SOLUTION: If sin(t)=2/3, and t is in quadrant II, find cos(t), sec(t), csc(t), tan(t), cot(t).

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Question 1203332: If sin(t)=2/3, and t is in quadrant II, find
cos(t), sec(t), csc(t), tan(t), cot(t).

Found 3 solutions by Alan3354, ikleyn, math_tutor2020:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
If sin(t)=2/3, and t is in quadrant II, find
cos(t), sec(t), csc(t), tan(t), cot(t).
-----------------
cos = sqrt(1 - sin^2) = sqrt(5/9)
---
tan = sin/cos
cot = cos/sin
sec = 1/cos
csc = 1/sin

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
If sin(t)=2/3, and t is in quadrant II, find
cos(t), sec(t), csc(t), tan(t), cot(t).
-----------------
cos = -sqrt(1 - sin^2) = -sqrt(5)/3,   since in quadrant II cosine function is negative.
---
tan = sin/cos
cot = cos/sin
sec = 1/cos
csc = 1/sin

Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

I'll use the greek letter theta (pronounced "theta") in place of t because theta is often used in trig.

sin(theta) = 2/3
sine = opposite/hypotenuse

We could say
opposite = 2
hypotenuse = 3

Draw a right triangle in quadrant Q2 (northwest quadrant)
The reference angle theta is placed in the bottom right corner of the triangle.

The unknown side adjacent is labeled x.

Use the pythagorean theorem
a%5E2%2Bb%5E2+=+c%5E2

2%5E2%2Bx%5E2+=+3%5E2

4%2Bx%5E2+=+9

x%5E2+=+9-4

x%5E2+=+5

x+=+sqrt%285%29 or x+=+-sqrt%285%29
I'll go with the negative version of x to indicate we're to the left of the y axis.
x < 0 in Q2


Here's the updated diagram

opposite = 2
adjacent = -sqrt%285%29
hypotenuse = 3

Then,
cos%28theta%29+=+adjacent%2Fhypotenuse+=+-sqrt%285%29%2F3

tan%28theta%29+=+opposite%2Fadjacent+=+2%2F%28-sqrt%285%29%29+=+-2sqrt%285%29%2F5

csc%28theta%29+=+hypotenuse%2Fopposite+=+3%2F2 (this is the reciprocal of sine)

sec%28theta%29+=+hypotenuse%2Fadjacent+=+3%2F%28-sqrt%285%29%29+=+-3sqrt%285%29%2F5 (this is the reciprocal of cosine)

cot%28theta%29+=+adjacent%2Fopposite+=+-sqrt%285%29%2F2 (this is the reciprocal of tangent)