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

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?

• I think it is $$arctanh(\cos(2x))$$ – Dr. Sonnhard Graubner Feb 6 at 20:50
• Could you explain the reason? – Rithik Kapoor Feb 6 at 20:52
• For $|x|<1$, $arctanh(x)=\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$. – DiegoMath Feb 6 at 20:53
• it is $$\sin(2(x+\frac{\pi}{4}))=\sin(2x+\frac{\pi}{2})=\cos(2x)$$ – Dr. Sonnhard Graubner Feb 6 at 20:54

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