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 $\sinh x={\displaystyle \frac{e^x-e^{-x}}{2}}$ and $\cosh x={\displaystyle \frac{e^x+e^{-x}}{2}}$. 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,
$\Rightarrow \sinh x={\displaystyle \frac{e^x-e^{-x}}{2}}$ ….. (1)
The hyperbolic cosine is defined in terms of the exponential function $e^x$ as,
$\Rightarrow \cosh x={\displaystyle \frac{e^x+e^{-x}}{2}}$ ….. (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),
$\Rightarrow {\displaystyle \frac{e^x-e^{-x}}{2}}+{\displaystyle \frac{e^x+e^{-x}}{2}}$
Since the denominator of both terms is the same. So, add the numerator of both terms,
$\Rightarrow {\displaystyle \frac{e^x-e^{-x}+e^x+e^{-x}}{2}}$
On simplifying the terms, we get
$\Rightarrow {\displaystyle \frac{2e^x}{2}}$
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.