cos(π/9)*cos(2π/9)*cos(3π/9)*cos(4π/9)

=sin(π/9)*cos(π/9)*cos(2π/9)*cos(3π/9)*cos(4π/9)/sin(π/9)
=sin(2π/9)cos(2π/9)*cos(3π/9)*cos(4π/9)/[2*sin(π/9)]
=sin(4π/9)*cos(3π/9)*cos(4π/9)/[4*sin(π/9)]
=sin(8π/9)*cos(3π/9)/[8*sin(π/9)]
=sin(π/9)*cos(3π/9)/[8*sin(π/9)]
=cos(3π/9)/8
=1/16

In LaTex with MathML :

  \(\displaystyle= { \sin\frac{\pi}{9}\times\cos\frac{\pi}{9}\times\cos\frac{2\pi}{9}\times\cos\frac{3\pi}{9}\times\cos\frac{4\pi}{9}\over\sin\frac{\pi}{9} } \)
\(\displaystyle= { \sin\frac{2\pi}{9}\times\cos\frac{2\pi}{9}\times\cos\frac{3\pi}{9}\times\cos\frac{4\pi}{9}\over 2\times\sin\frac{\pi}{9} } \) \( ( \because \sin(x)\cos(y)=\frac{1}{2}\left[\sin(x+y)+\sin(x-y)\right] )\)
\(\displaystyle= { \sin\frac{4\pi}{9}\times\cos\frac{3\pi}{9}\times\cos\frac{4\pi}{9}\over 4\times\sin\frac{\pi}{9} } \) \( ( \because \sin(x)\cos(y)=\frac{1}{2}\left[\sin(x+y)+\sin(x-y)\right] )\)
\(\displaystyle= { \sin\frac{8\pi}{9}\times\cos\frac{3\pi}{9}\over 8\times\sin\frac{\pi}{9} } \) \( ( \because \sin(x)\cos(y)=\frac{1}{2}\left[\sin(x+y)+\sin(x-y)\right] )\)
\(\displaystyle= { \sin\frac{\pi}{9}\times\cos\frac{3\pi}{9}\over 8\times\sin\frac{\pi}{9} } \)
\(\displaystyle= { \cos\frac{3\pi}{9}\over 8 } \)
\(\displaystyle= { \frac{1}{16} } \)