My book defines the Fourier transform of a function $f(t)$ as:
$$ F(w)= \frac{1}{\sqrt{2\pi}} \int\limits_{-\infty}^{\infty}f(t)e^{-iwt}\ dt$$ I am stuck on this question:
Show that the Fourier transform of $g(t)= f(t)cos(bt)$ is related to the Fourier transform of $f(t)$ through:
$G(w) = \frac{1}{2}(F(w + b) + F(w-b))$
So I tried to work it out in the following way using the translation properties of the Fourier transform:
$$F(w+b) = e^{ibw}F(w)\\ F(w-b) = e^{-ibw}F(w)\\ \therefore F(w+b) + F(w-b) = 2cos(bw)F(w)$$
From this relation I then should obtain:
$$G(w) = \frac{2cos(bw)F(w)}{2} = cos(bw)F(w) = \frac{cos(bw)}{\sqrt{2\pi}} \int\limits_{-\infty}^{\infty}f(t)e^{-iwt}\ dt$$
This however doesn't seem to be the Fourier transform of $g(t)$:
$$G(w) = \frac{1}{\sqrt{2\pi}} \int\limits_{-\infty}^{\infty}f(t)cos(bt)e^{-iwt}\ dt\ne \frac{cos(bw)}{\sqrt{2\pi}} \int\limits_{-\infty}^{\infty}f(t)e^{-iwt}\ dt$$
I am struggling to understand where I went wrong.
I believe where you went wrong is the first step - what you wrote as the translation properties are not actually what the true translation properties are.
Hint:$$\frac1{\sqrt{2\pi}}\int_\Bbb R f(t)\cos(bt)e^{-iwt}\,\mathrm dt =\frac1{2\sqrt{2\pi}}\int_\Bbb R f(t)e^{-iwt}\left(e^{ibt}+e^{-ibt}\right)\,\mathrm dt\\ =\frac12\left\{\frac1{\sqrt{2\pi}}\int_\Bbb R f(t)e^{-i(w-b)t}\,\mathrm dt+\frac1{\sqrt{2\pi}}\int_\Bbb R f(t)e^{-i(w+b)t}\,\mathrm dt\right\}\\$$
It is always best to go back to the definition of the Fourier transform in terms of an integral when you are stuck, since you can't go wrong. All the shift properties follow from this definition anyway, so if you make a false step, it is clear if you use this method.