Q. If $A + B + C = \pi$, then $\sin{2A + \sin{2B + \sin{2C}}}$ is equal to
A) $4\sin{A\sin{B\sin C}}$
B) $4\cos{A\cos{B\cos C}}$
C) $2\cos{A\cos{B\cos C}}$
D) $2\sin{A\ \sin\ B\sin C}$
Explanation
(a)
$B + C = \pi - A$
$\Rightarrow \ \sin{(B + C) = \sin{(\pi - A) = \sin A}}$
$\therefore\ \sin{2A + \sin{2B + \sin{2C}}}$
$= 2\sin{A\cos\ A + 2\sin\ (B + C)}\cos(B - C)$
$= 2\sin{A\left\lbrack \cos{A + \cos(B - C)} \right\rbrack}$
=$2\sin{A\lbrack\cos{(B - C) - \cos{(B + C)}}\rbrack}$
$= 2\sin{A\lbrack 2\sin{B\sin C}\rbrack}$
$= 4\sin{A\sin{B\sin C}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.