What are the coefficients of these trigonometric sums? I have two functions that I'm working on.  The first is:
$$
\begin{align}
\cos x &= (\cos 1)^3 \cos(3-x) \\
&{}+ 3 (\cos 1)^2 (\sin 1) \sin(3-x) \\
&{}- 3 (\cos 1) (\sin 1)^2 \cos(3-x) \\
&{}- (\sin 1)^3 \sin(3-x)
\end{align}
$$
...In other words, we can represent cosine as a series involving powers of cosines and sines.  We can call the above function $f_3(x)$, since the maximum power of either sine or cosine is 3.  In general, then, we can represent cosine as:
$$f_k(x) = \cos x = \sum_{i=0}^k c_{(k,i)} (\cos 1)^i (\sin 1)^{k-i}\cdot\left\{
   \begin{array}{lr}
     \cos{(k-x)} & : i \text{ odd}\\
     \sin{(k-x)} & : i \text{ even}
   \end{array}
\right.
$$
I've been trying to derive the coefficients ($c_{(k,i)} $) of this function.  In general, I believe they are binomial coefficients, but I can't seem to figure out the pattern of signs (i.e. $+1$ or $-1$) for the coefficients.  I'm wondering if anyone can get the formula for the coefficients of this function.  I would greatly appreciate it.
Please note:  $k$ is a natural.
 A: You have effectively decomposed $\cos x$ as
$$\cos x = \cos( k - ( k - x ) ) = \cos k \cos(k-x) + \sin k \sin(k-x)$$
while expressing $\cos k$ and $\sin k$ using "multiple angle" expansions in terms of $\sin 1$ and $\cos 1$.
The binomial-like formulas for multiple angle formulas are well-known:
$$\begin{align}
\sin k\theta \quad&=\quad \sum_{j=0}^k {k \choose j} \; \cos^j\theta \; \sin^{k-j}\theta \;\cdot\; \sin\frac{(k-j)\pi}{2} \\
\cos k\theta \quad&=\quad \sum_{j=0}^k {k \choose j} \; \cos^j\theta \; \sin^{k-j}\theta \;\cdot\; \cos\frac{(k-j)\pi}{2}
\end{align}$$
where those final factors ---$\sin\frac{(k-j)\pi}{2}$ and $\cos\frac{(k-j)\pi}{2}$--- are cleverly generating the $\pm 1$s (and term-killing $0$s) that seem to have stumped you.

The best way to understand the binomial-like nature of these formulas is to introduce DeMoivre's Formula from complex analysis:
$$\cos k\theta + i \sin k \theta = \left(\; \cos\theta + i \sin\theta \;\right)^k$$
Expanding the right-hand side of DeMoivre's Formula using the Binomial Theorem, then equating real and imaginary parts of the left- and right-hand sides of the equation, gives the multiple-angle formulas.
