Is $\tanh^{-1}\left(\sin 2 \left(x +\dfrac {\pi}{4}\right) \right) = \dfrac {1}{\pi} \ln \left( \cot ^2 x \right) $?

Question

https://www.desmos.com/calculator/av124c6vix

As you see the above graph that $artanh\left(\sin 2 \left(x +\frac {\pi}{4}\right) \right)$ is similar to $ \frac {1}{\pi} \ln \left( \cot ^2 x \right) $.

Does this mean that by some minor alterations in the functions, they would be equal?

Why are both of functions so similar?

Answer 1

We have \begin{align} \operatorname{artanh}\left(\sin 2 \left(x +\frac {\pi}{4}\right) \right)=\operatorname{artanh}(\cos2x)&=\frac{1}{2}\ln\left(\frac{1+\cos2x}{1-\cos2x}\right)\\[4px] &=\frac{1}{2}\ln\left(\frac{1+\cos^2x-\sin^2x}{1-\cos^2x+\sin^2x}\right)\\[4px] &=\frac{1}{2}\ln\left(\frac{2\cos^2x}{2\sin^2x}\right)\\[4px] &=\frac{1}{2}\ln(\cot^2x) \end{align}

Thus, the "error" is only in the $\pi$ factor