Đáp án:
a) $\sqrt 2.\sin\left(x+\dfrac{\pi}{4}\right)$
b) $\sqrt 2.\sin\left(x-\dfrac{\pi}{4}\right)$
Giải thích các bước giải:
a) $\sin x+\cos x$
$=\sqrt 2.\left(\dfrac{\sqrt 2}{2}\sin x+\dfrac{\sqrt 2}{2}\cos x\right)$
$=\sqrt 2.\left(\sin x.\cos \dfrac{\pi}{4}+\sin\dfrac{\pi}{4}.\cos x\right)$
$=\sqrt 2.\sin\left(x+\dfrac{\pi}{4}\right)$
b) $\sin x-\cos x$
$=\sqrt 2.\left(\dfrac{\sqrt 2}{2}\sin x-\dfrac{\sqrt 2}{2}\cos x\right)$
$=\sqrt 2.\left(\sin x.\cos \dfrac{\pi}{4}-\sin\dfrac{\pi}{4}.\cos x\right)$
$=\sqrt 2.\sin\left(x-\dfrac{\pi}{4}\right)$.
