Question

\(\sin^4x+\cos^4x=1-\frac{1}{2}\sin^22x\)

Answer 1

\(sin^4x+cos^4x=\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\)

\(=1-\frac{1}{2}\left(2sinx.cosx\right)^2=1-\frac{1}{2}sin^22x\)