Question
Without using tables, evaluate the following: $4(\sin^430^\circ + \cos^460^\circ ) - 3(\cos^245^\circ - \sin^290^\circ )$.
✓
Answer
$4\left(\sin ^4 30^{\circ}+\cos ^4 60^{\circ}\right)-3\left(\cos ^2 45^{\circ}-\sin ^2 90^{\circ}\right)$
$\sin 30^{\circ}=\frac{1}{2}$
$\sin 90^{\circ}=1$
$\cos 45^{\circ}=\frac{1}{\sqrt{2}}$
$\cos 60^{\circ}=\frac{1}{2}$
$4\left(\sin ^4 30^{\circ}+\cos ^4 60^{\circ}\right)-3\left(\cos ^2 45^{\circ}-\sin ^2 90^{\circ}\right)$
$=4\left(\left(\frac{1}{2}\right)^4+\left(\frac{1}{2}\right)^4\right)-3\left(\left(\frac{1}{\sqrt{2}}\right)^2-(1)^2\right)$
$=4\left(\frac{1}{16}+\frac{1}{16}\right)-3\left(\frac{1}{2}-1\right)$
$=4 \times \frac{2}{16}+3 \times \frac{1}{2}$
$=\frac{1}{2}+\frac{3}{2}$
$=\frac{4}{2}$
$=2$
$\sin 30^{\circ}=\frac{1}{2}$
$\sin 90^{\circ}=1$
$\cos 45^{\circ}=\frac{1}{\sqrt{2}}$
$\cos 60^{\circ}=\frac{1}{2}$
$4\left(\sin ^4 30^{\circ}+\cos ^4 60^{\circ}\right)-3\left(\cos ^2 45^{\circ}-\sin ^2 90^{\circ}\right)$
$=4\left(\left(\frac{1}{2}\right)^4+\left(\frac{1}{2}\right)^4\right)-3\left(\left(\frac{1}{\sqrt{2}}\right)^2-(1)^2\right)$
$=4\left(\frac{1}{16}+\frac{1}{16}\right)-3\left(\frac{1}{2}-1\right)$
$=4 \times \frac{2}{16}+3 \times \frac{1}{2}$
$=\frac{1}{2}+\frac{3}{2}$
$=\frac{4}{2}$
$=2$
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