Download MTH 1120 – Quiz 1 w Solutions (2011) – College Trigonometry

MTH 1120, College Trigonometry, Quiz 1
Name:
Instructions: Work the following problems; give your reasoning where appropriate. Do not give
decimal approximations.
1. If t is a real number,
√
1
1
=√ .
15, find sin t. Solution: sin t =
csc t
15
√
5
1
3
(b) and cot t = −
, find tan t. Solution: tan t =
= −√ .
3
cot t
5
(a) and csc t =
2. Find the values of all of the trigonometric functions for the real number t given that tan t =
15
−
and P (t), the image of t under the action of the wrapping function, is in quadrant II.
8
Solution: If P (t) is in quadrant II and tan t = −15/8, then P (t) must have the form (x, y),
where x2 +y 2 = 1 and y/x = −15/8. From the latter of these equations, we see that y = −15x/8.
Replacing y with −15x/8 in the equation for the unit circle gives
15
1=x +y =x + − x
8
2
2
2
2
225
289 2
= 1+
x2 =
x .
64
64
Hence x = ±8/17. But P (t) is in the second quadrant, so x = −8/17. From y = −15x/8, we
8
15
17
17
8
have y = 15/17. It follows that cos t = − ; sin t =
; sec t = − ; csc t =
; cot t = − .
17
17
8
15
15
3. You are given that P (t), the image of the real number t under the action of the wrapping
function, lies in quadrant II, and that the point (−5, 12) lies on the half-line that extends
from the origin through P (t). Find the values of the six trigonometric functions of t.
Solution: We know that P (t) lies on the unit circle and on the line through (0, 0) and (−5, 12),
whose equation is y = −12x/5. Thus, the coordinates of P (t) have the form (x, y), where
x2 + y 2 = 1 and y = −12x/5. Substituting the latter into the former, we find that
12
1=x +y =x + − x
5
2
2
2
2
144
169 2
= 1+
x2 =
x ,
25
25
so that x = ±5/13. But P (t) lies in the second quadrant, so x = −5/13. Then y = −12x/5 =
5
12
12
13
13
12/13, and it follows that cos θ = − ; sin θ =
; tan θ = − ; sec θ = − ; csc θ =
;
13
13
5
5
12
5
cot θ = − .
12