Question

$4\sin \left({420}^0-a\right)\cos \left(60+a\right)=$

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

Answer 1

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

$=4\sin (420-a)\cos (60+a)=4[\sin 420\cos a-\cos 420\sin a][\cos 60\cos a-\sin 60\sin a]=4[\frac{\sqrt{3}}{2}\cos a-\frac{1}{2}\sin a][\frac{1}{2}\cos a-\frac{\sqrt{3}}{2}\sin a]=(2\sqrt{3}\cos a-2\sin a)(2\cos a-2\sqrt{3}\sin a)=\sqrt{3}(\cos {}^{2}a+\sin {}^{2}a)-4\sin a\cos a=\sqrt{3}-2\sin 2A$