If Sinθ=4/5, find the exact value cos2θ

Question

I don’t know exactly how to do this problem.. please help

Answer 1

sinx = 4/5, x is in quadrant I or II

cosx =3/5 or -3/5, cosx = + or - sqr(1-sin^2x) = sqr(1-16/25) = sqr(9/25 = 3/5

use one of the double angle formula for cosines

cos2x = cos^2 - sin^2= 9/25 -16/25 = - 7/25

it's negative because 2x is in quadrant II or III where cosines are negative

or use cos2x = 1-2sin^2x = 1 - 2(4/5)^2 = 1-2(16/25) = 1-32/25 = 25/25-32/25 =-7/25