Simplify leaving answers in exact values$(\frac{1-cos(\frac{pi}{7})+isin\frac{pi}{7}}{1-cos\frac{pi}{7}-isin\frac{pi}{7}})^7$

I do not understand the solution in Terry Lee so thank you in advance.

2. ## Re: Complex help please

I'm assuming you're having difficulty with the first part of his solution. What he does is use double angles to get rid of the 1 at the front of the numerator and the denominator. I'll do the angles in the numerator as an example and then you should be able to do the rest:

<a href="http://www.codecogs.com/eqnedit.php?latex=cos2\theta =1-2sin^2\theta \\ \therefore cos(\frac{\pi}{7})=1- 2sin^2(\frac{\pi}{14}) \\ sin2\theta =2cos\theta sin\theta \\ \therefore isin(\frac{\pi}{7})=2icos(\frac{\pi}{14})sin(\frac {\pi}{14})" target="_blank"><img src="http://latex.codecogs.com/gif.latex?cos2\theta =1-2sin^2\theta \\ \therefore cos(\frac{\pi}{7})=1- 2sin^2(\frac{\pi}{14}) \\ sin2\theta =2cos\theta sin\theta \\ \therefore isin(\frac{\pi}{7})=2icos(\frac{\pi}{14})sin(\frac {\pi}{14})" title="cos2\theta =1-2sin^2\theta \\ \therefore cos(\frac{\pi}{7})=1- 2sin^2(\frac{\pi}{14}) \\ sin2\theta =2cos\theta sin\theta \\ \therefore isin(\frac{\pi}{7})=2icos(\frac{\pi}{14})sin(\frac {\pi}{14})" /></a>

When you sub those values in, the 1 will neatly disappear and the same thing will happen with the numerator. Good luck

3. ## Re: Complex help please

Originally Posted by deswa1
I'm assuming you're having difficulty with the first part of his solution. What he does is use double angles to get rid of the 1 at the front of the numerator and the denominator. I'll do the angles in the numerator as an example and then you should be able to do the rest:

$cos2\theta =1-2sin^2\theta \\ \therefore cos(\frac{\pi}{7})=1- 2sin^2(\frac{\pi}{14}) \\ sin2\theta =2cos\theta sin\theta \\ \therefore isin(\frac{\pi}{7})=2icos(\frac{\pi}{14})sin(\frac{\pi}{14})$

When you sub those values in, the 1 will neatly disappear and the same thing will happen with the numerator. Good luck
Thank you <3 This was exactly what I needed.

4. ## Re: Complex help please

I copy pasted the question and put it into google with the smallest hope that i will find the answer, and by some small miracle we were both looking for the solution to the exact same question, so thank you

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