Chứng minh rằng:

a) $\sin \alpha + \cos \alpha = \sqrt 2 \cos \left( {\alpha - \frac{\pi }{4}} \right) = \sqrt 2 \sin \left( {\alpha + \frac{\pi }{4}} \right)$;

a) Ta có:

$\sqrt 2 \cos \left( {\alpha - \frac{\pi }{4}} \right) = \sqrt 2 \left( {\cos \alpha \cos \frac{\pi }{4} + \sin \alpha \sin \frac{\pi }{4}} \right)$

$= \sqrt 2 \left( {\cos \alpha .\frac{{\sqrt 2 }}{2} + \sin \alpha .\frac{{\sqrt 2 }}{2}} \right)$

$= \sqrt 2 \frac{{\sqrt 2 }}{2}\left( {\cos \alpha + \sin \alpha } \right)$

= cosα + sin α        (1)

$\sqrt 2 \sin \left( {\alpha + \frac{\pi }{4}} \right) = \sqrt 2 \left( {\sin \alpha \cos \frac{\pi }{4} + \cos \alpha \sin \frac{\pi }{4}} \right)$

$= \sqrt 2 \left( {\sin \alpha .\frac{{\sqrt 2 }}{2} + \cos \alpha .\frac{{\sqrt 2 }}{2}} \right)$

$= \sqrt 2 \frac{{\sqrt 2 }}{2}\left( {\sin \alpha + \cos \alpha } \right)$

= sin α + cosα       (2)

Từ (1) và (2) $\Rightarrow \sin \alpha + \cos \alpha = \sqrt 2 \cos \left( {\alpha - \frac{\pi }{4}} \right) = \sqrt 2 \sin \left( {\alpha + \frac{\pi }{4}} \right)$ (đccm);

b) Ta có: $\sqrt 2 \sin \left( {\alpha - \frac{\pi }{4}} \right) = \sqrt 2 \left( {\sin \alpha \cos \frac{\pi }{4} - \cos \alpha \sin \frac{\pi }{4}} \right)$

$= \sqrt 2 \left( {\cos \alpha .\frac{{\sqrt 2 }}{2} - \sin \alpha .\frac{{\sqrt 2 }}{2}} \right)$

$= \sqrt 2 \frac{{\sqrt 2 }}{2}\left( {\cos \alpha - \sin \alpha } \right)$

= cosα – sin α (3)

$- \sqrt 2 \cos \left( {\alpha - \frac{\pi }{4}} \right) = - \sqrt 2 \left( {\sin \alpha \cos \frac{\pi }{4} - \cos \alpha \sin \frac{\pi }{4}} \right)$

$= - \sqrt 2 \left( {\sin \alpha .\frac{{\sqrt 2 }}{2} - \cos \alpha .\frac{{\sqrt 2 }}{2}} \right)$

\( = - \sqrt 2 \frac{{\sqrt 2 }}{2}\left( {\sin \alpha  - \cos \alpha } \right)\)

= – (sin α – cosα) = cosα – sin α (4)

Từ (3) và (4) $\Rightarrow \sin \alpha - \cos \alpha = \sqrt 2 \sin \left( {\alpha - \frac{\pi }{4}} \right) = - \sqrt 2 \cos \left( {\alpha + \frac{\pi }{4}} \right)$ (đpcm).