Question

The value of ${\sin }^2{1}^{\circ }+{\sin }^2{2}^{\circ }+........+{\sin }^2{89}^{\circ }$ is $............$

• A. $1$
• B. $44$
• C. $44\frac{1}{2}$
• D. $45$

Answer 1

Answer: (C)

$(44\frac{1}{2})$

Consider ${\sin }^2\theta +{\sin }^2({90}^0-\theta )$
$={\sin }^2\theta +{\cos }^2\theta$
$\therefore {\sin }^2\theta +{\sin }^2({90}^0-\theta )=1$
Applying the above identity, we get
$({\sin }^2{1}^0+{\sin }^2{89}^0)+({\sin }^2{2}^0+{\sin }^2{82}^0)+...({\sin }^2{44}^0+{\sin }^2{46}^0)+{\sin }^2{45}^0$
$=1+1+\mathrm{1....}+1+\frac{1}{2}$
$=\frac{89-1}{2}+\frac{1}{2}$
$=44+\frac{1}{2}$
$=44\frac{1}{2}$
Therefore, the value of the given expression is $=44\frac{1}{2}$
Hence, the correct option is B.