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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.

Bumblebee
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Mr. Y
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3 Answers3

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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?

Chinny84
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  • Well,not really.What should be the best way to simplify $\left(2 \cos (\pi/20)\right)^6 $ ? Do I transform it in $\cos (\cfrac{\pi/10}{2})$ and then apply the identity for $\cos(x/2)$ ?Btw this approach is pretty neat ,thank you ! – Mr. Y Dec 21 '15 at 10:45
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    You can do whatever is required to simplify it. It is a simpler to reduce from here. Alternatively you could expand extract one term completely (at the point where I obtain the $\cos $ relation) and use standard Binomal expansion. Which you could also have achieved from the original statement. I hope it helps. – Chinny84 Dec 21 '15 at 10:49
  • Note: $\mathrm{e}^{a} = \mathrm{e}^{i\pi/30}$ then we can see that $$\left(\mathrm{e}^a+\mathrm{e}^{4a}\right)^6 = \mathrm{e}^{6a}\left(1+\mathrm{e}^{4a-a}\right)^6$$ hope this helps. – Chinny84 Dec 21 '15 at 11:30
  • ah you are right. – Chinny84 Dec 21 '15 at 11:35
  • Note to self $\frac{2\pi}{30}\neq \frac{\pi}{30}$! – Chinny84 Dec 21 '15 at 11:38
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Let me try.

$$2^6\cos^6 18(\cos 30 + i\sin 30 )^6 = 64\cos^6 18 (\cos 180 + i^6\sin 180) = -64\cos^6 18.$$

GAVD
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HINT: $$\Im \left(\left(\cos 12^\circ +i \sin 12^\circ +\cos 48^\circ+i\sin 48 ^\circ \right)^6\right)$$ $$=\large{\Im\left(\left(e^{i\frac{\pi}{15}}+e^{i\frac{4\pi}{15}}\right)^6\right)}$$

K. Rmth
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