Question

Chứng minh:

\(cos\left(\frac{17\pi}{4}+x\right).cos\left(\frac{\pi}{4}-x\right)+sin^2x=\frac{1}{2}\)

Answer 1

\(cos\left(\frac{17\pi}{4}+x\right)cos\left(\frac{\pi}{4}-x\right)+sin^2x\)

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

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

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

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

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