So you have so far:
2z1−2n(z2−1z4n−1)Using a similar method to how you would have established
cosnθ=0.5(zn+z−n), you can work out that
isinnθ=0.5(zn−z−n). This is useful because
0.5(zn−z−n)=0.5(zn−zn1)=0.5znz2n−1 so
2isinnθ=znz2n−1.
Your answer so far can be re-written as
21((z2−1)/z(z4n−1)/z2n) and you can put this in terms of theta using the above formula.
I'll give you an idea of the thought process that I went through in answering this question. My first instinct was to try and use the formula
cosnθ=0.5(zn+z−n) and perhaps replacing n with some other number as appropriate. I noticed that
cosnθ=0.5(zn+z−n)=0.5(z2n+1)/zn which looked very similar to the terms at the top and bottom of the fraction in the sum of the g.p. but the problem here was the minus sign so I wouldn't have been able to use these workings to make a substitution. So then I thought, if the only difference here is a minus sign then perhaps I might have more luck getting a formula for the g.p. in terms of sin? So I put together a similar formula in terms of sin n theta and it turns out that I can use it to replace the terms involving z.