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I met this problem in Ratcliffe's Foundations of Hyperbolic Manifolds. Please help me prove this.

In an equilateral triangle with side length $a$ and angle $\alpha$, $$\cosh \left(\frac{1}{2} a\right) \sin \left(\frac {1}{2} \alpha\right) = \frac{1}{2}.$$

Thank you.

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You can use the hyperbolic law of sines, applied to half your equilateral triangle:

\begin{align*} \frac{\sin\frac\pi2}{\sinh a} &= \frac{\sin\frac\alpha2}{\sinh\frac a2} \\ \sinh\frac a2 &= \sin\frac\alpha2\sinh a \\ \sinh\frac a2 &= \sin\frac\alpha22\sinh\frac a2\cosh\frac a2 \\ \frac12 &= \sin\frac\alpha2\cosh\frac a2 \\ \end{align*}

This uses the identity $\sinh(2x)=2\sinh(x)\cosh(x)$ from the Wikipedia article on hyperbolic functions.

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