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?

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


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