SOLUTIONS

RTD’s Math 30-1
SOLUTIONS
TRIG WORKSHOP To β€œQuestion Set” items
Question Set 1
1. Convert the following angles to radians: (a) 310°
11
2. Convert the following angles to degrees: (a)
5πœ‹
6
(b) βˆ’110°
(b) βˆ’
11πœ‹
2
(c) 118.38° (nearest hundredth)
(c) 3.2 (nearest degree)
3. Find each missing value: provide any angles in both radians (nearest hundredth) and degrees (nearest degree)
(a)
ο€± SE Solve multi-step problems based on the relationship πœƒ = π‘Žπ‘Ÿ
(c)
Given that the two circles have the same radius,
determine the measure of πœƒ in the second circle.
Question Set 2
(b)
(angle conversion is not considered a step)
(d)
The minute hand on the clock shown
measures 18 cm. Assuming it
represents the radius of the circle,
determine the length of the arc
formed by the hour and minute hand
at 10:10am.
Question Set 2
4. State one positive and one negative co-terminal angle (in both degrees and radians) for each of the following:
(a)
(b)
5.
For each angle in #5, state an expression for all co-terminal angles. (in both degrees and radians)
6.
State the principal angle for each of the following:
(a)
(b)
7. For each angle in standard position, state the quadrant which the terminal arm lies in, and the measure of the
reference angle. (Draw a diagram)
(a) πœƒ = 261°
(b)
5πœ‹
7
(c) 920°
(d)
20πœ‹
3
8. An angle in standard position πœƒ passes through a point (-4, -5). Determine the exact value of all six trigonometric
ratios of πœƒ, and the measure of πœƒ. (Correct to the nearest degree and hundredth of a radian)
9. Given that 0 ≀ πœƒ < 2πœ‹, π‘π‘œπ‘ πœƒ =
5
√61
and π‘‘π‘Žπ‘›πœƒ < 0, determine:
(a) The exact value of π‘π‘ π‘πœƒ and π‘π‘œπ‘‘πœƒ.
(b) The approximate value of πœƒ. (Correct to the nearest degree and hundredth of a radian)
10. An angle in standard position passes through a point 𝑃(βˆ’π‘, 2𝑏). Determine:
(a) The exact value of π‘π‘œπ‘ πœƒ.
(b) The approximate value of πœƒ. (Correct to the nearest degree and hundredth of a radian)
4
11. If π‘‘π‘Žπ‘›πœƒ = βˆ’ 5, determine the largest possible value of πœƒ on 0 ≀ πœƒ < 2πœ‹. (nearest hundredth)
3
12. If π‘ π‘’π‘πœƒ = βˆ’ 2 and π‘‘π‘Žπ‘›πœƒ > 0, determine (a) the exact value of π‘ π‘–π‘›πœƒ, and (b) The approximate value of πœƒ in degrees
(nearest whole number) and radians (nearest hundredth).
13. An angle in standard position πœƒ passes through a point 𝑃(π‘Ž, 𝑏). If the measure of πœƒ is
2πœ‹
3
and the distance from 𝑃 to
the origin is 5 units, determine the exact values of π‘Ž and 𝑏.
14. An angle in standard position πœƒ passes through a point 𝑃(0, 4). State an expression for all angles co-terminal to πœƒ.
(in both degrees and radians) (nearest degree / hundredth of a radian)
Question Set 3
15. Determine
16. For each of the following unit circle diagrams, determine the missing coordinate (as an exact value), and determine the
value of πœƒ, in degrees and (for (a) only exact, in terms of πœ‹) radians.
(π‘₯,
12
)
13
1
(π‘₯, )
2
πœƒ
πœƒ
17. An
18. On the
πœ‹
19. In the diagram below, the reference angle is 𝛼 = 6 . State the value of πœƒ (degrees and exact radians).
20. A
21. Determine the exact value of each trigonometric ratio. Draw the terminal arm on each unit circle.
22. Determine quadrant of the terminal arm of the angle in standard position, and state the approximate value of each trig ratio:
(a) sin
7πœ‹
(b) sec(βˆ’295°)
5
(c) cot 17.5
23. State the solutions to each equation, for 0 ≀ πœƒ < 2πœ‹: (Where possible, exact solutions in terms of πœ‹)
(a) sin πœƒ =
1
2
(b) cos πœƒ = βˆ’
√3
2
(c) csc πœƒ = √2
24. An angle πœƒ has tan πœƒ = √3 and cos πœƒ < 0. Determine the value of πœƒ, in degrees and radians.
1
25. An angle πœƒ has sin πœƒ = βˆ’ 3 and sec πœƒ > 0. Determine the value of πœƒ, correct to the
26. If the point 𝑃(0.2, π‘˜) lies on a circle with a radius of 1, then the exact value of π‘˜ can be what two values?
πœ‹
7πœ‹
27. Determine the exact value of sin (βˆ’ 6 ) + π‘π‘œπ‘  ( 4 ).
28. P
𝒂 = πŸ•, 𝒃 = 𝟏𝟐
29.
Question Set 43
30. State the period, amplitude, domain, and range for the graphs of each of the following functions.
1
(a) 𝑦 = 3sin(4π‘₯)
(b) 𝑦 = βˆ’2 cos (4 π‘₯)
(c) 𝑦 = tan(3π‘₯)
31. Determine sine equation for each of the following graphs:
(a)
(b)
State period to nearest hundredth
32. State the period, amplitude, domain, and range for the graphs of each of the following functions.
3
(a) 𝑦 = βˆ’1.25 sin (2 π‘₯) βˆ’ 1.5
π‘₯
(b) 𝑦 = 200 cos (50) + 200
(c) 𝑦 = 20.4 sin(2.8π‘₯) βˆ’ 10.4
33. State the period, amplitude, domain, range, and horizontal phase shift for the graphs of each of the following
functions.
(b) 𝑦 = 4 sin[3(π‘₯ βˆ’ 75°)] βˆ’ 5
1
πœ‹
(b) 𝑦 = 5 cos[8 (π‘₯ + 6 )] + 1
(c) 𝑦 = cos(4π‘₯ βˆ’ πœ‹) βˆ’ 1
ο€± Determine the complete equation, finding all 4 parameters, for a sinusoidal curve given the graph, the characteristics, or
a real-world situation SE
34. Determine both a sine and cosine equation for each of the following graphs:
(a)
(b)
ο€± Provide a complete explanation of how the characteristics of the graph of a trigonometric function relate to the
conditions in a contextual situation. SE
35. The
36. The graph shows how the height of a bicycle pedal changes as the bike is pedalled at a constant speed.
(a) Determine a cosine equation to model the height of the pedal, β„Ž, as a
function of time in seconds, 𝑑. (In the form 𝑦 = π‘Žcos 𝑏[(π‘₯ βˆ’ 𝑐)] + 𝑑.)
(b) Use your equation to predict how many seconds (nearest tenth) the pedal is
above 40cm in the first 10 s.
(c) Which equation parameter(s) (a, b, c, or d) would change if the bicycle were
pedalled at a greater constant speed?
5A
Question Set 4a
𝑠𝑖𝑛π‘₯
37. Given the identity π‘‘π‘Žπ‘›π‘₯ = π‘π‘œπ‘ π‘₯,
πœ‹
(a) Verify using the angles 10° and 6
(b) Verify graphically. Sketch resulting graphs here
(b) Prove algebraically
Left Side
Right Side
(c) State any non-permissible values
38. Simplify each of the following trig expressions to π‘ π‘–π‘›πœƒ, π‘π‘œπ‘ πœƒ, or 1.
(a) π‘ π‘–π‘›πœƒπ‘π‘œπ‘‘πœƒ
(b) π‘‘π‘Žπ‘›πœƒπ‘ π‘’π‘πœƒπ‘π‘œπ‘  2 πœƒ
(c) π‘π‘œπ‘‘πœƒπ‘π‘œπ‘ πœƒπ‘π‘ π‘πœƒπ‘‘π‘Žπ‘›2 πœƒ
39. Simplify into a single trigonometric expression.
(a)
π‘π‘œπ‘‘π‘₯π‘‘π‘Žπ‘›π‘₯
𝑐𝑠𝑐π‘₯
(b)
𝑠𝑖𝑛π‘₯
1βˆ’π‘π‘œπ‘ 2 π‘₯
π‘π‘œπ‘ π‘₯
40. Consider the equation 𝑠𝑒𝑐π‘₯ + π‘‘π‘Žπ‘›π‘₯ = 1βˆ’π‘ π‘–π‘›π‘₯.
(a) State the non-permissible values, in radians.
(c) PROVE that this is an identity.
41. ANS: A
42. ANS: A
43.
(c)
𝑠𝑖𝑛π‘₯ 𝑐𝑠𝑐 2 π‘₯
𝑠𝑒𝑐π‘₯
(d)
1βˆ’π‘ π‘–π‘›2 π‘₯
π‘π‘œπ‘‘ 2 π‘₯
βˆ— 𝑐𝑠𝑐π‘₯
πœ‹
4
(b) Verify that this equation is true using an angle π‘₯ = .
Question Set 5B
4b #43
1. Use the appropriate formula to evaluate: cos(90ο‚° ο€­ 30ο‚°)
Use cos(𝛼 βˆ’ 𝛽) = π‘π‘œπ‘ π›Όπ‘π‘œπ‘ π›½ + 𝑠𝑖𝑛𝛼𝑠𝑖𝑛𝛽
cos(90° βˆ’ 30°) = π‘π‘œπ‘ 90°π‘π‘œπ‘ 30° + 𝑠𝑖𝑛90°π‘ π‘–𝑛30°
1
√3
= (0) ( ) + (1) ( )
2
2
=0+
1

2
𝟏
=
𝟐
2. Use the Pythagorean Identity to express the identity cos(2𝛼) = π‘π‘œπ‘  2 𝛼 βˆ’ 𝑠𝑖𝑛2 𝛼 entirely in terms of 𝑠𝑖𝑛𝛼.
Pyth. Identity is π‘π‘œπ‘  2 𝛼 + 𝑠𝑖𝑛2 𝛼 = 1
So we can re-arrange to get π‘π‘œπ‘  2 𝛼 = 1 βˆ’ 𝑠𝑖𝑛2 𝛼
So now cos(2𝛼) = π‘π‘œπ‘  2 𝛼 βˆ’ 𝑠𝑖𝑛2 𝛼
This is 1 βˆ’ 𝑠𝑖𝑛2 𝛼
cos(2π‘Ž) = (1 βˆ’ 𝑠𝑖𝑛2 𝛼) βˆ’ 𝑠𝑖𝑛2 𝛼
 𝐜𝐨𝐬(𝟐𝜢) = 𝟏 βˆ’ πŸπ’”π’Šπ’πŸ 𝜢
3. Express as a single trig function:
(a) cos 50ο‚° cos 20ο‚°  sin 50ο‚° sin 20ο‚°
Use cos(𝛼 βˆ’ 𝛽) = π‘π‘œπ‘ π›Όπ‘π‘œπ‘ π›½ + 𝑠𝑖𝑛𝛼𝑠𝑖𝑛𝛽 as what we’re given β€œfits that pattern”!
=𝐜𝐨𝐬(πŸ‘πŸŽ°)
𝛼 = 50°, 𝛽 = 20° … So we have cos(50° βˆ’ 20°)
πœ‹
πœ‹
πœ‹
πœ‹
(b) 𝑠𝑖𝑛 π‘π‘œπ‘  + π‘π‘œπ‘  𝑠𝑖𝑛
5
8
5
8
Use sin(𝛼 + 𝛽) = π‘ π‘–π‘›π›Όπ‘π‘œπ‘ π›½ + π‘π‘œπ‘ π›Όπ‘ π‘–π‘›π›½
πœ‹
πœ‹
πœ‹
πœ‹
5
8
πŸπŸ‘π…
)
πŸ’πŸŽ
𝛼 = , 𝛽 = … So we have sin( + ) =𝐬𝐒𝐧(
(c)
5
8
π‘‘π‘Žπ‘›20°+π‘‘π‘Žπ‘›15°
1βˆ’π‘‘π‘Žπ‘›20°π‘‘π‘Žπ‘›15°
Use tan(𝛼 + 𝛽) =
π‘‘π‘Žπ‘›π›Ό+π‘‘π‘Žπ‘›π›½
1βˆ’π‘‘π‘Žπ‘›π›Όπ‘‘π‘Žπ‘›π›½
𝛼 = 20°, 𝛽 = 15° … So we have tan(20° + 15°)
πœ‹
4. Simplify sin( βˆ’ πœƒ)
2
πœ‹
πœ‹
= sin ( ) π‘π‘œπ‘ πœƒ βˆ’ cos ( ) π‘ π‘–π‘›πœƒ
2
2
= (1)(π‘π‘œπ‘ πœƒ) βˆ’ (0)(π‘ π‘–π‘›πœƒ) = π’„π’π’”πœ½
=𝐭𝐚𝐧(πŸ‘πŸ“°)
5. Express as a single trigonometric function: 2π‘π‘œπ‘  2
2
Use cos(2𝛼) = 2π‘π‘œπ‘  𝛼 βˆ’ 1, with 𝛼 =
= cos(2 βˆ—
6. Use the appropriate double-angle formula to evaluate:
(a) sin(2 ο€ͺ 330ο‚°)
Use sin(2𝛼) = 2π‘ π‘–π‘›π›Όπ‘π‘œπ‘ π›½, with 𝛼 = 330°
= 2𝑠𝑖𝑛330°π‘π‘œπ‘ 330°
1
√3
2
2
= 2(βˆ’ )( ) 
=βˆ’
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βˆšπŸ‘
𝟐
πœ‹
(b)
2π‘‘π‘Žπ‘›12
πœ‹
1βˆ’π‘‘π‘Žπ‘›2
12
Use tan(2𝛼) =
= tan(2 βˆ—
πœ‹
12
2π‘‘π‘Žπ‘›π›Ό
1βˆ’π‘‘π‘Žπ‘›2 𝛼
, with 𝛼 =
πœ‹
)  = tan( )
6
πœ‹
12
=
βˆšπŸ‘
πŸ‘
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5πœ‹
12
10πœ‹
)  = cos(
12
5πœ‹
12
5πœ‹
12
πŸ“π…
)
πŸ”
)  = 𝐜𝐨𝐬(
βˆ’1
7. Express as a single trig function:
(a) 2 sin 60ο‚° cos 60ο‚°
Use sin(2𝛼) = 2𝑠𝑖𝑛𝛼 π‘π‘œπ‘ π›Ό, with 𝛼 = 60°
= sin(2 βˆ— 60°)
(b) cos 2

6
ο€­ sin2

6
πœ‹
= cos(2 βˆ— )
6
Use cos(2𝛼) = π‘π‘œπ‘  2 ∝ βˆ’π‘ π‘–π‘›2 𝛼, with 𝛼 =
πœ‹
6
𝝅
= 𝐜𝐨𝐬( )
πŸ‘
= 𝐬𝐒𝐧(𝟏𝟐𝟎°)
8. Simplify the function π‘π‘œπ‘ 2π‘₯ π‘π‘œπ‘ π‘₯ + 𝑠𝑖𝑛2π‘₯ 𝑠𝑖𝑛π‘₯
Use cos(𝛼 βˆ’ 𝛽) = π‘π‘œπ‘ π›Όπ‘π‘œπ‘ π›½ + 𝑠𝑖𝑛𝛼𝑠𝑖𝑛𝛽, with 𝛼 = "2π‘₯" and 𝛽 = "π‘₯"
= cos(2π‘₯ βˆ’ π‘₯)
= 𝒄𝒐𝒔𝒙
9. Given cos A ο€½
4
7
3
3
and sin B ο€½ ο€­
, where
, find the exact value of sin(A ο€­ B ) .
ο‚£ A ο‚£ 2 and  ο‚£ B ο‚£
5
25
2
2
10. Find the exact value of sin

12
First, change the angle to degrees so it’s easier to come up with two unit circle angles that add or subtract to it.
= 𝑠𝑖𝑛15°
Next, come up with two unit circle angles that add or subtract to 15°.
= sin(45° βˆ’ 30°)
Finally, use sin(𝛼 βˆ’ 𝛽) = π‘ π‘–π‘›π›Όπ‘π‘œπ‘ π›½ βˆ’ π‘π‘œπ‘ π›Όπ‘ π‘–π‘›π›½
sin(45° βˆ’ 30°) = 𝑠𝑖𝑛45°π‘π‘œπ‘ 30° βˆ’ π‘π‘œπ‘ 45°π‘ π‘–𝑛30°
√2
√3
√2
1
2
2
2
2
= ( )( )βˆ’ ( )( )
=
√6
4
βˆ’
√2
4
 =
βˆšπŸ”βˆ’βˆšπŸ
πŸ’
11. Find the exact value of π‘‘π‘Žπ‘›
17πœ‹
12
< Express in terms of unit circle angles
Convert to degrees… = tan(255°)
< This is just one option
= tan(225° + 30°)
tan(𝛼 + 𝛽) =
π‘‘π‘Žπ‘›π›Ό + π‘‘π‘Žπ‘›π›½
1 βˆ’ π‘‘π‘Žπ‘›π›Όπ‘‘π‘Žπ‘›π›½
Here, 𝛼 = 225° and 𝛽 = 30°
=
=
=
π‘‘π‘Žπ‘›225° + π‘‘π‘Žπ‘›30°
1 βˆ’ π‘‘π‘Žπ‘›225°π‘‘π‘Žπ‘›30°
1+√3/3
1βˆ’(1)(√3/3)
𝟏𝟐+πŸ”βˆšπŸ‘
πŸ”


=
3+√3
3
3βˆ’βˆš3
3

=
3+√3
3
βˆ— 3βˆ’ 3
3
√
3+√3
√3
 = 3βˆ’
3+√3
√3
βˆ— 3+
Rationalize denominator
= βˆšπŸ‘ + 𝟐
Prove the identity
2π‘‘π‘Žπ‘›π‘₯
1βˆ’π‘‘π‘Žπ‘›2 π‘₯
=
sin(2π‘₯)
π‘π‘œπ‘  2 π‘₯βˆ’π‘ π‘–π‘›2 π‘₯
(state any variable restrictions)
LS
RS
2𝑠𝑖𝑛π‘₯
π‘π‘œπ‘ π‘₯
=
π‘π‘œπ‘  2 π‘₯ 𝑠𝑖𝑛2 π‘₯
βˆ’
π‘π‘œπ‘  2 π‘₯ π‘π‘œπ‘  2 π‘₯
=
=
=
2𝑠𝑖𝑛π‘₯π‘π‘œπ‘ π‘₯
π‘π‘œπ‘  2 π‘₯ βˆ’ 𝑠𝑖𝑛2 π‘₯
οƒΌ
2𝑠𝑖𝑛π‘₯
π‘π‘œπ‘ π‘₯
π‘π‘œπ‘  2 π‘₯ βˆ’ 𝑠𝑖𝑛2 π‘₯
π‘π‘œπ‘  2 π‘₯
2𝑠𝑖𝑛π‘₯
π‘π‘œπ‘  2 π‘₯
βˆ—
π‘π‘œπ‘ π‘₯ π‘π‘œπ‘  2 π‘₯ βˆ’ 𝑠𝑖𝑛2 π‘₯
=
2𝑠𝑖𝑛π‘₯π‘π‘œπ‘ π‘₯
π‘π‘œπ‘  2 π‘₯ βˆ’ 𝑠𝑖𝑛2 π‘₯
3
οƒΌ
44. Given that π‘‘π‘Žπ‘›π‘₯ = 4, where 180° < π‘₯ < 270°, the exact value of cos(π‘₯ βˆ’ 30°) is:
A.
B.
C.
D.
3√3+4
10
βˆ’3√3βˆ’4
10
3+4√3
10
βˆ’3βˆ’4√3
10
45. Simplify each of the following to a single numerical value:
(a) π‘π‘œπ‘‘ 2 π‘₯ βˆ’ 𝑐𝑠𝑐 2 π‘₯
(b) 𝑠𝑒𝑐 2 π‘₯ βˆ’ π‘‘π‘Žπ‘›2 π‘₯
π‘‘π‘Žπ‘›π‘₯
(c) 𝑠𝑖𝑛π‘₯ βˆ’ 𝑠𝑒𝑐π‘₯
(d)
1
π‘π‘œπ‘  2 π‘₯
7
1
+ 7 𝑠𝑖𝑛2 π‘₯
46. (Added in class)
TRIGONOMETRIC EQUATIONS
 I.E.E.#1 Simplify Consider the solution to the equation 𝑠𝑖𝑛2π‘₯ βˆ’ π‘π‘œπ‘ π‘₯ = 0 ; 0 ≀ π‘₯ < 2πœ‹
Solve algebraically:
 I.E.E.#2 Determine a general solution to the equation 2π‘π‘œπ‘  2 π‘₯ βˆ’ π‘π‘œπ‘ π‘₯ βˆ’ 1 = 0
 I.E.E.#3
Answer: β€œB”
 I.E.E.#3 ANS: β€œB”
 I.E.E.#4
οƒΌ
=
2π‘‘π‘Žπ‘›π‘₯
1 + π‘‘π‘Žπ‘›2 π‘₯