Question

The value of tthe expression
${\cos }^2{1}^{\circ }+{\cos }^2{2}^{\circ }+{\cos }^2{3}^{\circ }+....+{\cos }^2{90}^{\circ }=$
is equal to

Answer 1

Given
${\cos }^2{1}^{\circ }+{\cos }^2{2}^{\circ }+{\cos }^2{3}^{\circ }+....+{\cos }^2{90}^{\circ }$
$=({\cos }^2{1}^{\circ }+{\cos }^2{89}^{\circ })+({\cos }^2{2}^{\circ }+{\cos }^2{88}^{\circ })+...+({\cos }^2{44}^{\circ }+{\cos }^2{46}^{\circ })+{\cos }^2{45}^{\circ }$

As we know $\cos \theta =\sin (90-\theta )$

$=({\cos }^2{1}^{\circ }+{\sin }^2{1}^{\circ })+({\cos }^2{2}^{\circ }+{\sin }^2{2}^{\circ })+...+({\cos }^2{44}^{\circ }+{\sin }^2{44}^{\circ })+{\cos }^2{45}^{\circ }$
$=44+\frac{1}{2}=\frac{89}{2}$
The term ${\cos }^2\theta +{\sin }^2\theta =1$ appear $44$ times and ${\cos }^2{45}^{\circ }$ is middle term.So Hence the correct answer is Option D.