Question

Prove that $4\left[{\sin }^4{30}^{\circ }+{\cos }^4{60}^{\circ }\right]-3\left[{\cos }^2{45}^{\circ }-{\sin }^2{90}^{\circ }\right]=2$

Answer 1

$\sin {30}^0=\frac{1}{2}\Rightarrow {\sin }^4{30}^0=\frac{1}{16}$

$\cos {60}^0=\frac{1}{2}\Rightarrow {\cos }^4{60}^0=\frac{1}{16}$

$\cos {45}^0=\frac{1}{2}{\sin }^2{90}^0=1$

$\Rightarrow 4(\frac{1}{16}+\frac{1}{16})-3(\frac{1}{2}-1)$

$\Rightarrow \frac{1}{2}-\frac{-3}{2}=2$