Name
January, 2015
Honors Math 3 review problem set
page 1
Review: Trigonometry (Chapter 4)
This week we will have a test on Trigonometry, Chapter 4. There are two sets of problems to
help you prepare: the review problems in this packet, and the textbook chapter review on page
389.
Here is a list of some of the key skills and concepts from the chapter.
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Understand and apply the unit circle definitions of the trigonometric functions
Find trigonometric function values using the unit circle, using special triangles (for 30°, 45°,
60°, and angles related to these), and in general using a calculator
Prove and apply the Pythagorean identity (sin2  + cos2 )
Find trigonometric function values when given other values using quadrant relationships, the
Pythagorean identity, and other identities
Solve trigonometric equations by hand, using inverses on the calculator, and graphically on
the calculator
Graph the functions sin x, cos x, and tan x and identify the periods of the graphs
Prove and apply the angle sum identities (formulas for sin( + ) and cos ( + ) will be
given on the test)
Prove and apply the Law of Sines and Law of Cosines.
Solve triangles (find unknown sides and angles) when given SSS, SAS, ASA, or AAS.
Solve triangles in the potentially ambiguous case SSA.
Review problems
1. Given that sin(39.05°) = 0.6300, answer the following without using the sin, cos, and tan
keys on your calculator.
a. Find cos(39.05°).
b. Find tan(39.05°).
c. Find an angle between 0° and 90° whose cosine equals 0.6300.
d. Find an angle between 90° and 360° whose cosine equals 0.6300.
e. Find all the solutions to the equation sin(x) = –0.6300.
Directions for problems 2–6: For the triangles given below (each with labels arranged in the
usual way): solve for the remaining angles and sides, and also find the triangle’s area.
2. In ABC, A = 20°, B = 60°, and c = 10.
3. In XYZ, x = 17, z = 8, and Y = 52°.
4. In ABC, a = 3, A = 120°, and B = 20°.
5. In XYZ, x = 3, y = 7, and z = 11.
6. In ABC, a = 6, b = 12, and c = 16.
Name
January, 2015
Honors Math 3 review problem set
page 2
7. Two trains leave a station on different straight
tracks. The tracks make an angle of 125° with the
station as the vertex. The first train travels at an
average speed of 100 kilometers per hour, and the
second train travels at an average speed of 65
kilometers per hour. How far apart are the trains
after 2 hours?
B
c
125°
C
station
A
8. For each of the following sets of SSA measurements for ABC, find all unknown sides and
angles. If there are two possible sets of measures, find both.
a. a  7, b  5, A  25.8
b. a  6, b  10, A  31.2
c. a  3, b  10, A  31.2
9. Prove using two different methods (1. with a unit circle diagram, 2. with an angle-sum
identity) that sin 270      cos
10. Solve this equation, finding solutions in the interval 0° to 360°:
sin 2   sin   cos 2 
11. Solve this equation, finding all solutions: sin  cos   3 sin 
12. Find sin  , cos , and tan  if   315
13. Which of the following angles does not have the same sin, cos, and tan values as 45°?
(a)  315
(b) 225
(c) 405
(d)  675
14. Answer the questions using the following information: sin 20° ≈ 0.342, cos 20° ≈ 0.940.
a. What are the angles in Quadrants II, III, and IV whose sine and cosine values are either
equal or opposite to those of 20°?
b. Find sine and cosine for each of the angles you listed in part a.
15. Using your work from page 330 exercise 13 and without using your calculator, find an exact
value for sin 162°.
16. What degree angle does the line y = 2x make with the x-axis? Hint: Draw this line on a
unit-circle diagram, as on page 340.
17. Take it further. Applying the angle-sum identities, find a formula for cos(4) in terms of
cos() and sin(). Hint: 4 = 2 + 2.
Name
January, 2015
Answers
Honors Math 3 review problem set
page 3
(first draft; if any answers don’t look right, check with teacher)
1. a. Use the Pythagorean Identity. Answer: 0.7766.
b. Divide the sine by the cosine. Answer: 0.8112.
c. 50.95°
d. 309.05°
e. 219.05° + 360° n and 320.95° + 360° n for any integer n
2. (ASA case) C = 100°, a = 3.47, b = 8.79; Area 15.04
3. (SAS case) X = 100.43°, Z = 27.57°, y = 13.62; Area 53.58
4. (AAS case) C = 40°, b = 1.18, c = 2.23; Area 1.14
5. No such triangle exists because 11  3 + 7.
6. (SSS case) A = 18.57°, B = 39.57°, C = 121.86°; Area 30.58
7. 294.5 kilometers
8. a. c  11.2, B  18.1 , C  136.1
b.
c. no triangle possible
or
9. sin(270 + q ) = sin270cosq + cos270sinq = -1× cosq + 0× sinq = -cosq (see teacher for unit
circle proof)
10. 90°, 210°, 330°
11. 180° n where n is any integer
-1 - 2
1
2
=
,cosq =
=
,tanq = -1
2
2
2
2
13. Answer: (b). 45° and 225° do have equal tan values, but their sin’s and cos’s are opposite.
cos70 = 0.342
14. sin160 = 0.342 cos160 = -0.940
sin70 = 0.940
12. sinq =
sin200 = -0.342 cos200 = -0.940
sin340 = -0.342 cos340 = 0.940
cos110 = -0.342
cos250 = -0.342
cos290 = 0.342
sin110 = 0.940
sin250 = -0.940
sin290 = -0.940
5 1
4
–1
16. tan (2) ≈ 63.435°
17. cos(4) = cos(2)cos(2) – sin(2)sin(2)
= (cos(+))2 – (sin(+))2
= (cos2  – sin2)2 – (2 sin  cos )2.
If you expand in the first term and then collect like terms, this simplifies to
sin4() + cos4() – 6 sin2() cos2().
15.