Q. In a $\mathrm{\Delta}ABC,\cos\left( \frac{B + 2C + 3A}{2} \right) + \cos\left( \frac{A - B}{2} \right)$ is equal to
A) $- 1$
B) 0
C) 1
D) 2
Explanation
(b)
In a $\mathrm{\Delta}ABC$
$A + B + C = \pi$
$\therefore\ \cos\left( \frac{B + 2C + 3A}{2} \right) + \cos\left( \frac{A - B}{2} \right)$
$= 2\cos\left( \frac{2C + 4A}{4} \right)\cos\left( \frac{2A + 2B + 2C}{4} \right)$
$= 2\cos{\left( \frac{C + 2A}{2} \right)\cos{\left( \frac{\pi}{2} \right) = 0}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.