Sin ^{2} 1+sin ^{2} 3+sin ^{2} 87+sin ^{2} 89=ldots ldots ldots ldots

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8. Introduction to Trigonometry

easy

$\sin ^{2} 1+\sin ^{2} 3+\sin ^{2} 87+\sin ^{2} 89=\ldots \ldots \ldots \ldots$

A

$0$

B

$1$

C

$2$

D

$4$

Solution

$\sin ^{2} 1+\sin ^{2} 3+\sin ^{2} 87+\sin ^{2} 89$

$=\sin ^{2} 1+\sin ^{2} 3+\sin ^{2}(90-3)+\sin ^{2}(90-1)$

$=\sin ^{2} 1+\sin ^{2} 3+\cos ^{2} 3+\cos ^{2} 1$

$=\left(\sin ^{2} 1+\cos ^{2} 1\right)+\left(\sin ^{2} 3+\cos ^{2} 3\right)$

$=1+1=2$

Standard 10

Mathematics

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