Question

The value of ${\sin }^{2}6{}^{\circ }+{\sin }^{2}12{}^{\circ }+{\sin }^{2}18{}^{\circ }+\cdots +{\sin }^{2}84{}^{\circ }+{\sin }^{2}90{}^{\circ }$
is

• A. $0$
• B. $7$
• C. $8$
• D. $9$

Answer 1

The correct option is C $8$

Given,
${\sin }^{2}6{}^{\circ }+{\sin }^{2}12{}^{\circ }+{\sin }^{2}18{}^{\circ }+\cdots +{\sin }^{2}84{}^{\circ }+{\sin }^{2}90{}^{\circ }=({\sin }^{2}6{}^{\circ }+{\sin }^{2}84{}^{\circ })+({\sin }^{2}12{}^{\circ }+{\sin }^{2}78{}^{\circ })+({\sin }^{2}18{}^{\circ }+{\sin }^{2}72{}^{\circ })+\dots +({\sin }^{2}42{}^{\circ }+{\sin }^{2}48{}^{\circ })+{\sin }^{2}90{}^{\circ }=({\sin }^{2}6{}^{\circ }+{\cos }^{2}6{}^{\circ })+({\sin }^{2}12{}^{\circ }+{\cos }^{2}12{}^{\circ })+({\sin }^{2}18{}^{\circ }+{\cos }^{2}18{}^{\circ })+\dots +({\sin }^{2}42{}^{\circ }+{\cos }^{2}42{}^{\circ })+1=7\left(1\right)+1=8$