Question

How do you prove $\sinh x + \cosh x = {e^x}$?

Answer 1

Hint: We recall the definition of sine and cosine hyperbolic function as $\sinh x = \dfrac{{{e^x} - {e^{ - x}}}}{2}$ and $\cosh x = \dfrac{{{e^x} + {e^{ - x}}}}{2}$. We begin from the left-hand side of the given statement and substitute the value and simplify them to arrive at the right-hand side.

Complete step-by-step answer:
We know that hyperbolic functions are functions analogous to ordinary trigonometric functions defined for the hyperbola, rather than the circle which is means just $\left( {\cos t,\sin t} \right)$ with parameter $t$ represents a circle with unit radius, the point $\left( {\cosh t,\sinh t} \right)$ represents form the right half of the equilateral parabola.
The basic hyperbolic functions are the sine hyperbolic function $\left( {\sinh x:R \to R} \right)$ and cosine hyperbolic function $\left( {\cosh x:R \to R} \right)$. All the other hyperbolic functions are derived from hyperbolic sine and hyperbolic cosine.
The hyperbolic sine is defined in terms of the exponential function ${e^x}$ as,
$ \Rightarrow \sinh x = \dfrac{{{e^x} - {e^{ - x}}}}{2}$ ….. (1)
The hyperbolic cosine is defined in terms of the exponential function ${e^x}$ as,
$ \Rightarrow \cosh x = \dfrac{{{e^x} + {e^{ - x}}}}{2}$ ….. (2)
We are asked to prove the following statement
$ \Rightarrow \sinh x + \cosh x = {e^x}$
We shall begin simplifying from the left-hand side that is
$ \Rightarrow \sinh x + \cosh x$
Substitute the values from equation (1) and (2),
$ \Rightarrow \dfrac{{{e^x} - {e^{ - x}}}}{2} + \dfrac{{{e^x} + {e^{ - x}}}}{2}$
Since the denominator of both terms is the same. So, add the numerator of both terms,
$ \Rightarrow \dfrac{{{e^x} - {e^{ - x}} + {e^x} + {e^{ - x}}}}{2}$
On simplifying the terms, we get
$ \Rightarrow \dfrac{{2{e^x}}}{2}$
Cancel out common factors from the numerator and denominator,
$\therefore {e^x}$
The above result is the same as the right-hand side of the statement.

Hence it is proved.

Note:
Be careful about the calculation and the signs while opening the brackets. The general mistake that a student can make is \[1 + x - \left( {x - 1} \right) = 1 + x - x - 1\]. Also, you need to remember the values of the hyperbolic functions \[\sinh x\] and $\cosh x$ to be able to solve such problems and make sure that you are not confused between their values, as they differ by only a sign.