B= sin(5π/2 – anpha) – sin( apha-6π) + cos(apha+5π)-sin(apha-π) 30/10/2021 Bởi Melanie B= sin(5π/2 – anpha) – sin( apha-6π) + cos(apha+5π)-sin(apha-π)
$B= sin(2\pi+\frac{\pi}{2}-\alpha)-sin(\alpha-6\pi)+ cos(\alpha+\pi+4\pi)+sin(\pi-\alpha)$ $= cos\alpha-sin\alpha-cos\alpha-sin\alpha$ $=0$ Bình luận
$B= sin(2\pi+\frac{\pi}{2}-\alpha)-sin(\alpha-6\pi)+ cos(\alpha+\pi+4\pi)+sin(\pi-\alpha)$
$= cos\alpha-sin\alpha-cos\alpha-sin\alpha$
$=0$