I am asked to find the Fourier Series of $\sin(x-\frac{\pi}{6})$
The question tells me that integration is not necessary and to use Trigonemetric Identities but i can't seem to figure out what trigonometric identity will help.
I have tried comparing this function $\sin(x-\frac{\pi}{6})$ to $\sin(x)$ since the only difference is their period but i'm not quite sure if that's the right approach.
Secondly, is $\sin(x)$ and $\cos(x)$ odd and even respectively regardless of what's inside the paranthesis(as long as there's a variable $x$) ?
Recall the Fourier series of a periodic function $f(x)$ is
$$ f(x) = \frac{1}{2}a_0 + \sum_{n=1}^{+\infty} \left(a_n \cos \frac{2\pi n x}{L} + b_n \sin \frac{2\pi n x}{L}\right) \tag{1} $$
where $L$ is the period of $f$. Now, use the fact that
$$ \sin(x - \pi/6) = \sin x \cos\frac{\pi}{6} - \cos x \sin \frac{\pi}{6} = \frac{\sqrt{3}}{2}\sin x - \frac{1}{2}\cos x \tag{2} $$
If you compare (2) against (1) you will see that the series of $f(x) = \sin(x - \pi/6)$ is
\begin{eqnarray} a_0 &=& 0 \\ a_1 &=& -1/2 \\ a_k &=& 0 ~~~\mbox{for}~~~ k=2,3,\cdots\\ b_1 &=& \sqrt{3}/2 \\ b_k &=& 0 ~~~\mbox{for}~~~ k=2,3,\cdots \tag{3} \end{eqnarray}
and $L = 2\pi$