Đáp án:
a) $sin(\pi-\dfrac{\pi}{\alpha})=0$
b) $cos(\dfrac{3\pi}{2}-\alpha)=1$
Giải thích các bước giải:
$cot\alpha =-3$
Ta có: $\dfrac{3\pi}{2}<\alpha<2\pi$
⇒ $\begin{cases} cos\alpha>0 \\ sin\alpha<0\\ tan\alpha<0 \end{cases}$
$tan\alpha.cot\alpha=1$
⇒ $tan\alpha=-\dfrac{1}{3}$
Mà: $tan\alpha=\dfrac{sin\alpha}{cos\alpha}$
⇔ $\dfrac{sin\alpha}{cos\alpha}=-\dfrac{1}{3}$
⇔ $\begin{cases} sin\alpha=-1 \\ cos\alpha=3 \end{cases}$
a) $sin(\pi-\dfrac{\pi}{\alpha})=sin\pi .cos\dfrac{\pi}{\alpha}-cos\pi.sin\dfrac{\pi}{\alpha}$
= $sin\pi .cos\dfrac{\pi}{3}-cos\pi.sin\dfrac{\pi}{-1}$
= $0$
b) $cos(\dfrac{3\pi}{2}-\alpha)=cos\dfrac{3\pi}{2}.cos\alpha+sin\dfrac{3\pi}{2}.sin\alpha$
= $cos\dfrac{3\pi}{2}.3+sin\dfrac{3\pi}{2}.(-1)$
= $1$
