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Note. The answer that inspired this question had a sign error. That error has been corrected; I'm making the corresponding correction to the formula shown here. (Existing answers acknowledge the error.) --Blue

How can I prove that $$ \frac{\tanh a \tanh b \sinh^2 c\sinh a \sinh b \tanh^2c}{\sinh^2 c \tanh^2 c} = \frac{\sinh a \sinh b}{1+\cosh a \cosh b} $$ where $\cosh c = \cosh a \cosh b$?

I know $\tanh a = \frac{\sinh a}{\cosh a}$ and $\tanh b = \frac{\sinh b}{\cosh b}$, but where do I go from there?


This relation appears in an answer to this question about the area $K$ of a hyperbolic right triangle with legs $a$, $b$ and hypotenuse $c$. The expression on the right is a target formula for $\sin K$; the expression on the left is an intermediate step at which the answer stops.

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    $\begingroup$ Consider the case when $a=b=c\neq 0$. The left-hand side becomes zero, but the right-hand side does not. The expressions are not equal in general. Is there an unspoken relationship between $a$, $b$, $c$? $\endgroup$ – Blue Apr 25 at 1:27
  • $\begingroup$ @Blue (math.stackexchange.com/questions/1259520/…) I'm going based off of this question asked. $\endgroup$ – mcobe Apr 25 at 1:35
  • $\begingroup$ Ah, so there is a relation. Since $a$, $b$, $c$ are sides of a right triangle ($c$ being the hypotenuse), we have that $\cosh c = \cosh a \cosh b$. (That's the hyperbolic counterpart of the Pythagorean Theorem. It's mentioned in the linked question.) Under that condition, the identity holds. Do you still need help demonstrating that? (The only other thing you need to know is that $\cosh^2x-\sinh^2x=1$. With that, the task is just a bit of straightforward algebraic manipulation.) $\endgroup$ – Blue Apr 25 at 1:51
  • $\begingroup$ @Blue yes please. I'm trying to get the left hand equation in terms of sides "a" and "b." $\endgroup$ – mcobe Apr 25 at 1:55
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It's a bit of a symbolic slog, but let's take things step-by-step.


Note. I've edited this answer to match the corrected version of the question.


$\require{cancel}$ Expressing the tangents in terms of sine and cosine, the denominator on the left becomes $$\sinh^2c\;\frac{\sinh^2c}{\cosh^2c} = \frac{\sinh^4c}{\cosh^2c} \tag{1}$$ Dividing by that fraction is the same as multiplying by its reciprocal, so we have $$\frac{\cosh^2c}{\sinh^4c}\left(\frac{\sinh a}{\cosh a}\frac{\sinh b}{\cosh b}\;\sinh^2 c-\sinh a \sinh b \;\frac{\sinh^2c}{\cosh^2 c}\right) \tag{2}$$ Factoring-out $\sinh a\sinh b\sinh^2 c$ (and canceling that last factor appropriately), we have $$\frac{\sinh a \sinh b \cosh^2 c}{\sinh^2 c}\left(\frac{1}{\cosh a\cosh b}-\frac{1}{\cosh^2c}\right) \tag{3}$$ Using the hyperbolic Pythagorean relation $\cosh c = \cosh a \cosh b$, this is $$\begin{align} \frac{\sinh a \sinh b \cosh^2 c}{\sinh^2 c}\left(\frac{1}{\cosh c}-\frac{1}{\cosh^2c}\right) &=\frac{\sinh a\sinh b\cancel{\cosh^2 c}}{\sinh^2 c}\;\frac{\cosh c-1}{\cancel{\cosh^2 c}} \tag{4} \\[6pt] &=\frac{\sinh a\sinh b}{\sinh^2 c}\;(\cosh c-1) \tag{5} \end{align}$$ Now, recall that $\cosh^2 x-\sinh^2x=1$, we can rewrite $\sinh^2 c$: $$\frac{\sinh a\sinh b (\cosh c-1)}{\cosh^2c-1}=\frac{\sinh a\sinh b \cancel{(\cosh c-1)}}{\cancel{(\cosh c-1)}(\cosh c+1)}=\frac{\sinh a\sinh b}{\cosh c+1} \tag{6}$$ Finally, the Pythagorean relation allows us to rewrite the denominator $$\frac{\sinh a \sinh b}{1+\cosh a \cosh b} \tag{$\star$}$$

$\square$

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  • $\begingroup$ just to clarify, if i start with $$\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$$ , it still equates to your end solution (star) but with the sign flipped? sinh(a)sinh(b) / 1 +cosh(a)cosh(b) ? $\endgroup$ – mcobe Apr 25 at 3:49
  • $\begingroup$ Yes. The steps are just as I've outlined them, but the terms are reversed: "this minus that" should be "that minus this". (I'm going to make that edit, but got distracted. :) So, at step (5), there'll be a factor of $\cosh c-1$ to cancel instead of $1-\cosh c$, so that there'll be no stray negative sign. $\endgroup$ – Blue Apr 25 at 3:53
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    $\begingroup$ thank you so much $\endgroup$ – mcobe Apr 25 at 3:54
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The original answer you link to has a sign error. The step $$ \sin [\frac 12 \pi -(\alpha + \beta )] = \sin\alpha\sin\beta-\cos\alpha\cos\beta $$ is almost correct, except the two sides are additive inverses. You can check this easily. If $\ \alpha = \beta = 0,\ $ then the left side is $\ \sin(\pi/2) = 1\ $ and the right side is $\ 0 - 1 = -1.\ $ The answer is now corrected.

I have an algebraic method that I have used with good results. Let $$ \sinh(a) = (A - 1/A)/2,\ \cosh(a) = (A + 1/A)/2\ \textrm{ where } A := e^a.$$ Similarly with $\ b,c.\ $ The equation $\ \cosh(c) = \cosh(a) \cosh(b) \ $ when written as a difference is $$ D := \cosh(c) - \cosh(a) \cosh(b) = 2 A B(1 + C^2) - C\ (1+A^2)(1+B^2). $$ Now the (sign corrected) equation to prove is $$ \frac{\tanh a \tanh b \sinh^2 c - \sinh a \sinh b \tanh^2c}{\sinh^2 c \tanh^2 c} = \frac{\sinh a \sinh b}{1+\cosh a \cosh b}. $$ When the right side is subtracted from the left side, the resulting expression is a rational function with $\ D\ $ as a factor in the numerator. Thus the two sides are equal. For this approach it helps to have a computer algebra system that can factor rational functions.

For the curious, the difference is expressed as

$$ \frac{(1-A^2)(1-B^2)((2AB(1+C^2))^2 - (C(1+A^2)(1+B^2))^2)} {AB(1+A^2)(1+B^2)(1-C^2)^2(4AB+(1+A^2)(1+B^2))} $$ and $\ D\ $ divides the third difference of two squares factor in the numerator.

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