The value of $4 ({sin}^{4} 30^{0} + {cos}^{4} 30^{0}) - 3 ({cos}^{2} 45^{0} - {sin}^{2} 90^{0}) =$

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Solution

To Prove: $4 ({sin}^{4} 30^{0} + {cos}^{4} 30^{0}) - 3 ({cos}^{2} 45^{0} - {sin}^{2} 90^{0}) = 2$
L.H.S.: $4 ({sin}^{4} 30^{0} + {cos}^{4} 30^{0}) - 3 ({cos}^{2} 45^{0} - {sin}^{2} 90^{0})$
= $4 ((\frac{1}{2})^{4} + (\frac{\sqrt{3}}{2})^{4}) - 3 ((\frac{1}{\sqrt{2}})^{2} - (1)^{2})$
= $4 (\frac{1}{16} + \frac{9}{16}) - 3 (\frac{1}{2} - 1)$
= $4 (\frac{10}{16}) + \frac{3}{2}$
= $\frac{5}{2} + \frac{3}{2}$
= $4$

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4 ({sin}^{4} 30^{0} + {cos}^{4} 60^{0}) - 3 ({cos}^{2} 45^{0} - {sin}^{2} 90^{0})

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