Question
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
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