Question

the hyperbolic cosine function, designated by $\cosh x$, is defined as $\cosh x=\frac{1}{2}(e^{x}+e^{-x})$ (a) show that $f(x)=\cosh x$ is an even function. (b) graph $f(x)=\cosh x$ using a graphing utility. (c) the hyperbolic sine function, designated by $\sinh x$, is defined as $\sinh x=\frac{1}{2}(e^{x}-e^{-x})$. show that, for every $x$, $(\cosh x)^{2}-(\sinh x)^{2}=1$

Explanation:

Step1: Substitute \(-x\) into \(\cosh x\)

Given \(\cosh x=\frac{1}{2}(e^{x}+e^{-x})\), substitute \(x\) with \(-x\):
\(\cosh(-x)=\frac{1}{2}(e^{-x}+e^{-(-x)})\)

Step2: Simplify the expression

Simplify \(e^{-(-x)}\) to \(e^{x}\), so \(\cosh(-x)=\frac{1}{2}(e^{-x}+e^{x})\)
Since addition is commutative, \(\frac{1}{2}(e^{-x}+e^{x})=\frac{1}{2}(e^{x}+e^{-x})\)
And \(\frac{1}{2}(e^{x}+e^{-x})=\cosh x\)

Answer:

\(\frac{1}{2}(e^{-x}+e^{x})\)