I was working through my physics book when I came upon the following:

cos<sup>m</sup>(x) = 1/2<sup>m-1</sup>(cos(mx) + m*cos((m-2)x) + ... + B(m,n)*cos((m-2*n)x) + ... + L)

Where B(m,n) is the binomial coefficient [m!/(n!(m-n)!)] and

L = m!/(2((m/2)!)<sup>2</sup>) for m even

L = m!/(2((m+1)/2)!((m-1)/2)!) for m odd

It seems to me like L should be 0 for m odd. Also, can anyone help me out with a proof of it? I have cos(mx) expanded with the Chebyshev Polynomials, but I am having trouble explicitly expressing an inverse to find cos<sup>m</sup>(x) in terms of cos(mx),cos((m-1)x,...

Thanks,

Vroom