# How to find the Fourier components

I've got a sinusodial signal:

$$\Delta I=\cos(\Delta \omega t + \varphi)).$$

and I would like to rewrite it as the sum of a cosine and a sine signal(without a phaseterm):

$$\Delta I = \mathcal{M}* \sin(\Delta \omega t) + \mathcal{N}* \cos(\Delta \omega t)$$

My book ("Quantum Optics" by J.C. Garrison p. 293) does it, but I can't see how they got there.

If I perform a sin/cos transform I get the period of the signal as a factor: \begin{align} M(t) &= \mathcal{S}(\Delta I(\Delta\omega)) \\ &= 4E_L E_P \int_{-T/2+t}^{T/2+t}cos(\Delta \omega \tau + \varphi))\sin(\Delta\omega)d\tau\\ &= -4E_L E_P T \sin(\varphi) \\ \end{align}

I'm fairly certain, that I have to integrate over $$\cos(\Delta \omega \tau + \varphi))\sin(\Delta\omega) = \frac{1}{2}\left[ \sin(\varphi) + \sin(2\Delta\omega + \varphi) \right]$$ and $\cos(\Delta \omega \tau + \varphi))\cos(\Delta\omega)$ in order to remove the rapidly oscillating term $\sin(2\Delta\omega t)$

Why don't you just use the relation $\cos(x+y)=\cos x \cos y - \sin x \sin y$? So
$$\cos(\Delta \omega t + \varphi))=\cos(\Delta\omega t)\cos\varphi-\sin(\Delta\omega t)\sin\varphi$$