cos alpha + sin alpha < hoac = can 2 lam ho vs thankkkkkkkkkkkkkkkkkk 17/07/2021 Bởi aikhanh cos alpha + sin alpha < hoac = can 2 lam ho vs thankkkkkkkkkkkkkkkkkk
Giải thích các bước giải: Ta có: \(\begin{array}{l}\sin \alpha + \cos \alpha \\ = \sqrt 2 .\left( {\dfrac{{\sqrt 2 }}{2}.\sin \alpha + \dfrac{{\sqrt 2 }}{2}.\cos \alpha } \right)\\ = \sqrt 2 .\left( {\sin \alpha .\cos \dfrac{\pi }{4} + \cos \alpha .\sin \dfrac{\pi }{4}} \right)\\ = \sqrt 2 .\sin \left( {\alpha + \dfrac{\pi }{4}} \right)\\ – 1 \le \sin \left( {\alpha + \dfrac{\pi }{4}} \right) \le 1\\ \Rightarrow \sqrt 2 \sin \left( {\alpha + \dfrac{\pi }{4}} \right) \le \sqrt 2 \\ \Rightarrow \sin \alpha + \cos \alpha \le \sqrt 2 \end{array}\) Bình luận
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\sin \alpha + \cos \alpha \\
= \sqrt 2 .\left( {\dfrac{{\sqrt 2 }}{2}.\sin \alpha + \dfrac{{\sqrt 2 }}{2}.\cos \alpha } \right)\\
= \sqrt 2 .\left( {\sin \alpha .\cos \dfrac{\pi }{4} + \cos \alpha .\sin \dfrac{\pi }{4}} \right)\\
= \sqrt 2 .\sin \left( {\alpha + \dfrac{\pi }{4}} \right)\\
– 1 \le \sin \left( {\alpha + \dfrac{\pi }{4}} \right) \le 1\\
\Rightarrow \sqrt 2 \sin \left( {\alpha + \dfrac{\pi }{4}} \right) \le \sqrt 2 \\
\Rightarrow \sin \alpha + \cos \alpha \le \sqrt 2
\end{array}\)