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EXAM 1 PRACTICE SOLUTIONS
1. Given that the point P = (
2
,
5
p
21
)
5
is on the unit circle what are the exact values of all six
Cos = -2/5, Sin = sqrt(21)/5, Tan = -sqrt(21)/2, Cot = -2/sqrt(21)
p
p
p
21
• tan = 2 2121 • csc = 52 • sec = 5 2121 • cot =
. Sec = -5/2, Csc = 5/sqrt(21)
2
trigonometric functions?
2
5
• sin =
p
21
5
• cos =
2. Find the exact value of the following
a. sin( 11⇡
) = sin(3 · 2⇡
2
d. cos(405 ) =
=
sin( ⇡2 ) =
1
3
2
b. sin( 300 ) =
c. cos( 10⇡
)=
3
⇡
)
2
p
1
2
p
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.
p
p
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 ✓ given that
cos ✓ =
3
5
and
3⇡
2
< ✓ < 2⇡.
Solution cos ✓ =
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 ✓ < 0. Hence
4
, tan ✓
5
• sin ✓ =
=
4
, sec ✓
3
= 53 , csc ✓ =
5
, cot ✓
4
=
3
.
4
6. Find the exact value of each of the remaining trigonometric functions of ✓ given that
tan ✓ =
1
3
and sin ✓ > 0.
Solution Since tan ✓ < 0 and sin ✓ > 0 it must be that ✓ 2 ( ⇡2 , ⇡) and if we write tan ✓ =
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
p
p
p
p
10
• sin ✓ = 1010 , • cos ✓ = 3 1010 , • cot ✓ = 3, • csc ✓ = 10, sec ✓ =
3
7. For the function f (x) =
3
4
cos(3x
⇡
)
4
p
10. Hence
+1
a. What is the amplitude? 3/4
b. What is the period?
2⇡
3
c. What is the phase shift?
⇡
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 + ⇡) + 1 or 3 cos(3(x + ⇡2 )) + 1 or
3 cos(3(x
⇡
))
2
+1
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