Now, the first problem we encounter, is that we have 1 + sin(pi/7) + icos(pi/7), and 1 + sin (pi/7) - icos(pi/7) on the bottom. Evidently, the cos is imaginary, and the sin is real, which is a bit of a problem. But, we can use complementary angles to help us out here.
(1 + cos (pi/2 - pi/7) + isin (p/2 - pi/7))/(1 + cos (pi/2 - pi/7) - isin (pi/2 - pi/7)), at the moment ignoring the 7th power.
Simplifying this, we get:
(1 + cos (5pi/14) + isin (5pi/14)) / (1 + cos (5pi/14) - isin (5pi/14))
Now, the next problem is the "1", as lyounamu correctly pointed out. This took me a little while to get rid of, but it's really very simple. Lets use cos(5pi/14) and isin (5pi/14) as double angles.
We know that cos2x = 2cos^2x - 1
and that cos2x + 1 = 2cos^2x. So we assume that 5pi/14 is the double angle, so we can change that to:
[2cos^2(5pi/28) + 2isin (5pi/28) cos (5pi/28)]/[2cos^2(5pi/28) - 2isin (5pi/28) cos (5pi/28)]
now, we can factor out 2cos (5pi/28) from both numerator and denominator:
{2cos(5pi/28) [cos (5pi/28) + i sin (5pi/28)]} / {2cos (5pi/28) [ cos (5pi/28) - isin (5pi/28)]}
The 2cos (5pi/28) will cancel out, and you're left with:
[cos (5pi/28) + i sin (5pi/28]/[cos (5pi/28) - isin (5pi/28)]
Now, we re-introduce the power of 7, and using de moivres theorem:
[cos (5pi/4) + i sin (5pi/4)] / [cos (5pi/4) - isin (5pi/4)]
Simplifying,
[-1/root2 - (1/root2)i] / [-1/root2 + (1/root2) i]
= [(-1-i)/root2] / [(-1 + i)/root2]
= -(1 + i)/ -(1-i)
= (1 + i) / (1 - i), after dividing top and bottom by -1
= i or/ cis pi/2
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