Q. The value of$\ \sin{A\sin{(60{^\circ} - A)\sin(60{^\circ} + A)}}$ is equal to
A) $\sin{3A}$
B) $\frac{\sin{3A}}{2}$
C) $\frac{\sin{3A}}{4}$
D) None of these
Explanation
(c)
$\sin{A\sin{(60{^\circ} - A)\sin(60{^\circ} + A)}}$
$= \sin{A(\sin^{2}{60{^\circ} - \sin^{2}{A)}}}$
$= \sin{A\left( \frac{3}{4} - \sin^{2}A \right)}$
$= \frac{3\sin{A - 4\sin^{3}A}}{4} = \frac{\sin{3A}}{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.