# Evaluate each of the following

Question:

Evaluate each of the following

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

Solution:

We have to find

$\sin ^{2} 30^{\circ}+\sin ^{2} 45^{\circ}+\sin ^{2} 60^{\circ}+\sin ^{2} 90^{\circ} \ldots \ldots$ (1)

Now,

$\sin 45^{\circ}=\frac{1}{\sqrt{2}}, \sin 30^{\circ}=\frac{1}{2}, \sin 60^{\circ}=\frac{\sqrt{3}}{2}, \sin 90^{\circ}=1$

So by substituting above values in equation (1)

We get,

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

$=\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}+(1)^{2}$

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

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

Now by taking denominator 4 together and simplifying

We get,

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

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

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

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

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

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

Now by taking LCM

We get,

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

$=\frac{2 \times 2}{1 \times 2}+\frac{1}{2}$

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

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

$=\frac{5}{2}$

Therefore,

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