mn cho e hỏi xét tính chẵn lẻ hàm số y=cos(2x+ pi/4) +sin(2x-pi/4)

\(y=\cos(2x+ \dfrac{\pi}{4}) +\sin(2x-\dfrac{\pi}{4})\)

\(=\dfrac{\sqrt 2}{2}\cos 2x-\dfrac{\sqrt 2}{2}\sin 2x+\dfrac{\sqrt 2}{2}\sin 2x-\dfrac{\sqrt 2}{2}\cos 2x\)

\(=0\)

TXĐ: \(D=\mathbb R\)

Khi đó \(x\in D\) \(\Rightarrow \exists (-x)\in D\)

Xét: \(y(-x)=\cos(-2x+ \dfrac{\pi}{4}) +\sin(-2x-\dfrac{\pi}{4})\)

\(=\dfrac{\sqrt 2}{2}\cos (-2x)-\dfrac{\sqrt 2}{2}\sin (-2x)+\dfrac{\sqrt 2}{2}\sin (-2x)-\dfrac{\sqrt 2}{2}\cos (-2x)=0=y(x)\)

Vậy hàm đã cho là hàm chẵn.