Question

Evaluate $4({\sin }^4{30}^{\circ }+{\cos }^4{60}^{\circ }-3({\cos }^2{45}^{\circ }-{\sin }^2{90}^{\circ })$

Answer 1

$4({\sin }^{4}30°+{\cos }^{4}60°-3({\cos }^{2}45°-{\sin }^{2}90°))$

$\Rightarrow 4({\left(\frac{1}{2}\right)}^4+{\left(\frac{1}{2}\right)}^4-3({\left(\frac{1}{\sqrt{2}}\right)}^2-1^2))=4(\frac{1}{16}+\frac{1}{16}-3(\frac{1}{2}-1))$

$\Rightarrow 4(\frac{1}{8}-3(-\frac{1}{2}))=4(\frac{1}{8}+\frac{3}{2})=4\left(\frac{1+12}{8}\right)$

$\Rightarrow 4\times \frac{13}{8}=\frac{13}{2}$

Hence, the answer is $=\frac{13}{2}.$