TITLE: Can a "difference formula" be a "sum formula" too?
QUESTION [5 upvotes]: Consider the following "product-to-difference" formula:
$$\sin p\sin q=\frac{1}{2}[\cos(p-q)-\cos(p+q)].$$
I wonder whether the right-hand side can be expressed as a weighted sum of cosine functions with positive coefficients, i.e., whether a "difference" formula can be a sum formula.
For concreteness I formulate the following question: Given $a_0>0$, do there exist $a_i\ge 0$ and $b_i$ such that 
$$\sin (a_0x)\sin x=\sum_{i=1}^{\infty}a_i\cos b_i x?$$
With a restriction on $a_0$ and $b_i$, namely that $a_0, b_i\in \mathbb{N}$, the uniqueness of Fourier series implies that the problem has no solutions. Even more simply, we can evaluate both sides at $x=0$ to see that the equality cannot hold. To fix this, I ask the following modified question:
Given $a_0>0$, do there exist $a_i\ge 0$ and $b_i$ and $\mathrm{err}_{a_0}=\mathrm{err}$ such that 
$$\sin (a_0x)\sin x=\sum_{i=1}^{\infty}a_i\cos b_i x +\mathrm{err}(x)$$
where the function err is an arbitrary sine series, namely,
$$\mathrm{err}(x)=\sum_{j=1}^{\infty}c_i\sin d_ix$$
where $c_i$ and $d_i$ are arbitrary real numbers.
Does the problem have a solution in general, without any restrictions?

REPLY [0 votes]: Note that ${\rm err}$ is odd, being a series of sines. Assume that there exist coefficients $c_i, \space d_j$ as desired. The formula ${\rm err}(x) = \sin (a_0 x) \sin x - \sum \limits _{i=1} ^\infty a_i \cos (b_i x)$ show that ${\rm err}$ must also be even, since its right-hand side is so. Being odd and even, ${\rm err}$ must be identically $0$, which is what you were trying to avoid.
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