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)
If $g(x)= f(N-1-x)$ then $$\begin{aligned} (\operatorname{DCT}\:g)(u) =& c(u)\cdot\sum_{x=0}^{N-1}g(x)\cdot\cos\bigl((2x+1)u\cdot\tfrac\pi{2N}\bigr) \\=& c(u)\cdot\sum_{x=0}^{N-1}f(N-1-x) \cdot\cos\bigl((2N-2N+1+2x)u\cdot\tfrac\pi{2N}\bigr) \\=& c(u)\cdot\sum_{(N-1-x)=(N-1)}^{0}f(N-1-x) \cdot\cos\Bigl(\bigl(2N-1-2(N-1-x)\bigr)u\cdot\tfrac\pi{2N}\Bigr) \end{aligned}$$ substitute now $N-1-x=:\xi$, $$\begin{aligned} (\operatorname{DCT}\:g)(u) =& c(u)\cdot\sum_{\xi=0}^{N-1}f(\xi) \cdot\cos\Bigl(\bigl(2N-1-2\xi\bigr)u\cdot\tfrac\pi{2N}\Bigr) \\=& c(u)\cdot\sum_{\xi=0}^{N-1}f(\xi) \cdot\cos\Bigl(\pi\bigl(\tfrac{2N}{2N}-\tfrac{2\xi+1}{2N}\bigr)u\Bigr). \end{aligned}$$ In the cosine, we have now a shift by an integer multiple ($u$) of $\pi$. Such a shift in a sine or cosine function is equivalent to multiplying the function with $(-1)^u$, So the result is $$\begin{aligned} (\operatorname{DCT}\:g)(u) =& c(u)\cdot\sum_{\xi=0}^{N-1}f(\xi) \cdot\cos\bigl(\pi\tfrac{2\xi+1}{2N}u\bigr)\cdot(-1)^u \\=& (-1)^u\cdot(\operatorname{DCT}\:f)(u). \end{aligned}$$ Not quite the expression you wanted to show, but this should be the correct one.
