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

Any trick for evaluating ${\left(\frac{\sqrt{3}}{2}\mathrm{cos}\theta +\frac{i}{2}\mathrm{sin}\left(\theta \right)\right)}^{7}$?
Expressions of the form ${\left(a\mathrm{cos}\left(\theta \right)+bi\mathrm{sin}\left(\theta \right)\right)}^{n}$ come up from time to time in applications of complex analysis, but to my knowledge the De Moivre's formula can only be applied with a=b. Is there some trick to deal with the case of $a\ne b$, for example when the expression is ${\left(\frac{\sqrt{3}}{2}\mathrm{cos}\theta +\frac{i}{2}\mathrm{sin}\theta \right)}^{7}$?
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nick1337

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

user_27qwe

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

karton

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