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Einstein says $$\cos\mathrm{i}x=\frac1{\sqrt{1-\left(\mathrm{i}\tan\mathrm{i}x\right)^2}},$$ but WolframAlpha says that this isn't true for $x=\pm2$ and $x=\pm9/5$. What's happening?

From page 34 of The Meaning of Relativity:

$$v=\mathrm{i}\tan\mathrm{i}\theta,\quad\mathrm{i}\sin\mathrm{i}\theta=\dfrac{v}{\sqrt{1-v^2}}\quad\text{and}\quad \cos\mathrm{i}\theta=\dfrac1{\sqrt{1-v^2}}.$$

Otherwise, the functions do appear to be identical.

Ice Tea
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Directions In Physics
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    I wonder why Einstein didn’t like the hyperbolic functions cosh and tanh. All those $i$’s are awful. – G. Smith Feb 18 '21 at 06:27
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    compare your query with https://www.wolframalpha.com/input/?i=Is+cos+%28ix%29sqrt%281-%28itan%28ix%29%29%5E2%29+%3D+1 and https://www.wolframalpha.com/input/?i=cos+%28ix%29sqrt%281-%28itan%28ix%29%29%5E2%29+ – robphy Feb 18 '21 at 04:38
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    On general formatting, note that plus/minus is \pm and it generally looks nicer to use \sqrt{ ... }. – Richard Myers Feb 18 '21 at 04:45
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    Mathematica says that that equality is true, symbolically. At $x=2$ both sides are numerically equal to 3.7622. I would ignore what Wolfram Alpha is saying. – G. Smith Feb 18 '21 at 04:59
  • There is a branch cut for the function on the right hand side in the complex plane due to the square root. I bet that wolfram alpha has some default prescription for handling the branch cut that caused it to make a silly choice when answering your question in the case where it does not assume $x$ is real. On the other hand, the box "assuming $x$ is real" (a) is the special case you are interested in so you should use this, and (b) I bet wolfram alpha makes a different choice for the branch cut so it doesn't hit the real axis in this case. –  Feb 18 '21 at 11:36
  • Please provide the reference where Einstein used this. – robphy Feb 18 '21 at 19:36
  • @robphy https://www.gutenberg.org/files/36276/36276-pdf.pdf page 35 – Directions In Physics Feb 18 '21 at 21:07
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    WolframAlpha is math. Physics is NOT math. I would tend to believe Einstein. – David White Feb 18 '21 at 06:43
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    Without working thru the example I would take it that Wolfram is considering points where the denominator goes to zero. This is mathematically possible but maybe not physically possible. –  Feb 18 '21 at 04:20

1 Answers1

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Einstein is right assuming real x. $$\cos (ix) = \frac{1}{\sqrt{1-(i\tan(ix))^2}} = \frac{1}{\sqrt{1+\tan^2(ix)}} = \frac{1}{\frac{1}{\cos(ix)}}~.$$

As $\cos(ix)\gt 0$ this is an identity for all real x.

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