# Evaluate each of the following:$\sin ^{2} 30^{\circ}+\sin ^{2} 45^{\circ}+\sin ^{2} 60^{\circ}+\sin ^{2} 90^{\circ}$

Given:

$\sin ^{2} 30^{\circ}+\sin ^{2} 45^{\circ}+\sin ^{2} 60^{\circ}+\sin ^{2} 90^{\circ}$

To do:

We have to evaluate $\sin ^{2} 30^{\circ}+\sin ^{2} 45^{\circ}+\sin ^{2} 60^{\circ}+\sin ^{2} 90^{\circ}$.

Solution:

We know that,

$\sin 30^{\circ}=\frac{1}{2}$

$\sin 45^{\circ}=\frac{1}{\sqrt2}$

$\sin 60^{\circ}=\frac{\sqrt3}{2}$

$\sin 90^{\circ}=1$

Therefore,

$\sin ^{2} 30^{\circ}+\sin ^{2} 45^{\circ}+\sin ^{2} 60^{\circ}+\sin ^{2} 90^{\circ}=\left(\frac{1}{2}\right)^{2} +\left(\frac{1}{\sqrt{2}}\right)^{2} +\left(\frac{\sqrt{3}}{2}\right)^{2} +( 1)^{2}$

$=\frac{1}{4} +\frac{1}{2} +\frac{3}{4} +1$

$=\frac{1+1( 2) +3+1( 4)}{4}$

$=\frac{1+2+3+4}{4}$

$=\frac{10}{4}$

$=\frac{5}{2}$

Hence, $\sin ^{2} 30^{\circ}+\sin ^{2} 45^{\circ}+\sin ^{2} 60^{\circ}+\sin ^{2} 90^{\circ}=\frac{5}{2}$.

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Updated on: 10-Oct-2022

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