# Equivalence of two methods to obtain $\sin \theta \geq 1$ with a complex $\theta$

When evaluating the function $$\sin \theta$$ with a complex angle $$\theta$$, a real value $$A \geq 1$$ can be obtained in two ways.

1. Considering $$\theta = i \log \left[ -i \left( A + \sqrt{A^2 - 1} \right) \right]$$, as stated in this page. After the substitution of $$\theta$$,

$$\sin \theta = \frac{e^{i \theta} - e^{-i \theta}}{2i} = A$$

1. Considering $$\theta = \frac{\pi}{2} - iB$$. This way,

$$\sin \theta = \cosh B$$

I want to prove that the two methods are equivalent.

My attempt:

I applied the inverse hyperbolic cosine definition:

$$\cosh B = A \Rightarrow B = \mathrm{arccosh} A = \log \left[ A + \sqrt{A^2 - 1} \right]$$ $$\theta = \frac{\pi}{2} - i\log \left[ A + \sqrt{A^2 - 1} \right] \label{a} \tag{1}$$

In method 1, $$\theta$$ can be rewritten as

$$\theta = i \log \left[ -i \left( A + \sqrt{A^2 - 1} \right) \right] =\\ = i \log \left( \frac{A + \sqrt{A^2 - 1}}{j} \right) = i \left[ \log \left( A + \sqrt{A^2 - 1} \right) - \log i \right] = \\ = i \left[ \log \left( A + \sqrt{A^2 - 1} \right) - i \frac{\pi}{2} \right] =$$ $$=i \log \left( A + \sqrt{A^2 - 1} \right) + \frac{\pi}{2} \label{b} \tag{2}$$

My question:

In my attempt, \ref{b} doesn't match \ref{a}. Why? What is wrong?

This question is related to a Physics question about Snell's law.

Yes, they match. Just note that$$(\forall z\in\mathbb{C}):\sin\left(\frac\pi2-z\right)=\sin\left(\frac\pi2+z\right)=\cos(z).$$