Question

Chứng minh
a) \(sin^4x=\frac{3}{8}-\frac{1}{2}cos2x+\frac{1}{8}cos4x\)

b) \(\frac{cos\left(a+b\right)cos\left(a-b\right)}{cos^2a.cos^2b}=1-tan^2a.tan^2b\)

Answer 1

\(sin^4x=\left(sin^2x\right)^2=\left(\frac{1}{2}-\frac{1}{2}cos2x\right)^2=\frac{1}{4}-\frac{1}{2}cos2x+\frac{1}{4}cos^22x\)

\(=\frac{1}{4}-\frac{1}{2}cos2x+\frac{1}{4}\left(\frac{1}{2}+\frac{1}{2}cos4x\right)=\frac{3}{8}-\frac{1}{2}cos2x+\frac{1}{8}cos4x\)

\(\frac{cos\left(a+b\right)cos\left(a-b\right)}{cos^2a.cos^2b}=\frac{\left(cosa.cosb-sina.sinb\right)\left(cosa.cosb+sina.sinb\right)}{cos^2a.cos^2b}\)

\(=\frac{cos^2a.cos^2b-sin^2a.sin^2b}{cos^2a.cos^2b}=1-\frac{sin^2a.sin^2b}{cos^2a.cos^2b}=1-tan^2a.tan^2b\)