Question

${\sin }^2{5}^{\circ }+{\sin }^{2}10^{\circ }+{\sin }^{2}15^{\circ }+....+{\sin }^{2}90^{\circ }$ =

• A. $8$
• B. $9$
• C. $10$
• D. $9.5$

Answer 1

Answer: (D)

${\sin }^2{5}^{\circ }+{\sin }^{2}10^{\circ }+{\sin }^{2}15^{\circ }+...+{\sin }^{2}90^{\circ }$
$={\sin }^2{5}^{\circ }+{\sin }^{2}10^{\circ }+....+{\sin }^{2}35^{\circ }+{\sin }^{2}40^{\circ }+{\sin }^{2}45^{\circ }+{\sin }^2\left(90^{\circ }-40^{\circ }\right)+{\sin }^2\left(90^{\circ }-35^{\circ }\right)+....+{\sin }^2\left(90^{\circ }-15\right)+{\sin }^2\left(90^{\circ }-10^{\circ }\right)+{\sin }^2\left(90^{\circ }-5^{\circ }\right)+{\sin }^{2}90^{\circ }$
$={\sin }^2{5}^{\circ }+{\sin }^{2}10^{\circ }+....+{\sin }^{2}35^{\circ }+{\sin }^{2}40^{\circ }+{\sin }^{2}45^{\circ }+{\cos }^{2}40^{\circ }+{\cos }^{2}35^{\circ }+....+{\cos }^{2}10^{\circ }+{\cos }^2{5}^{\circ }+{\sin }^{2}90^{\circ }$
$=\left[\left({\sin }^2{5}^{\circ }+{\cos }^2{5}^{\circ }\right)+\left({\sin }^{2}10^{\circ }+{\cos }^{2}10^{\circ }\right)+......+{\sin }^{2}40^{\circ }+{\cos }^{2}40^{\circ }\right]+{\sin }^{2}45^{\circ }+{\sin }^{2}90^{\circ }$
$=[1+1+.....+1]+{\left(\frac{1}{\sqrt{2}}\right)}^2+1$
$=8+\frac{1}{2}+1=\frac{19}{2}=9.5$