If Ѳ is in Quadrant II, and cos Ѳ=-3/4 , find an exact value for sin 2Ѳ.

Answer 1

Make a sketch of your triangle in quadrant II to see that
x = -3, r = 4 and from Pythagoras that y = √7 in II

so sinØ = √7/4

sin 2Ø = 2sinØcosØ
= 2((√7/4)(-3/4)
= -3√7/8

Answer 2

sin2A= 2sinAcosA

now we know that cosA=-3/4, and theta is in the second quadrant so you sketch a triangle in the second quadrant, remember the bowtie method

-3 is a leg and 4 is the hypotenuse based on sohcahtoa

plug these two into the pythagorean theorem to find the missing leg.

then you know that the missing leg is sqrt of 7

so sinA is sqrt7/4

2sinAcosA= sin2A

2(sqrt7/4)(-3/4)= sin2A

= -6(sqrt7)/16
= -3(sqrt7)/8