Question

Simplify $4\sin ({420}^{\circ }-\alpha )\cos ({60}^{\circ }+\alpha )$.

• A. $\sqrt{3}-2\sin 2\alpha$
• B. $\sqrt{3}+2\sin 2\alpha$
• C. $\sqrt{3}-2\cos 2\alpha$
• D. $\sqrt{3}+2\cos 2\alpha$

Answer 1

The correct option is A $\sqrt{3}-2\sin 2\alpha$

$4\sin ({420}^{\circ }-\alpha )\cos ({60}^{\circ }+\alpha )=2[\sin \left({480}^{\circ }\right)+\sin ({360}^{\circ }-2\alpha )]$
$=2[\frac{\sqrt{3}}{2}-\sin 2\alpha ]$
$=\sqrt{3}-2\sin 2\alpha$
$[2\sin (A+B)\cos (A-B)=\sin A+\sin B]$
Hence, option 'A' is correct.