Question 825698
If secθ= 4/5 and θ terminates in Quadrant II, find the exact value of sin2θ.

cos^2θ + sin^2θ = 1 because of Pythagorean identities

Therefore sin^2θ = 1- cos^2θ

cosθ = 1/secθ

So,

sin^2θ = 1- cos^2θ
sin^2θ = 1- (1/secθ)^2

sin^2θ = 1- {1/ (4/5)}^2
sin^2θ = 1- (5/4)^2
sin^2θ = 1- (25/16)
sin^2θ = 1- (25/16)
sin^2θ = 16/16 - (25/16)
sin^2θ = 16-25 /16
sin^2θ = -9/16

Tht's the answer :)