Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\sin \alpha + \cos \alpha \\
= \sqrt 2 .\left( {\dfrac{{\sqrt 2 }}{2}.\sin \alpha + \dfrac{{\sqrt 2 }}{2}.\cos \alpha } \right)\\
= \sqrt 2 .\left( {\sin \alpha .\cos \dfrac{\pi }{4} + \cos \alpha .\sin \dfrac{\pi }{4}} \right)\\
= \sqrt 2 .\sin \left( {\alpha + \dfrac{\pi }{4}} \right)\\
- 1 \le \sin \left( {\alpha + \dfrac{\pi }{4}} \right) \le 1\\
\Rightarrow \sqrt 2 \sin \left( {\alpha + \dfrac{\pi }{4}} \right) \le \sqrt 2 \\
\Rightarrow \sin \alpha + \cos \alpha \le \sqrt 2
\end{array}\)
