Question

Verify that the given equations are identities. \(\sinh (x-y)=\sinh x \cosh y-\cosh x \sinh y\)

Short answer

The equation \(\sinh(x-y)=\sinh x \cosh y-\cosh x \sinh y\) is an identity.

Step-by-step solution

Review Hyperbolic Identities

Recall the definitions for hyperbolic sine and cosine: \(\sinh{x} = \frac{e^x - e^{-x}}{2}\) and \(\cosh{x} = \frac{e^x + e^{-x}}{2}\). Also, remember the subtraction formula for the hyperbolic sine function: \(\sinh(x-y) = \sinh{x}\cosh{y} - \cosh{x}\sinh{y}\). We will now verify this identity.

Express Hyperbolic Functions in Exponential Form

Replace \(\sinh{x-y}\), \(\sinh{x}\), \(\cosh{y}\), \(\cosh{x}\), and \(\sinh{y}\) using their exponential definitions: - \(\sinh{x-y} = \frac{e^{x-y} - e^{-(x-y)}}{2}\) - \(\sinh{x} = \frac{e^x - e^{-x}}{2}\)- \(\cosh{y} = \frac{e^y + e^{-y}}{2}\)- \(\cosh{x} = \frac{e^x + e^{-x}}{2}\)- \(\sinh{y} = \frac{e^y - e^{-y}}{2}\)

Substitute into Identity

Substitute the exponential forms into the identity: \[\sinh{x-y} = \left(\frac{e^x - e^{-x}}{2}\right) \left(\frac{e^y + e^{-y}}{2}\right) - \left(\frac{e^x + e^{-x}}{2}\right) \left(\frac{e^y - e^{-y}}{2}\right)\]

Simplify Exponential Expression

Expand both terms:1. \(\left(\frac{e^x - e^{-x}}{2}\right) \left(\frac{e^y + e^{-y}}{2}\right)\) becomes \(\frac{e^{x+y} + e^{-(x+y)} - e^{x-y} - e^{-(x-y)}}{4}\)2. \(\left(\frac{e^x + e^{-x}}{2}\right) \left(\frac{e^y - e^{-y}}{2}\right)\) becomes \(\frac{e^{x+y} + e^{-(x+y)} - e^{-(x-y)} - e^{x-y}}{4}\)

Combine and Cancel Terms

Combine both expanded terms:\[\frac{(e^{x+y} + e^{-(x+y)} - e^{x-y} - e^{-(x-y)}) - (e^{x+y} + e^{-(x+y)} - e^{-(x-y)} - e^{x-y})}{4}\]They simplify to:\[\frac{-2(e^{x-y}) + 2(e^{-(x-y)})}{4}\]is reduced to\[\frac{e^{x-y} - e^{-(x-y)}}{2}\] which is exactly \(\sinh(x-y)\) in exponential form.

Key concepts

Hyperbolic Functions

Hyperbolic functions are analogs of the familiar trigonometric functions but they are based on hyperbolas rather than circles. The two most common hyperbolic functions are hyperbolic sine, denoted as \( \sinh{x} \), and hyperbolic cosine, denoted as \( \cosh{x} \). These functions are defined using exponential functions, setting them apart from the circular functions:

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

Hyperbolic functions have various applications in physics, engineering, and mathematics through modeling different phenomena such as wave equations and relativity theories.
One interesting property of hyperbolic functions is their relationships, similar to trigonometric identities. These relations, or identities, help simplify problems and solve complex equations. For example, the identity \( \cosh^2{x} - \sinh^2{x} = 1 \) is reminiscent of \( \cos^2{x} + \sin^2{x} = 1 \).

Subtraction Formula

The subtraction formula for hyperbolic functions is akin to those of trigonometric functions. It's particularly vital for expanding expressions and verifying identities. In our identity check, the subtraction formula for hyperbolic sine is:\[\sinh(x-y) = \sinh{x} \cosh{y} - \cosh{x} \sinh{y}\]
This formula is essential for proving identities like the one in our exercise by expressing the hyperbolic sine of a difference in terms of products of hyperbolic sines and cosines. Understanding this formula helps simplify expressions and solve equations involving hyperbolic functions efficiently.
The subtraction formula can be derived from the definitions of \( \sinh{x} \) and \( \cosh{x} \) using the exponential functions. This connection between the trigonometric and hyperbolic subtraction formulae provides deeper insight into the nature of these functions.

Exponential Form

The exponential form of hyperbolic functions is fundamental for understanding and utilizing hyperbolic identities. By expressing \( \sinh{x} \) and \( \cosh{x} \) using exponentials:

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

these functions are both simplified and made more robust in proofs and applications.
We used these exponential definitions in the problem to convert \( \sinh(x-y) \), \( \sinh{x} \), \( \cosh{y} \), and others into an exponential form. This conversion allows for straightforward algebraic manipulation, critical in verifying identities.
Applying exponential expressions of hyperbolic functions enables complex operations and transformations to be handled more directly. The exponential perspective reveals not only the symmetry of hyperbolic functions but also their distinct role in analytical problems.

Identities Verification

Verifying identities is vital for confirming the relationships and structures in mathematical expressions. This ensures that algebraic transformations and manipulations are valid.
In this exercise, we verified:

• \( \sinh(x-y) = \sinh{x} \cosh{y} - \cosh{x} \sinh{y} \)

by substituting their exponential forms for a detailed breakdown.
Through expanding and simplifying, we found that both the left- and right-hand sides of this equation match perfectly, confirming our identity. Verification like this not only validates the formula but enhances your understanding of how these expressions interact.
Learning how to verify identities is crucial for problem-solving, offering confidence in your manipulations and intelligence in handling both theoretical and practical cases involving hyperbolic functions. Knowing and seeing these identities verified is a powerful tool in mathematics, opening avenues to new results and applications.