DFT N-point circular convolution of $\cos(2 \pi n/N)$ and $\sin(2 \pi n/N)$
HINT : $\cos(2\pi n/N) = 0.5e^{j2\pi n/N}+0.5e^{-j2\pi n/N}$. This itself is a IDFT equation, where $X_1[k] = 0.5\delta[k-1] + 0.5\delta[k+1]$. 

Similarly for $\sin(2\pi n/N)$, $X_2[k] = (0.5/j)\delta[k-1]-(0.5/j)\delta[k+1]$. 

So $Y[k] = (0.25/j)\delta[k-1] + (0.25/j)\delta[k+1]$, and $y[n] = (1/2) \times (0.5/j)e^{j2\pi n/N} -(1/2) \times (0.5/j)e^{-j2\pi n/N} = (1/2) \sin(2\pi n/N)$ 

To expand mathematically
$$
X_1[k] = \sum (0.5e^{j2\pi n/N}+0.5e^{-j2\pi n/N})e^{-j2\pi kn/N}$$ $$= \sum (0.5e^{j2\pi (1-k)n/N }) +(0.5e^{j2\pi (-1-k)n/N })$$ $$= 0.5\frac{(e^{j2\pi (1-k)N/N} - 1)}{(e^{j2\pi (1-k)/N} - 1)} + 0.5\frac{(e^{j2\pi (-1-k)N/N} - 1)}{(e^{j2\pi (-1-k)/N} - 1)}$$ $$= 0.5\frac{e^{j\pi(1-k)}\sin(\pi(1-k))}{(e^{j\pi/N(1-k)}\sin(\pi/N(1-k))} + 0.5\frac{e^{j\pi(-1-k)}\sin(\pi(-1-k))}{(e^{j\pi/N(-1-k)}\sin(\pi/N(-1-k))}
$$
You can use similar steps to compute $X_2[k]$. You need NOT proceed to multiply from here. Look at the numerator term of each fraction. $\sin(\pi(m-k))/\sin(\pi/N(m-k))$ is zero when $k \ne m$. For $k=m$, even before you reach the geometric series sum, you can see all terms in summation are 1. So $X[k] = N$ when $k=m$. So $X_1[k] = N\times 0.5\delta[k-1] + N \times 0.5\delta[k+1]$ . Though I am bit confused about the $N$ factor.