Question
Prove that:
$\sin\frac{5\pi}{18}-\cos\frac{4\pi}{9}=\sqrt3\sin\frac{\pi}{9}$
$\sin\frac{5\pi}{18}-\cos\frac{4\pi}{9}=\sqrt3\sin\frac{\pi}{9}$
✓
Answer
$\sin\frac{5\pi}{18}-\cos\frac{4\pi}{9}=\sqrt3\sin\frac{\pi}{9}$
$\text{LHS}=\sin\frac{5\pi}{18}-\cos\frac{4\pi}{9}$
$=\ \sin50^\circ-\cos80^\circ$
$=\ \sin50^\circ-\sin10^\circ$
$=\ 2\sin\Big(\frac{50^\circ-10^\circ}{2}\Big)\cos\Big(\frac{50^\circ+10^\circ}{2}\Big)$
$=\ 2\sin20^\circ\cos30^\circ$
$=\ 2\sin20^\circ\times\frac{\sqrt3}{2}$
$=\ \sqrt3\sin\frac{\pi}{9}=\text{RHS}$
$\text{LHS}=\sin\frac{5\pi}{18}-\cos\frac{4\pi}{9}$
$=\ \sin50^\circ-\cos80^\circ$
$=\ \sin50^\circ-\sin10^\circ$
$=\ 2\sin\Big(\frac{50^\circ-10^\circ}{2}\Big)\cos\Big(\frac{50^\circ+10^\circ}{2}\Big)$
$=\ 2\sin20^\circ\cos30^\circ$
$=\ 2\sin20^\circ\times\frac{\sqrt3}{2}$
$=\ \sqrt3\sin\frac{\pi}{9}=\text{RHS}$
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