Honors Math 3
Name:
Date:
Final Exam Review Problems Part 2
Chapter 4 and 5: Trigonometry
22. Without a calculator, evaluate sin(cos–1(3/5)).
23. For any angle , how are cos( p2 – ) and sin( – ) related? Justify your answer using
a circle diagram.
cos3x = 4cos3 x - 3cos x
24. Prove the identity:
25. Suppose that sin  =
3
5
and sin  =
24
25
, where 0 <  < p2 <  < . Find cos( + ).
26. Find the general solution to each equation (in radians):
a.
27.
cos x = -0.82
b. sin x = 0.812
Find the general solution to each equation (in degrees):
a. cos x = 0.756
b. sin x = -0.155
28. Using non-graphical methods, find all solutions to the equations in the interval 0 ≤ x
< 2. Check your answer graphically.
a. sin(3x) = –0.5
b. sin2 x – sin x = cos2 x
c. 3sin ( x + 2 ) -10 = -11
d.
1 æp ö
cosç x ÷ - 3 = 4
2 è4 ø
29. Sketch two periods of the following sinusoids. Be accurate and label appropriately.
a. f (x) = -2cos( p2 x ) -1
6
( ( x - ))
b. g(x) = 3sin
p
1
2
2
30. Find 2 equations of the sinusoid, one
using the sine function and one using the
4
cosine function.
2
2π
π
π
2π
3π
2
4
6
8
10
31. a. Write a function formula for a sinusoidal function f(x) having the following
properties:
o Two adjacent maximum points of f(x) are located at (3, 5) and (7, 5).
o The graph of f(x) is tangent to the x-axis.
b. Suppose that the graph of g(x) is formed by compressing f(x) horizontally by a
factor of 8. Write a function formula for g(x).
32. Assume that you are aboard a submarine, submerged in the Pacific Ocean. At time t
= 0 minutes you spot an enemy destroyer. Immediately, you start diving lower to
avoid detection. At t = 4 minutes, you are at your greatest depth of -1000 miles. At
time t = 9 minutes, you have ascended to your minimum depth of -200 miles.
Assume that the path of the submarine varies sinusoidally with respect to t for t ³ 0.
a. Sketch two of these dives for the submarine (2 cycles).
b. Find a possible formula for f(t) which represents the depth of the submarine.
c. Your submarine is safe when it is below a depth of -300 miles. At time t = 0
minutes, was your submarine safe? Explain your answer.
d. Between what two (non-negative) times is your submarine first safe?
33.
Solve the following triangles below (find all possible missing angles and sides).
Not draw to scale.
a.
b.
B
E
10
7
60°
40°
A
12
c.
C
D
12
Y
6
32°
X
34.
10
Z
Find the area of quadrilateral ABCD below. (Hint: Start by connecting points A
and C)
D
55º
A
8
10.8
B
120º C
10
F
Chapter 6: Complex Numbers and Polynomials
35. Find a cubic polynomial, C(x), with real coefficients has the following properties:
 1- 2i is a complex root.
36.
 the graph of C(x) has an x-intercept at
x = 2.

C(1) = -12
Prove that the quotient of two complex numbers is complex. In other words,
show that the quotient can be expressed in the form a + bi. State any restrictions that
are necessary.
37. Solve each equation. Find all solutions.
a. x 3 = x 2 + 23x + 42
b. x 3 = -8
38. You are given this information about F(x):


F(x) is a polynomial with real coefficients.
The graph of F(x) for real numbers x is given on the
grid. All zeros and extrema can be seen from this
graph. The intercepts are at (2, 0), (4, 0), and (0, 5).
 F(x) has the smallest degree that it could possibly have
based on the number of zeros and the number of
extrema shown on the graph.
 In the complex numbers, F(–3 + i) = 0.
Find the factorization of F(x) in the real number system.
39.
Given: z = 3+ 4i and w =1+ 2i. Find the magnitude and argument of zw.
Answers
22. A right triangle with angle cos–1( 35 ) will have sides of 3, 4, and 5. This gives
sin(cos–1( 35 )) = 45 .
23. cos( p2 – ) = sin  because the x-coordinate at angle p2 –  is the same as the ycoordinate at angle  (complementary angles). sin( – ) = sin  because the ycoordinates at the same at angle  –  and angle . Therefore, cos( p2 – ) = sin( – ).
24.
cos3x = cos(2x + x) = cos2x cos x - sin2x sin x = (cos2 x - sin2 x )× cos x - (2sin x cos x )× sin x =
cos3 x - sin2 x× cos x - 2sin2 x cos x = cos3 x - 3sin2 x cos x =
cos3 x - 3(1- cos2 x)× cos x = cos3 x - 3cos x + 3cos3 x = 4cos3 x - 3cos x
25. The right triangle with angle sin–1( 35 ) will have sides of 3, 4, and 5 in Q1. The right
triangle with angle sin–1( 24
) will have sides of 7, 24, and 25 in Q2.
25
So cos( + ) = cos  cos  – sin  sin  = 45 ×
-7
25
- 35 ×
24
25
=
-100
125
=
-4
5
.
26. a. x = 2.532 + 2p k , x = 3.751+ 2p k for any integer k
b. x = 0.948+ 2p k , x = 2.194 + 2p k for any integer k
27.
a. x = 40.887°+360°k , x = 319.113°+360°k for any integer k
b. x = 351.083°+360°k , x =188.917°+360°k for any integer k
28. a. 3x = 76p + 2pk or 3x = 116p + 2pk
7p
7p 11p 19p 23p 31p 35p
+ 23 pk or x = 1118p + 23 pk In the specified interval: { 18
, 18 , 18 , 18 , 18 , 18 }.
x = 18
b. sin2 x – sin x = 1 – sin2 x
2 sin2 x – sin x – 1 = 0
(2 sin x + 1)(sin x – 1) = 0
sin x = - 12 or sin x = 1
In the specified interval: { p2 , 76p , 116p }.
c. sin(x + 2) = - 13 so x + 2 = 5.943 or 3.481, x = 3.943 or 1.481
d. cos( p4 x) =14 so no solution
29. check graphs on your calculator
æ 2 æ p öö
æ 2 æ p öö
30. possible answers: f (x) = 5sin ç ç x + ÷÷ - 4
f (x) = 5cos ç ç x - ÷÷ - 4
4 øø
2 øø
è3è
è3è
31. a. possible answers: f (x) = 2.5sin( 24p (x - 2))+ 2.5 or 2.5cos( 24p (x - 3))+ 2.5
b. period 1/8 as big now so c = 4p : g(x) = 2.5sin(4p (x - 2))+ 2.5
æp
ö
32. b. possible answer: f (t) = -400cos ç ( t - 4) ÷ - 600
c. not safe f (0) = -276.39 ft
è5
ø
d. Solve f (t) = -300 and find both solutions. Between 0.150 and 7.850 min
33. a. ASA – one triangle
Use triangle sum to find last angle ÐB = 80°
Use Law of Sines to find each of the other two sides
12
a
c
=
=
c =10.553
a = 7.832
sin80 sin 40 sin 60
b. SSS – one triangle
Use Law of Cosines to find two of the angles
ÐE = 87.95°
ÐD = 56.39°
Use triangle sum to find the third angle ÐF = 35.66°
122 =102 + 72 - 2(10)(7)cosE
102 =122 + 72 - 2(12)(7)cosD
c. SSA – could be 0, 1, or 2 triangles possible
Can use either LOS or LOC to start…
If use LOS to find another angle
6
10
=
ÐY = 62.03° OR ÐY =117.97°
sin32 sinY
By triangle sum…
ÐX = 85.97° OR ÐX = 30.03°
Using LOS or LOC x =11.294 OR x = 5.666
34.
Area = 108.365 sq units
35.
C must have 3 zeros (it is a cubic). Since 1- 2i is a zero, 1+ 2i must also be a
zero (real coefficients, complex conjugates theorem). C must have the form:
C(x) = a(x - 2)(x - (1- 2i))(x - (1+ 2i)), where a will be determined by the condition
f (1) = -12
C(x) = a(x 3 - 4x 2 + 9x -10)
36.
C(1) = a(1- 4 + 9 -10) = -12 Þ a = 3
Let our two complex #s be a + bi and c + di with a,b,c,d real #s and c,d ≠ 0
)i
cb-ad
Divide to get: a+bi
× c-di = (ac+bd2)+(cb-ad
= ac+db
2
2 2 + 2 2 i
c+di c-di
c +d
( ) ( )
c +d
c +d
the last expression is complex: (real number) + (real number)i .
37. a. x 3 - x 2 - 23x - 42 = 0 Checking factors of -42 see that x = 6 is a zero. Use
division to get: (x - 6)(x 2 + 5x + 7) = 0 use the quadratic formula to find the other two
-5 ± i 3
solutions. Solutions: x = 6,
.
2
b. x 3 +8 = 0 ; same method can be used as above. Alternatively, you can use the sum of
cubes to get (x + 2)(x 2 - 2x + 4) = 0 first. Solutions are: x = -2, 1± i 3 .
38. Since -3+ i is a zero, -3- i must also be a zero (real coefficients, complex
conjugates theorem). F must have the form:
F(x) = a(x - 2)(x - 4)2 (x - (-3+i))(x - (-3- i)) , where a will be determined by the
condition f ( 0) = 5
f (x) =
-1
( x - 2)( x - 4) 2 (x 2 + 6x +10)
64
39. magnitude of zw = zw = z × w = 5 5
arg(zw) = arg(z)+arg(w) =