For a length- DFT with even, split the input into its even and odd entries. Two length- DFTs determine the result, followed by at most multiplications by twiddle factors. Hence the multiplication count satisfies
so induction gives for powers of two. This proves the needed fast Fourier transform bound rather than assuming it.
Part (a) computes the length- DFT of the reflected vector and then recovers
Forming uses no multiplications and the final recovery uses only . Therefore the discrete cosine transform costs at most

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