Prove that: 10 3 13 6 + 8 3 5 6 =-1

Question

Prove that:
$\sin\frac{10\pi}{3}\cos\frac{13\pi}{6}+\cos\frac{8\pi}{3}\sin\frac{5\pi}{6}=-1$

✓

Answer

$\text{L.H.S}=\sin\frac{10\pi}{3}\cos\frac{13\pi}{6}+\cos\frac{8\pi}{3}\sin\frac{5\pi}{6}$
$=\sin600^\circ\cos390^\circ+\cos480^\circ\sin150^\circ$
$=\sin\Big(3\pi+\frac{\pi}{3}\Big)\cos\Big(2\pi+\frac{\pi}{6}\Big)+\cos\Big(3\pi-\frac{\pi}{3}\Big)\sin\Big(\pi-\frac{\pi}{6}\Big)$
$=-\sin\frac{\pi}{3}\cos\frac{\pi}{6}-\cos\frac{\pi}{3}-\sin\frac{\pi}{6}$$\Big(\because\sin\Big(3\pi+\frac{\pi}{3}\Big)=-\sin\frac{\pi}{3 }\&\cos\Big(3\pi-\frac{\pi}{3}\Big)=-\cos\frac{\pi}{3}\Big)$
$=\frac{-\sqrt3}{2}\times\frac{-\sqrt3}{2}-\frac{1}{2}\times\frac{1}{2}$
$=\frac{-3}{4}-\frac{1}{4}$
$=\frac{-4}{4}$
$=-1$
$=\text{R.H.S}$
$\text{Proved}$

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