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$$\frac{1+\cos 5x+i\sin 5x}{1+\cos 5x-i\sin 5x}=\cos 5x+i\sin 5x$$

When I attempted this I first tried multiplying top and bottom of the LHS by the complex conjugate of what's on the bottom, $1+\cos 5x+i\sin 5x$. After simplification I got:

$$LHS=\frac{1+2\cos 5x+2i\sin 5x+\cos^25x+2i\sin 5x \cos5x-\sin^2x}{2\cos 5x+sin 5x}$$

I cannot see a way of simplifying further to give the RHS, where have I gone wrong?

Also, I know that since $\cos 5x+i\sin 5x=(\cos x+i\sin x)^5$ I could do an expansion but after doing that I could also see no way of getting the LHS. Please help.

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    $\begingroup$ The $5$ looks irrelevant. I think you can safely delete all of them. $\endgroup$ – DanielV Mar 19 '15 at 18:56
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Let $$ s = e^{5ix} $$ Then $$\text{LHS } = \frac{1+s}{1+1/s} = \frac{ s+s^2}{s+1} = \frac{s(s+1)}{s+1} = s = e^{5ix} $$

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You wish to prove $$\frac{1+e^{5ix}}{1+e^{-5ix}}=e^{5ix}$$ You multiplied top and bottom by $1+e^{5ix}$. I think you'll have better luck using $e^{5ix}$ instead.

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Another way: multiply both sides of the equation by the denominator of the LHS. Then on the right hand side use $\cos^2 + \sin^2 = 1$.

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