Find $\Im((\cos 12^\circ +i \sin 12^\circ +\cos 48^\circ+i\sin 48
^\circ )^6)$.
I've solved this problem but I think I've taken the long way to do this, so I am asking if there's some slick way to solve this.
That's how I solved it:
I've applied the identity $\cos 48 +\cos 12=2\cos\left(\cfrac{48+12}{2}\right)\cdot \cos\left(\cfrac{48-12}{2}\right)=2 \cos 30 \cdot \cos18 =\sqrt{3}\cdot \cos 18 $
and the same for $\sin (12)+\sin(48)=2\sin (30)\cdot \cos(18)=\cos(18) $
Thus I have $\Im((\cos 12^\circ +i \sin 12^\circ +\cos 48^\circ+i\sin 48 ^\circ )^6)=\Im((\sqrt{3} \cos 18 +i \cos 18)^6)$
In the end I've applied the Binomial Theorem
(if it's necessary, I will edit including this step ), however this was really a painfull process.
So is there some slick way to solve this ?
My thought:
I think there must be some way to turn the above expression into something like $(\cos\theta +i \sin \theta )^6$, that would be pretty neat.
Best Answer
You were close. The argument of the expression is equivalent to $$ \left(\mathrm{e}^{i\frac{\pi}{15}}+\mathrm{e}^{i\frac{4\pi}{15}}\right)^6 = \mathrm{e}^{i\frac{2\pi}{5}}\left(1+\mathrm{e}^{i\frac{\pi}{5}}\right)^6 $$ then you can extract $\mathrm{e}^{i\pi/10}$ to find $$ \mathrm{e}^{i\frac{2\pi}{5}+i\frac{6\pi}{10}}\left(\mathrm{e}^{-i\frac{\pi}{10}}+\mathrm{e}^{i\frac{\pi}{10}}\right)^6 $$ then knowing that $$ \cos (x) =\frac{\mathrm{e}^{ix}+\mathrm{e}^{-ix}}{2}\implies 2\cos (x) = \mathrm{e}^{ix}+\mathrm{e}^{-ix} $$ I can re-write as $$ \mathrm{e}^{i\frac{2\pi}{5}+i\frac{3\pi}{5}}\left(2\cos (\frac{\pi}{10}) \right)^6 $$ you should be good to go right?