Prove that: ${sin}^{2} 5^{o} + {sin}^{2} 10^{o} + {sin}^{2} 15^{o} . . . . + {sin}^{2} 85^{o} + {sin}^{2} 90^{o} = 9 \frac{1}{2}$

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Solution

$8 + 1 + \frac{1}{2} = 9 \frac{1}{2}$ . The angles are in A.P of 18 terms of which ${sin}^{2} 90^{o} = 1$
${sin}^{2} 45^{o} = \frac{1}{2}$ and 16 terms form 8 pairs like
${sin}^{2} 5^{o} + {sin}^{2} 85^{o} = {sin}^{2} 5^{o} + {cos}^{2} 5^{o} = 1$

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