4.1: TrigonometryI can Solve a Triangle.I know the 6 Trig Ratios.I know the sides of 30-60-90 & 45-45-90 Triangles.

12, 42, 47, 704.2: Degrees & RadiansI can convert Radians to Degrees & Visa-Versa. I can find DMS formI can find the Reference Angle.I can find Co-Terminal Angles.I can find Arc Length.I can determine the Linear & Angular Velocity.I can find the Area of a Sector.

8, 10, 22, 28, 34, 444.3: Unit CircleI know the values of the 6 Trig Functions on the Unit Circle.I can write the Coordinates of a Point on a using (Sine, Cosine).I can determine the Slope of a point and the Origin using Tangent.4, 8, 34, 48, 684.4: Graphs of Sine & CosineI can Graph Sine & Cosine.I can Determine the Period, Frequency, Amplitude, Vertical & Phase Shifts of graph from its Equation. Given the above parts, I can determine the Equation.

4, 8, 15, 34, 38Look at 31,32 in book

4.5: Graphs of the 4 OthersI can determine all of the parts listed in 4.4 with the other 4 Trig Functions.I can find the Dampening Factor and write an Equation Modeling it

16, 284.6: Inverse Trig FunctionsI can Find the Inverse of Trig Functions.I can Find the Domain & Range of Inverse Trig Functions.I Can Graph Inverse Trig Functions.I can Simplify Compound Expressions.

10, 20, 38, 46

4.7: Law of Sines and Cosines:I can Use the Law of Sines & Cosines to Solve a non-right triangle.I can determine if a Triangle has zero, one, or two possible solutions.I can use ½ a*b*sin© to find the Area of a Triangle.

6, 20, 46, 56

4.1: Trigonometry

I can Solve a Triangle.

I know the 6 Trig Ratios.

I know the sides of 30-60-90 & 45-45-90 Triangles.

12, 42, 47, 70

12. Use the given trigonometric function value of the acute angle θ to find the exact values of the five remaining trigonometric function values of θ. sec θ = 8

Draw a right triangle and label one acute angle θ. Because

sec θ = = 8 = , label the hypotenuse 8 and the adjacent side 1.

By the Pythagorean Theorem, the length of the side adjacent

to θ is .

42. SKI LIFT A company is installing a new ski lift on a 225-meter-high mountain that will ascend at a 48º angle of elevation.

a. Draw a diagram to represent the situation.

b. Determine the length of cable the lift requires to extend from the base to the peak of the mountain.

a. Draw a diagram of a right triangle. The height of the mountain is give. Label the vertical side 225. Then label the acute angle opposite the 225 m side as 48º.

b. Because an acute angle and the length of the side opposite the angle are given, you can use the sine function to find the length of the hypotenuse.

47. Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

Use trigonometric functions to find b and c.

Because the measures of two angles are given, B can be found by subtracting A from

So, the company will need about 303 meters of cable.

Therefore, b 16.5, c 17.5.

70. Without using a calculator, determine the value of x.

SOLUTION:

Because the triangle is a triangle, the legs are the same length.

4.2: Degrees & RadiansI can convert Radians to Degrees & Visa-Versa. I can find DMS formI can find the Reference Angle.I can find Co-Terminal Angles.I can find Arc Length.I can determine the Linear & Angular Velocity.I can find the Area of a Sector.

8, 10, 22, 28, 34, 44

8. Write each decimal degree measure in DMS form.

  

 301° 42′ 8  ″    

  SOLUTION:  

Each minute is   of a degree and each

second is   of a minute, so each second

is   of a degree.

Therefore, 301° 42′ 8″ can be written as about 301.702°.

28. Find the length of the intercepted arc with the given central angle measure in a circle with the given radius. Round to the nearest tenth.

  

   R=3 inches  

 

10. Write each degree measure in radians as a multiple of π and each radian measure in degrees.

  

  30°   

  SOLUTION:  To convert a degree measure

to radians, multiply

by  

  

22. Identify all angles that are coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

  

     

 

  SOLUTION:  

All angles measuring  are coterminal with

a   angle.  Sample answer: Let n = 1 and −1. 

 

  SOLUTION:  

   34. Find the rotation in revolutions per minute given the angular speed and the radius given the linear speed and the rate of rotation.

  

 

  =   rad/s

SOLUTION: The angular speed is radians per second. Use the angular speed equation to find the angle of rotation, θ.

Each revolution measures 2π radians. 40π ÷ 2π = 20

  

44. Find the area of each sector.

  

  

The measure of the sector’s central angle θ is and the radius is 3.4 meters.

Therefore, the area of the sector is about 24.2 square meters.

SOLUTION:

The measure of the sector’s central angle θ is and the radius is 3.4 meters.

Therefore, the area of the sector is about 24.2 square meters.

4.3: Unit CircleI know the values of the 6 Trig Functions on the Unit Circle.I can write the Coordinates of a Point on a using (Sine, Cosine).I can determine the Slope of a point and the Origin using Tangent.4, 8, 34, 48, 68

4. Find the exact values of the six trigonometric functions of θ.

  

     

 

  SOLUTION:  The length of the side opposite θ is 12,

the length of the side adjacent to θ is 35, and the length of the hypotenuse is 37.

8. Find the exact values of the six trigonometric functions of θ.

  

     

 

  SOLUTION:  The length of the side opposite θ is

8 and the length of the side adjacent to θ is 32. By the Pythagorean Theorem, the length of the

hypotenuse is  .

34. Find the measure of angle θ. Round to the nearest degree, if necessary. 

  

     

   SOLUTION:  Because the lengths of the sides

opposite and adjacent to θ  are given, the tangent function can be used to find θ .

48. Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

  

     

   SOLUTION:  Use trigonometric functions to find y and z.

 Because the measures of two angles are given, X can be found by subtracting Z from 

 Therefore, X = 29°, y   37.1, z   32.5.

68. Without using a calculator, find the measure of the acute angle θ  in a right triangle that satisfies each equation.

  

   sec θ = 2

  

   SOLUTION:  Because sec θ = 2 and sec θ  =

, it follows that   = 2

=  . In the  triangle below, the hypotenuse is 2 and the side length that is adjacent to the

60º angle is 1. So,   =   = 2. 

 Therefore, θ = 60°.

4.4: Graphs of Sine & CosineI can Graph Sine & Cosine.I can Determine the Period, Frequency, Amplitude, Vertical & Phase Shifts of graph from its Equation. Given the above parts, I can determine the Equation.

4, 8, 15, 34, 38Look at 31,32 in book

4. Describe how the graphs of f(x) and g(x) are related. Then find the amplitude of g(x), and sketch two periods of both functions on the same coordinate axes.

  

   f(x) = sin x; g(x) = −8 sin x

  

 

  SOLUTION:  The graph of g(x) is the graph of f(x) expanded

vertically. The amplitude of g(x) is   or 8. Create a table listing the coordinates of the x-intercepts and extrema for f(x) = sin x for one period, 2 , on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph. 

Function f(x) = sin x

x-int (0, 0)

Max

x-int

Min

x-int

8. Describe how the graphs of f(x) and g(x) are related. Then find the period of g(x), and sketch at least one period of both functions on the same coordinate axes.

  

   f(x) =

sin x; 

  

 

  SOLUTION:  The graph of g(x) is the graph of f(x) expanded horizontally. The

period of g(x) is  . To find corresponding points on the graph of g(x), change the x-coordinates of those key points on f(x) so that they range from 0 to  , increasing by increments

of  or   . 

Functions f(x) = sin x

x-int (0, 0) (0, 0)

Max

x-int

Min

x-int

 Sketch the curve through the indicated points for each function.

  

Function g(x) = −8 sin x

x-int (0, 0)

Min

x-int

Max

x-int

 Sketch the curve through the indicated points for each function. Then repeat the pattern to complete a second period.

Then repeat the pattern to complete a second period.

15. State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

   34. Write an equation that corresponds to each graph.

  

     

 

  

44. Write a sinusoidal function with the given period and amplitude that passes through the given point.

  

   period: 3π;

amplitude:  ;

point: 

  

 

  

     

 

  SOLUTION:  In this function, a = 1, b = 

, c =  , and d = 0. Because d = 0, there is no vertical shift. 

SOLUTION:  Sample answer: There is an x-intercept at 0, so one

equation that corresponds to this graph is y = a sin (bx + c) + d. Half of the distance from the maximum to the minimum value of the function is 8 ÷ 2 or 4. So, the amplitude is 4. It appears that the function completes one period on [0, 4π]. Find b.

SOLUTION:  Use the period to find b.

 Sample answer: One sinusoidal function in which a = 0.5

and b =   is y =   sin  x. Evaluate the function for 

4.5: Graphs of the 4 OthersI can determine all of the parts listed in 4.4 with the other 4 Trig Functions.I can find the Dampening Factor and write an Equation Modeling it

2, 8, 16, 18, 28

2. Locate the vertical asymptotes, and sketch the graph of each function.

  

     

 

  SOLUTION:  

The graph of   is the graph of y =

tan x shifted   units to the left. The period is   or  .  

y = a tan (bx + c), so a = 1, b = 1, and c =  . Use the tangent asymptote equations to find the location of the asymptotes.

     

    Create a table listing the coordinates of key points

for  for one period on  .

8. Locate the vertical asymptotes, and sketch the graph of each function.

  

     

 

  SOLUTION:  

The graph of   is the graph of y =

cot x expanded horizontally. The period is   or 2 .  

y = a cot (bx + c), so a = 1, b =  , and c = 0. Use the tangent asymptote equations to find the location of the asymptotes. Find the location of two consecutive vertical asymptotes.

      

     Create a table listing the coordinates of key points

 Function

  y = tan x y = tan (x + 4)

VerticalAsymptote

IntermediatePointx-int

  (0, 0)

IntermediatePoint (0, 1)

Vertical Asymptote

 Sketch the curve through the indicated key points for the function. Then repeat the pattern.

for  for one period on  . 

Function 

y = cot x

VerticalAsympt

otex = 0 x = 0

IntermediatePointx-int

 Interme

diatePoint

VerticalAsympt

ote Sketch the curve through the indicated key points for the function. Then repeat the pattern.

16. Locate the vertical asymptotes, and sketch the graph of each function.

  

  y = –sec 

  

18. Identify the damping factor f(x) of each function. Then use a graphing calculator to sketch the graphs of f(x), −f(x), and the given function in the same viewing window. Describe

    28. DIVING  The end of a diving board is 20.3 centimeters above its resting position at the moment a diver leaves the board. Two seconds later, the board has moved down and up 12 times. The damping constant for the board is 0.901.

  

 

  

SOLUTION:  

The graph of   is the graph of y = sec x expanded horizontally and reflected in the x-

axis. The period is   or 16 .  

y = a sec (bx + c), so a = –1, b =  , and c = 0. Use the asymptote equations to find the location of two vertical asymptotes. 

      

     Create a table listing the coordinates of key points

for   for one period on [−4 , 12 ]. 

the behavior of the graph.   y =

4x cos x

  

 

  

SOLUTION:  The function y = 4x cos x is the product of the

functions y = 4x and y = cos x, so f(x) = 4x. Use a graphing calculator to graph f(x), –f(x), and y = 4x cos x in the same viewing window.

 The amplitude of the function is decreasing as x approaches 0 from both directions, and increasing as x approaches positive and negative infinity.

  

 

a. Write a trigonometric function that models the motion of the diving board y as a function of time t.b. Determine the amount of time t that it takes the diving board to be damped so that –0.5 ≤ y ≤ 0.5.

 

  SOLUTION:  a. The maximum displacement of the diving

board occurs when t = 0, so y = ke−ct cos ωt can be used to model the motion of the diving board because the graph of y = cos t has a y-intercept other than 0. The maximum displacement occurs when the diving board is 20.3 centimeters above its resting position. The total displacement k is the maximum displacement minus the minimum displacement. So, the total displacement is 20.3 centimeters. The damping constant c for the diving board is 0.901. Since after 2 seconds the board has moved down and up 12 times, it will have moved down and up 6 times after 1 second. Use the value of the frequency to find  .

Write a function using the values of k,   , and c.y = 20.3e−0.901t cos 12πt is one model that describes the motion of the diving board. b. Use a graphing calculator to determine the value of t when the graph of y = 20.3e −0.901t cos

Function 

y = sec x

VerticalAsympt

oteInterme

diatePoint

(0, 1) (0, –1)

x-int 

IntermediatePoint

VerticalAsympt

ote Sketch the curve through the indicated key points for the function. Then repeat the pattern.

12πt is oscillating between y =−0.5 and y = 0.5.Graph y = 20.3e −0.901t cos 12πt, y = 0.5, and y = −0.5 on the same coordinate plane.

From this window, it is unclear as to when the graph of y = 20.3e −0.901t cos 12πt oscillates within the interval −0.5 ≤ y ≤ 0.5. Adjust the window.

Using the intersect function from the CALC menu, it can be found that the graph of y = 20.3e −0.901t cos 12πt intersects y = −0.5 for the last time when x ≈ 4.09. When x > 4.09, the graph of y = 20.3e −0.901t cos 12πt will stay between y = −0.5 and y = 0.5. So, it takes approximately 4.09 seconds for the graph of y = 20.3e −0.901t cos 12πt to oscillate within the interval −0.5 ≤ y ≤ 0.5.

4.6: Inverse Trig FunctionsI can Find the Inverse of Trig Functions.I can Find the Domain & Range of Inverse Trig Functions.I Can Graph Inverse Trig Functions.I can Simplify Compound Expressions.

2, 10, 20, 38, 46