Using Euler's Formula, e^ix = cos(x) + i*sin(x), or otherwise, derive the Fourier transform of the function:
g(x) = cos(ax)f(x)
in terms of a, x and F(w), the Fourier transform of f(x).
Simply stuck on this question, where do you simplify down. My initial idea was substituting in (e^axi + e^-axi)/2 for cos(ax) as that formula was shown in one of our lectures but I just get stuck with:
F[g]= 1/(2 * ROOT(2 * pi)) * ∫ f(x) * e^((a-w)ix) dx + ∫ f(x) * e^(-(a+w)ix) dx
and don't know what to do from here, I'm very new to Fourier transforms and Euler's formula and the only examples given to us are like for F[Af(x) + BG(x)] not when they're multiplied together so I'm struggling with this. Am I suppose to integrate or substitute back to trig or am I completely off?
The basic Fourier Transform example they gave us is in the lecture is:
F[f] (w) = 1/(ROOT(2 * pi)) * ∫ f(x) * e^(-iwx) dx
You were on the right track and very close. You just missed that you can treat $\omega\pm a$ as a constant during the integration.
Here's the follow-through of what you started, using intermediate substitutions for $\omega\pm a$ to hopefully make things more obvious:
$$\begin{align*}\mathcal{F}\left\{g(x)\right\} &= \mathcal{F}\left\{\cos(ax)f(x)\right\}\\ \\ &=\dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \cos(ax)f(x)e^{-i\omega x}dx\\ \\ &=\dfrac{1}{2}\left[\dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-iax}f(x)e^{-i\omega x}dx+\dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{iax}f(x)e^{-i\omega x}dx\right]\\ \\ &=\dfrac{1}{2}\left[\dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(x)e^{-i(\omega+a) x}dx+\dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(x)e^{-i(\omega-a)x}dx\right]\\ \\ &=\dfrac{1}{2}\left[\dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(x)e^{-iu x}dx+\dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(x)e^{-ivx}dx\right]\\ \\ &=\dfrac{1}{2}\left[F(u)+F(v)\right]\\ \\ &=\dfrac{1}{2}\left[F(\omega+a)+F(\omega-a)\right]\\ \end{align*}$$