Question

Show that $(1+\cos \frac{\pi }{9})(1+\cos \pi \frac{3\pi }{9})(1+\cos \frac{5\pi }{9})(1+\cos \frac{7\pi }{9})=\frac{9}{16}$.

Answer 1

$\left(1+\cos \frac{\pi }{9}\right)\left(1+\cos \frac{3\pi }{9}\right)\left(1+\cos \frac{5\pi }{9}\right)\left(1+\cos \frac{7\pi }{9}\right)$

Using formulae.

$\begin{array}{c}1+\cos 2\theta =2{\cos }^2\theta \\ 2{\cos }^2\frac{\pi }{18}.2{\cos }^2\frac{3\pi }{18}.2{\cos }^2\frac{5\pi }{18}.2{\cos }^2\frac{7\pi }{18}\\ =16{\cos }^2{10}^{\circ }.{\cos }^2{30}^{\circ }.{\cos }^2{50}^{\circ }.{\cos }^{2-}{70}^{\circ }\\ =16{\left(\frac{\sqrt{3}}{2}\right)}^2\left[{\cos }^{2}10.{\cos }^{2}50.{\cos }^{2-}70\right]\end{array}$

Using formulae.

$\begin{array}{c}\cos x.\cos \left(60+x\right)\cos \left(60-x\right)=\frac{\cos 3x}{4}\\ 16\times \frac{3}{4}{\left[\cos 10.\cos 50.\cos 70\right]}^2\\ =12{\left[\cos \frac{3\times 10}{4}\right]}^2\\ =\frac{12}{16}\times \frac{3}{4}=\frac{9}{16}\end{array}$