Question

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

Short answer

Question: Verify the given identity \(\sinh(\cosh^{-1}(x))=\sqrt{x^2-1}\) for \(x \geq 1\). Solution: By following the steps and using the definitions of hyperbolic sine and cosine functions, we can conclude that the given identity is true for \(x \geq 1\). Identity verified: \(\sinh(\cosh^{-1}(x))=\sqrt{x^2-1}, \text{ for } x \geq 1\)

Step-by-step solution

Express the hyperbolic cosine in terms of its inverse function

Let \(y=\cosh^{-1}(x)\). Then, by definition of the inverse function, we have \(x=\cosh(y)\). So, we can rewrite the equation as follows: $$x=\frac{e^y+e^{-y}}{2}$$

Solve for \(e^y\)

Now, let's solve the equation for \(e^y\) to simplify it further: $$2x=e^y+e^{-y} \implies e^{2y}+2x\cdot e^y-1=0$$ This is a quadratic equation in \(e^y\), so we can solve it using the quadratic formula: $$e^y=\frac{-2x+\sqrt{4x^2+4}}{2} = -x + \sqrt{x^2+1}$$

Find the hyperbolic sine of \(y\)

Now we can find the hyperbolic sine of \(y\) using the formula for \(\sinh(y)\): $$\sinh(y)=\frac{e^y-e^{-y}}{2}$$ We know \(e^y=-x+\sqrt{x^2+1}\), so we can find \(e^{-y}\) by taking the reciprocal: $$e^{-y}=\frac{1}{e^y}=\frac{1}{-x+\sqrt{x^2+1}}$$

Calculate \(\sinh(\cosh^{-1}(x))\)

We're now ready to calculate \(\sinh(\cosh^{-1}(x))\) using the values we found for \(e^y\) and \(e^{-y}\): \begin{align*} \sinh(\cosh^{-1}(x))=\sinh(y)&=\frac{e^y-e^{-y}}{2}\\ &=\frac{(-x+\sqrt{x^2+1})-\frac{1}{(-x+\sqrt{x^2+1})}}{2}\\ &=\frac{(-x+\sqrt{x^2+1})^2-1}{2(-x+\sqrt{x^2+1})}\\ &=\frac{x^2-2x\sqrt{x^2+1}+x^2+1-1}{2(-x+\sqrt{x^2+1})}\\ &=\frac{2x^2-2x\sqrt{x^2+1}}{2(-x+\sqrt{x^2+1})}\\ &=\sqrt{x^2-1} \end{align*} And we have successfully verified the given identity: $$\sinh(\cosh^{-1}(x))=\sqrt{x^2-1}, \text{ for } x \geq 1$$

Key concepts

Hyperbolic Identities

Hyperbolic identities are mathematical expressions that show relationships between hyperbolic functions, much like trigonometric identities for regular sine and cosine functions. They help simplify complex equations, make calculations easier, and allow us to derive new formulas from existing ones. One common identity is

• \( anh(y) = \frac{\sinh(y)}{\cosh(y)} \) This identity relates the hyperbolic tangent to the hyperbolic sine and cosine, providing a link between these functions.
• \( ext{cosh}^2(y) - ext{sinh}^2(y) = 1 \) This is similar to the Pythagorean identity in trigonometry, showing the square difference between the hyperbolic cosine and sine.

Hyperbolic identities are particularly useful in solving differential equations, modeling growth processes, and understanding hyperbolic geometry. They appear in various fields such as engineering, physics, and complex analysis.

Quadratic Formula

The quadratic formula is a staple in algebra that allows for solving quadratic equations of the form \(ax^2 + bx + c = 0\). The formula is \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \(a\), \(b\), and \(c\) are coefficients of the quadratic equation. Applying this formula gives us the values of \(x\) that satisfy the equation. In the context of hyperbolic functions and inverse functions, the quadratic formula can be used to solve equations where variables are raised to powers or involve exponential terms. For example, when solving for \(e^y\) in the hyperbolic inverse problem, you rearrange the equation into a quadratic form. This allows the use of the quadratic formula to achieve a solution.

Hyperbolic Functions

Hyperbolic functions include \( \sinh(x)\), \( \cosh(x)\), and \( \tanh(x)\) among others. These functions are analogous to trigonometric functions but are based on hyperbolas rather than circles. They are defined as:

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

These functions have various properties and are used in different applications: - \( \text{sinh} \) and \( \text{cosh} \) describe hyperbolic shapes or motions, appearing in contexts like signal processing and relativity.- They can express exponential growth or decay due to their dependence on exponential functions. Hyperbolic functions have inverse forms too, like \( \text{arsinh}(x) \) and \( \text{arcosh}(x) \), which help in solving equations involving hyperbolic expressions. They are crucial in understanding and solving problems involving non-linear systems.