When I was playing around with the formula for $\sin(\theta)$, I found out that $\ln(2i\sin(\theta)) = \ln(e^{2i\theta}-1) - i\theta$. Using common sense, I could derive that this would equal a little under $i\theta$. But I would like to know the exact value, so I was wondering if there was any way to compare $\ln(x)$ to $\ln(x - 1)$. I have tried asking Wolfram Alpha and Microsoft Math Solver, but their output did not help.
Raw Wolfram Alpha input:
ln(e^(2iθ)-1)-iθ
Raw Microsoft Maths Solver input (translated into raw input very weirdly):
\log_{ e }({ { e }^{ 2i \theta } -1 }) -i \theta
(They're both the same equation.)
I found out you can use a Taylor series. The Taylor series relies on the fact that: $$f(x+h)=\sum_{n=0}^\infty \frac{h^n f'^n(x)}{n!}$$ where $f'^n(x)$ is the $n^\text{th}$ derivative of $f(x)$. To solve this, $x$ will be $x$ and $h$ will be $-1$. The Taylor series now becomes: $$\ln([x+-1]=[x-1])=\sum_{n=0}^\infty \frac{(-1)^n\ln'^n(x)}{n!}$$ Writing out the sum gets: $$\ln(x)-\frac{1}{x}-\frac{1}{2x^2}-\frac{1}{3x^3}-\frac{1}{4x^4}...$$ Noticing the pattern makes the sum: $$\ln(x)-\sum_{n=1}^\infty \frac1{nx^n}$$ Therefore: $$\ln(x-1)=\ln(x)+\sum_{n=1}^\infty -\frac{1}{nx^n}$$ Adapting this to the problem I had: $$\ln(e^{2i\theta}-1)=2i\theta+\sum_{n=1}^\infty -\frac{1}{n \left(e^{2i\theta n}\right)}$$ So that is your answer. If anyone else has a better one, they should feel free to post theirs!