Đáp án:
Câu 35: $B=0$
Cau 30: $A=-\sin x-2\cos x$
Giải thích các bước giải:
Câu 35:
$B=\cos\left(\dfrac{3\pi}{2}-\alpha\right)+\sin\left(\dfrac{3\pi}{2}-\alpha\right)-\cos\left(\dfrac{3\pi}{2}-\alpha\right)-\sin\left(\dfrac{3\pi}{2}+\alpha\right)$
$=\sin\left(\dfrac{3\pi}{2}-\alpha\right)-\sin\left(\dfrac{3\pi}{2}+\alpha\right)$
$=\sin\left(\pi+\dfrac{\pi}{2}-\alpha\right)-\sin\left(\pi+\dfrac{\pi}{2}+\alpha\right)$
$=-\sin\left(\dfrac{\pi}{2}-\alpha\right)+\sin\left(\dfrac{\pi}{2}+\alpha\right)$
$=-\cos\alpha+\cos(-\alpha)$
$=\cos\alpha-\cos\alpha$
$=0$
Câu 30:
$A=\cos\left(\dfrac{\pi}{2}+x\right)-\cos(2\pi-x)+\cos(3\pi+x)$
$=\cos\left[\dfrac{\pi}{2}-(-x)\right]-\cos(-x)+\cos(\pi+x)$
$=\sin(-x)-\cos x-\cos x$
$=-\sin x-2\cos x$.
