sin 2B using given information and double angle formula?

Question

GIVEN:
tan A= -3/5, 3rd quad. (π
3π/2sin B= -1/4, 2nd quad.(π/2
cot C= 1/2, 2nd quad.(π/2

Double-angle formula for sine:

sin2x = 2sinxcosx



THANK YOU!!!!!

Answer 1

The only info you need to worry about is the part pertaining to angle B

I'm not sure what 3π/2 is doing in front of sin B. I'll assume that is a typo.
So sin B = -1/4

You are given the double angle formula: sin (2B) = 2sinBcosB
You already know sinB = -1/4, so you need to find cosB

cosB = adjacent / hypotenuse
sinB = opposite / hypotenuse = -1 / 4
You can use the pythagorean theorem to find the adjacent side.
1^2 + (adjacent side)^2 = 4^2
1 + (adjacent side)^2 = 16
(adjacent side)^2 = 15
adjacent side = sqrt(15) or -sqrt(15)
You are given that the angle is in the second quadrant (because π/2 < B < π), and since cosine is negative there, then the adjacent will be negative.

So sin(2B)
= 2sinBcosB
= 2(-1/4)(-sqrt(15))
= sqrt(15) / 2

Looking back on the problem, I'm going to assume there is another typo. It says second quadrant for x. Is that x supposed to be B? If it is B, then something is wrong because sine is positive in the second quadrant. So if sinB is -1/4, then B has to be in the third or fourth quadrant.
You'll just have to figure out the signs.