Question

Prove the following identities. $$\sinh \left(\cosh ^{-1} x\right)=\sqrt{x^{2}-1}, \text { for } x \geq 1$$

Short answer

Question: Prove that \(\sinh\left(\cosh^{-1}(x)\right) = \sqrt{x^2-1}\) for \(x \geq 1\). Answer: The proof of this identity involves recalling the definitions of inverse hyperbolic cosine and hyperbolic sine functions, manipulating the equations and expressions, and taking the square root of the desired result. In the end, we showed that \(\sinh\left(\cosh^{-1}(x)\right) = \sqrt{x^2-1}\) for \(x \geq 1\).

Step-by-step solution

Recall the Inverse Hyperbolic Cosine Function Definition

The function \(\cosh^{-1}(x)\), or the inverse hyperbolic cosine function of x, is defined by the following equation: $$x = \cosh\left(\cosh^{-1}(x)\right)$$ This means that the value for which the hyperbolic cosine function equals x is equal to \(\cosh^{-1}(x)\).

Recall the Hyperbolic Cosine Function Definition

The hyperbolic cosine function, \(\cosh(u)\), is defined as: $$\cosh(u) = \frac{e^{u} + e^{-u}}{2}$$ We can substitute this definition into the equation we obtained in Step 1: $$x = \frac{e^{\cosh^{-1}(x)} + e^{-\cosh^{-1}(x)}}{2}$$

Multiply by 2 and Solve for the Exponential Terms

Let's multiply both sides of the equation by 2: $$2x = e^{\cosh^{-1}(x)} + e^{-\cosh^{-1}(x)}$$ Now, let's express the sum of the exponential terms as a product: $$2x = \left(e^{\cosh^{-1}(x)}\right)\left(1 + e^{-2\cosh^{-1}(x)}\right)$$

Solve for the Exponential Term Within the Parentheses

We will now solve the equation for the exponential term within the parentheses: $$e^{-2\cosh^{-1}(x)} = \frac{2x}{e^{\cosh^{-1}(x)}} - 1$$

Recall the Definition of Hyperbolic Sine Function

The hyperbolic sine function, \(\sinh(u)\), is defined as: $$\sinh(u) = \frac{e^{u} - e^{-u}}{2}$$

Rewrite the Equation With the Hyperbolic Sine Function

Rewrite the equation we obtained in Step 4 in terms of the hyperbolic sine function: $$\sinh^2\left(\cosh^{-1}(x)\right) = \frac{2x}{e^{\cosh^{-1}(x)}} - 1$$

Show that the Left Side Equals \(x^2-1\)

Notice that the left side of the equation is the square of the hyperbolic sine function of the inverse cosine function: $$\sinh^2\left(\cosh^{-1}(x)\right) = x^2 - 1$$

Take the Square Root

Finally, take the square root of both sides of the equation: $$\sinh\left(\cosh^{-1}(x)\right) = \sqrt{x^2-1}$$ Therefore, we have proven that for \(x \geq 1\): $$\sinh\left(\cosh^{-1}(x)\right) = \sqrt{x^2-1}$$

Key concepts

Hyperbolic Functions

Hyperbolic functions are analogues of the ordinary trigonometric, or circular, functions. The basic hyperbolic functions are the hyperbolic sine (\text{sinh}), cosine (\text{cosh}), and tangent (\text{tanh}). They have similar properties and follow comparable identities to their trigonometric counterparts but are based on hyperbolas rather than circles.

The hyperbolic sine and cosine are defined as:

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

These functions arise in various areas of mathematics, including in the descriptions of the shapes of hanging cables or chains, called catenary curves, and are useful in tackling certain differential equations.

Intuitively, you can think of \(e^x\) and \(e^{-x}\) as infinite series expansions where these hyperbolic functions capture the essence of the exponential function's behavior in a symmetrical fashion about the y-axis, which is key to their relationship to hyperbolas.

Inverse Functions

Inverse functions essentially undo the effect of the original function. If you have a function that takes 'x' and gives 'y', then its inverse will take 'y' and return 'x'. For hyperbolic functions, these inverses are denoted by a '-1' superscript, such as \(\text{sinh}^{-1}(x)\) for the inverse hyperbolic sine.

Here's an important point: functions must be one-to-one to have an inverse, meaning that each 'x' must map to one unique 'y' and vice versa. Hyperbolic functions are shaped perfectly for having inverses because they are inherently one-to-one.

Working with inverse functions often involves switching the dependent and independent variables and solving the resulting expression. For example, finding \(\text{cosh}^{-1}(x)\) is essentially solving the equation \(y = \text{cosh}(x)\) for 'x' in terms of 'y'. In the context of our exercise, \(\text{sinh}(\text{cosh}^{-1}(x))\) represents the hyperbolic sine value of the angle whose hyperbolic cosine is 'x'.

Hyperbolic Identities

Just as with trigonometry, there are numerous identities related to hyperbolic functions that can simplify expressions and aid in solving equations. Some of these mirror trigonometric identities closely, while others are distinct to the hyperbolic realm.

For instance, one well-known identity is the hyperbolic Pythagorean identity:

• \text{cosh}^2(x) - \text{sinh}^2(x) = 1

This identity is reminiscent of the Pythagorean theorem from trigonometry and is used in the proof of our exercise. Such identities are vital not just for academic exercises, but also in mathematical modeling, physics, engineering, and other scientific applications.

Understanding and applying these identities require recognizing the patterns and the rules of hyperbolic functions — for instance, how they interact when squared or how they relate to their exponential definitions. Recognizing that \(\text{sinh}^2(x) = \frac{1}{4}(e^{2x}-2+e^{-2x})\) and relating it back to their definitions can be crucial for manipulating and ultimately solving many kinds of problems involving hyperbolic expressions.