# Evaluate each of the following

Question:

Evaluate each of the following

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

Solution:

We have,

$4\left(\sin ^{4} 30^{\circ}+\cos ^{2} 60^{\circ}\right)-3\left(\cos ^{2} 45^{-}-\sin ^{2} 90^{\circ}\right)-\sin ^{2} 60^{\circ} \ldots \ldots$ (1)

Now,

$\sin 30^{\circ}=\cos 60^{\circ}=\frac{1}{2}, \sin 90^{\circ}=1, \cos 45^{\circ}=\frac{1}{\sqrt{2}}, \sin 60^{\circ}=\frac{\sqrt{3}}{2}$

So by substituting above values in equation (1)

We get,

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

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

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

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

LCM of 16 and 4 in the first term of above expression is 16 and

Similarly LCM of 2 and 1 in the second term of above expression is 2

Therefore,

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

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

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

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

$=4\left(\frac{5}{16}\right)-3\left(\frac{-1}{2}\right)-\frac{3}{4}$

Now in the second term of the above expression $-3 \times-1=+3$

Therefore,

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

$=4\left(\frac{5}{16}\right)+\frac{3}{2}-\frac{3}{4}$

Now, in the above expression 4 cancels 16 and 4 remains in the denominator of first term

Therefore,

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

$=\frac{5}{4}+\frac{3}{2}-\frac{3}{4}$

Now by taking LCM =4 in the above expression

We get,

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

$=\frac{5}{4}+\frac{3 \times 2}{2 \times 2}-\frac{3}{4}$

$=\frac{5}{4}+\frac{6}{4}-\frac{3}{4}$

$=\frac{5+6-3}{4}$

$=\frac{11-3}{4}$

$=\frac{8}{4}$

Now, in the above expression $\frac{8}{4}$ gets reduced to 2

Therefore,

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