Cosine Last Year, Sine Next Year

The solution of the problem lies in complex numbers and binomial theorem. Let me show only for $\cos nx$ and I believe you can do for $\sin$ .

By De-Moivre's theorem, we know that $(\cos x + i\sin x)^n = \cos nx + i \sin nx$ , and similarly $(\cos x - i \sin x)^n = \cos nx - i \sin nx$

Hence, $2 \cos nx = (\cos x + i\sin x)^n + (\cos x - i\sin x)^n$

Expand it binomially to get:

$\displaystyle 2 \cos nx = 2 \sum_{2r \leq n} {n \choose 2r} (i \sin x)^{2r} (\cos x)^{n-2r}$

$\displaystyle \Rightarrow \cos nx = \sum_{2r \leq n} (-1)^r {n \choose 2r} (\sin^2 x)^r (\cos x)^{n-2r}$

$\displaystyle \Rightarrow \cos nx = \sum_{2r \leq n} (-1)^r {n \choose 2r} (1 - \cos^2 x)^r (\cos x)^{n-2r}$ .

Now, as we have to find the coefficient of $(\cos x)^n$ , we take only $(-1)^r (\cos x)^{2r}$ from the binomial expansion of $(1- \cos^2 x)^r$ . Note that we can also obtain whole expression, if we consider all terms of its expansion, but that is not needed for this problem.

Hence, we get the coefficient of $(\cos x)^n$ as $\displaystyle \sum_{2r \leq n} (-1)^{r} {n \choose 2r} (-1)^r = \displaystyle \sum_{2r \leq n} {n \choose 2r}$

= $\boxed{2^{n-1}}$

In our case it is $2^{2013-1} = 2^{2012}$ .

By similar method you can expand $\sin nx$ , for odd $n$ . You get the coefficient of $(\sin x)^{2015}$ in $\sin (2015 x)$ as $-2^{2014}$

Hence, $A=1,B=2012,C=-1,D=2014$

Thus, $A+B - C - D = 0$