Question

Find value of ${\sin }^2{10}^0+{\sin }^2{20}^0+{\sin }^2{40}^0+{\sin }^2{50}^0+{\sin }^2{70}^0+{\sin }^2{80}^0$

Answer 1

$si{n}^2{10}^{\circ }+si{n}^2{20}^{\circ }+si{n}^2{40}^{\circ }+si{n}^2(90-40{)}^{\circ }+si{n}^2(90-20{)}^{\circ }+si{n}^2(90-10{)}^{\circ }=si{n}^2{10}^{\circ }+si{n}^2{20}^{\circ }+si{n}^2{40}^{\circ }+co{s}^2{10}^{\circ }+co{s}^2{20}^{\circ }+co{s}^2{40}^{\circ }=(si{n}^2{10}^{\circ }+co{s}^2{10}^{\circ })+(si{n}^2{20}^{\circ }+co{s}^2{20}^{\circ })+(si{n}^2{40}^{\circ }+co{s}^2{40}^{\circ })=1+1+1=3$