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

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