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

Given:

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

To do:

We have to evaluate \( 4\left(\sin ^{4} 30^{\circ}+\cos ^{4} 60^{\circ}\right)-\frac{2}{3}\left(\sin ^{2} 60^{\circ}-\cos ^{2} 45^{\circ}\right)+\frac{1}{2} \tan ^{2} 60^{\circ} \).

Solution:  

We know that,

$\sin 30^{\circ}=\frac{1}{2}$

$\cos 60^{\circ}=\frac{1}{2}$

$\sin 60^{\circ}=\frac{\sqrt3}{2}$

$\cos 45^{\circ}=\frac{1}{\sqrt2}$

$\tan 60^{\circ}=\sqrt3$

$4\left(\sin ^{4} 30^{\circ}+\cos ^{4} 60^{\circ}\right)-\frac{2}{3}\left(\sin ^{2} 60^{\circ}-\cos ^{2} 45^{\circ}\right)+\frac{1}{2} \tan ^{2} 60^{\circ}=4\left[\left(\frac{1}{2}\right)^{4} +\left(\frac{1}{2}\right)^{4}\right] -\frac{2}{3}\left[\left(\frac{\sqrt{3}}{2}\right)^{2} -\left(\frac{1}{\sqrt{2}}\right)^{2}\right] +\frac{1}{2}\left(\sqrt{3}\right)^{2}$

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

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

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

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

$=\left(\frac{1}{2}\right) -\frac{1}{6} +\frac{3}{2}$

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

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

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

$=\frac{12-1}{6}$

$=\frac{11}{6}$

Hence, $4\left(\sin ^{4} 30^{\circ}+\cos ^{4} 60^{\circ}\right)-\frac{2}{3}\left(\sin ^{2} 60^{\circ}-\cos ^{2} 45^{\circ}\right)+\frac{1}{2} \tan ^{2} 60^{\circ}=\frac{11}{6}$.   

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Updated on: 10-Oct-2022

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