Question

Prove the identity\(sinh2x = 2sinhxcoshx\).

Short answer

The identity \(\sinh 2x = 2\sinh x\cosh x\) is proved.

Step-by-step solution

Given identity

The identity is \(\sinh 2x = 2\sinh x\cosh x\).

Formula of hyperbolic function and sine cosine functions

Hyperbolic function and sine cosine function:

\(\begin{array}{c}sinhx = \frac{{{e^x} - {e^{ - x}}}}{2}\\\cosh x = \frac{{{e^x} + {e^{ - x}}}}{2}\end{array}\)

Use the formula and substitute the value

Consider, \(\sinh 2x\).

Substitute 2x for \(x\)in\(\sin hx = \frac{{{e^x} - {e^{ - x}}}}{2}\).

\(\sinh 2x = \frac{{{e^{2x}} - {e^{ - 2x}}}}{2}\)

Add and subtract the value \({e^0}\) on numerator.

\(\begin{array}{c}\sinh 2x = \frac{{{e^{2x}} - {e^{ - 2x}} + {e^0} - {e^0}}}{2}\\ = \frac{{{e^{2x}} - {e^{ - 2x}} + {e^{x - x}} - {e^{ - x + x}}}}{2}\\ = \frac{{{e^{x + x}} + {e^{x - x}} - {e^{ - x - x}} - {e^{ - x + x}}}}{2}\\ = \frac{{{e^x}\left( {{e^x} + {e^{ - x}}} \right) - {e^{ - x}}\left( {{e^x} + {e^{ - x}}} \right)}}{2}\end{array}\)

Multiply the numerator and the denominator by \(2\).

\(\begin{array}{c}\sinh 2x = 2\frac{{{e^x}\left( {{e^x} + {e^{ - x}}} \right) - {e^{ - x}}\left( {{e^x} + {e^{ - x}}} \right)}}{4}\\ = 2\frac{{\left( {{e^x} + {e^{ - x}}} \right)\left( {{e^x} - {e^{ - x}}} \right)}}{{2 \cdot 2}}\\ = 2\left( {\frac{{{e^x} - {e^{ - x}}}}{2}} \right)\left( {\frac{{{e^x} + {e^{ - x}}}}{2}} \right)\\ = 2\sinh x\cosh x\quad \end{array}\)

Hence, the required identity \(\sinh 2x = 2\sinh x\cosh x\) is proved.