Math 102 Test # 4 Spring 2016
Chapter 9.4, 9.5, 10
Instructor: Smith-Subbarao
______________________________
Name
Raw Score: _______________
Percent: _________________
Directions:
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Show all work and circle/box your answers.
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Partial credit may be given, even if the answer is incorrect, if your work is clear – attach additional scratch
pages you wish to be considered.
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If you do not show your work, you may not get credit.
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Leave fractions as fractions, square roots as square roots – need not evaluate.
You may use a calculator on this test
Suggestions:
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Choose the problems you understand best to work first.
If you get stuck, write down what you do understand for partial credit and move on
Show your work clearly
Check your solutions
Evaluate your solutions for “reasonableness”
1+cos 𝑎
cos (2a) = cos2 – sin2 a
cos(a/2) = ±√
sin (2a) = 2sin a cos a
sin (a/2) = ±√
2
1−cos 𝑎
2
1. Complete the following table
arcsin
arcos
artan
-1 -pi/2, pi/2 -pi/2
1 0, pi
pi
all reals -pi/2, pi/2 -pi/4
0
pi/2
0
Evaluate the following:
2. sin−1 (sin
5𝜋
6
)
/6; must be in the range -/2 to /6=2
4
3. tan (cos−1 (− 5))
-3/4
4. Cot-1(sec(-2))
This was not supposed to be this problem, but you had a calculator
Sec(-2) = -2.4, arcot(-2.4) = arctan(-1/2.4) = -0.39
This has been scored as a bonus problem. You may have received some credit for an attempt.
The problem cot(arcsec(-2)) is solvable, however.
5. Evaluate each expression without using a calculator and write your answer using radians.
a. cos-1 (-1/2 )
b. cot-1(-1/√3 )
2/3
- /3
Again, you needed to be in the range of the inverse trig functions
6. Solve by drawing a triangle and labeling its sides:
tan(cos-1(x/√𝑥 2 + 9 )
sides are x, 3, √𝑥 2 + 9, tan = 3/x
Solve (a) for the principal root, (b) for all solutions in the interval [0,2π), and (c) all real roots.
7. 12 cos x = -6 √3
cos x= -√3 /2; reference angle is /6, in Q2
Principle root is 5/6
interval is 5/6, 7/6
all 5/6 + 2n, 7/6 + 2n
8. √2 csc (4x) = 2
csc (4x) = 2/√2 or sin(4x) = √2/2
4x = /4 + 2n, 3/4 + 2n, so x = /16 + n/2, 3/16 + n/2
Principle: /16, 3/16, 9/16, 11/16
Interval: /16, 3/16, 9/16, 11/16, 17/16, 19/16, 20/16, 25/16, 28/16
all: /16 + n/2, 3/16 + n/2
Solve in [0,2π)
9. csc2x – sec2x = 0
multiply by cos2x sin2x giving cos2x – sin2x = 0
cos 2x = cos2x – sin2x, so cos 2x = 0
2x = /2, 3/2, + 2n, x = /4, 3/4 + n
x= /4, 3/4, 5/4, 7/4
10. 2 sin 2𝑥 = 4 sin 𝑥
sin 2x = 2 sinx cos x
4 sin x cos x = 4 sinx , sinx = 0 or cos x = 1
x = 0, 
11. Find all the missing parts of the triangle if C  50o , a  18 cm, and b  20 cm, solve the triangle. Round all
answers to two decimal places.
Law of cosines: c2 = a2 + b2 – 2 ab cos C= 324 + 400 - 720 (0.642) = 724 - 462.81 = 261.119, c = 16.16 cm
Law of sines sin A = a sin C/c = 18 (0.766)/16.16 = 0.40, C = 58.57 deg; 180 - 50 - 58.57 = 121.43 B = 71.43 deg
A = 58.57, B = 71.43, c = 16.16 cm
12. Find two triangles for which A  34o , b  2.3 mm , and a  1.9 mm . Round all answers to two decimal places.
Law of sines: Sin B = b sin A/a = (2.3/1.9) (0.559)=0.677, B = 42.60 deg , or B = 180-42.60 = 137.40 deg
C = 180 – 34 – 42.6 = 103.4 deg, or C = 180 – 34 – 137.4 = 8.6 deg
c = a (sin C/sin A) = 1.9 (0.677/0.559) = 3.31 mm, using 103.4 deg for C or
c = 1.9(0.15/0.559)= 0.51
Check: triangle 1: 0.2943= sin A/a = sin B/b =0.294 = sin C/c = 0.294
Check triangle 2: 0.2943 = sinA/a = sin B/b = 0.294 = sin C/c = 0.294
Check triangle 1 legs: 2.3, 1.9, 3.31: sum of any two is longer than the third
Check triangle 2 legs: 2.3, 1.9. 0.51; sum of any two is longer than the third
TRIANGLE 1: B = 42.6, C = 103.4, c = 3.31
TRIANGLE 2: B = 137.4, C = 8.6, c = 0.51
1
13. Convert the given polar equation to rectangular form: 𝑟 2 = 3+cos2 𝜃
Multiply both sides by 3 + cos2 
3r2 + r2cos2  = 1
r2 = x2 + y2
3(x2 + y2 ) + x2 = 1
4 x2 + 3 y2 = 1
14. Convert to polar form: 𝑥 2 + 𝑦 2 = 36
r2 = 36, r = 6
15. Graph the given parametric equations for the indicated values of t. Be sure to indicate the direction of the
graph. Then write the equation in terms of 𝑥 and 𝑦 by eliminating the parameter.
𝑥 = 5 cos 𝑡
𝑦 = −3 sin 𝑡
0 ≤ 𝑡 ≤ 2𝜋
To find the direction, put in t = 0, t = /2, etc.
cos t = x/5, sin t = -y/3
cos2t + sin2t = 1
(x/5)2 + (y/3)2 = 1
4
3
2
1
-6
-4
-2
0
-1 0
2
4
6
-2
-3
-4
16. Write the equation in terms of 𝑥 and 𝑦 by eliminating the parameter.
x = 3t
y = t2 + 5
t>0
t = x/3, y = (x/3)2 + 5 = x2/9 + 5
17. Write the following rectangular coordinates in terms of polar coordinates:
a. (-5/2, -5 √3/2 )
b. ( 3, 3)
r2 = 25/4 + 75/4 = 100/4
r = 10/2,  = arctan(√3) =/3,in Q3
r = 3√2,  = /4
(10/2, 4/3)
(3√2, /4)
18. Write the following polar coordinates in terms of rectangular coordinates:

)
6
a. (4, /3)
b. (3,
cos(/3) = 1/2
sin(/3) = √3/2
(3√3/2, 3/2)
(2, 2√3)