Let $H_1$ be the branch (connected component) of the unit hyperbola $x^2-y^2=1$ with $x>0$ .

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Let $\alpha>0$ . Then $(\cosh\alpha,\sinh\alpha)$ is the point on $H_1$ with hyperbolic angle $\alpha$ .

In order to draw the hyperbolic angle, the line passing through $(0,0)$ and $(\cosh\alpha,\sinh\alpha)$ must be drawn. Recall that $\tanh\alpha$ is defined by$$ \tanh\alpha :=\frac{\sinh\alpha}{\cosh\alpha}.$$ Thus, the equation of the line passing through $(0,0)$ and $(\cosh\alpha,\sinh\alpha)$ is $y=(\tanh\alpha)x$ .

Below is the graph of $H_1$ and the line $y=(\tanh\alpha)x$ .

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Observe also that $\sinh\alpha>0$ and $\cosh\alpha>1$ .

In the hyperbolic angle entry, it is discussed that $\alpha$ is twice the area bounded by the $x$ axis, $H_1$ , and the line $y=(\tanh\alpha)x$ . We will use this fact to obtain formulas for $\cosh\alpha$ and $\sinh\alpha$ .

In the calculations below, the following integration formula will be used: $$ \int\sqrt{x^2-1} \, dx=\frac{x}{2}\sqrt{x^2-1}-\frac{1}{2}\ln\left|x+\sqrt{x^2-1}\right|+C$$

Thus, we have