Let a = pi/11 Use the identity that cos x = (e^(i x) + e^(- i x))/2 Let p = e^(i a) So cos(pi/11) = cos(a) = (p^1 + p^-1)/2 The value you are looking for is then: (p^1 + p^-1)(p^2 + p^-2)(p^3 + p^-3)(p^4 + p^-4)(p^5 + p^-5)/32 Multiplying that out, a term at a time, from the right: (p^5 + p^-5)/32 (p^9+p^1+p^-1+p^-9)/32 (p^12+p^6+p^4+p^2+p^-2+p^-4+p^-6+p^-12)/32 (p^14+p^12+p^8+2p^4+p^2+2p^0+p^-2+2p^-4+p^-8+p^-12+p^-14)/32 (p^15+p^13+p^11+2p^9+2p^7+3p^5+3p^3+3p^1+3p^-1+3p^-3+3p^-5+2p^-7+2p^-9+p^-11+p^-13+p^-15)/32 p^11 is e^(i pi) which by Euler is -1 So p^22 is 1 The negative powers can be made postive by multiplying by p^22 = 1 (3p^21+3p^19+3p^17+3p^15+3p^13+2p^11+3p^9+3p^7+3p^5+3p^3+3p^1)/32 The coefficient of p^11 is a touch small, I'll add and subtract p^11: (3p^21+3p^19+3p^17+3p^15+3p^13+3p^11+3p^9+3p^7+3p^5+3p^3+3p^1-p^11)/32 Take out a common factor of 3p (3p(p^20+p^18+p^16+p^13+p^12+p^10+p^8+p^6+p^4+p^2+p^0) - p^11)/32 I claim that p^20+p^18+...+p^0 is zero: p^11+1 is zero so (p^11+1)(p^10+p^9+p^8+...+p^0) is zero. so p^21+p^20+...+p^0 is zero so (1+p)(p^20+p^18+p^16+...+p^0) is zero and since (1+p) is NOT zero, (p^20+p^18+...+p^0) is zero. So that leads to: -p^11/32 which is 1/32