$sinx+cosx= sinx+sin(\frac{\pi}{2}-x)= 2sin\frac{\pi}{4}cos(x+\frac{\pi}{4})= \sqrt{2}cos(x+\frac{\pi}{4})$ $sinx-cosx= sinx-sin(\frac{\pi}{2}-x)= 2cos\frac{\pi}{4}sin(x+\frac{\pi}{4})=\sqrt{2}sin(x+\frac{\pi}{4})$ Bình luận
*sinx + cosx = √2 sin(x+$\pi$ /4) = √2 cos( x – $\pi$ /4) *sinx – cosx = √2 sin(x – $\pi$ /4) = -√2 cos(x + $\pi$ /4) Bình luận
$sinx+cosx= sinx+sin(\frac{\pi}{2}-x)= 2sin\frac{\pi}{4}cos(x+\frac{\pi}{4})= \sqrt{2}cos(x+\frac{\pi}{4})$
$sinx-cosx= sinx-sin(\frac{\pi}{2}-x)= 2cos\frac{\pi}{4}sin(x+\frac{\pi}{4})=\sqrt{2}sin(x+\frac{\pi}{4})$
*sinx + cosx = √2 sin(x+$\pi$ /4) = √2 cos( x – $\pi$ /4)
*sinx – cosx = √2 sin(x – $\pi$ /4) = -√2 cos(x + $\pi$ /4)