Question
Evaluate:
$4(\sin^430^\circ+\cos^460^\circ)-\frac{2}{3}(\sin^260^\circ-\cos^245^\circ)+\frac{1}{2} \ \tan^260^\circ$
$4(\sin^430^\circ+\cos^460^\circ)-\frac{2}{3}(\sin^260^\circ-\cos^245^\circ)+\frac{1}{2} \ \tan^260^\circ$
✓
Answer
$4(\sin^430^\circ+\cos^460^\circ)-\frac{2}{3}(\sin^260^\circ-\cos^245^\circ)+\frac{1}{2} \ \tan^260^\circ$
$= 4\bigg\{\Big(\frac{1}{2}\Big)^4+\Big(\frac{1}{2}\Big)^4\bigg\}-\frac{2}{3}\bigg\{\Big(\frac{\sqrt{3}}{2}\Big)^2-\Big(\frac{1}{\sqrt{2}}\Big)^2\bigg\}+\frac{1}{2}\times\big(\sqrt{3}\big)^2$
$ =4\Big(\frac{1}{16}+\frac{1}{16}\Big)-\frac{2}{3}\Big(\frac{3}{4}-\frac{1}{2}\Big)+\frac{3}{2}$
$ =4\Big(\frac{2}{16}\Big)-\frac{2}{3}\Big(\frac{3-2}{4}\Big)+\frac{3}{2}$
$ =\frac{8}{16}-\frac{2}{3}\times\frac{1}{4}+\frac{3}{2}$
$ =\frac{1}{2}-\frac{1}{6}+\frac{3}{2}$
$ = \frac{3-1+9}{6}$
$=\frac{11}{6}$
$= 4\bigg\{\Big(\frac{1}{2}\Big)^4+\Big(\frac{1}{2}\Big)^4\bigg\}-\frac{2}{3}\bigg\{\Big(\frac{\sqrt{3}}{2}\Big)^2-\Big(\frac{1}{\sqrt{2}}\Big)^2\bigg\}+\frac{1}{2}\times\big(\sqrt{3}\big)^2$
$ =4\Big(\frac{1}{16}+\frac{1}{16}\Big)-\frac{2}{3}\Big(\frac{3}{4}-\frac{1}{2}\Big)+\frac{3}{2}$
$ =4\Big(\frac{2}{16}\Big)-\frac{2}{3}\Big(\frac{3-2}{4}\Big)+\frac{3}{2}$
$ =\frac{8}{16}-\frac{2}{3}\times\frac{1}{4}+\frac{3}{2}$
$ =\frac{1}{2}-\frac{1}{6}+\frac{3}{2}$
$ = \frac{3-1+9}{6}$
$=\frac{11}{6}$
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