Đề bài

Rút gọn các biểu thức sau:

a, 🅷\(\sqrt 2 \sin \left( {\alpha  + \frac{\pi }{4}} \right) - cos\alpha \),

b,♋ \({\left( {cos\alpha  + \sin \alpha } \right)^2} - \sin 2\alpha \)

Lời giải chi tiết

a, Ta có:\(\begin{array}{l}\sqrt 2 \sin \left( {\alpha  + \frac{\pi }{4}} \right) - cos\alpha  = \sqrt 2 .\left( {\sin \alpha \cos \frac{\pi }{4} + \cos \alpha \sin \frac{\pi }{4}} \right) - cos\alpha \\ = \sqrt 2 .\left( {\sin \alpha .\frac{{\sqrt 2 }}{2} + \cos \alpha .\frac{{\sqrt 2 }}{2}} \right) - cos\alpha \\ =  \sqrt 2 .{\sin \alpha .\frac{{\sqrt 2 }}{2} + \sqrt 2 .\cos \alpha .\frac{{\sqrt 2 }}{2}} - cos\alpha \\ =\sin \alpha  + \cos \alpha  - cos\alpha \\ = \sin \alpha \end{array}\)b, Ta  có:\(\begin{array}{l}{\left( {cos\alpha  + \sin \alpha } \right)^2} - \sin 2\alpha \\ = co{s^2}\alpha  + {\sin ^2}\alpha  + 2cos\alpha \sin \alpha  - 2\sin \alpha cos\alpha \\ = {\sin ^2}\alpha + co{s^2}\alpha = 1\end{array}\)