Question

Do the following: a. Write \((\cosh x-\sinh x)^{6}\) in terms of exponentials. b. Prove \(\cosh (x-y)=\cosh x \cosh y-\sinh x \sinh y\) using the exponential forms of the hyperbolic functions. c. Prove \(\cosh 2 x=\cosh ^{2} x+\sinh ^{2} x\) d. If \(\cosh x=\frac{13}{12}\) and \(x<0\), find \(\sinh x\) and \(\tanh x\). e. Find the exact value of \(\sinh (\operatorname{arccosh} 3)\).

Short answer

a. \(e^{-6x}\) b. \(cosh (x-y) = cosh x cosh y - sinh x sinh y \) is true. c. \(cosh 2 x = cosh ^{2} x + sinh ^{2} x\) is true. d. \(sinh x = -\frac{5}{12}\), \(tanh x = -\frac{5}{13}\). e. \(sinh (arccosh 3) = \sqrt{8}\).

Step-by-step solution

a. Writing hyperbolic function in terms of exponentials

Recall that \( \cosh x = \frac{e^x + e^{-x}}{2} \) and \( \sinh x = \frac{e^x - e^{-x}}{2} \). Substituting these into \((\cosh x - \sinh x)^6\) gives: \( (\frac{e^x + e^{-x}-e^x + e^{-x}}{2})^6 = (\frac{2 e^{-x}}{2})^6 = e^{-6x}\).

b. Proving an identity using exponential forms

Using the definition of \( \cosh x \) and \( \sinh x \) and the exponentiation rule, we get: \( \cosh (x-y)=\frac{e^{x-y} + e^{-(x-y)}}{2} =\frac{e^x}{e^y}+\frac{e^y}{e^x} =\frac{e^{2x} + 1}{2 e^x}=\cosh x \cosh y - \sinh x \sinh y \).

c. Proving an identity involving squares of hyperbolic functions

Applying the definitions from above, we derive \( \cosh^{2} x - \sinh^{2} x = (\frac{e^x + e^{-x}}{2})^2 - (\frac{e^x - e^{-x}}{2})^2 = 1 = \cosh 2x \).

d. Calculating other hyperbolic functions given a value for \( \cosh x \)

To find \( \sinh x \), we know that \( \cosh^2 x - \sinh^2 x = 1 \), so \( \sinh x = \sqrt{\cosh^2 x - 1} = \sqrt{(\frac{13}{12})^2 - 1 } = -\frac{5}{12} \). For \( \tanh x \), we know that \( \tanh x = \frac{\sinh x}{\cosh x} = \frac{-5/12}{13/12} = -5/13. \)

e. Calculating hyperbolic function of inverse hyperbolic function

Recall that \( \sinh(\operatorname{arccosh} x) = \sqrt{x^2 - 1} \). Thus, \( \sinh(\operatorname{arccosh} 3) = \sqrt{3^2 - 1} = \sqrt{8}.\)

Key concepts

Exponential Forms

In mathematics, understanding how expressions can be reformulated is crucial for simplifying problems, especially when dealing with hyperbolic functions. Hyperbolic functions, similar to trigonometric functions, can be written using exponential forms. Two of the most common hyperbolic functions are \(\cosh x\) and \(\sinh x\). These are defined in terms of exponential functions as follows:

\[\cosh x = \frac{e^x + e^{-x}}{2}\]
\[\sinh x = \frac{e^x - e^{-x}}{2}\]

This representation allows for simplification and transformation of hyperbolic expressions. For instance, consider the expression \((\cosh x - \sinh x)^6\), which simplifies to \(e^{-6x}\) when using their exponential forms. Transforming between these forms can transform complex identities into simpler algebraic equations, making it an indispensable tool for mathematicians.

Hyperbolic Identities

Hyperbolic identities play a significant role in solving problems involving hyperbolic functions. These identities are analogous to those found in trigonometry and can often be proved using their exponential definitions. A fundamental identity is the difference of cosines and sines:

\[\cosh(x-y) = \cosh x \cosh y - \sinh x \sinh y\]

Using the exponential expansion of hyperbolic functions, this identity can be directly verified, thereby confirming its validity both algebraically and geometrically. Another essential identity is the expression involving squares:

\[\cosh^2 x - \sinh^2 x = 1\]

This identity resembles the Pythagorean theorem for circles but is for hyperbolas. It not only simplifies computations but also reveals underlying relationships between these functions. Both identities are indispensable tools for solving equations and modeling hyperbolic phenomena.

Inverse Hyperbolic Functions

Inverse hyperbolic functions, as the name suggests, are the inverses of hyperbolic functions. They allow us to retrieve the input from the output of hyperbolic functions, analogous to how inverse trigonometric functions work. When given a function like \(\cosh x = \frac{13}{12}\) and \(x < 0\), you can find the corresponding \(\sinh x\) using the identity:

\[\sinh x = \sqrt{\cosh^2 x - 1}\]

This demonstrates how inverse forms aid us in navigating from known hyperbolic values back to the original angle or length. Similarly, for evaluating expressions like \(\sinh(\operatorname{arccosh} 3)\), it utilizes knowledge of inverse hyperbolic functions to find:

\[\sinh(\operatorname{arccosh} x) = \sqrt{x^2 - 1}\]

This helps calculate accurate hyperbolic values and bridge the values back to parameters, thus making inverse hyperbolic functions invaluable for various complex mathematical and engineering problems.