This was a past year paper question. Not sure how to answer it.
Question: The 1-D discrete cosine transform(DCT) of a sequence f(x), x =0,1,...,N-1 is
F(u) = c(u)* Summation of f(x) cos ( (2x+1)u*pi/2N)
show that 1-D DCT of the sequence
g(x) = f(N-1-x), x = 0,1,...,N-1
can be expressed as G(u) = (-1)*F(u), u=0,1..., N-1
My solution (If wrong please correct me)
G(u) = c(u)* Summation of f(N-1-x) cos ((2(N-1-x)+1)u*pi/2N) = c(u)* Summation of f(N-1-x) cos (2N-2x-1)u*pi/2N)