# Fourier transformation of sin, cos, sinh and cosh

I am trying to solve the following exercise

Use $\mathcal{F}(e^{xb}) = 2\pi \delta_{ib}$ to calculate the Fourier-Transformation of $\sin x$, $\cos x$, $\sinh x$ and $\cosh x$

Now I am a little bit confused, because the fourier transformation of $\sin x$ is simply $\sin x$, of $\cos x$ is $\cos x$, of $\sinh x$ it is $-i \sin(ix)$ and of $\cosh(x)$ it is $\cos(ix)$ simply by definition, so why should I use the relation $\mathcal{F}(e^{xb}) = 2\pi \delta_{ib}$?

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"The Fourier transform of $\sin(x)$ is $\sin(x)$"? What definition are you using? – Cocopuffs Jul 9 '12 at 13:01
that it is an infinity sum of sin- and cos terms – Stefan Jul 9 '12 at 13:15
You're thinking of the Fourier series. – Cocopuffs Jul 9 '12 at 13:19
ah, ok you mean the integral with $e^{-i\xi x}$ ok, then i think i know how to calculate it! – Stefan Jul 9 '12 at 14:22
@Stefan is an infinity sum of polynomials still a polynomial? – chaohuang Jul 9 '12 at 14:23

Because there are formulae: $$\sin x=\frac{e^{ix}-e^{-ix}}{2i}\quad\cos x=\frac{e^{ix}+e^{-ix}}2$$ $$\sinh x=\frac{e^x-e^{-x}}2\quad\cosh x=\frac{e^x+e^{-x}}2$$ Use them and the Fourier transform of $e^{bx}$