Question

${\cos }^2{1}^{\circ }+{\cos }^2{2}^{\circ }+{\cos }^2{3}^{\circ }+....+{\cos }^2{90}^{\circ }=$

• A. \N
• B. 1
• C. 45
• D. $\frac{89}{2}$

Answer 1

The correct option is D $\frac{89}{2}$

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