Question

use the double angle identities to rewrite sin^2 (x) in terms of linear powers of common trig functions

Answer 1

\[\cos{2x}=\cos^2{x}-\sin^2{x}=1-\sin^2{x}-\sin^2{x}=1-2\sin^2{x}\]\[\Rightarrow\quad\sin^2{x}=\frac{1-\cos{2x}}{2}\]

Answer 2

@nikvist wait i just dont get how this solves the problem.. what are the common trig functions?