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EXAM 1 PRACTICE SOLUTIONS
1. Given that the point P = (− 52 ,
√
21
)
5
is on the unit circle what are the exact values of all six
trigonometric functions?
• sin = − 52 • cos =
√
21
5
√
• tan = − 2 2121 • csc = − 52 • sec =
√
5 21
21
√
• cot = −
21
.
2
2. Find the exact value of the following
) = sin(3 · 2π − π2 ) = − sin( π2 ) = −1
a. sin( 11π
2
√
b. sin(−300◦ ) =
3
2
c. cos( 10π
) = − 12
3
d. cos(405◦ ) =
√
2
2
The following illustration is the labeling we used for the following two questions
3. Solve the right triangle if a = 6, b = 4.
√
√
Solution c = 52 = 2 13, A = 56.3◦ , B = 33.7◦ .
4. Solve the right triangle if B = 25◦ , c = 13.
Solution A = 65◦ , b = 5.49, a = 11.78
5. Find the exact value of each of the remaining trigonometric functions of θ given that
cos θ =
3
5
and
3π
2
Solution cos θ =
< θ < 2π.
x
hyp
=
3
5
implying that x = 3 and hyp = 5 and so by the Pythagorean theorem y = 4. Since the
angle is in the fourth quadrant sin θ < 0. Hence
• sin θ = − 54 , tan θ = − 43 , sec θ = 53 , csc θ = − 54 , cot θ = − 34 .
6. Find the exact value of each of the remaining trigonometric functions of θ given that
tan θ = − 31 and sin θ > 0.
Solution Since tan θ < 0 and sin θ > 0 it must be that θ ∈ ( π2 , π) and if we write tan θ =
1
y
x
we have that y = 1
2
EXAM 1 PRACTICE SOLUTIONS
and x = −3. By the Pythagorean theorem the hypotenuse of the right triangle is
√
√
√
√
• sin θ = 1010 , • cos θ = − 3 1010 , • cot θ = −3, • csc θ = 10, sec θ = − 310
√
10. Hence
7. For the function f (x) = − 43 cos(3x − π4 ) + 1
a. What is the amplitude? 3/4
b. What is the period?
2π
3
c. What is the phase shift?
π
12
d. What is the maximum value f (x) can take? 3/4+1
e. What is the minimum value f (x) can take? -3/4+1
8. For the graph below, write two equations for the graph, one using the function sine and the
other using the function cosine.
−3 sin(3x + π) + 1 or 3 cos(3(x + π2 )) + 1 or −3 cos(3(x − π2 )) + 1
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