Find the value of ${sin}^{2} 9^{o} + {sin}^{2} 18^{o} + . . . . . + {sin}^{2} 90^{o}$ .

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Solution

Now,
${sin}^{2} 9^{o} + {sin}^{2} 18^{o} + . . . . . + {sin}^{2} 90^{o}$
$= {sin}^{2} 9^{o} + {sin}^{2} 18^{o} + . . . + {sin}^{2} 45^{o} + {sin}^{2} 54^{o} + . . . + {sin}^{2} 81^{o} + {sin}^{2} 90^{o}$
$= {sin}^{2} 9^{o} + {sin}^{2} 18^{o} + . . . + {sin}^{2} 36^{o} + {sin}^{2} 45^{o} + {cos}^{2} 36^{o} + . . . + {cos}^{2} 9^{o} + {sin}^{2} 90^{o}$ [ Since $sin 81^{o} = cos 9^{o}$ and so on]
$= ({sin}^{2} 9^{o} + {cos}^{2} 9^{o}) + ({sin}^{2} 18^{o} + {cos}^{2} 18^{o}) + . . . + ({sin}^{2} 36^{o} + {cos}^{2} 36^{o}) + {sin}^{2} 45^{o} + {sin}^{2} 90^{o}$
$= 1 + 1 + 1 + 1 + \frac{1}{2} + 1$ [ Since ${sin}^{2} θ + {cos}^{2} θ = 1$ ]
$= \frac{11}{2}$ .

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