Question

The value of ${\sin }^2{1}^{\circ }+{\sin }^2{2}^{\circ }+{\sin }^2{2}^{\circ }+...+{\sin }^2{89}^{\circ }+{\sin }^2{90}^{\circ }$

• A. 1
• B. 0
• C. 45.5
• D. 44

Answer 1

As we know that ${\sin }^2\theta +{\cos }^2\theta =1$ .....(i)
and also $\cos \theta =\sin (90-\theta )$........(ii)
Substitute equation(ii) in equation(i), we get
${\sin }^2\theta +{\sin }^2(90-\theta )=1$.........(iii)
Given ${\sin }^2{1}^{\circ }+{\sin }^2{2}^{\circ }+{\sin }^2{2}^{\circ }+...+{\sin }^2{89}^{\circ }+{\sin }^2{90}^{\circ }$
$=({\sin }^{2}1+{\sin }^{2}89)+({\sin }^{2}2+{\sin }^{2}88)+....+({\sin }^{2}44+{\sin }^{2}46)+({\sin }^{2}45+{\sin }^{2}90)$
$=({\sin }^{2}1+{\sin }^2(90-1\left)\right)+({\sin }^2+{\sin }^2(90-2\left)\right)+....+({\sin }^{2}44+{\sin }^2(90-44\left)\right)+({\sin }^{2}45+{\sin }^{2}90)$
$=1+1......\left(44\text{times}\right)+(\frac{1}{\sqrt{2}}{)}^2+1$
[From equation(iii)]
$=45+\frac{1}{2}$
$=45.5$
Therefore, the correct option is D $\left(45.5\right)$