sin ^2 5^{circ}+sin ^2 10^{circ}+sin ^2 15^{c | vedclass

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(C) The given expression is $S = \sin ^2 5^{\circ}+\sin ^2 10^{\circ}+\ldots+\sin ^2 85^{\circ}+\sin ^2 90^{\circ}$.There are $18$ terms in total from

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$9 \frac{1}{2}$

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(C) The given expression is $S = \sin ^2 5^{\circ}+\sin ^2 10^{\circ}+\ldots+\sin ^2 85^{\circ}+\sin ^2 90^{\circ}$.There are $18$ terms in total from $5^{\circ}$ to $90^{\circ}$ in steps of $5^{\circ}$.We can pair terms using the identity $\sin ^2 	heta + \sin ^2 (90^{\circ} - 	heta) = \sin ^2 	heta + \cos ^2 	heta = 1$.Pairing the terms: $(\sin ^2 5^{\circ} + \sin ^2 85^{\circ}) + (\sin ^2 10^{\circ} + \sin ^2 80^{\circ}) + \ldots + (\sin ^2 40^{\circ} + \sin ^2 50^{\circ}) + \sin ^2 45^{\circ} + \sin ^2 90^{\circ}$.There are $8$ such pairs,each equal to $1$.So,$S = 8 	imes 1 + \sin ^2 45^{\circ} + \sin ^2 90^{\circ}$.Since $\sin 45^{\circ} = \frac{1}{\sqrt{2}}$ and $\sin 90^{\circ} = 1$,we have $S = 8 + (\frac{1}{\sqrt{2}})^2 + 1^2 = 8 + \frac{1}{2} + 1 = 9 \frac{1}{2}$.

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

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