# How to prove such a hyperbolic sine cosine related equality? [closed]

$$\ln \left(\frac{\left(1+\sqrt{5}\right)^2 \left(2+\sqrt{5}\right)}{4}\right)=\text{arcsinh }(2)+2 \text{ arccsch }(2)$$

## closed as off-topic by Claude Leibovici, Watson, user91500, Chill2Macht, Daniel W. FarlowJul 23 '16 at 16:39

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$$\ln \left(\frac{\left(1+\sqrt{5}\right)^2 \left(2+\sqrt{5}\right)}{4}\right)=\ln\left(\left(1+\sqrt{5}\right)^2\left(2+\sqrt{5}\right)\right)-\ln (4)$$
$$=\ln\left(1+\sqrt{5}\right)^2+\underbrace{\ln\left(2+\sqrt{5}\right)}_{=\,\text{arcsinh}\, 2}-\ln 4$$
$$=\ln\frac{\left(1+\sqrt{5}\right)^2}{4}+\underbrace{\ln\left(2+\sqrt{5}\right)}_{=\, \text{arcsinh} \,2}$$
$$=\underbrace{2\ln\frac{\left(1+\sqrt{5}\right)}{2}}_{=\,2\,\text{arccsch}\, 2}+\underbrace{\ln\left(2+\sqrt{5}\right)}_{=\,\text{arcsinh} \,2}$$