Q. $\sin^{2}{5{^\circ} + \sin^{2}{10{^\circ} + \sin^{2}{15{^\circ} + ... + \sin^{2}{90{^\circ}}}}}$ is equal to

A) $8\frac{1}{2}$
B) 9
C) $9\frac{1}{2}$
D) $4\frac{1}{2}$
Explanation

(c)
$\sin^{2}{5{^\circ} + \sin^{2}{10{^\circ} + \sin^{2}{15{^\circ} + ... + \sin^{2}{90{^\circ}}}}}$
$= \sin^{2}{5{^\circ} + \sin^{2}{10{^\circ} + {\ldots + \sin^{2}}{45{^\circ} + ... + \sin^{2}{80{^\circ} + \sin^{2}{85{^\circ}}} + \sin^{2}{90{^\circ}}}}}$
$= \sin^{2}{5{^\circ} + \sin^{2}{10{^\circ} + ... + \frac{1}{2} + {\ldots + \cos^{2}}{10{^\circ}\mathbf{+}cos^{2}5{^\circ} + \sin^{2}{90{^\circ}}}}}$
$= \left( \sin^{2}{5{^\circ} + \cos^{2}{5{^\circ}}} \right) + (\sin^{2}{10{^\circ} + \cos^{2}{10{^\circ}}) + \ldots} +$
$\left( \sin^{2}{40{^\circ} + \cos^{2}{40{^\circ}}} \right) + \sin^{2}{45{^\circ}} + \sin^{2}{90{^\circ}}$
$= 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + \frac{1}{2} + 1 = 9\frac{1}{2}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.