Question

Prove the identity.

19. \(\sinh 2x = 2\sinh x\cosh x\)

Short answer

It is proved that \(\sinh 2x = 2\sinh x\cosh x\).

Step-by-step solution

Definition of Hyperbolic function

The formulas of the hyperbolic function as shown below:

\(\begin{aligned}\sinh x = \frac{{{e^x} - {e^{ - x}}}}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\mathop{\rm csch}\nolimits} x = \frac{1}{{\sinh x}}\\\cosh x = \frac{{{e^x} + {e^{ - x}}}}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,{\mathop{\rm sech}\nolimits} x = \frac{1}{{\cosh x}}\\\tanh x = \frac{{\sinh x}}{{\cosh x}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\coth x = \frac{{\cosh x}}{{\sinh x}}\end{aligned}\)

Prove the identity \(\sinh 2x = 2\sinh x\cosh x\)

Use the identity\(\sinh \left( {x + y} \right) = \sinh x\cosh y + \cosh x\sinh y\).

Substitute \(y = x\) in the above result to obtain as shown below:

\(\begin{aligned}\sinh 2x &= \sinh \left( {x + x} \right)\\ &= \sinh x\cosh x + \cosh x\sinh x\\ &= 2\sinh x\cosh x\end{aligned}\)

Thus, it is proved that \(\sinh 2x = 2\sinh x\cosh x\).