Áp dụng BĐT cauchy:
\(\left(\sin\alpha+\cos\alpha\right)^2=\sin^2\alpha+\cos^2\alpha+2\sin\alpha.\cos\alpha=1+2\sin\alpha.\cos\alpha\le1+\sin^2\alpha+\cos^2\alpha=1+1=2\\ \Leftrightarrow\sin\alpha+\cos\alpha\le\sqrt{2}\)
Dấu \("="\Leftrightarrow\sin\alpha=\cos\alpha=\dfrac{\sqrt{2}}{2}\) hay \(\alpha=45^0\)