The value of {sin ^2}{5^o} + {sin ^2}{10^o} + | vedclass

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Question

(d) Given expression is

${\sin ^2}{5^o} + {\sin ^2}{10^o} + {\sin ^2}{15^o} + ..... + {\sin ^2}{85^o} + {\sin ^2}{90^o}.$

We know that $\sin {90

Vedclass · Accepted Answer

$9\frac{1}{2}$

Vedclass · Answer

(d) Given expression is

${\sin ^2}{5^o} + {\sin ^2}{10^o} + {\sin ^2}{15^o} + ..... + {\sin ^2}{85^o} + {\sin ^2}{90^o}.$

We know that $\sin {90^o} = 1$ or ${\sin ^2}{90^o} = 1$. 

Similarly, $\sin {45^o} = \frac{1}{{\sqrt 2 }}{m{or}}\,{m{si}}{{m{n}}^{m{2}}}{45^o} = \frac{1}{2}$ 

and the angles are in $A.P. $ of $18$ terms. 

We also know that ${\sin ^2}{85^o} = {[\sin ({90^o} - {5^o})]^2}$$ = {\cos ^2}{5^o}.$ 

Therefore from the complementary rule, we find 

$	herefore$ ${\sin ^2}{5^o} + {\sin ^2}{85^o} = {\sin ^2}{5^o} + {\cos ^2}{5^o} = 1.$ 

Therefore, 

${\sin ^2}{5^o} + {\sin ^2}{10^o} + {\sin ^2}{15^o} + ... + {\sin ^2}{85^o} + {\sin ^2}{90^o}$ 

$ = (1 + 1 + 1 + 1 + 1 + 1 + 1 + 1) + 1 + \frac{1}{2} = 9\frac{1}{2}$.

The value of ${\sin ^2}{5^o} + {\sin ^2}{10^o} + {\sin ^2}{15^o} + ... + $ ${\sin ^2}{85^o} + {\sin ^2}{90^o}$ is equal to

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