Question

Find the value of ${\cos }^2{5}^0+{\cos }^2{10}^0+{\cos }^2{15}^0+\dots .+{\cos }^2{360}^0$.

• A. $18$
• B. $27$
• C. $36$
• D. $45$

Answer 1

The correct option is C $36$

Given

${\cos }^2{5}^{\circ }+{\cos }^2{10}^{\circ }+{\cos }^2{15}^{\circ }+\cdots +{\cos }^2{360}^{\circ }$

${\cos }^2{5}^{\circ }+{\cos }^2{10}^{\circ }+{\cos }^2{15}^{\circ }+\cdots +{\cos }^2{295}^{\circ }+1$

${\cos }^2{5}^{\circ }+{\cos }^2{10}^{\circ }+{\cos }^2{15}^{\circ }+\cdots +{\cos }^2(360-5{)}^{\circ }+1$

${\cos }^2{5}^{\circ }+{\cos }^2{10}^{\circ }+{\cos }^2{15}^{\circ }+\cdots +{\cos }^2(5{)}^{\circ }+1$

$(\cos (2\pi -\theta )=\cos \theta )$

$2({\cos }^2{5}^{\circ }+{\cos }^2{10}^{\circ }+\cdots +{\cos }^2{175}^{\circ })+{\cos }^2{180}^{\circ }+1$

$2({\cos }^2{5}^{\circ }+{\cos }^2{10}^{\circ }+\cdots +{\cos }^2(180-5{)}^{\circ })+2$

$2({\cos }^2{5}^{\circ }+{\cos }^2{10}^{\circ }+\cdots +{\cos }^2(5{)}^{\circ })+2$

$(\cos (\pi -\theta )=-\cos \theta )$

$2\left(2\right({\cos }^2{5}^{\circ }+{\cos }^2{10}^{\circ }+\cdots +{\cos }^2{85}^{\circ })+{\cos }^2{90}^{\circ })+2$

$4({\cos }^2{5}^{\circ }+{\cos }^2{10}^{\circ }+\cdots +{\cos }^2{85}^{\circ })+2$

$4({\cos }^2{5}^{\circ }+{\cos }^2{10}^{\circ }+\cdots +{\cos }^2(90-5{)}^{\circ })+2$

$4({\cos }^2{5}^{\circ }+{\cos }^2{10}^{\circ }+\cdots +{\sin }^{2}5\circ )+2$

$(\cos (90-\theta )=\sin (90-\theta )$

$4({\cos }^2{5}^{\circ }{\sin }^2{5}^{\circ }+{\cos }^2{10}^{\circ }+{\sin }^2{10}^{\circ }\cdots +{\cos }^{2}45\circ )+2$

$4(1+1+1+1+1+(8times)+{\cos }^{2}45\circ )+2$

$4\times 8+4\times \frac{1}{2}+2$

$32+2+2=36$