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

Question

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

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.

Answer 1

Let $$ s = e^{5ix} $$ Then $$\text{LHS } = \frac{1+s}{1+1/s} = \frac{ s+s^2}{s+1} = \frac{s(s+1)}{s+1} = s = e^{5ix} $$

Answer 2

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.

Answer 3

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

Answer 4

$$ \begin{aligned} \frac{1+\cos 5 x+i \sin 5 x}{1+\cos 5 x-i \sin 5 x} = & \frac{2 \cos ^2 \frac{5 x}{2}+2 i \sin \frac{5 x}{2} \cos \frac{5 x}{2}}{2 \cos ^2 \frac{5 x}{2}-2 i \sin \frac{5 x}{2} \cos \frac{5 x}{2}} \\ = & \frac{\cos \frac{5 x}{2}+i \sin \frac{5 x}{2}}{\cos \frac{5 x}{2}-i \sin \frac{5 x}{2}} \\ = & \left(\cos \frac{5 x}{2}+i \sin \frac{5 x}{2}\right)^2 \\ = & \cos ^2 \frac{5 x}{2}-\sin \frac{5 x}{2}+2 i \cos \frac{5 x}{2} \sin \frac{5 x}{2} \\ = & \cos 5 x+i \sin 5 x . \end{aligned} $$