Question

Evaluate :
$4({\sin }^4{30}^{\circ }+{\cos }^4{60}^{\circ })-\frac{2}{3}({\sin }^2{60}^{\circ }-{\cos }^2{45}^{\circ })+\frac{1}{2}{\tan }^2{60}^{\circ }$

Answer 1

Given $4({\sin }^4{30}^{\circ }+{\cos }^4{60}^{\circ })-\frac{2}{3}({\sin }^2{60}^{\circ }-{\cos }^2{45}^{\circ })+\frac{1}{2}{\tan }^2{60}^{\circ }$

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

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

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

$=\frac{1}{2}-\frac{1}{6}+\frac{3}{2}$

$=\frac{4}{2}-\frac{1}{6}$

$=\frac{2}{1}-\frac{1}{6}$

$=2-\frac{1}{6}$

$=\frac{11}{6}$