Answer
\[\sinh 2x=2\sinh x\cosh x\]
Work Step by Step
To show that:-\[\sinh 2x=2\sin hx\;\cos hx\]
We know that \[\sinh x=\frac{e^x-e^{-x}}{2}\;\;\;...(1)\]
and \[\cosh x=\frac{e^x+e^{-x}}{2}\;\;\;...(2)\]
Using (1)
\[\sinh 2x=\frac{e^{2x}-e^{-2x}}{2}\;\;\;...(3)\]
Consider \[I=2\sin hx\;\cos hx\]
Using (1) and (2)
\[I=2\left(\frac{e^x-e^{-x}}{2}\right)\left(\frac{e^x+e^{-x}}{2}\right)\]
\[I=\frac{(e^x-e^{-x})(e^{x}+e^{-x})}{2}\]
\[\Rightarrow I=\frac{e^{2x}-e^{-2x}}{2}\;\;\;...(4)\]
Using (3) and (4)
\[\sinh 2x=I\]
\[\Rightarrow \sinh 2x=2\sinh x\cosh x\]
Hence, \[\sinh 2x=2\sinh x\cosh x\]
