Question

How do you prove $\cos \left(\frac{\mathrm{\pi }}{7}\right)\cos \left(\frac{2\mathrm{\pi }}{7}\right)\cos \left(\frac{3\mathrm{\pi }}{7}\right)=\frac{1}{8}$?

Answer 1

Simplify the expression:

Consider,

$$\begin{gathered} \mathrm{L}.\mathrm{H}.\mathrm{S}=\cos \left(\frac{\mathrm{\pi }}{7}\right)\cos \left(\frac{2\mathrm{\pi }}{7}\right)\cos \left(\frac{3\mathrm{\pi }}{7}\right) \\ =\frac{1}{2}\left[2\cos \left(\frac{\mathrm{\pi }}{7}\right)\cos \left(\frac{2\mathrm{\pi }}{7}\right)\right]\cos \left(\frac{3\mathrm{\pi }}{7}\right) \\ =\frac{1}{2}\left[\cos \left(\frac{3\mathrm{\pi }}{7}\right)+\cos \left(\frac{-\mathrm{\pi }}{7}\right)\right]\cos \left(\frac{3\mathrm{\pi }}{7}\right)\left[\because 2\cos A\cos B=\cos \left(A+B\right)+\cos \left(A-B\right)\right] \\ =\frac{1}{4}\left[2\cos \left(\frac{3\mathrm{\pi }}{7}\right)\cos \left(\frac{3\mathrm{\pi }}{7}\right)+2\cos \left(\frac{-\mathrm{\pi }}{7}\right)\cos \left(\frac{3\mathrm{\pi }}{7}\right)\right] \\ =\frac{1}{4}\left[\cos \left(\frac{6\mathrm{\pi }}{7}\right)+\cos \left(0\right)+\cos \left(\frac{2\mathrm{\pi }}{7}\right)+\cos \left(\frac{-4\mathrm{\pi }}{7}\right)\right]\left[\because 2\cos A\cos B=\cos \left(A+B\right)+\cos \left(A-B\right)\right] \\ =\frac{1}{4}\left[1-\frac{1}{2}\right] \\ =\frac{1}{8} \\ =\mathrm{R}.\mathrm{H}.\mathrm{S} \end{gathered}$$

Hence, $\mathbf{cos}\mathbf{\left(}\frac{\mathbf{\pi }}{\mathbf{7}}\mathbf{\right)}\mathbf{}\mathbf{cos}\mathbf{\left(}\frac{\mathbf{2}\mathbf{\pi }}{\mathbf{7}}\mathbf{\right)}\mathbf{}\mathbf{cos}\mathbf{\left(}\frac{\mathbf{3}\mathbf{\pi }}{\mathbf{7}}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{8}}$ has been proved.