Fractions and Decimals

Discrete Mathematics

Differential Equations

Solved on Jan 17, 2024

Prove the identity $\cosh(-x) = \cosh(x)$ , showing that cosh is an even function.

STEP 1

Assumptions
1. The function $\cosh(x)$ is defined as $\cosh(x) = \frac{1}{2}(e^x + e^{-x})$ .
2. We need to prove that $\cosh(-x) = \cosh(x)$ , which would show that $\cosh(x)$ is an even function.
3. An even function is one for which $f(-x) = f(x)$ for all $x$ in the domain of $f$ .

STEP 2

Start by writing the definition of $\cosh(-x)$ using the definition of the hyperbolic cosine function.
$\cosh(-x) = \frac{1}{2}(e^{-x} + e^{-(-x)})$

STEP 3

Simplify the expression inside the parentheses by recognizing that $e^{-(-x)}$ is equivalent to $e^x$ .
$\cosh(-x) = \frac{1}{2}(e^{-x} + e^x)$

STEP 4

Notice that the expression for $\cosh(-x)$ is now in the same form as the definition of $\cosh(x)$ .
$\cosh(-x) = \frac{1}{2}(e^x + e^{-x})$

STEP 5

By comparing this with the definition of $\cosh(x)$ , we can see that the expressions are identical.
$\cosh(-x) = \cosh(x)$

STEP 6

Conclude that since $\cosh(-x)$ is equal to $\cosh(x)$ , the hyperbolic cosine function is indeed an even function.
The identity $\cosh(-x) = \cosh(x)$ has been proven, which confirms that $\cosh$ is an even function.

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