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)=\fra...