Math 104, Sample Test 3
Smith-Subbarao, Spring 2017
1. Solve: 𝑠𝑖𝑛2 𝑥 + 𝑐𝑜𝑠 2 𝑥 = 0
Since sin2x + cos2x = 1, this has no solution.
2. Solve: sin(t/2) = ½
let y = t/2, sin(y ) = ½, y = /6 + 2n or 5/6 + 2n. t = /3+4n or 5/3 + 4n
3. What is the exact value of:
 
2   
  sin   ?
 24 
 24 
a.) cos 2 
(Hint: Each of these can be evaluated by an identity.)
1, Pythag
b.) sin23t + cos2 3t - 3? cos2a = 1-sin2a, so we have sin23t + (1-sin23t)-3 = -2
4. Evaluate each expression without using a calculator and write your answer using radians.

a. sin 1  


3
 (4 points)
2 

b. cos 1  


2
 (4 points)
2 

c. tan 1  

Note: since these are inverse trig functions, need to restrict range
a. works for 4/3, 5/3
b. works for 3/4, 5/4
but only 5/3 (-/3) in range
but only 3/4 in range
1 
 (4 points)
3
c. works for 5/6 or -/6
but only -/6 in range
5. Evaluate the expression by drawing a right triangle and labeling its sides.

sin  tan 1

Op = x, adj = sqrt(x2 +4), hyp = sqrt(2x2 +4), sin = x/sqrt(2x2 +4)


x2  4 
x
x
Sqrt(2x2+4)
Sqrt(x2+4)
6. Solve for all solutions in the interval [0,2π)
8sinx = -4 √2
Sin x = -sqrt(2)/2, x = 5 /4, 7/4
7. Solve for all real roots.
 3 sec 3x   2
sec (3x) = -2/√3. Same as cos(3x) = −√3/2. If y = 3x, y = 5/6, 7/6; need all reals, so y = 5/6+2n and
7/6+2n; Now solve for x by dividing by 3, x = 5/18+2n/3 and 7/18+2n/3
8. Solve the equation in [0,2π) by factoring
4sin 3 x  4sin 2 x  sin x  1  0
4 sin2x(sinx – 1) – (sinx -1) = (4 sin2x -1) (sinx -1); sinx = 1, x = /2, sin2x = ¼, sinx = ± ½, x = /6, 5/6, 7/6, 11/6
9. State the period P of the function and find all solutions in [0,P).


150 cos  x    75  0
6
4
The period is 2/(/4) = 8;
Cos(x/4 + /6) = ½, x/4 + /3- /6 or x/4 = /6 +2n. For the next value, x/4 + /6 = 5/3 or x/4 = 3/2 + 2n
Now solve for x: x/4 = 1/6 + 2n or x/4 = 3/2 + 2n. x = 2/3 +8n or x = 6 + 8n. Try values of n: n = 0, x =6 or 2/3, both in
the period.
If n = 1, x is too large. We have x = {2/3, -1/4}
10. Graph one period of the function f(x) = -3 sin (2t + /4)
the period P = 2/2 = . The amplitude is -3. Need to shift horizontally, as t=0 gives -3sin(/4)
11. Graph the function f(x) = 2cot(x)
period is 1, amplitude is 2 – means that the value at ¼ the period, or at 0.25 is 2, not 1
12. Complete the following table;
f(x)
Domain
Range
arcsin(x) [-1, 1] [-pi/2, pi/2]
arcos(x) [-1, 1]
[0, pi]
arctan(x)
[
[-pi/2, pi/2]
f(-1)
-pi/2
pi
-pi/4
For problems 13-15, evaluate the following:
f(0)
0
pi/2
0
13. cos−1 (cos 2𝜋)
cos 2 𝜋 = 1, arcos(1) = 2 𝜋
4
14. tan (cos−1 (− 5))
if cos = 4/5, we have adj/hyp = 4/5. By Pythagoras, we can find opp = 3. Tangent is ¾. Now, get the sign
right. Cos is negative in the 2nd or 3rd quadrant, however arcos is only a function in quadrants 1 and 2.
4
Thus we have tan (cos −1 (− 5)) = -3/4
𝜋
15. tan−1 (2 sin 3 )
sin /3 = √3 /2, so the argument is √3 . Where is tan = √3? Need sin/cos = √3. This happens for 30 deg
or /3, where sin is ½ and cos is √3 /2.
16. 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
17. 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,
18. 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), and ( – 4 3 , 4)
For (4, -4), radius is √32 = 4√2 . Angle is arctan of -4/4; angle is 7/4 since is in Q4,
Polar is (4√2, 7/4),
For (-4√3 , 4), radius is √16 + 48 = √64 = 8, angle is arctan ( -1/√3) = 5/6, since is in Q2.
Polar is (8, 5/6).

)
6
For (-4, π), radius is 4, angle is π, so along x axis, but, since we have -4, is positive x axis. Hence we have
rectangular coordinates of (4, 0)




For (4, ), radius is 4, tan( ) =√3 /2. So we have x = 4 cos
, y = 4 sin
.
6
6
6
6
The rectangular coordinates are (2√3 , 2)
19. Convert from polar to rectangular coordinates: (-4, π) and (4,