Question
Prove that:
$2\sin\frac{5\pi}{12}\sin\frac{\pi}{12}=\frac{1}{2}$

Answer

$\text{LHS}=2\sin\frac{5\pi}{12}\sin\frac{\pi}{12}$
$\because\ 2\sin\text{A}\sin\text{B}=\cos(\text{A}-\text{B})-\cos(\text{A+B})$
$\Rightarrow\ 2\sin\frac{5\pi}{12}\sin\frac{\pi}{12}=\cos\Big(\frac{5\pi}{12}-\frac{\pi}{12}\Big)-\cos\Big(\frac{5\pi}{12}+\frac{\pi}{12}\Big)$
$=\ \cos\Big(\frac{4\pi}{12}\Big)-\cos\Big(\frac{6\pi}{12}\Big)$
$=\ \cos\Big(\frac{\pi}{3}\Big)-\cos\Big(\frac{6\pi}{12}\Big)$
$=\ \frac{1}{2}-0=\frac{1}{2}=\text{RHS}$

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