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Prove that: \(\sin A\sin (\frac {\pi }{3}-A)\sin (\frac {\pi }{3}+A)=\frac {1}{4}\sin 3A\)
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Prove that: \(\sin A\sin (\frac {\pi }{3}-A)\sin (\frac {\pi }{3}+A)=\frac {1}{4}\sin 3A\)

Answer

\(\text{LHS}=\sin A \sin \left(\frac{\pi}{3}-A\right) \sin \left(\frac{\pi}{3}+A\right)\)
\(=\frac{1}{2} \sin A\left\{2 \sin \left(\frac{\pi}{3}-A\right) \sin \left(\frac{\pi}{3}+A\right)\right\}\)
\(=\frac{1}{2} \sin A\left[\cos \left\{\left(\frac{\pi}{3}-A\right)-\left(\frac{\pi}{3}+A\right)\right\}-\cos \left\{\left(\frac{\pi}{3}-A\right)+\left(\frac{\pi}{3}+A\right)\right\}\right]\)
\(=\frac{1}{2} \sin A\left\{\cos (-2 A)-\cos \frac{2 \pi}{3}\right\}\)
\(=\frac{1}{2} \sin A\left\{\cos 2 A+\frac{1}{2}\right\}\)
\(=\frac{1}{2} \sin A \cos 2 A+\frac{1}{4} \sin A\)
\(=\frac{1}{4}(2 \sin A \cos 2 A)+\frac{1}{4} \sin A\)
\(=\frac{1}{4}\{\sin (A+2 A)+\sin (A-2 A)\}+\frac{1}{4} \sin A\)
\(=\frac{1}{4}\{\sin 3 A+\sin (-A)\}+\frac{1}{4} \sin A\)
\(=\frac{1}{4} \sin 3 A-\frac{1}{4} \sin A+\frac{1}{4} \sin A\)
\(=\frac{1}{4} \sin 3 A\)
\(=\mathrm{RHS}\)
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