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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,
$\sin 30^{\circ}=\frac{1}{2}$
$\sin 45^{\circ}=\frac{1}{\sqrt2}$
$\sin 60^{\circ}=\frac{\sqrt3}{2}$
$\sin 90^{\circ}=1$
$ \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, $\sin ^{2} 30^{\circ}+\sin ^{2} 45^{\circ}+\sin ^{2} 60^{\circ}+\sin ^{2} 90^{\circ}=\frac{5}{2}$.