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).