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== 7.4#h(0.6em)Chapter 7 Summary and Review


    
      
=== Key Concepts


      + Changes to the amplitude, period, and midline of the basic sine and cosine graphs are called #strong[transformations].  Changing the midline shifts the graph vertically, changing the amplitude stretches or compresses the graph vertically, and changing the period stretches or compresses the graph horizontally.
+ The order in which we apply transformations to a function makes a difference in the graph.
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Amplitude, Period, and Midline of Sinusoidal Functions]

+ The graph of
      #math.equation(block: true, alt: "y equals A cos ⁡ open parenthesis x close parenthesis or y equals A sin ⁡ open parenthesis x close parenthesis")[$y = A cos ( x ) " " " " " " " " "or" " " " " " " " " y = A sin ( x )$]
    has #strong[amplitude] #math.equation(block: false, alt: "vertical bar A vertical bar")[$| A |$].
+ The graph of
      #math.equation(block: true, alt: "y equals cos ⁡ open parenthesis B x close parenthesis or y equals sin ⁡ open parenthesis B x close parenthesis")[$y = cos ( B x ) " " " " " " " " "or" " " " " " " " " y = sin ( B x )$]
    has #strong[period] #math.equation(block: false, alt: "the fraction 2 π over B")[$display(frac(2 π, B))$].
+ The graph of
      #math.equation(block: true, alt: "y equals k plus cos ⁡ open parenthesis x close parenthesis or y equals k plus sin ⁡ open parenthesis x close parenthesis")[$y = k + cos ( x ) " " " " " " " " "or" " " " " " " " " y = k + sin ( x )$]
    has #strong[midline] #math.equation(block: false, alt: "y equals k")[$y = k$].
]
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Horizontal Shifts]

The graphs of

#math.equation(block: true, alt: "y equals sin ⁡ open parenthesis x minus h close parenthesis and t equals cos ⁡ open parenthesis x minus h close parenthesis")[$y = sin ( x − h ) " " " " " " " " "and" " " " " " " " " t = cos ( x − h )$]are #emph[shifted horizontally] compared to the graphs of #math.equation(block: false, alt: "y equals sin ⁡ open parenthesis x close parenthesis")[$y = sin ( x )$] and #math.equation(block: false, alt: "y equals cos ⁡ open parenthesis x close parenthesis")[$y = cos ( x )$].

- If #math.equation(block: false, alt: "h greater than 0")[$h > 0$], the graph is shifted to the right.
- If #math.equation(block: false, alt: "h less than 0")[$h < 0$], the graph is shifted to the left.
]
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Standard Form for Sinusoidal Functions]

The graphs of the functions

#math.equation(block: true, alt: "y equals A sin ⁡ B open parenthesis x minus h close parenthesis plus k and y equals A cos ⁡ B open parenthesis x minus h close parenthesis plus k")[$y = A sin B ( x − h ) + k " " " " " " " " "and" " " " " " " " " y = A cos B ( x − h ) + k$]are transformations of the sine and cosine graphs.

+ The amplitude is #math.equation(block: false, alt: "vertical bar A vertical bar")[$| A |$].
+ The midline is #math.equation(block: false, alt: "y equals k")[$y = k$].
+ The period is #math.equation(block: false, alt: "the fraction 2 π over vertical bar B vertical bar , B not equal to 0")[$display(frac(2 π, | B |)) , " " " " B ≠ 0$].
+ The horizontal shift is #math.equation(block: false, alt: "h")[$h$] units to the right if #math.equation(block: false, alt: "h")[$h$] is positive, and #math.equation(block: false, alt: "h")[$h$] 
     units to the left if #math.equation(block: false, alt: "h")[$h$] is negative.
]
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Solutions of Trigonometric Equations]

+ The equation #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals k , minus 1 less than k less than 1")[$cos ( θ ) = k , " " " " − 1 < k < 1$], has two solutions between #math.equation(block: false, alt: "0")[$0$] and #math.equation(block: false, alt: "2 π")[$2 π$]:
      #math.equation(block: true, alt: "θ sub 1 equals cos to the power minus 1 ⁡ open parenthesis k close parenthesis and θ sub 2 equals 2 π minus θ sub 1")[$θ_(1) = cos^(− 1) ( k ) " " " " "and" " " " " θ_(2) = 2 π − θ_(1)$]
+ The equation #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals k , minus 1 less than k less than 1")[$sin ( θ ) = k , " " " " − 1 < k < 1$], has two solutions between #math.equation(block: false, alt: "0")[$0$] and #math.equation(block: false, alt: "2 π")[$2 π$]:
        #math.equation(block: true, alt: "If k greater than 0 : θ sub 1 equals sin to the power minus 1 ⁡ open parenthesis k close parenthesis and θ sub 2 equals π minus θ sub 1")[$"If" " " k > 0 : " " " " θ_(1) = sin^(− 1) ( k ) " " " " "and" " " " " θ_(2) = π − θ_(1)$]
        #math.equation(block: true, alt: "If k less than 0 : θ sub 1 equals sin to the power minus 1 ⁡ open parenthesis k close parenthesis plus 2 π and θ sub 2 equals π minus sin to the power minus 1 ⁡ open parenthesis k close parenthesis")[$"If" " " k < 0 : " " " " θ_(1) = sin^(− 1) ( k ) + 2 π " " " " "and" " " " " θ_(2) = π − sin^(− 1) ( k )$]
+ The equation #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals k")[$tan ( θ ) = k$], has two solutions between #math.equation(block: false, alt: "0")[$0$] and #math.equation(block: false, alt: "2 π")[$2 π$]:
        #math.equation(block: true, alt: "If k greater than 0 : θ sub 1 equals tan to the power minus 1 ⁡ open parenthesis k close parenthesis and θ sub 2 equals π plus θ sub 1")[$"If" " " k > 0 : " " " " θ_(1) = tan^(− 1) ( k ) " " " " "and" " " " " θ_(2) = π + θ_(1)$]
        #math.equation(block: true, alt: "If k less than 0 : θ sub 1 equals tan to the power minus 1 ⁡ open parenthesis k close parenthesis plus π and θ sub 2 equals π plus θ sub 1")[$"If" " " k < 0 : " " " " θ_(1) = tan^(− 1) ( k ) + π " " " " "and" " " " " θ_(2) = π + θ_(1)$]
]
+ If #math.equation(block: false, alt: "n")[$n$] is a positive integer, the equations #math.equation(block: false, alt: "sin ⁡ open parenthesis n θ close parenthesis equals k")[$sin ( n θ ) = k$] and #math.equation(block: false, alt: "cos ⁡ open parenthesis n θ close parenthesis equals k")[$cos ( n θ ) = k$] each have #math.equation(block: false, alt: "2 n")[$2 n$] solutions between #math.equation(block: false, alt: "0")[$0$] and #math.equation(block: false, alt: "2 π")[$2 π$], for #math.equation(block: false, alt: "minus 1 less than k less than 1")[$− 1 < k < 1$].
+ The equation #math.equation(block: false, alt: "tan ⁡ open parenthesis n θ close parenthesis equals k")[$tan ( n θ ) = k$] has one solution in each cycle of the graph.
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Using a Substitution to Solve Trigonometric Equations]

To solve the equation #math.equation(block: false, alt: "sin ⁡ open parenthesis B x plus C close parenthesis equals k")[$sin ( B x + C ) = k$]  or  #math.equation(block: false, alt: "cos ⁡ open parenthesis B x plus C close parenthesis equals k")[$cos ( B x + C ) = k$]:

+ Substitute #math.equation(block: false, alt: "θ equals B x plus C")[$θ = B x + C$],  and find two solutions for #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals k")[$sin ( θ ) = k$]  or  #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals k")[$cos ( θ ) = k$].
+ Replace #math.equation(block: false, alt: "θ")[$θ$] by #math.equation(block: false, alt: "B x plus C")[$B x + C$] in each solution, and solve for #math.equation(block: false, alt: "x")[$x$].
+ Find the other solutions by adding multiples of #math.equation(block: false, alt: "the fraction 2 π over B")[$display(frac(2 π, B))$] to the first two solutions.

To solve the equation #math.equation(block: false, alt: "tan ⁡ open parenthesis B x plus C close parenthesis equals k")[$tan ( B x + C ) = k$]:

+ Substitute #math.equation(block: false, alt: "θ equals B x plus C")[$θ = B x + C$],  and find one solution for #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals k")[$tan ( θ ) = k$].
+ Replace #math.equation(block: false, alt: "θ")[$θ$] by #math.equation(block: false, alt: "B x plus C")[$B x + C$] and solve for #math.equation(block: false, alt: "x")[$x$].
+ Find the other solutions by adding multiples of #math.equation(block: false, alt: "the fraction π over B")[$display(frac(π, B))$] to the first solution.
]


    
    
      
=== Review Problems


      For Problems 1–4, state the amplitude, period, and midline of the graph.


      
        #math.equation(block: true, alt: "y equals 4 minus 2 sin ⁡ open parenthesis 3 x close parenthesis")[$y = 4 − 2 sin ( 3 x )$]


        
          amp: #math.equation(block: false, alt: "2")[$2$], period: #math.equation(block: false, alt: "the fraction 2 π over 3")[$display(frac(2 π, 3))$]; mid: #math.equation(block: false, alt: "y equals 4")[$y = 4$]


        
      
      
        #math.equation(block: true, alt: "y equals 1 plus 5 cos ⁡ open parenthesis the fraction x over 2 close parenthesis")[$y = 1 + 5 cos ( display(frac(x, 2)) )$]


      
      
        #math.equation(block: true, alt: "y equals 2.5 cos ⁡ open parenthesis π x close parenthesis minus 2")[$y = 2.5 cos ( π x ) − 2$]


        
          amp: #math.equation(block: false, alt: "2.5")[$2.5$], period: #math.equation(block: false, alt: "2")[$2$]; mid: #math.equation(block: false, alt: "y equals minus 2")[$y = − 2$]


        
      
      
        #math.equation(block: true, alt: "y equals 0.8 sin ⁡ open parenthesis the fraction π x over 6 close parenthesis plus 0.3")[$y = 0.8 sin ( display(frac(π x, 6)) ) + 0.3$]


      
      For Problems 5–8, use transformations to sketch graphs of the functions.


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


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


        
      
      
        #math.equation(block: true, alt: "g open parenthesis t close parenthesis equals minus 4 minus 2 sin ⁡ open parenthesis t close parenthesis")[$g ( t ) = − 4 − 2 sin ( t )$]


      
      
        #math.equation(block: true, alt: "h open parenthesis w close parenthesis equals minus 4 sin ⁡ open parenthesis π w close parenthesis")[$h ( w ) = − 4 sin ( π w )$]


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


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


      
      For Problems 9–12, write an equation for the graph using sine or cosine.


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


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


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


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


        
          #math.equation(block: true, alt: "y equals 4 minus 3 sin ⁡ open parenthesis the fraction x over 4 close parenthesis")[$y = 4 − 3 sin ( display(frac(x, 4)) )$]


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


      
      For Problems 13–16, complete the table of values and sketch a graph of the function.


      
        #math.equation(block: true, alt: "y equals sin ⁡ open parenthesis the fraction x over 2 plus the fraction π over 6 close parenthesis")[$y = sin ( display(frac(x, 2)) + display(frac(π, 6)) )$]


        + What are the period and the horizontal shift? #linebreak() (Hint: Factor out #math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$] from #math.equation(block: false, alt: "the fraction x over 2 plus the fraction π over 6")[$display(frac(x, 2)) + display(frac(π, 6))$].)
+ Fill in the table of values.#figure(table(
  columns: 4, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "the fraction x over 2")[$display(frac(x, 2))$]], [#math.equation(block: false, alt: "the fraction x over 2 plus the fraction π over 6")[$display(frac(x, 2)) + display(frac(π, 6))$]], [#math.equation(block: false, alt: "sin ⁡ open parenthesis the fraction x over 2 plus the fraction π over 6 close parenthesis")[$sin ( display(frac(x, 2)) + display(frac(π, 6)) )$]]),
  [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction minus π over 6")[$display(frac(− π, 6))$]], [$#hide($0000$)$],
  [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "0")[$0$]], [$#hide($0000$)$],
  [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction π over 6")[$display(frac(π, 6))$]], [$#hide($0000$)$],
  [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [$#hide($0000$)$],
  [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$]], [$#hide($0000$)$],
  [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [$#hide($0000$)$],
  [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction 2 π over 3")[$display(frac(2 π, 3))$]], [$#hide($0000$)$],
))
+ Sketch the graph.#figure(figph[grid], alt: "grid", caption: none)
+ Solve #math.equation(block: false, alt: "sin ⁡ open parenthesis the fraction x over 2 plus the fraction π over 6 close parenthesis equals 1 , for the fraction minus 2 π over 3 less than or equal to x less than or equal to the fraction 2 π over 3")[$" " " " sin ( display(frac(x, 2)) + display(frac(π, 6)) ) = 1 , " " " " "for" " " " " display(frac(− 2 π, 3)) ≤ x ≤ display(frac(2 π, 3))$]
+ Solve #math.equation(block: false, alt: "sin ⁡ open parenthesis the fraction x over 2 plus the fraction π over 6 close parenthesis equals 0 , for the fraction minus 2 π over 3 less than or equal to x less than or equal to the fraction 2 π over 3")[$" " " " sin ( display(frac(x, 2)) + display(frac(π, 6)) ) = 0 , " " " " "for" " " " " display(frac(− 2 π, 3)) ≤ x ≤ display(frac(2 π, 3))$]


        
          + period: #math.equation(block: false, alt: "4 π")[$4 π$], shift: #math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$] left
+ #figure(table(
  columns: 4, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "the fraction x over 2")[$display(frac(x, 2))$]], [#math.equation(block: false, alt: "the fraction x over 2 plus the fraction π over 6")[$display(frac(x, 2)) + display(frac(π, 6))$]], [#math.equation(block: false, alt: "sin ⁡ open parenthesis the fraction x over 2 plus the fraction π over 6 close parenthesis")[$sin ( display(frac(x, 2)) + display(frac(π, 6)) )$]]),
  [#math.equation(block: false, alt: "the fraction minus 2 π over 3")[$display(frac(− 2 π, 3))$]], [#math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$]], [#math.equation(block: false, alt: "the fraction minus π over 6")[$display(frac(− π, 6))$]], [#math.equation(block: false, alt: "the fraction minus 1 over 2")[$display(frac(− 1, 2))$]],
  [#math.equation(block: false, alt: "the fraction minus π over 3")[$display(frac(− π, 3))$]], [#math.equation(block: false, alt: "the fraction minus π over 6")[$display(frac(− π, 6))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "0")[$0$]],
  [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 6")[$display(frac(π, 6))$]], [#math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$]],
  [#math.equation(block: false, alt: "the fraction π over 6")[$display(frac(π, 6))$]], [#math.equation(block: false, alt: "the fraction π over 12")[$display(frac(π, 12))$]], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [#math.equation(block: false, alt: "the fraction 1 over the square root of 2")[$display(frac(1, sqrt(2)))$]],
  [#math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$]], [#math.equation(block: false, alt: "the fraction π over 6")[$display(frac(π, 6))$]], [#math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$]], [#math.equation(block: false, alt: "the fraction the square root of 3 over 2")[$display(frac(sqrt(3), 2))$]],
  [#math.equation(block: false, alt: "the fraction 2 π over 3")[$display(frac(2 π, 3))$]], [#math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "1")[$1$]],
  [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "the fraction 2 π over 3")[$display(frac(2 π, 3))$]], [#math.equation(block: false, alt: "the fraction the square root of 3 over 2")[$display(frac(sqrt(3), 2))$]],
))
+ #figure(figph[sinusoidal graph], alt: "sinusoidal graph", caption: none)
+ #math.equation(block: false, alt: "the fraction 2 π over 3")[$display(frac(2 π, 3))$]
+ #math.equation(block: false, alt: "the fraction minus π over 3")[$display(frac(− π, 3))$]


        
      
      
        #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 2 cos ⁡ open parenthesis 3 x minus the fraction π over 2 close parenthesis plus 5")[$f ( x ) = 2 cos ( 3 x − display(frac(π, 2)) ) + 5$]


        + What are the midline, period, horizontal shift, and amplitude?
+ Fill in the table of values.#figure(table(
  columns: 5, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "3 x")[$3 x$]], [#math.equation(block: false, alt: "3 x minus the fraction π over 2")[$3 x − display(frac(π, 2))$]], [#math.equation(block: false, alt: "cos ⁡ open parenthesis 3 x minus the fraction π over 2 close parenthesis")[$cos ( 3 x − display(frac(π, 2)) )$]], [#math.equation(block: false, alt: "2 cos ⁡ open parenthesis 3 x minus the fraction π over 2 close parenthesis plus 5")[$2 cos ( 3 x − display(frac(π, 2)) ) + 5$]]),
  [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "0")[$0$]], [$#hide($0000$)$], [$#hide($0000$)$],
  [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [$#hide($0000$)$], [$#hide($0000$)$],
  [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "π")[$π$]], [$#hide($0000$)$], [$#hide($0000$)$],
  [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [$#hide($0000$)$], [$#hide($0000$)$],
  [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "2 π")[$2 π$]], [$#hide($0000$)$], [$#hide($0000$)$],
))
+ Sketch the graph.#figure(figph[grid], alt: "grid", caption: none)
+ Solve #math.equation(block: false, alt: "2 cos ⁡ open parenthesis 3 x minus the fraction π over 2 close parenthesis plus 5 equals 7 , for 0 less than or equal to x less than or equal to 2 π")[$" " " " 2 cos ( 3 x − display(frac(π, 2)) ) + 5 = 7 , " " " " "for" " " " " 0 ≤ x ≤ 2 π$]
+ Solve #math.equation(block: false, alt: "2 cos ⁡ open parenthesis 3 x minus the fraction π over 2 close parenthesis plus 5 equals 5 , for 0 less than or equal to x less than or equal to 2 π")[$" " " " 2 cos ( 3 x − display(frac(π, 2)) ) + 5 = 5 , " " " " "for" " " " " 0 ≤ x ≤ 2 π$]


      
      
        #math.equation(block: true, alt: "y equals 20 minus 5 cos ⁡ open parenthesis the fraction π over 30 x close parenthesis")[$y = 20 − 5 cos ( display(frac(π, 30)) x )$]


        + What are the midline, period, horizontal shift, and amplitude?
+ Fill in the table of values.#figure(table(
  columns: 4, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "the fraction π over 30 x")[$display(frac(π, 30)) x$]], [#math.equation(block: false, alt: "cos ⁡ open parenthesis the fraction π over 30 x close parenthesis")[$cos ( display(frac(π, 30)) x )$]], [#math.equation(block: false, alt: "20 minus 5 cos ⁡ open parenthesis the fraction π over 30 x close parenthesis")[$20 − 5 cos ( display(frac(π, 30)) x )$]]),
  [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction minus π over 6")[$display(frac(− π, 6))$]], [$#hide($0000$)$], [$#hide($0000$)$],
  [$#hide($0000$)$], [#math.equation(block: false, alt: "0")[$0$]], [$#hide($0000$)$], [$#hide($0000$)$],
  [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction π over 6")[$display(frac(π, 6))$]], [$#hide($0000$)$], [$#hide($0000$)$],
  [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$]], [$#hide($0000$)$], [$#hide($0000$)$],
  [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [$#hide($0000$)$], [$#hide($0000$)$],
  [$#hide($0000$)$], [#math.equation(block: false, alt: "π")[$π$]], [$#hide($0000$)$], [$#hide($0000$)$],
))
+ Sketch the graph.#figure(figph[grid], alt: "grid", caption: none)
+ Solve #math.equation(block: false, alt: "20 minus 5 cos ⁡ open parenthesis the fraction π over 30 x close parenthesis equals 25 , for 0 less than or equal to x less than or equal to 60")[$" " " " 20 − 5 cos ( display(frac(π, 30)) x ) = 25 , " " " " "for" " " " " 0 ≤ x ≤ 60$]
+ Solve #math.equation(block: false, alt: "20 minus 5 cos ⁡ open parenthesis the fraction π over 30 x close parenthesis equals 20 , for 0 less than or equal to x less than or equal to 60")[$" " " " 20 − 5 cos ( display(frac(π, 30)) x ) = 20 , " " " " "for" " " " " 0 ≤ x ≤ 60$]


        
          + mid: #math.equation(block: false, alt: "y equals 20")[$y = 20$], period: 0, amp: 5
+ Fill in the table of values.#figure(table(
  columns: 4, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "the fraction π over 30 x")[$display(frac(π, 30)) x$]], [#math.equation(block: false, alt: "cos ⁡ open parenthesis the fraction π over 30 x close parenthesis")[$cos ( display(frac(π, 30)) x )$]], [#math.equation(block: false, alt: "20 minus 5 cos ⁡ open parenthesis the fraction π over 30 x close parenthesis")[$20 − 5 cos ( display(frac(π, 30)) x )$]]),
  [#math.equation(block: false, alt: "minus 5")[$− 5$]], [#math.equation(block: false, alt: "the fraction minus π over 6")[$display(frac(− π, 6))$]], [#math.equation(block: false, alt: "the fraction the square root of 3 over 2")[$display(frac(sqrt(3), 2))$]], [#math.equation(block: false, alt: "20 minus the fraction the square root of 3 over 2")[$20 − display(frac(sqrt(3), 2))$]],
  [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "15")[$15$]],
  [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "the fraction π over 6")[$display(frac(π, 6))$]], [#math.equation(block: false, alt: "the fraction the square root of 3 over 2")[$display(frac(sqrt(3), 2))$]], [#math.equation(block: false, alt: "20 minus the fraction the square root of 3 over 2")[$20 − display(frac(sqrt(3), 2))$]],
  [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$]], [#math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$]], [#math.equation(block: false, alt: "17.5")[$17.5$]],
  [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "20")[$20$]],
  [#math.equation(block: false, alt: "50")[$50$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "25")[$25$]],
))
+ #figure(figph[sinusoidal graph], alt: "sinusoidal graph", caption: none)
+ 30
+ 15, 45


        
      
      
        #math.equation(block: true, alt: "y equals 50 minus 50 cos ⁡ open parenthesis 2 π x close parenthesis")[$y = 50 − 50 cos ( 2 π x )$]


        + What are the midline, period, horizontal shift, and amplitude?
+ Fill in the table of values.#figure(table(
  columns: 4, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "2 π x")[$2 π x$]], [#math.equation(block: false, alt: "cos ⁡ open parenthesis 2 π x close parenthesis")[$cos ( 2 π x )$]], [#math.equation(block: false, alt: "50 minus 50 cos ⁡ open parenthesis 2 π x close parenthesis")[$50 − 50 cos ( 2 π x )$]]),
  [$#hide($0000$)$], [#math.equation(block: false, alt: "0")[$0$]], [$#hide($0000$)$], [$#hide($0000$)$],
  [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [$#hide($0000$)$], [$#hide($0000$)$],
  [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$]], [$#hide($0000$)$], [$#hide($0000$)$],
  [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [$#hide($0000$)$], [$#hide($0000$)$],
  [$#hide($0000$)$], [#math.equation(block: false, alt: "π")[$π$]], [$#hide($0000$)$], [$#hide($0000$)$],
  [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction minus π over 3")[$display(frac(− π, 3))$]], [$#hide($0000$)$], [$#hide($0000$)$],
))
+ Sketch the graph.#figure(figph[grid], alt: "grid", caption: none)
+ Solve #math.equation(block: false, alt: "50 minus 50 cos ⁡ open parenthesis 2 π x close parenthesis equals 50 , for minus 1 less than or equal to x less than or equal to 1")[$" " " " 50 − 50 cos ( 2 π x ) = 50 , " " " " "for" " " " " − 1 ≤ x ≤ 1$]
+ Solve #math.equation(block: false, alt: "50 minus 50 cos ⁡ open parenthesis 2 π x close parenthesis equals 0 , for minus 1 less than or equal to x less than or equal to 1")[$" " " " 50 − 50 cos ( 2 π x ) = 0 , " " " " "for" " " " " − 1 ≤ x ≤ 1$]


      
      For Problems 17–18, label the scales on the axes for the graph.


      
        #math.equation(block: true, alt: "y equals the fraction 1 over 4 sin ⁡ open parenthesis the fraction x over 6 close parenthesis plus the fraction 1 over 2")[$y = display(frac(1, 4)) sin ( display(frac(x, 6)) ) + display(frac(1, 2))$]


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


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


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


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


      
      For Problems 19–20,


      + Use a calculator to graph the function for #math.equation(block: false, alt: "0 less than or equal to x less than or equal to 2 π")[$0 ≤ x ≤ 2 π$].
+ Use the intersect feature to find all solutions between #math.equation(block: false, alt: "0")[$0$] and #math.equation(block: false, alt: "2 π")[$2 π$].  Round your answers to hundredths.


      
        + #math.equation(block: false, alt: "y equals minus 5 cos ⁡ open parenthesis 2 x minus 0.5 close parenthesis plus 3")[$y = − 5 cos ( 2 x − 0.5 ) + 3$]
+ #math.equation(block: false, alt: "minus 5 cos ⁡ open parenthesis 2 x minus 0.5 close parenthesis plus 3 equals minus 1")[$− 5 cos ( 2 x − 0.5 ) + 3 = − 1$]


        
          + #figure(figph[sinusoidal graph], alt: "sinusoidal graph", caption: none)
+ 0.57, 3.07, 3.71


        
      
      
        + #math.equation(block: false, alt: "y equals 2 minus 4 sin ⁡ 3 open parenthesis x plus 0.2 close parenthesis")[$y = 2 − 4 sin 3 ( x + 0.2 )$]
+ #math.equation(block: false, alt: "2 minus 4 sin ⁡ 3 open parenthesis x plus 0.2 close parenthesis equals 5")[$2 − 4 sin 3 ( x + 0.2 ) = 5$]


      
      For Problems 21–22, write a formula for the function.


      
        The average high temperature in Phoenix, Arizona is minimum in January at 66#math.equation(block: false, alt: "∘")[$∘$] and maximum in July at 105#math.equation(block: false, alt: "∘")[$∘$].  Write a sinusoidal function that models the average high temperature in Phoenix.


        
          #math.equation(block: true, alt: "y equals 85.5 minus 19.5 cos ⁡ open parenthesis the fraction π over 6 t close parenthesis")[$y = 85.5 − 19.5 cos ( display(frac(π, 6)) t )$]


        
      
      
        The average monthly rainfall in Hawaii reaches a maximum of 3.4 inches in December and a minimum of 0.4 inches in June.  Write a sinusoidal function that models the monthly rainfall in Hawaii.


      
      For Problems 23–24,


      + Estimate the amplitude, period, and midline of a circular function that fits the data.
+ Write a formula for the function.


      
        #figure(table(
  columns: 9, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "14")[$14$]]),
  [#math.equation(block: false, alt: "y")[$y$]], [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "13.4")[$13.4$]], [#math.equation(block: false, alt: "16.2")[$16.2$]], [#math.equation(block: false, alt: "18")[$18$]], [#math.equation(block: false, alt: "17")[$17$]], [#math.equation(block: false, alt: "14.1")[$14.1$]], [#math.equation(block: false, alt: "12.1")[$12.1$]], [#math.equation(block: false, alt: "12.7")[$12.7$]],
))


        
          + amp: 3, period: 12, midline: #math.equation(block: false, alt: "y equals 15")[$y = 15$]
+ #math.equation(block: false, alt: "y equals 15 minus 3 cos ⁡ open parenthesis the fraction π over 6 t close parenthesis")[$y = 15 − 3 cos ( frac(π, 6) t )$]


        
      
      
        #figure(table(
  columns: 10, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "0.05")[$0.05$]], [#math.equation(block: false, alt: "0.1")[$0.1$]], [#math.equation(block: false, alt: "0.15")[$0.15$]], [#math.equation(block: false, alt: "0.2")[$0.2$]], [#math.equation(block: false, alt: "0.25")[$0.25$]], [#math.equation(block: false, alt: "0.3")[$0.3$]], [#math.equation(block: false, alt: "0.35")[$0.35$]], [#math.equation(block: false, alt: "0.4")[$0.4$]]),
  [#math.equation(block: false, alt: "y")[$y$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "10.4")[$10.4$]], [#math.equation(block: false, alt: "11.8")[$11.8$]], [#math.equation(block: false, alt: "11.8")[$11.8$]], [#math.equation(block: false, alt: "10.4")[$10.4$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "5.6")[$5.6$]], [#math.equation(block: false, alt: "4.2")[$4.2$]], [#math.equation(block: false, alt: "4.2")[$4.2$]],
))


      
      For Problems 25–28, give exact values for the solutions between #math.equation(block: false, alt: "0")[$0$] and #math.equation(block: false, alt: "2 π")[$2 π$].


      
        #math.equation(block: true, alt: "10 sin ⁡ open parenthesis 2 θ close parenthesis equals minus 5")[$10 sin ( 2 θ ) = − 5$]


        
          #math.equation(block: true, alt: "the fraction 7 π over 12")[$display(frac(7 π, 12))$], #math.equation(block: true, alt: "the fraction 11 π over 12")[$display(frac(11 π, 12))$], #math.equation(block: true, alt: "the fraction 19 π over 12")[$display(frac(19 π, 12))$], #math.equation(block: true, alt: "the fraction 23 π over 12")[$display(frac(23 π, 12))$]


        
      
      
        #math.equation(block: true, alt: "the square root of 2 cos ⁡ open parenthesis 3 ϕ close parenthesis equals 1")[$sqrt(2) cos ( 3 ϕ ) = 1$]


      
      
        #math.equation(block: true, alt: "12 tan ⁡ open parenthesis 4 β close parenthesis equals 0")[$12 tan ( 4 β ) = 0$]


        
          #math.equation(block: true, alt: "0")[$0$], #math.equation(block: true, alt: "the fraction π over 4")[$display(frac(π, 4))$], #math.equation(block: true, alt: "the fraction π over 2")[$display(frac(π, 2))$], #math.equation(block: true, alt: "the fraction 3 π over 4")[$display(frac(3 π, 4))$], #math.equation(block: true, alt: "π")[$π$], #math.equation(block: true, alt: "the fraction 5 π over 4")[$display(frac(5 π, 4))$], #math.equation(block: true, alt: "the fraction 7 π over 4")[$display(frac(7 π, 4))$], #math.equation(block: true, alt: "2 π")[$2 π$]


        
      
      
        #math.equation(block: true, alt: "2 the square root of 3 tan ⁡ open parenthesis 2 α close parenthesis equals minus 6")[$2 sqrt(3) tan ( 2 α ) = − 6$]


      
      For Problems 29–32, find all solutions between #math.equation(block: false, alt: "0")[$0$] and #math.equation(block: false, alt: "2 π")[$2 π$].  Round your answers to three decimal places.


      
        #math.equation(block: true, alt: "5 tan ⁡ open parenthesis 3 x close parenthesis plus 2 equals 3")[$5 tan ( 3 x ) + 2 = 3$]


        
          0.066, 1.113, 2.160, 3.207, 4.255, 5.302


        
      
      
        #math.equation(block: true, alt: "minus 8 sin ⁡ open parenthesis 2 t close parenthesis minus 4 equals 3")[$− 8 sin ( 2 t ) − 4 = 3$]


      
      
        #math.equation(block: true, alt: "2.8 minus 3.6 cos ⁡ open parenthesis 2 s close parenthesis equals 5.2")[$2.8 − 3.6 cos ( 2 s ) = 5.2$]


        
          1.150, 1.991, 4.292, 5.133


        
      
      
        #math.equation(block: true, alt: "6.7 tan ⁡ open parenthesis 3 u close parenthesis plus 1.2 equals 28")[$6.7 tan ( 3 u ) + 1.2 = 28$]


      
      For Problems 33–36, use a substitution to find exact values for all solutions between #math.equation(block: false, alt: "0")[$0$] and #math.equation(block: false, alt: "2 π")[$2 π$].


      
        #math.equation(block: true, alt: "2 cos ⁡ open parenthesis 2 ϕ minus the fraction π over 4 close parenthesis equals the square root of 3")[$2 cos ( 2 ϕ − display(frac(π, 4)) ) = sqrt(3)$]


        
          #math.equation(block: true, alt: "the fraction π over 24")[$display(frac(π, 24))$], #math.equation(block: true, alt: "the fraction 5 π over 24")[$display(frac(5 π, 24))$], #math.equation(block: true, alt: "the fraction 25 π over 24")[$display(frac(25 π, 24))$], #math.equation(block: true, alt: "the fraction 29 π over 24")[$display(frac(29 π, 24))$]


        
      
      
        #math.equation(block: true, alt: "3 sin ⁡ open parenthesis 3 z plus π close parenthesis plus 2 equals minus 1")[$3 sin ( 3 z + π ) + 2 = − 1$]


      
      
        #math.equation(block: true, alt: "minus 4 sin ⁡ open parenthesis the fraction t over 2 plus the fraction π over 8 close parenthesis equals the square root of 8")[$− 4 sin ( display(frac(t, 2)) + display(frac(π, 8)) ) = sqrt(8)$]


        
          No solution


        
      
      
        #math.equation(block: true, alt: "7 cos ⁡ open parenthesis the fraction w over 2 minus the fraction π over 3 close parenthesis equals minus 3.5")[$7 cos ( display(frac(w, 2)) − display(frac(π, 3)) ) = − 3.5$]


      
      For Problems 37–40, use a substitution to find all solutions between #math.equation(block: false, alt: "0")[$0$] and #math.equation(block: false, alt: "2 π")[$2 π$]. Round your answers to hundredths.


      
        #math.equation(block: true, alt: "0.4 tan ⁡ open parenthesis 3 x plus 0.2 close parenthesis equals 1.6")[$0.4 tan ( 3 x + 0.2 ) = 1.6$]


        
          0.375, 1.422, 2.470, 3.517, 4.564, 5.611


        
      
      
        #math.equation(block: true, alt: "15 tan ⁡ open parenthesis 1.4 s minus 2 close parenthesis equals 20")[$15 tan ( 1.4 s − 2 ) = 20$]


      
      
        #math.equation(block: true, alt: "8 sin ⁡ open parenthesis the fraction π t over 6 minus the fraction π over 12 close parenthesis equals 6")[$8 sin ( display(frac(π t, 6)) − display(frac(π, 12)) ) = 6$]


        
          2.120, 4.880


        
      
      
        #math.equation(block: true, alt: "12 cos ⁡ open parenthesis the fraction π t over 2 minus the fraction 3 π over 5 close parenthesis equals 5")[$12 cos ( display(frac(π t, 2)) − display(frac(3 π, 5)) ) = 5$]