Question

Nếu cosa+sina=\(\sqrt{2}\) \(\left(0< a< \dfrac{\pi}{2}\right)\) thì a bằng

Answer 1

\(\sin\alpha+\cos\alpha=\sqrt{2}\) (1)

=> \(\left(\sin a+\cos a\right)^2=2\)

=> \(\sin^2\alpha+\cos^2\alpha+2\cdot\sin\alpha\cdot\cos\alpha=2\)

\(\Rightarrow1+2\cdot\sin\alpha\cdot\cos\alpha=2\Rightarrow2\cdot\sin\alpha\cdot\cos\alpha=1\)

\(\Rightarrow\sin\alpha\cdot\cos\alpha=\dfrac{1}{2}\)

Có: \(\left(\sin\alpha-\cos\alpha\right)^2=\left(\sin\alpha+\cos\alpha\right)^2-4\cdot\sin\alpha\cdot\cos a=2-2=0\)

=> \(\sin\alpha-\cos\alpha=0\) (2)

Từ (1),(2) => \(2\sin\alpha=\sqrt{2}\Rightarrow\sin a=\dfrac{\sqrt{2}}{2}\Rightarrow\alpha=45\)(ktm)

Vậy không có a nào t/m điều kiện