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Question

Prove that: $\mathrm{cos}36°\mathrm{cos}42°\mathrm{cos}60°\mathrm{cos}78°=\frac{1}{16}$

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Solution

$\mathrm{LHS}=\mathrm{cos}36°\mathrm{cos}42°\mathrm{cos}60°\mathrm{cos}78°\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\mathrm{cos}36°\mathrm{cos}60°\left(2\mathrm{cos}42°\mathrm{cos}78°\right)\phantom{\rule{0ex}{0ex}}\left[2\mathrm{cos}A\mathrm{cos}B=\mathrm{cos}\left(A+B\right)+\mathrm{cos}\left(A-B\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(\frac{\sqrt{5}+1}{4}\right)×\frac{1}{2}\left(\mathrm{cos}120°+\mathrm{cos}36°\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{\sqrt{5}+1}{16}\right)\left(-\frac{1}{2}+\frac{\sqrt{5}+1}{4}\right)\phantom{\rule{0ex}{0ex}}=\frac{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}{64}\phantom{\rule{0ex}{0ex}}=\frac{5-1}{64}\phantom{\rule{0ex}{0ex}}=\frac{1}{16}\phantom{\rule{0ex}{0ex}}=\mathrm{RHS}\phantom{\rule{0ex}{0ex}}\mathrm{Hence}\mathrm{proved}.$

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