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Question

# Prove that: $\mathrm{cos}78°\mathrm{cos}42°\mathrm{cos}36°=\frac{1}{8}$

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Solution

## $\mathrm{LHS}=\mathrm{cos}78°\mathrm{cos}42°\mathrm{cos}36°\phantom{\rule{0ex}{0ex}}=\frac{\left(2\mathrm{cos}78°\mathrm{cos}42°\right)}{2}\mathrm{cos}36°\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{cos}\left(78°+42°\right)+\mathrm{cos}\left(78°-42°\right)}{2}×\mathrm{cos}36°\phantom{\rule{0ex}{0ex}}\left[2\mathrm{cos}A\mathrm{cos}B=\mathrm{cos}\left(A+B\right)+\mathrm{cos}\left(A-B\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(\mathrm{cos}120°+\mathrm{cos}36°\right)\mathrm{cos}36°$ $=\frac{1}{2}\left(-\mathrm{cos}\left(180°-120°\right)+\mathrm{cos}36°\right)\mathrm{cos}36°\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(-\mathrm{cos}60°+\mathrm{cos}36°\right)\mathrm{cos}36°\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(-\frac{1}{2}+\frac{\sqrt{5}+1}{4}\right)\frac{\sqrt{5}+1}{4}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}×\frac{\sqrt{5}-1}{4}×\frac{\sqrt{5}+1}{4}\phantom{\rule{0ex}{0ex}}=\frac{1}{8}\phantom{\rule{0ex}{0ex}}=\mathrm{RHS}\phantom{\rule{0ex}{0ex}}\mathrm{Hence}\mathrm{proved}.$

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