Question

The value of ${\cos }^2{0}^0+{\cos }^2{30}^0+{\cos }^2{45}^0+{\cos }^2{60}^0+{\cos }^2{90}^0$ is

• A. $\frac{1}{2}$
• B. $\frac{3}{2}$
• C. $\frac{5}{2}$
• D. $\frac{7}{2}$

Answer 1

Answer: (C)

$\frac{5}{2}$

Consider the following identity
${\cos }^2\alpha +{\cos }^2({90}^0-\alpha )$
$={\cos }^2\alpha +{\sin }^2\alpha$
$=1$

Applying this identity in the above expression
${\cos }^{2}0+{\cos }^2{30}^0+{\cos }^2{45}^0+{\cos }^2{60}^0+{\cos }^2{90}^0$
$={\cos }^{2}0+{\cos }^2{90}^0+{\cos }^2{30}^0+{\cos }^2{60}^0+{\cos }^2{45}^0$
$={\cos }^{2}0+{\cos }^2({90}^0-0^0)+{\cos }^2{30}^0+{\cos }^2({90}^0-{30}^0)+\frac{1}{2}$
$=1+1+\frac{1}{2}$ ...(from the identity)
$=\frac{5}{2}$