Question
Prove that:
$\sin^2\frac{2\pi}{5}=\sin^2\frac{\pi}{3}=\frac{\sqrt{5}-1}{8}$
$\sin^2\frac{2\pi}{5}=\sin^2\frac{\pi}{3}=\frac{\sqrt{5}-1}{8}$
✓
Answer
We have,
$\sin^272^\circ-\sin^260^\circ,$
$=\sin^2(90^\circ-18^\circ)-\Big(\frac{\sqrt{3}}{2}\Big)^2$
$=\cos^218^\circ-\frac{3}{4}$
$\Bigg(\frac{\sqrt{10+2+\sqrt{5}}}{4}\Bigg)^2-\frac{3}{4}\Bigg[\because\cos18^\circ=\frac{\sqrt{10+2\sqrt{5}}}{4}\Bigg]$
$=\frac{10+2\sqrt{5}}{16}-\frac{3}{4}$
$=\frac{10+2\sqrt{5}-12}{16}$
$=\frac{2\sqrt{5}-2}{16}$
$=\frac{\sqrt{5}-1}{8}$
$\sin^272^\circ-\sin^260^\circ,$
$=\sin^2(90^\circ-18^\circ)-\Big(\frac{\sqrt{3}}{2}\Big)^2$
$=\cos^218^\circ-\frac{3}{4}$
$\Bigg(\frac{\sqrt{10+2+\sqrt{5}}}{4}\Bigg)^2-\frac{3}{4}\Bigg[\because\cos18^\circ=\frac{\sqrt{10+2\sqrt{5}}}{4}\Bigg]$
$=\frac{10+2\sqrt{5}}{16}-\frac{3}{4}$
$=\frac{10+2\sqrt{5}-12}{16}$
$=\frac{2\sqrt{5}-2}{16}$
$=\frac{\sqrt{5}-1}{8}$
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