1-D discrete cosine transform(DCT)
If $g(x)= f(N-1-x)$ then
$$\begin{aligned}
  (\operatorname{DCT}\:g)(u)
    =&amp; c(u)\cdot\sum_{x=0}^{N-1}g(x)\cdot\cos\bigl((2x+1)u\cdot\tfrac\pi{2N}\bigr)
  \\=&amp; c(u)\cdot\sum_{x=0}^{N-1}f(N-1-x)
        \cdot\cos\bigl((2N-2N+1+2x)u\cdot\tfrac\pi{2N}\bigr)
  \\=&amp; c(u)\cdot\sum_{(N-1-x)=(N-1)}^{0}f(N-1-x)
        \cdot\cos\Bigl(\bigl(2N-1-2(N-1-x)\bigr)u\cdot\tfrac\pi{2N}\Bigr)
\end{aligned}$$
substitute now $N-1-x=:\xi$,
$$\begin{aligned}
  (\operatorname{DCT}\:g)(u)
    =&amp; c(u)\cdot\sum_{\xi=0}^{N-1}f(\xi)
        \cdot\cos\Bigl(\bigl(2N-1-2\xi\bigr)u\cdot\tfrac\pi{2N}\Bigr)
  \\=&amp; c(u)\cdot\sum_{\xi=0}^{N-1}f(\xi)
        \cdot\cos\Bigl(\pi\bigl(\tfrac{2N}{2N}-\tfrac{2\xi+1}{2N}\bigr)u\Bigr).
\end{aligned}$$
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
$$\begin{aligned}
  (\operatorname{DCT}\:g)(u)
    =&amp; c(u)\cdot\sum_{\xi=0}^{N-1}f(\xi)
        \cdot\cos\bigl(\pi\tfrac{2\xi+1}{2N}u\bigr)\cdot(-1)^u
  \\=&amp; (-1)^u\cdot(\operatorname{DCT}\:f)(u).
\end{aligned}$$
Not quite the expression you wanted to show, but this should be the correct one.