Trig3.7

From Exampleproblems

Prove that $\sin A \sin [\frac{\pi}{3}+A] \sin [\frac{\pi}{3}-A]=\frac{1}{4} \sin 3A\,$ .Hence show that $\sin \frac{\pi}{9} \sin \frac{2\pi}{9} \sin \frac{3\pi}{9} \sin \frac{4\pi}{9}=\frac{3}{16}\,$

LHS= $\sin A \sin [\frac{\pi}{3}+A] \sin [\frac{\pi}{3}-A]=\sin A [\sin^2 \frac{\pi}{3}-\sin^2 A]=[\frac{3}{4}-\sin^2 A]\sin A\,$

$\frac{3\sin A-4\sin^3 A}{4}=\frac{1}{4} \sin 3A\,$

Putting $A=\frac{\pi}{9}\,$ in the above equation, we get

$\sin \frac{\pi}{9} \sin [\frac{\pi}{3}+\frac{\pi}{9}] \sin [\frac{\pi}{3}-\frac{\pi}{9}]=\frac{1}{4} \sin \frac{\pi}{3}\,$

$\sin \frac{\pi}{9} \sin \frac{4\pi}{9} \sin \frac{2\pi}{9}=\frac{1}{4} \sin \frac{\pi}{3}\,$

$\sin \frac{\pi}{9} \sin \frac{2\pi}{9} \sin \frac{3\pi}{9} \sin \frac{4\pi}{9}=\frac{1}{4} \sin^2 \frac{\pi}{3}\,$

$\sin \frac{\pi}{9} \sin \frac{2\pi}{9} \sin \frac{3\pi}{9} \sin \frac{4\pi}{9}=\frac{3}{16}\,$

Main Page:Trigonometry

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