Question

Prove the identity \(coshx + sinhx = {e^x}\).

Short answer

The identity \(\cosh x + \sinh x = {e^x}\)is proved.

Step-by-step solution

 Given

The identity is \(\cosh x + \sinh x = {e^x}\).

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

Since the formula of the hyperbolic sine and cosine functions are, \(\sin hx = \frac{{{e^x} - {e^{ - x}}}}{2}{\rm{ }}\)and \(\cosh x = \frac{{{e^x} + {e^{ - x}}}}{2}\).

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

Hence, the required identity \(\cosh x + \sinh x = {e^x}\) is proved.