SOLUTION: If θ is an angle in the third quadrant and csc θ=-(13/5), find the value of 3sin^2 θ

SOLUTION: If θ is an angle in the third quadrant and csc θ=-(13/5), find the value of 3sin^2 θ - 2cos θ as a fraction with no decimals.

Question 1197407: If θ is an angle in the third quadrant and csc θ=-(13/5), find the value of 3sin^2 θ - 2cos θ as a fraction with no decimals.
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

If you use the pythagorean theorem , you'll have a 5-12-13 right triangle.
leg1 = 5 units
leg2 = 12 units
hypotenuse = 13 units

Start at the origin of the xy coordinate axis.
Move 12 units to the left, then move 5 units down
This forms a right triangle with the 90 degree angle on the x axis.
Angle theta is shown in green below.

We start facing directly east along the positive x axis.
Then rotate until we are aiming in the southwest direction along the hypotenuse of length 13.

Csc = cosecant
csc is the reciprocal of sine

if , then

Then use the pythagorean trig identity with that sine value found to get .
Or you can use the reference diagram above.

sine = opposite/hypotenuse
cosine = adjacent/hypotenuse
In this case, the opposite side is the vertical leg labeled 5. The adjacent side is 12 units long.
The negative signs are there so we get the correct sign for each trig ratio.

Now we can compute what your teacher asked

Answer: 387/169

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