Question
Evaluate:
$\frac{2}{3}(\cos^430^\circ-\sin^445^\circ)-3(\sin^260^\circ-\sec^245^\circ)+\frac{1}{4}\cot^230^\circ$
$\frac{2}{3}(\cos^430^\circ-\sin^445^\circ)-3(\sin^260^\circ-\sec^245^\circ)+\frac{1}{4}\cot^230^\circ$
✓
Answer
$\cos30^\circ=\frac{\sqrt{3}}{2},\sin60^\circ=\frac{\sqrt{3}}{2},\cot30^\circ=\sqrt{3},$ $\sin45^\circ=\frac{1}{\sqrt{2}},\sec45^\circ=\sqrt{2}$
Substituting above value in question,
$=\frac{2}{3}\bigg[\Big(\frac{\sqrt{3}}{2}\Big)^4-\Big(\frac{1}{\sqrt{2}}\Big)^4\bigg]-3\bigg[\Big(\frac{\sqrt{3}}{2}\Big)^2-\Big(\sqrt{2}\Big)^2\bigg]+\frac{1}{4}\Big(\sqrt{3}\Big)^2$
$=\frac{2}{3}\Big[\frac{9}{16}-\frac{1}{4}\Big]-3\Big[\frac{3}{4}-2\Big]+\frac{1}{4}\times3$
$=\frac{2}{3}\Big[\frac{5}{16}\Big]-3\Big[\frac{-5}{4}\Big]+\frac{3}{4}$
$=\frac{5}{24}+\frac{15}{4}+\frac{3}{4}$
$=\frac{5+90+18}{24}$
$=\frac{133}{24}$
Substituting above value in question,
$=\frac{2}{3}\bigg[\Big(\frac{\sqrt{3}}{2}\Big)^4-\Big(\frac{1}{\sqrt{2}}\Big)^4\bigg]-3\bigg[\Big(\frac{\sqrt{3}}{2}\Big)^2-\Big(\sqrt{2}\Big)^2\bigg]+\frac{1}{4}\Big(\sqrt{3}\Big)^2$
$=\frac{2}{3}\Big[\frac{9}{16}-\frac{1}{4}\Big]-3\Big[\frac{3}{4}-2\Big]+\frac{1}{4}\times3$
$=\frac{2}{3}\Big[\frac{5}{16}\Big]-3\Big[\frac{-5}{4}\Big]+\frac{3}{4}$
$=\frac{5}{24}+\frac{15}{4}+\frac{3}{4}$
$=\frac{5+90+18}{24}$
$=\frac{133}{24}$
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