Question: Let (cosx)^6 = mcos6x+ncos5x+ocos4x+pcos3x+qcos2x+rcosx+a. What is the value of a?
cos(6x) = 32cos(x)^6 - 48cos(x)^4 + 18cos(x)^2 - 1 ---------------- (1)
cos(4x) = 8cos(x)^4 - 8*cos(x)^2 + 1 --------------------------------- (2)
cos(2x) = 2cos^2(x) - 1 ---------------------------------------------------- (3)
(1) + 6*(2)
cos(6x) + 6cos(4x) = 32cos(x)^6 - 30cos(x)^2 + 5 ------------------- (4)
(4) + 15*(3)
cos(6x) + 6cos(4x) + 15cos(2x) = 32cos^6(x) - 10
32cos^6(x) = cos(6x) + 6cos(4x) + 15cos(2x) + 10
cos^6(x) = (1/32)cos(6x) + (3/16)cos(4x) + (15/32)cos(2x) + 5/16
Hence a = 5/16