Here's what I can remember from 25 years ago:
I don't know how much you know so if anything below doesn't make sense then please write back.
A very effective way to prove most trig identities like the one above is to use complex numbers. To do this we use the formula
eix = cos(x) + isin(x)
So cos(x) = (1/2)(eix + e-ix)
and sin(x) = (1/2i)(eix - e-ix)
So if I set x = ip/11, then the question above is to show that:
(x4 - x-4)
------- + 2 (x1 - x-1) = i sqrt(11)
(x4 + x-4)
Now if we multiply both sides by (x4 + x-4) and then square each side and simplify then we get (after a little work which I'm certainly not typing out!):
4(x10 + x9 + x8 - x7 + x6 + 2x2 + 1 + x-10 + x-9 + x-8 - x-7 + x-6 + 2x-2) = 0
Now x11 = -1 (because x11 = eip = -1)
So -1 = x11 = x9+2 = x9 × x2 => x9 + x2 = 0
And similarly -x7 = x-4
So the long equation above becomes:
(x10 + x8 + x6 + x4 + x2 + x-10 + x-8 + x-6 + x-4 + x-2 + 1) = 0
So if we can prove this is true then the result you want will be true.
Now, x11 = -1 means that x22 = 1
So 0 = x22 - 1 = (x-1)(1 + x + x2 + x3 + ... + x21)
= (x-1)(1+x)(1 + x2 + x4 + x6 + ... + x20)
Now, clearly x isn't 1 or -1, so the above means that:
1 + x2 + x4 + ... + x20 = 0
Dividing by x10 (allowed as x isn't 0) gives:
x10 + x8 + x6 + x4 + x2 + x-10 + x-8 + x-6 + x-4 + x-2 + 1 = 0
Which is exactly what we wanted.
So tan(4p/11) + 4sin(p/11) = sqrt(11). QED
Whew!
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