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Evaluate each of the following:$ 4\left(\sin ^{4} 30^{\circ}+\cos ^{2} 60^{\circ}\right)-3\left(\cos ^{2} 45^{\circ}-\sin ^{2} 90^{\circ}\right)-\sin ^{2} 60^{\circ} $
Given:
\( 4\left(\sin ^{4} 30^{\circ}+\cos ^{2} 60^{\circ}\right)-3\left(\cos ^{2} 45^{\circ}-\sin ^{2} 90^{\circ}\right)-\sin ^{2} 60^{\circ} \)
To do:
We have to evaluate \( 4\left(\sin ^{4} 30^{\circ}+\cos ^{2} 60^{\circ}\right)-3\left(\cos ^{2} 45^{\circ}-\sin ^{2} 90^{\circ}\right)-\sin ^{2} 60^{\circ} \).
Solution:
We know that,
$sin 30^{\circ}=\frac{1}{2}$
$\cos 60^{\circ}=\frac{1}{2}$
$\cos 45^{\circ}=\frac{1}{\sqrt2}$
$\sin 90^{\circ}=1$
$\sin 60^{\circ}=\frac{\sqrt3}{2}$
Therefore,$4\left(\sin ^{4} 30^{\circ}+\cos ^{2} 60^{\circ}\right)-3\left(\cos ^{2} 45^{\circ}-\sin ^{2} 90^{\circ}\right)-\sin ^{2} 60^{\circ}=4\left[\left(\frac{1}{2}\right)^{4} +\left(\frac{1}{2}\right)^{2}\right] -3\left[\left(\frac{1}{\sqrt{2}}\right)^{2} -( 1)^{2}\right] -\left(\frac{\sqrt{3}}{2}\right)^{2}$
$=4\left(\frac{1}{16} +\frac{1}{4}\right) -3\left(\frac{1}{2} -1\right) -\frac{3}{4}$
$=4\left(\frac{1+1( 4)}{16}\right) -3\left(\frac{1-2( 1)}{2}\right) -\frac{3}{4}$
$=\frac{5}{4} -3\left(\frac{-1}{2}\right) -\frac{3}{4}$
$=\frac{5+3( 2) -3}{4}$
$=\frac{5+6-3}{4}$
$=\frac{8}{4}$
$=2$
Hence, $4\left(\sin ^{4} 30^{\circ}+\cos ^{2} 60^{\circ}\right)-3\left(\cos ^{2} 45^{\circ}-\sin ^{2} 90^{\circ}\right)-\sin ^{2} 60^{\circ}=2$.
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