Posted by
**Ashley** on
.

If secx = 8 and -pi/2 < x < 0, find the exact value of sin2x

Use the identity sin 2x = 2(sinx)(cosx)

if secx = 8, then cosx = 1/8 where x is in the fourth quadrant.

consider a right angled triangle with x=1, r=8, then y=??

by Pythagoras 1^2 + y^2 = 8^2

y = -√63 (it has to be negative in IV)

so sinx = -√63/8

so we have sin 2x = 2(-√63/8)(1/8)

=-√63/32