Sin 2B using given information and double angle formula?

GIVEN:


tan A= -3/5, 3rd quad. (π%26lt;x%26lt;3π/2)





3π/2sin B= -1/4, 2nd quad.(π/2%26lt;x%26lt;π)





cot C= 1/2, 2nd quad.(π/2%26lt;x%26lt;π)








Double-angle formula for sine:





sin2x = 2sinxcosx











THANK YOU!!!!!

Sin 2B using given information and double angle formula?
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 %26lt; B %26lt; 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.