Question

Single equation: if sin B = -12/13, Π<-B<3π/2 what is the answer for sin 2B or sin B/2

Answer 1

sin 2B is -120/169 using the double-angle formula, and sin B/2 is 3/√13 using the positive branch of the half-angle formula because B/2 is in the second quadrant.

Step-by-step explanation:

Since sin B is given as -12/13 and B is in the range between Π and 3π/2, B is in the third quadrant, where sine is negative and cosine is negative. We can find the cosine of B using the Pythagorean identity: cos2B = 1 - sin2B. This gives us:

cos2B = 1 - (-12/13)2

cos2B = 1 - 144/169

cos2B = 25/169

Since cosine is also negative in the third quadrant, cos B = -5/13.

For sin 2B, we use the double angle formula sin 2B = 2 sin B cos B.

sin 2B = 2(-12/13)(-5/13) = 24/13 * -5/13 = -120/169

To find sin B/2, we must use the half-angle formula. For sin B/2, we have two cases depending on the sign of the half angle. Since B/2 will be in the second quadrant, sin B/2 will be positive. We have:
sin B/2 = ±1 √(1 - cos B)/2

sin B/2 = √(1 - (-5/13))/2 = √(1 + 5/13)/2 = √(18/13)/2 = √(9/13) = 3/ √13

Answer 2

sin2B = 2sinBCosB (double angle formula)

sin(B/2) = 1/2 - 1/2 Cos(B)

Pi to 3pi/2 -> 3rd quadrant

cosB-> negative , sinB negative

CosB = -5/13

sin2B = 2 * -5/13 * -12/13 = 120/169

sinB/2 = 1/2 - 1/2 * * -5/13 = 9/13