# Let(theta) be in Q3 & csc(theta) = -5/3, find cos2(theta) and determine in which quadrant 2(theta) terminates. Draw 2 diagrams.

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`cosec(theta) = -5/3`

But we know, `sin(theta) = 1/(cosec(theta))`

Therefore, `sin(theta) = 1/(-5/3) = -3/5.`

`sin(theta) = -3/5`

Therefore `theta = sin^(-1)(-3/5) = -36.87` degrees.

But we know theta is in Q3, therefore we have to find the general solution for sine in Q3. The general solution for sine is given by,

`theta = n180+(-1)^n(-36.87)` where n is any integer.

n =1 would give the required answer,

`theta = 180+(-1)(-36.87)`

`theta = 180+36.87 = 216.87` degrees

Therefore `2theta = (2)(216.87)` degrees

` = 433.74` degrees

To find the qudrant, we can convert this to a basic angle by reducing 360 from it.

`2theta = 433.74 -360 = 73.74 ` degrees.

Therefore 2theta is in the first quadrant (Q1).

To find `cos(2theta)` we can us the following identity,

`cos(2theta) = 1 - 2sin^(2)(theta)`

`cos(2theta) = 1 - 2(-3/5)^2`

`= 0.28`

` cos(2theta) = 0.28`

I should also add that I swiched the values of sine and cosine in the calculation so the cosine of 2-theta is actually 7/25, and theta is about 74º.

csc(theta) = -5/3

sin (theta) = - 3/5

cos (theta) = (1 - sin^2 (theta)^.5

cos (theta) = (1 - (-3/5)^2)^.5

cos (theta) = (1 - (9/25))^.5

cos (theta) = (16/25)^.5

cos (theta) = + or - 4/5

cos (2theta) = cos^2 (theta) - sin^2 (theta)

cos (2theta) = (3/5)^2 - (4/5)^2

cos (2theta) = 9/25 - 16/25

cos (2theta) = -7/25

I don't know how to draw diagrams on Enotes.

As to the quadrant, I think the original angle is about 216º because the arcsin of (3/5) +180 is about 216º.

2*theta is about 432º.

432 - 360 = 72º, which is in the first quadrant.