Problem 46 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 41-52, verify that the given equations are identities. \(\sinh (x-y)=\sinh x \cosh y-\cosh x \sinh y\)

Step-by-Step Solution

Answer

The identity \(\sinh(x-y) = \sinh x \cosh y - \cosh x \sinh y\) is verified through expansion and simplification of hyperbolic functions.

Step 1: Understand Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but for a hyperbola. The hyperbolic sine and cosine functions are defined as follows:\[\sinh x = \frac{e^x - e^{-x}}{2}, \quad \cosh x = \frac{e^x + e^{-x}}{2}.\] These definitions will be used in the verification.

Step 2: Expand Left Side of Equation

Identify the left side of the given identity as \(\sinh(x-y)\). Using the definition of hyperbolic sine, expand it:\[\sinh(x-y) = \frac{e^{x-y} - e^{-(x-y)}}{2} = \frac{e^{x-y} - e^{-x+y}}{2}.\]

Step 3: Use Exponential Properties

Utilize the properties of exponents to separate terms:\[\frac{e^{x-y}}{2} = \frac{e^x e^{-y}}{2}, \quad \text{and} \quad \frac{e^{-x+y}}{2} = \frac{e^{-x} e^{y}}{2}.\] Therefore, \[\sinh(x-y) = \frac{e^x e^{-y} - e^{-x} e^{y}}{2}.\]

Step 4: Substitute Hyperbolic Identities

Now match this to the right side of the given equation \(\sinh x \cosh y - \cosh x \sinh y\):1. \(\sinh x = \frac{e^x - e^{-x}}{2}\) and \(\cosh y = \frac{e^y + e^{-y}}{2}\),2. \(\cosh x = \frac{e^x + e^{-x}}{2}\) and \(\sinh y = \frac{e^y - e^{-y}}{2}\).

Step 5: Multiply and Simplify Terms

Compute \(\sinh x \cosh y\):\[\left(\frac{e^x - e^{-x}}{2}\right)\left(\frac{e^y + e^{-y}}{2}\right) = \frac{e^{x+y} + e^{x-y} - e^{-x+y} - e^{-x-y}}{4}.\]Now compute \(\cosh x \sinh y\):\[\left(\frac{e^x + e^{-x}}{2}\right)\left(\frac{e^y - e^{-y}}{2}\right) = \frac{e^{x+y} - e^{x-y} + e^{-x+y} - e^{-x-y}}{4}.\]

Step 6: Subtract and Simplify Further

Subtract the second result from the first to get \(\sinh x \cosh y - \cosh x \sinh y\):\[\frac{e^{x+y} + e^{x-y} - e^{-x+y} - e^{-x-y}}{4} - \frac{e^{x+y} - e^{x-y} + e^{-x+y} - e^{-x-y}}{4}\]reduce to:\[\frac{2e^{x-y} - 2e^{-x+y}}{4} = \frac{e^{x-y} - e^{-x+y}}{2} = \sinh(x-y).\]

Step 7: Verify Conclusion

Through expansion, using exponent properties and simplifying, we've shown:\[\sinh(x-y) = \sinh x \cosh y - \cosh x \sinh y.\]Thus, the identity is verified.

Key Concepts

Hyperbolic Functions

Hyperbolic functions are similar to trigonometric functions but are related to hyperbolas instead of circles. They form the foundation of many areas of mathematics, especially when calculating idealized systems where circular trigonometry does not suffice. Unlike the circular functions like sine and cosine, which are periodic, hyperbolic functions like hyperbolic sine (\(\sinh x\)) and hyperbolic cosine (\(\cosh x\)) have a different pattern resembling exponential growth. The natural base of these functions is \(e\), which is the basis of exponential functions.

• Sinh: Does not loop, but rises and falls more like a wave.
• Cosh: Resembles a U-shaped curve that opens upwards.

Understanding these properties helps in solving equations involving hyperbolic functions, recognizing them as transformations of exponentials and vice versa.

Hyperbolic Sine

The hyperbolic sine function, denoted as \(\sinh x\), is a hyperbolic function that uses exponential functions to describe its behavior. It is defined by the equation:\[\sinh x = \frac{e^x - e^{-x}}{2}\]Here, the exponential terms \(e^x\) and \(e^{-x}\) showcase the function's reliance on exponential growth and decay, reflecting its sine-like wave along the real number line. The function plays a crucial role in scenarios where modeling cannot be neatly captured by regular trigonometric functions, such as in certain physics applications.

• The shape of \(\sinh(x)\) is useful in describing catenary shapes or in calculating distances in hyperbolic geometry.

Having a clear grasp of these definitions makes it easier to grasp further complex identities involving hyperbolic functions.

Exponential Properties

Understanding exponential properties is key to working with hyperbolic functions since these rely heavily on systems involving powers of e. When looking at expressions like \(e^{x-y}\) or \(e^{-x+y}\), it is essential to correctly apply the rules of exponents, such as:

• \(e^{a+b} = e^a \times e^b\)
• \(e^{-a} = \frac{1}{e^a}\)

This knowledge allows the manipulation of terms in a way that simplifies calculations and identity verifications. For instance, in the provided solution concerning hyperbolic identities, separating exponential components is a critical step in both defining and verifying the identity formula itself.Such manipulations demonstrate the flexibility of exponential forms and their importance in this mathematical scope.

Function Verification

Function verification involves demonstrating that both sides of an equation produce the same result under identical conditions. In the context of hyperbolic identities, this means breaking down \(\sinh(x-y)\) into its exponential components and showing it equal to \(\sinh x \cosh y - \cosh x \sinh y\). Essentially, it involves multiple steps:

• Translate each function using its exponential definition.
• Expand and simplify the equations leveraging exponential properties.
• Re-combine and reduce terms to confirm both sides are equivalent.

Such steps are not merely technical; they underpin the entire exercise of using identities \(\sinh x, \cosh y\) to transform one expression into another until they match perfectly, thus verifying the proposed identity.