sin(pi:2-x)sin(pi-x)-cos(pi:2-x)cos(pi-x) 30/09/2021 Bởi Daisy sin(pi:2-x)sin(pi-x)-cos(pi:2-x)cos(pi-x)
Đáp án: sin ( $\frac{\pi}{2}$-x).sin($\pi$ -x) – cos( $\frac{\pi}{2}$cos ($\pi$ -x) = cosx.sinx – sinx. (-cosx) = cosx.sinx + sinx.cosx= 2cosxsinx = sin2x Bình luận
$\sin\Bigg(\dfrac{\pi}{2}-x\Bigg).\sin(\pi-x)-\cos\Bigg(\dfrac{\pi}{2}-x\Bigg).\cos(\pi-x)\\=\cos x. \sin x – \sin x. (-\cos x)\\=\sin x .\cos x + \sin x. \cos x\\=2. \sin x. \cos x\\=\sin 2x$ Bình luận
Đáp án:
sin ( $\frac{\pi}{2}$-x).sin($\pi$ -x) – cos( $\frac{\pi}{2}$cos ($\pi$ -x)
= cosx.sinx – sinx. (-cosx) = cosx.sinx + sinx.cosx= 2cosxsinx = sin2x
$\sin\Bigg(\dfrac{\pi}{2}-x\Bigg).\sin(\pi-x)-\cos\Bigg(\dfrac{\pi}{2}-x\Bigg).\cos(\pi-x)\\=\cos x. \sin x – \sin x. (-\cos x)\\=\sin x .\cos x + \sin x. \cos x\\=2. \sin x. \cos x\\=\sin 2x$