5, 6)) , " " display(frac(π, 2)) < ρ < π ,$] and #math.equation(block: false, alt: "cos ⁡ open parenthesis μ close parenthesis equals the fraction minus 1 over 3 , the fraction π over 2 less than μ less than π ,")[$cos ( μ ) = display(frac(− 1, 3)) , " " display(frac(π, 2)) < μ < π ,$] find exact values for


        + #math.equation(block: false, alt: "cos ⁡ open parenthesis ρ minus μ close parenthesis")[$cos ( ρ − μ )$]
+ #math.equation(block: false, alt: "tan ⁡ open parenthesis ρ minus μ close parenthesis")[$tan ( ρ − μ )$]


      
      
        If #math.equation(block: false, alt: "tan ⁡ open parenthesis x plus y close parenthesis equals 2")[$" " tan ( x + y ) = 2$] and #math.equation(block: false, alt: "tan ⁡ open parenthesis y close parenthesis equals the fraction 1 over 3 ,")[$tan ( y ) = display(frac(1, 3)) ,$] find #math.equation(block: false, alt: "tan ⁡ open parenthesis x close parenthesis")[$tan ( x )$].


        
          #math.equation(block: true, alt: "1")[$1$]


        
      
      
        If #math.equation(block: false, alt: "tan ⁡ open parenthesis x minus y close parenthesis equals the fraction 1 over 4")[$" " tan ( x − y ) = display(frac(1, 4))$] and #math.equation(block: false, alt: "tan ⁡ open parenthesis x close parenthesis equals 4 ,")[$tan ( x ) = 4 ,$] find #math.equation(block: false, alt: "tan ⁡ open parenthesis y close parenthesis")[$tan ( y )$].


      
      For Problems 15-16, use the sum and difference formulas to expand each expression.


      
        #math.equation(block: true, alt: "tan ⁡ open parenthesis t minus the fraction 5 π over 3 close parenthesis")[$tan ( t − display(frac(5 π, 3)) )$]


        
          #math.equation(block: true, alt: "the fraction tan ⁡ open parenthesis t close parenthesis plus the square root of 3 over 1 minus the square root of 3 tan ⁡ open parenthesis t close parenthesis")[$display(frac(tan ( t ) + sqrt(3), 1 − sqrt(3) tan ( t )))$]


        
      
      
        #math.equation(block: true, alt: "cos ⁡ open parenthesis s plus the fraction 7 π over 4 close parenthesis")[$cos ( s + display(frac(7 π, 4)) )$]


      
      For Problems 17–18, use the figure to find the trigonometric ratios.


      #figure(figph[triangle], alt: "triangle", caption: none)


      
        + #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$]
+ #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$cos ( θ )$]
+ #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$]
+ #math.equation(block: false, alt: "sin ⁡ open parenthesis 2 θ close parenthesis")[$sin ( 2 θ )$]
+ #math.equation(block: false, alt: "cos ⁡ open parenthesis 2 θ close parenthesis")[$cos ( 2 θ )$]
+ #math.equation(block: false, alt: "tan ⁡ open parenthesis 2 θ close parenthesis")[$tan ( 2 θ )$]


        
          + #math.equation(block: false, alt: "the fraction 4 over 5")[$display(frac(4, 5))$]
+ #math.equation(block: false, alt: "the fraction 3 over 5")[$display(frac(3, 5))$]
+ #math.equation(block: false, alt: "the fraction 4 over 3")[$display(frac(4, 3))$]
+ #math.equation(block: false, alt: "the fraction 24 over 25")[$display(frac(24, 25))$]
+ #math.equation(block: false, alt: "the fraction minus 7 over 25")[$display(frac(− 7, 25))$]
+ #math.equation(block: false, alt: "the fraction minus 24 over 7")[$display(frac(− 24, 7))$]


        
      
      
        + #math.equation(block: false, alt: "sin ⁡ open parenthesis ϕ close parenthesis")[$sin ( ϕ )$]
+ #math.equation(block: false, alt: "cos ⁡ open parenthesis ϕ close parenthesis")[$cos ( ϕ )$]
+ #math.equation(block: false, alt: "tan ⁡ open parenthesis ϕ close parenthesis")[$tan ( ϕ )$]
+ #math.equation(block: false, alt: "sin ⁡ open parenthesis 2 ϕ close parenthesis")[$sin ( 2 ϕ )$]
+ #math.equation(block: false, alt: "cos ⁡ open parenthesis 2 ϕ close parenthesis")[$cos ( 2 ϕ )$]
+ #math.equation(block: false, alt: "tan ⁡ open parenthesis 2 ϕ close parenthesis")[$tan ( 2 ϕ )$]


      
      For Problems 19–24, use identities to simplify each expression.


      
        #math.equation(block: true, alt: "sin ⁡ open parenthesis 4 x close parenthesis cos ⁡ open parenthesis 5 x close parenthesis plus cos ⁡ open parenthesis 4 x close parenthesis sin ⁡ open parenthesis 5 x close parenthesis")[$sin ( 4 x ) cos ( 5 x ) + cos ( 4 x ) sin ( 5 x )$]


        
          #math.equation(block: true, alt: "sin ⁡ open parenthesis 9 x close parenthesis")[$sin ( 9 x )$]


        
      
      
        #math.equation(block: true, alt: "cos ⁡ open parenthesis 3 β close parenthesis cos ⁡ open parenthesis 1.5 close parenthesis minus sin ⁡ open parenthesis 3 β close parenthesis sin ⁡ open parenthesis 1.5 close parenthesis")[$cos ( 3 β ) cos ( 1.5 ) − sin ( 3 β ) sin ( 1.5 )$]


      
      
        #math.equation(block: true, alt: "the fraction tan ⁡ open parenthesis 2 ϕ close parenthesis minus tan ⁡ open parenthesis 2 close parenthesis over 1 plus tan ⁡ open parenthesis 2 ϕ close parenthesis tan ⁡ open parenthesis 2 close parenthesis")[$display(frac(tan ( 2 ϕ ) − tan ( 2 ), 1 + tan ( 2 ϕ ) tan ( 2 )))$]


        
          #math.equation(block: true, alt: "tan ⁡ open parenthesis 2 ϕ minus 2 close parenthesis")[$tan ( 2 ϕ − 2 )$]


        
      
      
        #math.equation(block: true, alt: "the fraction tan ⁡ open parenthesis the fraction 5 π over 9 close parenthesis minus tan ⁡ open parenthesis the fraction 2 π over 9 close parenthesis over 1 plus tan ⁡ open parenthesis the fraction 5 π over 9 close parenthesis tan ⁡ open parenthesis the fraction 2 π over 9 close parenthesis")[$display(frac(tan ( display(frac(5 π, 9)) ) − tan ( display(frac(2 π, 9)) ), 1 + tan ( display(frac(5 π, 9)) ) tan ( display(frac(2 π, 9)) )))$]


      
      
        #math.equation(block: true, alt: "2 sin ⁡ open parenthesis 4 θ close parenthesis cos ⁡ open parenthesis 4 θ close parenthesis")[$2 sin ( 4 θ ) cos ( 4 θ )$]


        
          #math.equation(block: true, alt: "sin ⁡ open parenthesis 8 θ close parenthesis")[$sin ( 8 θ )$]


        
      
      
        #math.equation(block: true, alt: "1 minus 2 sin squared ⁡ open parenthesis 3 ϕ close parenthesis")[$1 − 2 sin^(2) ( 3 ϕ )$]


      
      For Problems 25–26,


      + Use identities to rewrite the equation in terms of a single angle.
+ Solve.  Give exact solutions between #math.equation(block: false, alt: "0")[$0$] and #math.equation(block: false, alt: "2 π")[$2 π$].


      
        #math.equation(block: true, alt: "cos ⁡ open parenthesis 2 θ close parenthesis minus sin ⁡ open parenthesis θ close parenthesis equals 1")[$cos ( 2 θ ) − sin ( θ ) = 1$]


        
          + #math.equation(block: false, alt: "1 minus 2 sin squared ⁡ open parenthesis θ close parenthesis minus sin ⁡ open parenthesis θ close parenthesis equals 1")[$1 − 2 sin^(2) ( θ ) − sin ( θ ) = 1$]
+ #math.equation(block: false, alt: "0 , π , the fraction 7 π over 6 , the fraction 11 π over 6")[$0 , " " π , " " display(frac(7 π, 6)) , " " display(frac(11 π, 6))$]


        
      
      
        #math.equation(block: true, alt: "tan ⁡ open parenthesis 2 z close parenthesis plus tan ⁡ open parenthesis z close parenthesis equals 0")[$tan ( 2 z ) + tan ( z ) = 0$]


      
      For Problems 27–28, graph the function and decide if it has an inverse function.


      
        #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 4 x minus x cubed")[$f ( x ) = 4 x − x^(3)$]


        
          #figure(figph[graph], alt: "graph", caption: none)


          No


        
      
      
        #math.equation(block: true, alt: "g open parenthesis x close parenthesis equals 5 plus the cube root of x minus 2")[$g ( x ) = 5 + root(3, x − 2)$]


      
      For Problems 29–30, give exact values in radians.


      
        + #math.equation(block: false, alt: "tan to the power minus 1 ⁡ open parenthesis minus the square root of 3 close parenthesis")[$tan^(− 1) ( − sqrt(3) )$]
+ #math.equation(block: false, alt: "arccos ⁡ open parenthesis minus the fraction 1 over 2 close parenthesis")[$arccos ( − display(frac(1, 2)) )$]


        
          + #math.equation(block: false, alt: "the fraction minus π over 3")[$display(frac(− π, 3))$]
+ #math.equation(block: false, alt: "the fraction 2 π over 3")[$display(frac(2 π, 3))$]


        
      
      
        + #math.equation(block: false, alt: "arcsin ⁡ open parenthesis minus 1 close parenthesis")[$arcsin ( − 1 )$]
+ #math.equation(block: false, alt: "cos to the power minus 1 ⁡ open parenthesis minus 1 close parenthesis")[$cos^(− 1) ( − 1 )$]


      
      
        An IMAX movie screen is 52.8 feet high.


        + If your line of sight is level with the bottom of the screen, write an expression for the angle subtended by the screen when you sit #math.equation(block: false, alt: "x")[$x$] feet away.
+ Evaluate your expression for #math.equation(block: false, alt: "x equals 20")[$x = 20$] feet and for #math.equation(block: false, alt: "x equals 100")[$x = 100$] feet.


        
          + #math.equation(block: false, alt: "tan to the power minus 1 ⁡ open parenthesis the fraction 52.8 over x close parenthesis")[$tan^(− 1) ( display(frac(52.8, x)) )$]
+ #math.equation(block: false, alt: "69.25 degrees , 27.83 degrees")[$69.25^(∘) , " " 27.83^(∘)$]


        
      
      
        Rembrandt's painting The Night Watch measures 13 feet high by 16 feet wide.


        + Write an expression for the angle subtended by the width of the painting if you sit #math.equation(block: false, alt: "d")[$d$] feet back from the center of the painting.
+ Evaluate your expression for #math.equation(block: false, alt: "d equals 10")[$d = 10$] feet and for #math.equation(block: false, alt: "d equals 25")[$d = 25$] feet.


      
      For Problems 33–34, solve for #math.equation(block: false, alt: "θ")[$θ$].


      
        #math.equation(block: true, alt: "v sub y equals v sub 0 sin ⁡ open parenthesis θ close parenthesis minus g t")[$v_(y) = v_(0) sin ( θ ) − g t$]


        
          #math.equation(block: true, alt: "θ equals sin to the power minus 1 ⁡ open parenthesis the fraction v sub y plus g t over v sub 0 close parenthesis")[$θ = sin^(− 1) ( display(frac(v_(y) + g t, v_(0))) )$]


        
      
      
        #math.equation(block: true, alt: "Δ W equals minus q sub 0 E cos ⁡ open parenthesis π minus θ close parenthesis Δ l")[$upright(Δ) W = − q_(0) E cos ( π − θ ) upright(Δ) l$]


      
      For Problems 35–36, find exact values without using a calculator.


      
        #math.equation(block: true, alt: "cos ⁡ open bracket tan to the power minus 1 ⁡ open parenthesis the fraction minus the square root of 5 over 2 close parenthesis close bracket")[$cos [ tan^(− 1) ( display(frac(− sqrt(5), 2)) ) ]$]


        
          #math.equation(block: true, alt: "the fraction 2 over 3")[$display(frac(2, 3))$]


        
      
      
        #math.equation(block: true, alt: "tan ⁡ open bracket sin to the power minus 1 ⁡ open parenthesis the fraction 2 over 7 close parenthesis close bracket")[$tan [ sin^(− 1) ( display(frac(2, 7)) ) ]$]


      
      For Problems 37–38, simplify the expression.


      
        #math.equation(block: true, alt: "sin ⁡ open bracket cos to the power minus 1 ⁡ open parenthesis 2 t close parenthesis close bracket")[$sin [ cos^(− 1) ( 2 t ) ]$]


        
          #math.equation(block: true, alt: "the square root of 1 minus 4 t squared")[$sqrt(1 − 4 t^(2))$]


        
      
      
        #math.equation(block: true, alt: "tan ⁡ open bracket cos to the power minus 1 ⁡ open parenthesis m close parenthesis close bracket")[$tan [ cos^(− 1) ( m ) ]$]


      
      
        Explain why one of the expressions #math.equation(block: false, alt: "sin to the power minus 1 ⁡ open parenthesis x close parenthesis")[$sin^(− 1) ( x )$] or #math.equation(block: false, alt: "sin to the power minus 1 ⁡ open parenthesis the fraction 1 over x close parenthesis")[$sin^(− 1) ( display(frac(1, x)) )$] must be undefined.


        
          Because #math.equation(block: false, alt: "vertical bar sin ⁡ open parenthesis θ close parenthesis vertical bar less than or equal to 1 , sin to the power minus 1 ⁡ open parenthesis t close parenthesis")[$| sin ( θ ) | ≤ 1 , " " sin^(− 1) ( t )$] is undefined for #math.equation(block: false, alt: "vertical bar t vertical bar greater than 1")[$| t | > 1$]. If #math.equation(block: false, alt: "x not equal to 0")[$x ≠ 0$], then either #math.equation(block: false, alt: "vertical bar x vertical bar greater than 1")[$| x | > 1$] or #math.equation(block: false, alt: "vertical bar the fraction 1 over x vertical bar greater than 1")[$| display(frac(1, x)) | > 1$].  If #math.equation(block: false, alt: "x equals 0")[$x = 0$], then #math.equation(block: false, alt: "the fraction 1 over x")[$display(frac(1, x))$] is undefined.


        
      
      
        Does #math.equation(block: false, alt: "sin to the power minus 1 ⁡ open parenthesis minus x close parenthesis equals minus sin to the power minus 1 ⁡ open parenthesis x close parenthesis")[$sin^(− 1) ( − x ) = − sin^(− 1) ( x )$]?  Does #math.equation(block: false, alt: "cos to the power minus 1 ⁡ open parenthesis minus x close parenthesis equals minus cos to the power minus 1 ⁡ open parenthesis x close parenthesis")[$cos^(− 1) ( − x ) = − cos^(− 1) ( x )$]?


      
      For Problems 41–42, evaluate. Round answers to 3 decimal places if necessary.


      
        + #math.equation(block: false, alt: "csc ⁡ open parenthesis 27 degrees close parenthesis")[$csc ( 27^(∘) )$]
+ #math.equation(block: false, alt: "sec ⁡ open parenthesis 108 degrees close parenthesis")[$sec ( 108^(∘) )$]
+ #math.equation(block: false, alt: "cot ⁡ open parenthesis 245 degrees close parenthesis")[$cot ( 245^(∘) )$]


        
          + #math.equation(block: false, alt: "2.203")[$2.203$]
+ #math.equation(block: false, alt: "minus 3.236")[$− 3.236$]
+ #math.equation(block: false, alt: "0.466")[$0.466$]


        
      
      
        + #math.equation(block: false, alt: "csc ⁡ open parenthesis 5.3 close parenthesis")[$csc ( 5.3 )$]
+ #math.equation(block: false, alt: "cot ⁡ open parenthesis 0.98 close parenthesis")[$cot ( 0.98 )$]
+ #math.equation(block: false, alt: "sec ⁡ open parenthesis 2.17 close parenthesis")[$sec ( 2.17 )$]


      
      For Problems 43–50, find all six trigonometric ratios for the angle #math.equation(block: false, alt: "θ")[$θ$].


      
        #figure(figph[triangle], alt: "triangle", caption: none)


        
          #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction 13 over the square root of 313")[$sin ( θ ) = display(frac(13, sqrt(313)))$], #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction 12 over the square root of 313")[$" " cos ( θ ) = display(frac(12, sqrt(313)))$], #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction 13 over 12")[$" " tan ( θ ) = display(frac(13, 12))$], #math.equation(block: true, alt: "sec ⁡ open parenthesis θ close parenthesis equals the fraction the square root of 313 over 12")[$" " sec ( θ ) = display(frac(sqrt(313), 12))$], #math.equation(block: true, alt: "csc ⁡ open parenthesis θ close parenthesis equals the fraction the square root of 313 over 13")[$" " csc ( θ ) = display(frac(sqrt(313), 13))$], #math.equation(block: true, alt: "cot ⁡ open parenthesis θ close parenthesis equals the fraction 12 over 13")[$" " cot ( θ ) = display(frac(12, 13))$]


        
      
      
        #figure(figph[triangle], alt: "triangle", caption: none)


      
      
        #figure(figph[angle], alt: "angle", caption: none)


        
          #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction 1 over 3")[$sin ( θ ) = display(frac(1, 3))$], #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction minus 2 the square root of 2 over 3")[$" " cos ( θ ) = display(frac(− 2 sqrt(2), 3))$], #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction minus 1 over 2 the square root of 2")[$" " tan ( θ ) = display(frac(− 1, 2 sqrt(2)))$], #math.equation(block: true, alt: "sec ⁡ open parenthesis θ close parenthesis equals the fraction minus 3 over 2 the square root of 2")[$" " sec ( θ ) = display(frac(− 3, 2 sqrt(2)))$], #math.equation(block: true, alt: "csc ⁡ open parenthesis θ close parenthesis equals 3")[$" " csc ( θ ) = 3$], #math.equation(block: true, alt: "cot ⁡ open parenthesis θ close parenthesis equals minus 2 the square root of 2")[$" " cot ( θ ) = − 2 sqrt(2)$]


        
      
      
        #figure(figph[angle], alt: "angle", caption: none)


      
      
        #figure(figph[angle], alt: "angle", caption: none)


        
          #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction minus 9 over the square root of 106")[$sin ( θ ) = display(frac(− 9, sqrt(106)))$], #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction minus 5 over the square root of 106")[$" " cos ( θ ) = display(frac(− 5, sqrt(106)))$], #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction 9 over 5")[$" " tan ( θ ) = display(frac(9, 5))$], #math.equation(block: true, alt: "sec ⁡ open parenthesis θ close parenthesis equals the fraction minus the square root of 106 over 5")[$" " sec ( θ ) = display(frac(− sqrt(106), 5))$], #math.equation(block: true, alt: "csc ⁡ open parenthesis θ close parenthesis equals the fraction minus the square root of 106 over 9")[$" " csc ( θ ) = display(frac(− sqrt(106), 9))$], #math.equation(block: true, alt: "cot ⁡ open parenthesis θ close parenthesis equals the fraction 5 over 9")[$" " cot ( θ ) = display(frac(5, 9))$]


        
      
      
        #figure(figph[angle], alt: "angle", caption: none)


      
      
        #math.equation(block: true, alt: "6 cos ⁡ open parenthesis α close parenthesis equals minus 5")[$6 cos ( α ) = − 5$], #math.equation(block: true, alt: "180 degrees less than α less than 270 degrees")[$" " 180^(∘) < α < 270^(∘)$]


        
          #math.equation(block: true, alt: "sin ⁡ open parenthesis α close parenthesis equals the fraction minus the square root of 11 over 6")[$sin ( α ) = display(frac(− sqrt(11), 6))$], #math.equation(block: true, alt: "cos ⁡ open parenthesis α close parenthesis equals the fraction minus 5 over 6")[$" " cos ( α ) = display(frac(− 5, 6))$], #math.equation(block: true, alt: "tan ⁡ open parenthesis α close parenthesis equals the fraction the square root of 11 over 5")[$" " tan ( α ) = display(frac(sqrt(11), 5))$], #math.equation(block: true, alt: "sec ⁡ open parenthesis α close parenthesis equals the fraction minus 6 over 5 ,")[$" " sec ( α ) = display(frac(− 6, 5)) ,$]#math.equation(block: true, alt: "csc ⁡ open parenthesis α close parenthesis equals the fraction minus 6 over the square root of 11")[$" " csc ( α ) = display(frac(− 6, sqrt(11)))$], #math.equation(block: true, alt: "cot ⁡ open parenthesis α close parenthesis equals the fraction 5 over the square root of 11")[$" " cot ( α ) = display(frac(5, sqrt(11)))$]


        
      
      
        #math.equation(block: false, alt: "4 sin ⁡ open parenthesis θ close parenthesis equals 3 , θ")[$4 sin ( θ ) = 3 , " " θ$] is obtuse


      
      For Problems 51–56, write algebraic expressions for the six trigonometric ratios of the angle.


      
        #figure(figph[triangle], alt: "triangle", caption: none)


        
          #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction s over 4 ,")[$sin ( θ ) = display(frac(s, 4)) ,$] #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction the square root of 16 minus s squared over 4 ,")[$" " cos ( θ ) = display(frac(sqrt(16 − s^(2)), 4)) ,$] #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction s over the square root of 16 minus s squared ,")[$" " tan ( θ ) = display(frac(s, sqrt(16 − s^(2)))) ,$] #math.equation(block: true, alt: "sec ⁡ open parenthesis θ close parenthesis equals the fraction 4 over the square root of 16 minus s squared ,")[$" " sec ( θ ) = display(frac(4, sqrt(16 − s^(2)))) ,$] #math.equation(block: true, alt: "csc ⁡ open parenthesis θ close parenthesis equals the fraction 4 over s ,")[$" " csc ( θ ) = display(frac(4, s)) ,$] #math.equation(block: true, alt: "cot ⁡ open parenthesis θ close parenthesis equals the fraction the square root of 16 minus s squared over s")[$" " cot ( θ ) = display(frac(sqrt(16 − s^(2)), s))$]


        
      
      
        #figure(figph[triangle], alt: "triangle", caption: none)


      
      
        #figure(figph[triangle], alt: "triangle", caption: none)


        
          #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction w over the square root of w squared plus 144 ,")[$sin ( θ ) = display(frac(w, sqrt(w^(2) + 144))) ,$] #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction minus 12 over the square root of w squared plus 144 ,")[$" " cos ( θ ) = display(frac(− 12, sqrt(w^(2) + 144))) ,$] #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction minus w over 12 ,")[$" " tan ( θ ) = display(frac(− w, 12)) ,$] #math.equation(block: true, alt: "sec ⁡ open parenthesis θ close parenthesis equals the fraction minus the square root of w squared plus 144 over 12 ,")[$" " sec ( θ ) = display(frac(− sqrt(w^(2) + 144), 12)) ,$] #math.equation(block: true, alt: "csc ⁡ open parenthesis θ close parenthesis equals the fraction the square root of w squared plus 144 over w ,")[$" " csc ( θ ) = display(frac(sqrt(w^(2) + 144), w)) ,$] #math.equation(block: true, alt: "cot ⁡ open parenthesis θ close parenthesis equals the fraction minus 12 over w")[$" " cot ( θ ) = display(frac(− 12, w))$]


        
      
      
        #figure(figph[triangle], alt: "triangle", caption: none)


      
      
        #math.equation(block: true, alt: "2 sin ⁡ open parenthesis α close parenthesis minus k equals 0 , the fraction π over 2 less than α less than π")[$2 sin ( α ) − k = 0 , " " display(frac(π, 2)) < α < π$]


        
          #math.equation(block: true, alt: "sin ⁡ open parenthesis α close parenthesis equals the fraction k over 2 ,")[$sin ( α ) = display(frac(k, 2)) ,$] #math.equation(block: true, alt: "cos ⁡ open parenthesis α close parenthesis equals the fraction minus the square root of 4 minus k squared over 2 ,")[$" " cos ( α ) = display(frac(− sqrt(4 − k^(2)), 2)) ,$] #math.equation(block: true, alt: "tan ⁡ open parenthesis α close parenthesis equals the fraction minus k over the square root of 4 minus k squared ,")[$" " tan ( α ) = display(frac(− k, sqrt(4 − k^(2)))) ,$] #math.equation(block: true, alt: "sec ⁡ open parenthesis α close parenthesis equals the fraction minus 2 over the square root of 4 minus k squared ,")[$" " sec ( α ) = display(frac(− 2, sqrt(4 − k^(2)))) ,$] #math.equation(block: true, alt: "csc ⁡ open parenthesis α close parenthesis equals the fraction 2 over k ,")[$" " csc ( α ) = display(frac(2, k)) ,$] #math.equation(block: true, alt: "cot ⁡ open parenthesis α close parenthesis equals the fraction minus the square root of 4 minus k squared over k")[$" " cot ( α ) = display(frac(− sqrt(4 − k^(2)), k))$]


        
      
      
        #math.equation(block: true, alt: "h cos ⁡ open parenthesis β close parenthesis minus 3 equals 0 , the fraction 3 π over 2 less than β less than 2 π")[$h cos ( β ) − 3 = 0 , " " display(frac(3 π, 2)) < β < 2 π$]


      
      For Problems 57–58, find all six trigonometric ratios of the arc #math.equation(block: false, alt: "θ")[$θ$]. Round to two places.


      
        #figure(figph[circle], alt: "circle", caption: none)


        
          #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis equals 0.3")[$sin ( θ ) = 0.3$], #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals minus 0.4")[$cos ( θ ) = − 0.4$], #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis equals minus 0.75")[$tan ( θ ) = − 0.75$], #math.equation(block: true, alt: "sec ⁡ open parenthesis θ close parenthesis equals minus 2.5")[$sec ( θ ) = − 2.5$], #math.equation(block: true, alt: "csc ⁡ open parenthesis θ close parenthesis approximately equals 3.33")[$csc ( θ ) ≈ 3.33$], #math.equation(block: true, alt: "cot ⁡ θ open parenthesis approximately equals close parenthesis minus 1.33")[$cot θ ( ≈ ) − 1.33$]


        
      
      
        #figure(figph[circle], alt: "circle", caption: none)


      
      For Problems 59–62, evaluate exactly.


      
        #math.equation(block: true, alt: "4 cot ⁡ open parenthesis the fraction 3 π over 4 close parenthesis minus sec squared ⁡ open parenthesis the fraction π over 3 close parenthesis")[$4 cot ( display(frac(3 π, 4)) ) − sec^(2) ( display(frac(π, 3)) )$]


        
          #math.equation(block: true, alt: "minus 8")[$− 8$]


        
      
      
        #math.equation(block: true, alt: "the fraction 1 over 2 csc ⁡ open parenthesis the fraction 2 π over 3 close parenthesis plus tan squared ⁡ open parenthesis the fraction 5 π over 6 close parenthesis")[$display(frac(1, 2)) csc ( display(frac(2 π, 3)) ) + tan^(2) ( display(frac(5 π, 6)) )$]


      
      
        #math.equation(block: true, alt: "csc ⁡ open parenthesis the fraction 7 π over 6 close parenthesis cos ⁡ open parenthesis the fraction 5 π over 4 close parenthesis")[$csc ( display(frac(7 π, 6)) ) cos ( display(frac(5 π, 4)) )$]


        
          #math.equation(block: true, alt: "the square root of 2")[$sqrt(2)$]


        
      
      
        #math.equation(block: true, alt: "sec ⁡ open parenthesis the fraction 7 π over 4 close parenthesis cot ⁡ open parenthesis the fraction 4 π over 3 close parenthesis")[$sec ( display(frac(7 π, 4)) ) cot ( display(frac(4 π, 3)) )$]


      
      For Problems 63–64, find all solutions between #math.equation(block: false, alt: "0")[$0$] and #math.equation(block: false, alt: "2 π")[$2 π$]. Round your solutions to tenths.


      
        #math.equation(block: true, alt: "3 csc ⁡ open parenthesis θ close parenthesis plus 2 equals 12")[$3 csc ( θ ) + 2 = 12$]


        
          #math.equation(block: true, alt: "θ approximately equals 2.8 , θ approximately equals 0.30")[$θ ≈ 2.8 , " " θ ≈ 0.30$]


        
      
      
        #math.equation(block: true, alt: "5 cot ⁡ open parenthesis θ close parenthesis plus 15 equals minus 3")[$5 cot ( θ ) + 15 = − 3$]


      
      For Problems 65–70, sketch a graph of each function. Then choose the function or functions described by each statement.


      #math.equation(block: true, alt: "y equals sec ⁡ open parenthesis x close parenthesis y equals csc ⁡ open parenthesis x close parenthesis y equals cot ⁡ open parenthesis x close parenthesis")[$y = sec ( x ) " " " " " " " " " " " " " " " " " " " " " " y = csc ( x ) " " " " " " " " " " " " " " " " " " " " " " y = cot ( x )$]
      #math.equation(block: true, alt: "y equals cos to the power minus 1 ⁡ open parenthesis x close parenthesis y equals sin to the power minus 1 ⁡ open parenthesis x close parenthesis y equals tan to the power minus 1 ⁡ open parenthesis x close parenthesis")[$y = cos^(− 1) ( x ) " " " " " " " " " " " " " " " " y = sin^(− 1) ( x ) " " " " " " " " " " " " y = tan^(− 1) ( x )$]
      
        The graph has vertical asymptotes at multiples of #math.equation(block: false, alt: "π")[$π$].


        
          #math.equation(block: false, alt: "y equals csc ⁡ open parenthesis x close parenthesis")[$y = csc ( x )$] or #math.equation(block: false, alt: "y equals cot ⁡ open parenthesis x close parenthesis")[$y = cot ( x )$]


        
      
      
        The graph has a horizontal asymptote at #math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$].


      
      
        The function values are the reciprocals of #math.equation(block: false, alt: "y equals cos ⁡ open parenthesis x close parenthesis")[$y = cos ( x )$].


        
          #math.equation(block: true, alt: "y equals sec ⁡ open parenthesis x close parenthesis")[$y = sec ( x )$]


        
      
      
        The function is defined only for #math.equation(block: false, alt: "x")[$x$]-values between #math.equation(block: false, alt: "minus 1")[$− 1$] and #math.equation(block: false, alt: "1")[$1$], inclusive.


      
      
        None of the function values lie between #math.equation(block: false, alt: "minus 1")[$− 1$] and #math.equation(block: false, alt: "1")[$1$].


        
          #math.equation(block: false, alt: "y equals sec ⁡ open parenthesis x close parenthesis")[$y = sec ( x )$] or #math.equation(block: false, alt: "y equals csc ⁡ open parenthesis x close parenthesis")[$y = csc ( x )$]


        
      
      
        The graph includes the origin.


      
      For Problems 71–74,


      + Graph the function on the interval #math.equation(block: false, alt: "minus 2 π less than or equal to x less than or equal to 2 π")[$− 2 π ≤ " " x ≤ 2 π$], and use the graph to write the function in a simpler form.
+ Verify your conjecture algebraically.


      
        #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals tan ⁡ open parenthesis x close parenthesis open bracket cos ⁡ open parenthesis x close parenthesis minus cot ⁡ open parenthesis x close parenthesis close bracket")[$f ( x ) = tan ( x ) [ cos ( x ) − cot ( x ) ]$]


        
          #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals sin ⁡ open parenthesis x close parenthesis minus 1")[$f ( x ) = sin ( x ) − 1$]


        
      
      
        #math.equation(block: true, alt: "g open parenthesis x close parenthesis equals csc ⁡ open parenthesis x close parenthesis minus cot ⁡ open parenthesis x close parenthesis cos ⁡ open parenthesis x close parenthesis")[$g ( x ) = csc ( x ) − cot ( x ) cos ( x )$]


      
      
        #math.equation(block: true, alt: "G open parenthesis x close parenthesis equals sin ⁡ x open parenthesis sec ⁡ open parenthesis x close parenthesis minus csc ⁡ open parenthesis x close parenthesis close parenthesis")[$G ( x ) = sin x ( sec ( x ) − csc ( x ) )$]


        
          #math.equation(block: true, alt: "G open parenthesis x close parenthesis equals tan ⁡ open parenthesis x close parenthesis minus 1")[$G ( x ) = tan ( x ) − 1$]


        
      
      
        #math.equation(block: true, alt: "F open parenthesis x close parenthesis equals the fraction 1 over 2 open parenthesis the fraction cos ⁡ open parenthesis x close parenthesis over 1 plus sin ⁡ open parenthesis x close parenthesis plus the fraction 1 plus sin ⁡ open parenthesis x close parenthesis over cos ⁡ open parenthesis x close parenthesis close parenthesis")[$F ( x ) = display(frac(1, 2)) ( display(frac(cos ( x ), 1 + sin ( x ))) + display(frac(1 + sin ( x ), cos ( x ))) )$]


      
      For Problems 75–78, simplify the expression.


      
        #math.equation(block: true, alt: "1 minus the fraction sin ⁡ open parenthesis x close parenthesis over csc ⁡ open parenthesis x close parenthesis")[$1 − display(frac(sin ( x ), csc ( x )))$]


        
          #math.equation(block: true, alt: "cos squared ⁡ open parenthesis x close parenthesis")[$cos^(2) ( x )$]


        
      
      
        #math.equation(block: true, alt: "the fraction sin ⁡ open parenthesis x close parenthesis over csc ⁡ open parenthesis x close parenthesis plus the fraction cos ⁡ open parenthesis x close parenthesis over sec ⁡ open parenthesis x close parenthesis")[$display(frac(sin ( x ), csc ( x ))) + display(frac(cos ( x ), sec ( x )))$]


      
      
        #math.equation(block: true, alt: "the fraction 2 plus tan squared ⁡ open parenthesis B close parenthesis over sec squared ⁡ open parenthesis B close parenthesis minus 1")[$display(frac(2 + tan^(2) ( B ), sec^(2) ( B ))) − 1$]


        
          #math.equation(block: true, alt: "cos squared ⁡ open parenthesis B close parenthesis")[$cos^(2) ( B )$]


        
      
      
        #math.equation(block: true, alt: "the fraction csc ⁡ open parenthesis t close parenthesis over tan ⁡ open parenthesis t close parenthesis plus cot ⁡ open parenthesis t close parenthesis")[$display(frac(csc ( t ), tan ( t ) + cot ( t )))$]


      
      For Problems 79–82, use the suggested substitution to simplify the expression.


      
        #math.equation(block: true, alt: "the fraction the square root of 16 plus x squared over x , x equals 4 tan ⁡ open parenthesis θ close parenthesis")[$display(frac(sqrt(16 + x^(2)), x)) , " " " " x = 4 tan ( θ )$]


        
          #math.equation(block: true, alt: "csc ⁡ open parenthesis θ close parenthesis")[$csc ( θ )$]


        
      
      
        #math.equation(block: true, alt: "x the square root of 4 minus x squared , x equals 2 sin ⁡ open parenthesis θ close parenthesis")[$x sqrt(4 − x^(2)) , " " " " x = 2 sin ( θ )$]


      
      
        #math.equation(block: true, alt: "the fraction x squared minus 3 over x , x equals the square root of 3 sec ⁡ open parenthesis θ close parenthesis")[$display(frac(x^(2) − 3, x)) , " " " " x = sqrt(3) sec ( θ )$]


        
          #math.equation(block: true, alt: "the square root of 3 tan ⁡ open parenthesis θ close parenthesis sin ⁡ open parenthesis θ close parenthesis")[$sqrt(3) tan ( θ ) sin ( θ )$]


        
      
      
        #math.equation(block: true, alt: "the fraction x over the square root of x squared plus 2 , x equals the square root of 2 tan ⁡ open parenthesis θ close parenthesis")[$display(frac(x, sqrt(x^(2) + 2))) , " " " " x = sqrt(2) tan ( θ )$]


      
      
        This problem outlines a geometric proof of difference of angles formula for tangent.


        + In the figure below left, #math.equation(block: false, alt: "α equals ∠ A B C")[$α = upright(∠) A B C$] and #math.equation(block: false, alt: "β equals ∠ D B C")[$β = upright(∠) D B C$].  Write expressions in terms of #math.equation(block: false, alt: "α")[$α$] and #math.equation(block: false, alt: "β")[$β$] for the sides #math.equation(block: false, alt: "A C , D C ,")[$A C , " " D C ,$] and #math.equation(block: false, alt: "A D")[$A D$].#figure(figph[triangles], alt: "triangles", caption: none)
+ In the figure above right, explain why #math.equation(block: false, alt: "△ A B C")[$upright(△) A B C$] is similar to #math.equation(block: false, alt: "△ F B E")[$upright(△) F B E$].
+ Explain why #math.equation(block: false, alt: "∠ F D C equals α")[$upright(∠) F D C = α$].
+ Write an expression in terms of #math.equation(block: false, alt: "α")[$α$] and #math.equation(block: false, alt: "β")[$β$] for side #math.equation(block: false, alt: "C F")[$C F$].
+ Explain why #math.equation(block: false, alt: "△ F B E")[$upright(△) F B E$] is similar to #math.equation(block: false, alt: "△ A D E")[$upright(△) A D E$].
+ Justify each equality in the statement
        #math.equation(block: true, alt: "tan ⁡ open parenthesis α minus β close parenthesis equals the fraction D E over B E equals the fraction A D over B F equals the fraction tan ⁡ open parenthesis α close parenthesis minus tan ⁡ open parenthesis β close parenthesis over 1 plus tan ⁡ open parenthesis α close parenthesis tan ⁡ open parenthesis β close parenthesis")[$tan ( α − β ) = display(frac(D E, B E)) = display(frac(A D, B F)) = display(frac(tan ( α ) − tan ( β ), 1 + tan ( α ) tan ( β )))$]


        
          + #math.equation(block: false, alt: "A C equals tan ⁡ open parenthesis α close parenthesis , D C equals tan ⁡ open parenthesis β close parenthesis , A D equals tan ⁡ open parenthesis α close parenthesis minus tan ⁡ open parenthesis β close parenthesis")[$A C = tan ( α ) , " " D C = tan ( β ) , " " A D = tan ( α ) − tan ( β )$]
+ They are right triangles that share #math.equation(block: false, alt: "∠ B")[$upright(∠) B$].
+ #math.equation(block: false, alt: "∠ A equals ∠ F , ∠ B")[$upright(∠) A = upright(∠) F , " " upright(∠) B$] is the complement of #math.equation(block: false, alt: "∠ A ,")[$upright(∠) A ,$] and #math.equation(block: false, alt: "∠ F D C")[$upright(∠) F D C$] is the complement of #math.equation(block: false, alt: "∠ F")[$upright(∠) F$].
+ #math.equation(block: false, alt: "the fraction C F over C D equals tan ⁡ open parenthesis α close parenthesis ,")[$display(frac(C F, C D)) = tan ( α ) ,$] so #math.equation(block: false, alt: "C F equals tan ⁡ open parenthesis α close parenthesis tan ⁡ open parenthesis β close parenthesis")[$C F = tan ( α ) tan ( β )$]
+ They are right triangles with #math.equation(block: false, alt: "∠ A equals ∠ F")[$upright(∠) A = upright(∠) F$].
+ #math.equation(block: false, alt: "∠ E B D equals α minus β ,")[$upright(∠) E B D = α − β ,$] so #math.equation(block: false, alt: "tan ⁡ open parenthesis α minus β close parenthesis equals the fraction opp over adj equals the fraction D E over B E ; the fraction D E over B E")[$tan ( α − β ) = display(frac("opp", "adj")) = display(frac(D E, B E)) ; " " " " display(frac(D E, B E))$] and #math.equation(block: false, alt: "the fraction A D over B F")[$display(frac(A D, B F))$] are ratios of corresponding sides of similar triangles; 
    #math.equation(block: false, alt: "A D equals tan ⁡ open parenthesis α close parenthesis minus tan ⁡ open parenthesis β close parenthesis")[$A D = tan ( α ) − tan ( β )$] by part (a), #math.equation(block: false, alt: "B F equals B C plus C F equals 1 plus tan ⁡ open parenthesis α close parenthesis tan ⁡ open parenthesis β close parenthesis")[$B F = B C + C F = 1 + tan ( α ) tan ( β )$] by part (d).


        
      
      
        Let #math.equation(block: false, alt: "L sub 1")[$L_(1)$] and #math.equation(block: false, alt: "L sub 2")[$L_(2)$] be two lines with slopes #math.equation(block: false, alt: "m sub 1")[$m_(1)$] and #math.equation(block: false, alt: "m sub 2")[$m_(2)$], respectively, and let #math.equation(block: false, alt: "θ")[$θ$] be the acute angle formed between the two lines. Use an identity to show that


        #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction m sub 2 minus m sub 1 over 1 plus m sub 1 m sub 2")[$tan ( θ ) = display(frac(m_(2) − m_(1), 1 + m_(1) m_(2)))$]
      
      For Problems 85–86, use the fact that if #math.equation(block: false, alt: "θ")[$θ$] is one angle of a triangle and #math.equation(block: false, alt: "s")[$s$] is the length of the opposite side, then the diameter of the circumscribing circle is


      #math.equation(block: true, alt: "d equals s csc ⁡ open parenthesis θ close parenthesis")[$d = s csc ( θ )$]
      Round your answers to the nearest hundredth.


      #figure(figph[circle circumscribing a triangle], alt: "circle circumscribing a triangle", caption: none)


      
        In the figure above, find the diameter of the circumscribing circle, the angle #math.equation(block: false, alt: "α")[$α$], and the sides #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "b")[$b$].


        
          #math.equation(block: true, alt: "d equals 25 csc ⁡ open parenthesis 112 degrees close parenthesis , α equals 45 degrees , a approximately equals 19.07 , b approximately equals 10.54")[$d = 25 csc ( 112^(∘) ) , " " α = 45^(∘) , " " a ≈ 19.07 , " " b ≈ 10.54$]


        
      
      
        A triangle has one side of length 17cm and the angle opposite is #math.equation(block: false, alt: "26 degrees")[$26^(∘)$]. Find the diameter of the circle that circumscribes the triangle.