Any trick for evaluating (\frac{\sqrt3}{2}\cos \theta+\fr

In general, from de Moivre's theorem, we have:
$\cos x=\frac{e^{ix}+e^{-ix}}{2}$
and
$\sin x=\frac{e^{ix}-e^{-ix}}{2}$
Then,
$S=(a\cos x+ib\sin x{)}^n=\frac{((a+b)e^{ix}+(a-b)e^{-ix}{)}^n}{2^n}$
Hence, from the Binomial Theorem,
$S=\frac{\sum _{r=0}^n((n),(r))(a+b{)}^{n-r}(a-b{)}^r{e}^{i(n-2r)x}}{2^n}$
Since $e^{i(n-2r)x}=\cos ((n-2r)x)+i\sin ((n-2r)x)$ we have:
$S=(\frac{\sum _{r=0}^n((n),(r))(a+b{)}^{n-r}(a-b{)}^r\cos ((n-2r)x)}{2^n})+i(\frac{\sum _{r=0}^n((n),(r))(a+b{)}^{n-r}(a-b{)}^r\sin ((n-2r)x)}{2^n})$
Thus it is evident what this particular case shall simplify to.

Long Comment:
For your particular problem there is a small tiny tiny algebraic simplification as
$T=\sin (\frac{\pi }{3})\cos \theta -i\cos \frac{\pi }{3}\sin \theta =\frac{1}{2}\sqrt{3}\cos \theta -\frac{1}{2}i\sin \theta$ (1)
or
$T=\cos (\frac{\pi }{6})\cos \theta -i\sin (\frac{\pi }{6})\sin \theta =\frac{1}{2}\sqrt{3}\cos \theta -\frac{1}{2}i\sin \theta$ (2)
Rewriting (1) as
$S=(\sin \omega \cos \theta -i\cos \omega \sin \theta {)}^n$ (3)
we can get to
$S=(\frac{1}{2}+\frac{i}{2}{)}^n(\sin (\theta +\omega )+i\sin (\theta -\omega ){)}^n$ (4)
and likewise for (2)
$S=(\cos \theta \cos (\frac{\omega }{2})+i\sin \theta \sin (\frac{\omega }{2}){)}^n$
we get
$S=(\frac{1}{2}-\frac{i}{2}{)}^n(\cos (\theta +\frac{\omega }{2})+i\cos (\theta -\frac{\omega }{2}){)}^n$
Using @Buraian 's methods we have the interesting general result for (3)
$S=(\sin \omega \cos \theta -i\cos \omega \sin \theta {)}^n$ (5)
$=(\frac{1}{2}+\frac{i}{2}{)}^n(1-\cos 2\theta \cos 2\omega {)}^{\frac{n}{2}}{e}^{(in{\tan }^{-1}(\sin (\theta +\omega ),\sin (\theta -\omega )))}$
where ${\tan }^{-1}(x,y)={\tan }^{-1}(\frac{y}{x})$ gives the angle of the point [x,y] and where ${\sin }^2(\theta -\omega )+{\sin }^2(\theta +\omega )$ simplifies to $1-\cos (2\theta )\cos (2\omega )$

How about this:
$\frac{\sqrt{3}}{2}\cos \theta =\cos 30\cos \theta =\frac{1}{2}[\cos (30-\theta )+\cos (30+\theta )]=\frac{1}{2}(b+a)$
$\frac{1}{2}\sin \theta =\sin (30)\sin \theta =\frac{1}{2}[\cos (30-\theta )-\cos (30+\theta )]=\frac{1}{2}(b-a)$
We have:
$\frac{1}{2^7}[(a+b)+i(b-a){]}^7$
Call $a+b=u=\sqrt{3}\cos \theta ,b-a=v=\sin \theta$, then for the complicated part:
$[u+iv{]}^7=(u^2+v^2{)}^{\frac{7}{2}}[\frac{u+iv}{(u^2+v^2{)}^{\frac{1}{2}}}{]}^7=(u^2+v^2{)}^{\frac{7}{2}}{e}^{i7({\tan }^{-1}\frac{v}{u})}$
Now back substitute and simplify.