Question

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

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

Answer 1

The correct option is A $18$

${\cos }^{2}5°+{\cos }^{2}10°+{\cos }^{2}15°+......+{\cos }^{2}360°$

${\cos }^{2}5°+{\cos }^{2}10°+.....+{\cos }^{2}45°+......+{\cos }^{2}360°$

$\Rightarrow {\cos }^{2}5°+{\cos }^{2}85°+{\cos }^{2}10°+{\cos }^{2}80°+{\cos }^{2}15°+{\cos }^{2}75°+....8terms+{\left(\frac{1}{\sqrt{2}}\right)}^2+$

${\cos }^{2}90°+...+{\cos }^{2}135°+....+{\cos }^{2}150°+..+{\cos }^{2}225°+....{\cos }^{2}270°+....{\cos }^{2}245°+....{\cos }^{2}360°$

$\Rightarrow \left({\cos }^{2}5°+{\sin }^{2}5°\right)+({\cos }^{2}10°+{\sin }^{2}10°)+....8terms+\frac{1}{2}+\mathrm{0....}{\left(\frac{1}{\sqrt{2}}\right)}^2+....$

${(-1)}^2+.....{\left(\frac{-1}{\sqrt{2}}\right)}^2+{\left(0\right)}^2+....+{\left(\frac{-1}{\sqrt{2}}\right)}^2+....{\left(1\right)}^2$

$=1+1+1+.....8terms+(\frac{1}{2}+\frac{1}{2})+1+(\frac{1}{2}+\frac{1}{2})+1+.....6terms$

$=8+1+1+1+1+6$

$=18$

Hence, the answer is $18.$