Math 102, PALS Spring 2016
Sample Test 4; Smith-Subbarao
1. Complete the following table;
function f(x)
Domain
arcsin(x) [-1, 1]
arcos(x) [-1, 1]
arctan(x) , 
Range
f(-1)
[/2, /2] -/2
[0, ]

-/4
[-/2, /2]
f(0)
0
/2 1
0
Evaluate the following:
2. cos−1 (cos 2𝜋)= cos-1(0) = 0
4
3. tan (cos−1 (− 5)) If cos = -4/5, sin = 3/5, tan = -3/4
𝜋
4. tan−1 (2 sin 3 ) sin(/3) = √3 /2, tan-1(2√3 /2) = tan-1(√3 ) = /3
5. Evaluate each expression without using a calculator and write your answer using radians.

3
a. sin 1  
 = -/3
 2 

2
b. cos 1  
 = 3/4
 2 
 1 
c. tan 1  
 = -/6
3

6. Evaluate by drawing a right triangle and labeling its sides.


x
sin  tan 1
 sides are x, √𝑥 2 + 22 , and√2𝑥 2 + 4; sin = x/√2𝑥 2 + 4
2
x 4 

Solve (a) for the principal root, (b) for all solutions in the interval [0,2π), and (c) all real roots.
7. 8sin x  4 2 sinx = −√2 /2; reference angle is -/4; a) principle root is -/4, b) 5/4, 7/4,
c) 5/4 + 2n, 7/4 + 2n
8.
 3 sec  3x   2
cos (3x) = - √3 /2 so 3x = 5/6 + 2n, 7/6 + 2n; all roots: x = 5/18 + 2n/3, 7/18 +
2n/3; principle roots 5/18, 7/18,, in [0, 2): 5/18, 7/18, 11/18, 17/18, 19/18, 23/18. 25/18,
29/18, 31/18
Solve in [0,2π):
9. 4sin 3 x  4sin 2 x  sin x  1  0 , 4 sin2x(sin x – 1) – (sin x – 1) = 0 or (4 sin2x-1)(sin x – 1); sin x = 1/2,
sinx = 1, x = /6 + n, 5/6 + n or x = /2 + 2n. In interval, have /6 5/6, 7/6, 11/6, /2
10. tan2 𝑥 + tan 𝑥 = 0 , tanx (tanx + 1 ) = 0, tanx = 0 or tanx = -1, x = n or x = 3/4 + n; in interval, x =
0, , 3/4, 7/4
11.
1+cos 2𝑥
2
3
1+cos 2𝑥
= 4 Take square root of both sides: √
2
1+cos 2𝑥
=  √3/2, but √
2
= cos x. So cos x =
 √3/2; x = /6, 5/6, 7/6, 11/6 + 2n; Principle: /6, 5/6; in interval: /6, 5/6, 7/6, 11/6.
Alternatively, cos 2x /2= ¾-1/2 = ¼ or cos 2x = ½; so 2x = /3 or 5/3 + 2n  , x = /6 or 5/6+ n=
/6, 5/6, 7/6, 11/6
12. cos2 𝑥 − sin2 𝑥 = 0; cos 2x = 0, 2x = (2n+1)/2, x = (2n+1) /4
13. −2 sin 2𝑥 = √3 ; sin 2x = -√3/2, 2x = 4/3, 5/3 + 2n, x = 2/3, 5/6, 5/3, 11/6
14. sec2 𝑥 = 2 tan 𝑥; 1/cos2x = 2sinx/cosx; 1 = 2sinx cosx, 1 = sin2x, 2x = /2 , 3/2 + 2n; x = /4, 3/4,
5/4, 7/4
15. Find two triangles for which A  34o , b  2.3 mm , and a  1.9 mm . Round all answers to two decimal
places. sin A /a = 0.29 = sin B/b, so sin B = 0.68, B = 42.60 or 137.30. Since angles sum to 180, C =
103.4 or 8.7. For the first, c = (a/sin A)sin C = 3.35, and the second 0.52.
TRIANGLE 1: B = 42.6, C = 103.4, c = 3.35mm
TRIANGLE 2:
B = 137.30, C = 8.7, c = 0.52mm
16. Find all the missing parts of the triangle if a  12, b  8, and c  5 . Round all answers to two decimal
places.
a2 = b2 + c2 – 2bc cos A; cos A = [b2 + c2 –a2 ]/2bc, A=133.47.
sinA/a = sinB/b, B= 28.96, A + B + C = 180. C = 17.57; Check sinA/a=sinB/b=sinC/c = 0.060
17. Convert the given polar equation to rectangular form. 2 cos 𝜃 − 3 sin 𝜃 = 𝑟
multiply both sides by r: 2rcos  - 3rsin  = r2; 2x – 3y = x2+y2
18. Convert the given rectangular equation to polar form. 𝑥 2 + 4𝑥 + 𝑦 2 + 4𝑦 = 0 ; x2 + y2 + 4(x+y) = 0;
let x = r cos , etc.; r2 + 4r(cos  + sin ) = 0,
19. 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.
eliminate t: (x +3)/2 = t, put in 2nd equation: y = (x+3)2/2 – 4 or y = x2 /2 + 3x + 9/2 – 4 = x2 /2 + 3x +1/2
This is a parabola.
-4
𝑥 = 2𝑡 − 3
𝑦 = 𝑡2 − 4
𝑡≥0
-3
-2
5
4
3
2
1
0
-1 -1 0
-2
-3
-4
-5
1
2
20. Convert from rectangular to polar coordinates: (4, -4), ( – 4 3 , 4)
(4, 7/4), (8, 5/6); note that the 1st is in Q4 and the 2nd in Q2

)
6
For the 1st, the negative sign makes it point in the opposite direction
21. Convert from polar to rectangular coordinates: (-4, π), (4,
(4, 0), (2√3 , 2)