Rút gọn các biểu thức sau:
a, \(\sqrt 2 \sin \left( {\alpha + \frac{\pi }{4}} \right) - cos\alpha \),
b, \({\left( {cos\alpha + \sin \alpha } \right)^2} - \sin 2\alpha \)
Áp dụng công thức lượng giác
\(\sin \left( {a + b} \right) = \sin a\cos b + \cos a\sin b\)
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\({\sin ^2}\alpha + {\cos ^2}\alpha = 1\)
\(\sin 2a = 2\sin a\cos a\)
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:
\(\begin{array}{l}{\left( {cos\alpha + \sin \alpha } \right)^2} - \sin 2\alpha \\ = co{s^2}\alpha + {\sin ^2}\alpha + 2cos\alpha \sin \alpha - 2\sin \alpha cos\alpha \\ = {\sin ^2}\alpha + co{s^2}\alpha = 1\end{array}\)