Question

The value of ${\sin }^2{5}^{\circ }+{\sin }^2{10}^{\circ }+{\sin }^2{15}^{\circ }+$..............

$+{\sin }^2{85}^{\circ }+{\sin }^2{90}^{\circ }$ is equal to

• A. $7$
• B. $8$
• C. $9$
• D. $\frac{19}{2}$

Answer 1

The correct option is

D

$\frac{19}{2}$

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 $\sin {90}^{\circ }=1\text{or}{\sin }^2{90}^{\circ }=1$

The angles are in A.P. with common difference $5^{\circ }$ and has 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 ${\sin }^2{5}^{\circ }+{\sin }^2{85}^{\circ }={\sin }^2{5}^{\circ }+{\cos }^2{5}^{\circ }=1$
Similarly, for ${\sin }^2{10}^{\circ }+{\sin }^2{80}^{\circ }={\sin }^2{10}^{\circ }+{\sin }^2(90-10{)}^{\circ }$
$={\sin }^2{10}^{\circ }+{\cos }^2{10}^{\circ }=1$
So, all we can arrange all terms in pair wise except the middle terms ${\sin }^2{45}^{\circ }$ and last term $\sin {90}^{\circ }$ and its values are ${\sin }^2{45}^{\circ }=\frac{1}{2}$ and ${\sin }^2{90}^{\circ }=1$

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

$=(1+1+1+1+1+1+1+1)+1+\frac{1}{2}=9\frac{1}{2}=\frac{19}{2}$.