Question

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

• A. $9\frac{3}{4}$
• B. $8$
• C. $9$
• D. $9\frac{1}{2}$

Answer 1

The correct option is D $9\frac{1}{2}$

Given expression is:

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

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

Similarly, $\sin {45}^{\circ }=\frac{1}{\sqrt{2}}$ or ${\sin }^2{45}^{\circ }=\frac{1}{2}$ and the angles are in A.P. of 18 terms. We also know that

${\sin }^2{85}^{\circ }=\left[\sin \right({90}^{\circ }-5^{\circ }){]}^2={\cos }^2{5}^{\circ }$.

Therefore from the complementary rule, we find ${\sin }^2{5}^{\circ }+{\sin }^2{85}^{\circ }={\sin }^2{5}^{\circ }+{\cos }^2{5}^{\circ }$ = 1.

Therefore,

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

$=(1+1+1+1+1+1+1+1)+\frac{1}{2}+1=9\frac{1}{2}$