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Question

Prove that $\left|\mathrm{sin}x\mathrm{sin}\left(\frac{\pi }{3}-x\right)\mathrm{sin}\left(\frac{\pi }{3}+x\right)\right|\le \frac{1}{4}$ for all values of x

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Solution

$\frac{\pi }{3}=60°\phantom{\rule{0ex}{0ex}}\mathrm{We}\mathrm{have},\phantom{\rule{0ex}{0ex}}\left|\mathrm{sin}x\mathrm{sin}\left(60-x\right)\mathrm{sin}\left(60+x\right)\right|\phantom{\rule{0ex}{0ex}}=\left|\mathrm{sin}x\left({\mathrm{sin}}^{2}60-{\mathrm{sin}}^{2}x\right)\right|\phantom{\rule{0ex}{0ex}}\left[\because \mathrm{sin}\left(\mathrm{A}+\mathrm{B}\right)\mathrm{sin}\left(\mathrm{A}-\mathrm{B}\right)={\mathrm{sin}}^{2}\mathrm{A}-{\mathrm{sin}}^{2}\mathrm{B}\right]\phantom{\rule{0ex}{0ex}}=\left|\mathrm{sin}x\left(\frac{3}{4}-{\mathrm{sin}}^{2}x\right)\right|\phantom{\rule{0ex}{0ex}}=\left|\frac{1}{4}\mathrm{sin}x\left(3-4{\mathrm{sin}}^{2}x\right)\right|\phantom{\rule{0ex}{0ex}}=\left|\frac{1}{4}3\mathrm{sin}x-4{\mathrm{sin}}^{3}x\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\left|\mathrm{sin}3\mathrm{x}\right|\left(\because 3\mathrm{sin}x-4{\mathrm{sin}}^{3}x=\mathrm{sin}3x\right)\phantom{\rule{0ex}{0ex}}\le \frac{1}{4}\left(\because \left|\mathrm{sin}x\right|\le 1\mathrm{for}\mathrm{all}x\right)\phantom{\rule{0ex}{0ex}}\therefore \left|\mathrm{sin}x\mathrm{sin}\left(60-x\right)\mathrm{sin}\left(60+x\right)\right|\le \frac{1}{4}\phantom{\rule{0ex}{0ex}}\mathrm{Hence}\mathrm{proved}.$

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