Question

Prove that ${\sin }^2{12}^{\circ }+{\sin }^2{21}^{\circ }+{\sin }^2{39}^{\circ }+{\sin }^2{48}^{\circ }-{\sin }^2{9}^{\circ }-{\sin }^2{18}^{\circ }=1$.

Answer 1

${\sin }^2{12}^{\circ }+{\sin }^2{21}^{\circ }+{\sin }^2{39}^{\circ }+{\sin }^2{48}^{\circ }-{\sin }^2{9}^{\circ }-{\sin }^2{18}^{\circ }$
$=\frac{1-\cos {24}^{\circ }}{2}+\frac{1-\cos {42}^{\circ }}{2}+\frac{1-\cos {28}^{\circ }}{2}+\frac{1-\cos {96}^{\circ }}{2}-\left(\frac{1-\cos {18}^{\circ }}{2}\right)-\left(\frac{1-\cos {56}^{\circ }}{2}\right)=1-\cos {24}^{\circ }-\cos {42}^{\circ }-\cos {28}^{\circ }-\cos {96}^{\circ }+\cos {18}^{\circ }+\cos {36}^{\circ }=1-(\cos {24}^{\circ }+\cos {96}^{\circ })-\left(\cos {42}^{\circ }+\cos 8^{\circ }\right)+\cos {18}^{\circ }+\cos {36}^{\circ }=1-\left(2\cos {60}^{\circ }\cos {36}^{\circ }\right)-\left(2\cos {60}^{\circ }\cos {18}^{\circ }\right)+\cos {18}^{\circ }+\cos {36}^{\circ }=1-\cos {36}^{\circ }-\cos {18}^{\circ }+\cos {18}^{\circ }+\cos {36}^{\circ }=1$