Problem 91 - Precalculus

Question

In calculus the following two functions are studied: $$ \sinh x=\frac{e^{x}-e^{-x}}{2} \quad \text { and } \quad \cosh x=\frac{e^{x}+e^{-x}}{2} $$ $$\text { Show that } \cosh ^{2} x-\sinh ^{2} x=1$$

Step-by-Step Solution

Answer

\( \cosh^2 x - \sinh^2 x = 1 \) is true after simplifying both sides using algebra.

Step 1: Write Down the Identity to Prove

We need to prove the identity \(\cosh^2 x - \sinh^2 x = 1\). This can be verified by using the definitions of the hyperbolic sine and cosine functions.

Step 2: Write the Expressions for \( \cosh x \) and \( \sinh x \)

Start by writing the expressions for \( \cosh x \) and \( \sinh x \):\[\cosh x = \frac{e^x + e^{-x}}{2}\]\[\sinh x = \frac{e^x - e^{-x}}{2}\]

Step 3: Square the Expressions

Find \( \cosh^2 x \) and \( \sinh^2 x \) by squaring their expressions:\[\cosh^2 x = \left(\frac{e^x + e^{-x}}{2}\right)^2 = \frac{(e^x + e^{-x})^2}{4}\]\[\sinh^2 x = \left(\frac{e^x - e^{-x}}{2}\right)^2 = \frac{(e^x - e^{-x})^2}{4}\]

Step 4: Expand the Squared Expressions

Expand both expressions using the formula \((a+b)^2 = a^2 + 2ab + b^2\):\[(e^x + e^{-x})^2 = e^{2x} + 2 \cdot e^x \cdot e^{-x} + e^{-2x} = e^{2x} + 2 + e^{-2x}\]And for \((a-b)^2 = a^2 - 2ab + b^2\):\[(e^x - e^{-x})^2 = e^{2x} - 2 \cdot e^x \cdot e^{-x} + e^{-2x} = e^{2x} - 2 + e^{-2x}\]

Step 5: Subtract the Squared Terms

Subtract \( \sinh^2 x \) from \( \cosh^2 x \):\[\cosh^2 x - \sinh^2 x = \frac{(e^{2x} + 2 + e^{-2x})}{4} - \frac{(e^{2x} - 2 + e^{-2x})}{4}\]Simplify the expression:\[\cosh^2 x - \sinh^2 x = \frac{e^{2x} + 2 + e^{-2x} - e^{2x} + 2 - e^{-2x}}{4}\]The terms \( e^{2x} \) and \( e^{-2x} \) cancel each other out, leaving:\[\cosh^2 x - \sinh^2 x = \frac{4}{4} = 1\]

Step 6: Conclusion: Verify the Identity

After performing the algebraic operations, we verify that \( \cosh^2 x - \sinh^2 x = 1 \), confirming the given identity.

Key Concepts

Hyperbolic Sine

The hyperbolic sine function, denoted as \( \sinh x \), is an important concept in calculus, similar to its trigonometric counterpart, the sine function. Its definition is based on exponential functions, given by \( \sinh x = \frac{e^x - e^{-x}}{2} \). This formula uses the natural exponential function \( e \), providing an elegant symmetry between the exponential growth and decay. The hyperbolic sine function reflects properties of hyperbolas, analogous to how the sine function relates to circles.

• \( \sinh x \) is an odd function: \( \sinh(-x) = -\sinh x \).
• It's differentiable and smooth, critical for calculus applications.
• The function increases as \( x \) increases, reflecting exponential growth.

These properties make hyperbolic sine a powerful tool in understanding certain mathematical models, such as those describing hyperbolic geometry and special relativity.

Hyperbolic Cosine

Just like the hyperbolic sine, the hyperbolic cosine function has a key role in mathematics. It is expressed as \( \cosh x = \frac{e^x + e^{-x}}{2} \). Interestingly, it is based on the average of exponential growth and decay, which highlights its symmetry and provides useful properties.

• \( \cosh x \) is an even function: \( \cosh(-x) = \cosh x \).
• It is always non-negative, reflecting the positive values of exponential sums.
• As \( x \) approaches infinity, \( \cosh x \) grows exponentially.

The hyperbolic cosine function is instrumental in modeling real-world phenomena such as the shape of a hanging cable, known as a catenary. Its smoothness and symmetry make it a vital component in both theoretical and applied mathematics.

Identity Proof

In mathematics, proof of identities is essential to verifying mathematical facts and relationships. The identity \( \cosh^2 x - \sinh^2 x = 1 \) is one such relationship that shows a profound connection between hyperbolic functions. The proof demonstrates that hyperbolic functions, like their trigonometric relatives, have intrinsic relationships that are elegantly expressed using exponential functions.
To prove this:

• Start with the definitions: \( \cosh x = \frac{e^x + e^{-x}}{2} \) and \( \sinh x = \frac{e^x - e^{-x}}{2} \).
• Square both functions separately.
• Expand each using algebraic identities for squaring binomials.
• Subtract the squared terms to show all exponential terms cancel out, simplifying to 1.

The simplicity and elegance of this identity underline the powerful nature of hyperbolic functions and their applications.

Exponential Functions

Exponential functions are a cornerstone of both mathematics and applied sciences. Defined by the base of the natural logarithm \( e \), they exhibit dramatic growth and decay, with unique characteristics that make them indispensable. These functions form the backbone of hyperbolic functions with equations like \( e^x \) representing exponential growth and \( e^{-x} \) indicating decay.

• Exponential functions are always positive, never touching zero.
• They grow increasingly faster as \( x \) increases, or decrease slower as \( x \) decreases for negative exponents.
• They are the only functions with a rate of change directly proportional to the function's current value.

By expressing hyperbolic functions in terms of exponentials, one gains insight into their behavior and symmetries, bridging a gap between algebraic manipulation and geometric interpretation. Exponential growth models populations, finance, and even natural processes, demonstrating the concepts' vast relevance.