Q. If $A + B + C\ = \frac{3\pi}{2},\$then $\cos{2A + \cos{2B + \cos{2C}}}$ is equal to
A) $1 - 4\ \cos{A\cos{B\cos C}}$
B) $4\sin{A\sin{B\sin C}}$
C) $1 + 2\cos{A\cos{B\cos C}}$
D) $1 - 4\sin{A\sin{B\sin C}}$
Explanation
(d)
Since, $A + B + C = \frac{3\pi}{2}$
$\therefore\cos{2A + \cos{2B + \cos{2C}}}$
$= 2\cos{(\ A + B)\cos{(\ A - B) + \cos{2C}}}$
$= 2\cos{\left( \frac{3\pi}{2} - C \right)\cos{(\ A - B) + \ 1 - 2\sin^{2}C}}$
$= 1 - 2\sin{C\left\lbrack \cos{(A - B) + \sin\left( \frac{3\pi}{2} - (A + B) \right)} \right\rbrack}$
$= 1 - 2\sin{C\left\lbrack \cos{(A - B) - \cos{(A + B)}} \right\rbrack}$
$= 1 - 4\sin{A\sin{B\sin C}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.