What is the value of $4 ({sin}^{4} 30^{0} + {cos}^{4} 60^{0}) - 3 ({cos}^{2} 45^{0} - {sin}^{2} 90^{0})$ ?

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Solution

$4 ({sin}^{4} 30^{0} + {cos}^{4} 60^{0}) - 3 ({cos}^{2} 45^{0} - {sin}^{2} 90^{0})$
$= 4 [{(\frac{1}{2})}^{4} + {(\frac{1}{2})}^{4}] - 3 ⎡ ⎣ {(\frac{1}{\sqrt{2}})}^{2} - {(1)}^{2} ⎤ ⎦$
$= 4 (\frac{1}{16} + \frac{1}{16}) - 3 (\frac{1}{2} - 1) = \frac{1}{2} + \frac{3}{2} = 2$

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Similar questions

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

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

4 (s i n^{4} 30^{0} + c o s^{4} 60^{0}) (c o s^{2} 45^{0} - s i n^{2} 90^{0}) + t a n^{4} 45^{0} - c o t^{2} 45^{0}

{sin}^{2} 5^{0} + {sin}^{2} 10^{0} + {sin}^{2} 15^{0} + . . . + {sin}^{2} 90^{0} = a \frac{1}{2} .

a

\frac{4}{{tan}^{2} 60^{0}} + \frac{1}{{cos}^{2} 30^{0}} - {sin}^{2} 90^{0}

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