Question

[Bùi Bảo Trâm]

sina + cosa= \(\sqrt{2}\) sin(a+\(\dfrac{\pi}{4}\)) = \(\sqrt{2}\) cos( a-\(\dfrac{\pi}{4}\))

Answer 1

[Nguyễn Việt Lâm]

\(sina+cosa=\sqrt{2}\left(\dfrac{\sqrt{2}}{2}sina+\dfrac{\sqrt{2}}{2}cosa\right)\)

\(=\left[{}\begin{matrix}\sqrt{2}\left(sina.cos\dfrac{\pi}{4}+cosa.sin\dfrac{\pi}{4}\right)\\\sqrt{2}\left(sina.sin\dfrac{\pi}{4}+cosa.cos\dfrac{\pi}{4}\right)\end{matrix}\right.\)

\(=\left[{}\begin{matrix}\sqrt{2}sin\left(a+\dfrac{\pi}{4}\right)\\\sqrt{2}cos\left(a-\dfrac{\pi}{4}\right)\end{matrix}\right.\)