3. trignometrical ratios,functions and identities — ગણિત ધોરણ 11 સાયન્સ — Question

${\sin ^2}{5^o} + {\sin ^2}{10^o} + {\sin ^2}{15^o} + ..... + {\sin ^2}{85^o} + {\sin ^2}{90^o}.$

We know that $\sin {90^o} = 1$ or ${\sin ^2}{90^o} = 1$.

Similarly, $\sin {45^o} = \frac{1}{{\sqrt 2 }}{\rm{or}}\,{\rm{si}}{{\rm{n}}^{\rm{2}}}{45^o} = \frac{1}{2}$

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

$\therefore$ ${\sin ^2}{5^o} + {\sin ^2}{85^o} = {\sin ^2}{5^o} + {\cos ^2}{5^o} = 1.$

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}$.