Chapter 3.11, Problem 1PT

To determine

Whether the statement “$\sinh x=\frac{e^x+e^{-x}}{2}$” is true or false.

Answer to Problem 1PT

The given statement is $\underset{\_}{\text{false}}$.

Explanation of Solution

Check whether the value of $\sinh x=\frac{e^x+e^{-x}}{2}$ is true or not as follows.

It is known that $\sinh \left(-x\right)=-\sinh x$.

Suppose that $\sinh x=\frac{e^x-e^{-x}}{2}$. Then, prove that $\sinh \left(-x\right)=-\sinh x$.

Substitute $x=-x$ in $\sinh x=\frac{e^{x}s+e^{-x}}{2}$.

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

Here, $\sinh \left(-x\right)=\sinh x$. It is a contradiction to $\sinh \left(-x\right)=-\sinh x$.

That is, $\sinh x=\frac{e^x-e^{-x}}{2}$ is not true.

Therefore, the given statement is $\underset{\_}{\text{false}}$.