Problem 11 - Thomas Calculus

Question

Prove the identities $$\begin{aligned} \sinh (x+y) &=\sinh x \cosh y+\cosh x \sinh y \\ \cosh (x+y) &=\cosh x \cosh y+\sinh x \sinh y \end{aligned}$$ Then use them to show that \begin{equation}\begin{array}{l}{\text { a. } \sinh 2 x=2 \sinh x \cosh x} \\\ {\text { b. } \cosh 2 x=\cosh ^{2} x+\sinh ^{2} x}\end{array}\end{equation}

Step-by-Step Solution

Answer

The identities hold for both \( \sinh \) and \( \cosh \). \( \sinh(2x) = 2\sinh x\cosh x \) and \( \cosh(2x) = \cosh^2 x + \sinh^2 x \).

Step 1: Understand Hyperbolic Functions

The hyperbolic functions \( \sinh(x) \) and \( \cosh(x) \) are defined as: \( \sinh(x) = \frac{e^x - e^{-x}}{2} \) and \( \cosh(x) = \frac{e^x + e^{-x}}{2} \). These definitions are crucial for deriving the identities.

Step 2: Applying Addition Formulas

The identities to be proven are similar to the trigonometric addition formulas. Start by substituting the definitions of \( \sinh \) and \( \cosh \) into \( \sinh(x+y) \) and \( \cosh(x+y) \).

Step 3: Prove \( \sinh(x+y) = \sinh x \cosh y + \cosh x \sinh y \)

- Substitute \( \sinh(x+y) = \frac{e^{x+y} - e^{-(x+y)}}{2} \). - This becomes \( \frac{e^x e^y - e^{-x} e^{-y}}{2} \). - Distribute terms: \( \frac{(e^x - e^{-x})(e^y + e^{-y}) + (e^x + e^{-x})(e^y - e^{-y})}{4} \). - Recognize the first bracket as \( 2\sinh(x)\cdot\cosh(y) \) and the second as \( 2\cosh(x)\cdot\sinh(y) \). - Simplify to obtain \( \sinh x \cosh y + \cosh x \sinh y \).

Step 4: Prove \( \cosh(x+y) = \cosh x \cosh y + \sinh x \sinh y \)

- Substitute \( \cosh(x+y) = \frac{e^{x+y} + e^{-(x+y)}}{2} \). - This becomes \( \frac{e^x e^y + e^{-x} e^{-y}}{2} \). - Expand: \( \frac{(e^x + e^{-x})(e^y + e^{-y}) + (e^x - e^{-x})(e^y - e^{-y})}{4} \). - Identify these as \( 2\cosh(x)\cdot\cosh(y) \) plus \( 2\sinh(x)\cdot\sinh(y) \). - Simplify to get \( \cosh x \cosh y + \sinh x \sinh y \).

Step 5: Part a: Prove \( \sinh 2x = 2\sinh x\cosh x \)

- Use the identity \( \sinh(x+y) = \sinh x \cosh y + \cosh x \sinh y \) with \( y = x \). - This implies \( \sinh(2x) = \sinh x \cosh x + \cosh x \sinh x \). - Simplify to \( \sinh(2x) = 2\sinh x \cosh x \).

Step 6: Part b: Prove \( \cosh 2x = \cosh^2 x + \sinh^2 x \)

- Use the identity \( \cosh(x+y) = \cosh x \cosh y + \sinh x \sinh y \) with \( y = x \). - Thus \( \cosh(2x) = \cosh x \cosh x + \sinh x \sinh x \). - Rewrite this as \( \cosh^2 x + \sinh^2 x \).

Key Concepts

Hyperbolic Functions

Hyperbolic functions, much like their trigonometric counterparts, are built upon the exponential function. The two primary hyperbolic functions are the hyperbolic sine \( \sinh(x) \) and the hyperbolic cosine \( \cosh(x) \). These functions are defined using the exponential functions as follows: \[ \sinh(x) = \frac{e^x - e^{-x}}{2} \quad \text{and} \quad \cosh(x) = \frac{e^x + e^{-x}}{2}. \] They are particularly useful in various fields such as calculus, physics, and engineering.
Hyperbolic functions have unique properties that distinguish them from the standard sine and cosine functions:

• Both \( \sinh(x) \) and \( \cosh(x) \) are defined for all real numbers.
• Unlike \( \sin(x) \), the hyperbolic sine function is unbounded, meaning its range is all real numbers.
• The hyperbolic cosine function \( \cosh(x) \), like \( \cos(x) \), takes non-negative values and achieves a minimum at \( x = 0 \).

Understanding these properties allows you to explore deeper relationships and identities in mathematics.

Sinh and Cosh Properties

The properties of \( \sinh(x) \) and \( \cosh(x) \) are crucial for deriving and proving hyperbolic identities. These functions exhibit some relationships that are reminiscent of trigonometric identities. One fundamental identity involves the subtraction of their squares:

• The identity \( \cosh^2(x) - \sinh^2(x) = 1 \) is the hyperbolic equivalent of the Pythagorean identity in trigonometry.
• It is derived from the fact that \( (e^x + e^{-x})^2 - (e^x - e^{-x})^2 = 4 \), which simplifies using the basic definitions of \( \sinh(x) \) and \( \cosh(x) \).

This identity is vital in simplifying expressions and proving further hyperbolic identities.
Moreover, since the derivatives of these functions differ slightly from their trigonometric analogues, it's important to remember:

• The derivative of \( \sinh(x) \) is simply \( \cosh(x) \).
• The derivative of \( \cosh(x) \) is \( \sinh(x) \).

These derivatives will frequently appear in calculations and proofs involving hyperbolic functions.

Addition Formulas for Hyperbolic Functions

Addition formulas for hyperbolic functions are analogous to those in trigonometry and play a critical role in simplifying complex hyperbolic expressions. These formulas are derived directly from their exponential definitions.
The hyperbolic sine addition formula is given by:

• \( \sinh(x+y) = \sinh(x)\cosh(y) + \cosh(x)\sinh(y) \)

And for the hyperbolic cosine:

• \( \cosh(x+y) = \cosh(x)\cosh(y) + \sinh(x)\sinh(y) \)

These formulas are pivotal in proving further identities such as \( \sinh(2x) = 2\sinh(x)\cosh(x) \) and \( \cosh(2x) = \cosh^2(x) + \sinh^2(x) \).
To utilize these formulas, substitute the definitions of \( \sinh \) and \( \cosh \) into the expressions and simplify. For example:

• Using \( y = x \) in the formula for \( \sinh(x+y) \), we derive \( \sinh(2x) = 2\sinh(x)\cosh(x) \).
• Similarly, applying \( y = x \) in the \( \cosh(x+y) \) formula gives \( \cosh(2x) = \cosh^2(x) + \sinh^2(x) \).

Mastering these formulas enhances your problem-solving toolkit in calculus and analytic geometry.