Problem 8 - Essential Calculus Early Transcendentals

Question

Prove the identity. \(\cosh (-x)=\cosh x\) (This shows that cosh is an even function.)

Step-by-Step Solution

Answer

\(\cosh(-x) = \cosh(x)\), confirming \(\cosh\) is an even function.

Step 1: Recall the definition of cosh

The hyperbolic cosine function, \(\cosh(x)\), is defined as \(\cosh(x) = \frac{e^x + e^{-x}}{2}\). This definition will be used to verify the identity \(\cosh(-x) = \cosh(x)\).

Step 2: Substitute \(-x\) into the definition of \(\cosh(x)\)

Substitute \(-x\) for \(x\) in the definition of \(\cosh(x)\). This gives us:\[\cosh(-x) = \frac{e^{-x} + e^{x}}{2}\]

Step 3: Simplify \(\cosh(-x)\)

Notice that \(\cosh(-x) = \frac{e^{-x} + e^{x}}{2}\) can be rewritten as \(\frac{e^{x} + e^{-x}}{2}\) because addition is commutative. Thus,\[\cosh(-x) = \frac{e^{x} + e^{-x}}{2}\]

Step 4: Recognize the identity is satisfied

The expression \(\cosh(-x) = \frac{e^{x} + e^{-x}}{2}\) is identical to the original definition of \(\cosh(x)\). Thus, \(\cosh(-x) = \cosh(x)\), proving that \(\cosh\) is an even function.

Key Concepts

Even Function

An even function is a fascinating mathematical concept. It means a function that exhibits symmetry about the y-axis. In simpler terms, if you reflect its graph over the y-axis, it looks unchanged.
Mathematically, a function \( f(x) \) is even if it satisfies the equation \( f(-x) = f(x) \) for all values in its domain.
This characteristic implies that the values of the function for positive and negative inputs are the same.
When we say that the hyperbolic cosine function \( \cosh \) is even, we mean that the function behaves symmetrically with respect to the y-axis and follows this rule:

• For any real number \( x \), \( \cosh(-x) = \cosh(x) \).

This property is essential in proving many other identities and in understanding the symmetrical nature of hyperbolic functions.

cosh(x) Definition

The hyperbolic cosine function \( \cosh(x) \) is a crucial part of hyperbolic functions. This function, like its trigonometric counterpart, deals with hyperbolas instead of circles.
It is defined through the formula:

• \( \cosh(x) = \frac{e^x + e^{-x}}{2} \)

This representation uses the exponential function which is fundamental in calculus. It cleverly combines both the exponential and reciprocity, making it a pivotal concept in various mathematical fields.Unlike circular trigonometric functions, hyperbolic functions such as \( \cosh(x) \) model the geometry of hyperbolic shapes and motion.
Understanding \( \cosh(x) \) not only helps in academic exercises but also plays a vital role in areas like engineering and physics, involving wave mechanics and special relativity.

Hyperbolic Cosine Identity

The hyperbolic cosine identity is integral in proving that \( \cosh(x) \) is an even function. This identity is straightforward yet powerful:

• \( \cosh(-x) = \cosh(x) \)

To understand why this identity holds, we begin with the definition \( \cosh(x) = \frac{e^x + e^{-x}}{2} \). When you substitute \(-x\) into this formula, you get:

• \( \cosh(-x) = \frac{e^{-x} + e^x}{2} \)

Due to the commutative property of addition, this is simply the same as \( \cosh(x) \). Therefore, the identity \( \cosh(-x) = \cosh(x) \) is verified.
This illustrates the even nature of the \( \cosh \) function, emphasizing its incredible symmetry and utility in solving hyperbolic function problems.
Remembering these identities can simplify complex calculations and deepen your understanding of function characteristics.