${sin}^{2} 0 ° + {sin}^{2} 5 ° + {sin}^{2} 10 ° + . . . . {sin}^{2} 90 °$ is equal to

18

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9

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13

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\frac{19}{2}

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Solution

The correct option is D $\frac{19}{2}$
${sin}^{2} 0 ° + {sin}^{2} 5 ° + {sin}^{2} 10 ° + . . . + {sin}^{2} 85 ° + {sin}^{2} 90 °$

$= ({sin}^{2} 0 ° + {sin}^{2} 90 °) + ({sin}^{2} 5 ° + {sin}^{2} 85 °) + . . . + ({sin}^{2} 35 ° + {sin}^{2} 55 °) + ({sin}^{2} 40 ° + {sin}^{2} 50 °) + {sin}^{2} 45 °$

$= ({sin}^{2} 0 ° + {cos}^{2} 0 °) + ({sin}^{2} 5 ° + {cos}^{2} 5 °) + . . . + ({sin}^{2} 35 ° + {cos}^{2} 35 °) + ({sin}^{2} 40 ° + {cos}^{2} 40 °) + {sin}^{2} 45 °$

$= [1 + 1 + . . . . . + 1] + {(\frac{1}{\sqrt{2}})}^{2}$

where $0 °, 5 °, 10 °, . . .40 °$ total $9$ terms
Hence, $1 + 1 + . . . . . + 1$ for $9$ terms $= 9$

Final answer $= 9 + \frac{1}{2} = \frac{19}{2}$

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