Problem 29 - Advanced Engineering Mathematics

Question

Prove that \(\cosh ^{2} z-\sinh ^{2} z=1\)

Step-by-Step Solution

Answer

\(\cosh^2(z) - \sinh^2(z) = 1\) is verified using the definitions of hyperbolic functions.

Step 1: Review Hyperbolic Function Definitions

Recall the definitions of hyperbolic cosine and hyperbolic sine. The hyperbolic cosine function is defined as:\[ \cosh(z) = \frac{e^z + e^{-z}}{2} \]The hyperbolic sine function is defined as:\[ \sinh(z) = \frac{e^z - e^{-z}}{2} \]

Step 2: Compute \(\cosh^2 z\)

Calculate the square of the hyperbolic cosine function:\[ \cosh^2(z) = \left( \frac{e^z + e^{-z}}{2} \right)^2 = \frac{(e^z + e^{-z})^2}{4} \]

Step 3: Expand \((e^z + e^{-z})^2\)

Expand the square:\[ (e^z + e^{-z})^2 = e^{2z} + 2 + e^{-2z} \]

Step 4: Express \(\cosh^2 z\) in Expanded Form

Substitute the expanded form into the \(\cosh^2 z\) expression:\[ \cosh^2(z) = \frac{e^{2z} + 2 + e^{-2z}}{4} \]

Step 5: Compute \(\sinh^2 z\)

Calculate the square of the hyperbolic sine function:\[ \sinh^2(z) = \left( \frac{e^z - e^{-z}}{2} \right)^2 = \frac{(e^z - e^{-z})^2}{4} \]

Step 6: Expand \((e^z - e^{-z})^2\)

Expand the square:\[ (e^z - e^{-z})^2 = e^{2z} - 2 + e^{-2z} \]

Step 7: Express \(\sinh^2 z\) in Expanded Form

Substitute the expanded form into the \(\sinh^2 z\) expression:\[ \sinh^2(z) = \frac{e^{2z} - 2 + e^{-2z}}{4} \]

Step 8: Subtract \(\sinh^2(z)\) from \(\cosh^2(z)\)

Calculate the difference:\[ \cosh^2(z) - \sinh^2(z) = \frac{(e^{2z} + 2 + e^{-2z}) - (e^{2z} - 2 + e^{-2z})}{4} \]

Step 9: Simplify the Expression

Simplify the expression by subtracting:\[ \frac{4}{4} = 1 \]

Step 10: Conclusion

Thus, it is proven that \(\cosh^2(z) - \sinh^2(z) = 1\).

Key Concepts

Hyperbolic Functions

Hyperbolic functions are mathematical functions that share similarities with the trigonometric functions but are based on hyperbolas rather than circles. These functions commonly appear in various branches of mathematics, including calculus and complex numbers. Unlike trigonometric identities that deal with angles and circles, hyperbolic functions relate to hyperbolas and real exponential functions.
Understanding hyperbolic functions can enhance your mathematical toolkit, especially when dealing with differential equations and complex analysis. The two primary hyperbolic functions that are frequently discussed are hyperbolic cosine (\(\cosh\)) and hyperbolic sine (\(\sinh\)). Let's dive deeper into these definitions.

Cosh and Sinh Definitions

The hyperbolic cosine and sine functions are defined using exponential functions. They are analogous to the cosine and sine functions in trigonometry but involve exponential terms.

For hyperbolic cosine, the function is defined as follows:

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

This function averages the exponential growth and decay at a point \(z\).

For hyperbolic sine, the function is expressed as:

• \(\sinh(z) = \frac{e^z - e^{-z}}{2}\)

This function represents the difference in exponential growth and decay at a point \(z\).
Both functions are essential in describing hyperbolas, similar to how trigonometric functions describe circles. Understanding these definitions sets a good foundation for exploring the identities and properties of hyperbolic functions.

Mathematical Proofs

Mathematical proofs are essential tools in verifying equations and identities within mathematics. They provide a logical sequence of steps that lead from the assumptions of the problem to the desired conclusion. The proof we are focusing on involves verifying the identity \(\cosh^2 z - \sinh^2 z = 1\).

To prove this identity, we follow these steps:

• Start by using the definitions of \(\cosh(z)\) and \(\sinh(z)\).
• Calculate \(\cosh^2(z)\) by squaring its formula and expanding it, resulting in \(\cosh^2(z) = \frac{e^{2z} + 2 + e^{-2z}}{4}\).
• Similarly, find \(\sinh^2(z)\), which results in \(\sinh^2(z) = \frac{e^{2z} - 2 + e^{-2z}}{4}\).
• Subtract the two squared functions: \(\cosh^2(z) - \sinh^2(z)\).
• Upon simplifying, every exponential term cancels neatly, leaving us with \(1\).

This classic proof shows how hyperbolic identities link back to their exponential definitions, demonstrating the consistency and elegance present in mathematics.