Đáp án:
$A=-sinx$
$B=tanx$
Giải thích các bước giải:
$A=cos(\dfrac{\pi}{2}+x)+co(2\pi-x)+cos(3\pi+x)$
= $-sinx+cosx+cos(2\pi+\pi+x)$
= $-sinx+cosx+cos(\pi+x)$
= $-sinx+cosx-cosx$
= $-sinx$
$B=2cosx-3cos(\pi-x)+5sin(\dfrac{7\pi}{2}-x)+cot(\dfrac{3\pi}{2}-x)$
+) $3cos(\pi-x)=-3cosx$
+) $5sin(\dfrac{7\pi}{2})=5sin(3\pi+\dfrac{\pi}{2}-x)=-5sin(\dfrac{\pi}{2}-x)=-5cosx$
+) $cot(\dfrac{3\pi}{2}-x)=cot(\pi+\dfrac{\pi}{2}-x)=tanx$
⇒ $B=2cosx-(-3cosx)-5cosx+tanx=tanx$
