help!!

\\\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) \\=\left(2cos^2\frac{\pi}{18}\right)\left(2cos^2\frac{3\pi}{18}\right)\left(2cos^2\frac{5\pi}{18}\right)\left(2cos^2\frac{7\pi}{18}\right) \\=\left(4sin\frac{4\pi}{9}\times sin\frac{3\pi}{9}\times sin\frac{2\pi}{9}\times sin\frac{\pi}{9}\right)^2

now if you see ..

4sin(\pi/9)\times sin(\pi/9+\pi/3)\times sin(\pi/3-\pi/9)=sin(3\times\pi/9)

Hence the given expression becomes \left(sin\frac{3\pi}{9}\times sin\frac{3\pi}{9}\right)^2=9/16