Question

Inverse identity Show that \(\cosh ^{-1}(\cosh x)=|x|\) by using the formula \(\cosh ^{-1} t=\ln (t+\sqrt{t^{2}-1})\) and considering the cases \(x \geq 0\) and \(x<0\).

Short answer

Answer: The value of the inverse hyperbolic cosine function for an input of \(\cosh x\) is \(|x|\), which was proven in the step-by-step solution for both cases when \(x \geq 0\) and when \(x < 0\).

Step-by-step solution

Case 1: \(x \geq 0\)

We are given that \(\cosh^{-1} t = \ln(t + \sqrt{t^2 - 1})\). Let's consider the case when \(x \geq 0\) and we need to find \(\cosh^{-1}(\cosh x)\). By substituting \(t = \cosh x\) into the given formula, we get: \(\cosh^{-1}(\cosh x) = \ln(\cosh x + \sqrt{(\cosh x)^2 - 1})\) Now, recall the identity \(\cosh^2 x - \sinh^2 x = 1\). We can rewrite this identity as \((\cosh x)^2 - 1 = \sinh^2 x\). Substituting this identity into the previous expression, we get: \(\cosh^{-1}(\cosh x) = \ln(\cosh x + \sqrt{\sinh^2 x})\) Since \(\sinh x\) is non-negative when \(x \geq 0\), we can simplify the expression further: \(\cosh^{-1}(\cosh x) = \ln(\cosh x + \sinh x)\) By using the definition of the hyperbolic functions: \(\cosh x = \frac{e^x + e^{-x}}{2}\) and \(\sinh x = \frac{e^x - e^{-x}}{2}\), we can rewrite the expression as: \(\cosh^{-1}(\cosh x) = \ln\left(\frac{e^x + e^{-x}}{2} + \frac{e^x - e^{-x}}{2}\right) = \ln(e^x)\) Since \(x \geq 0\), we have \(|x| = x\), thus: \(\cosh^{-1}(\cosh x) = \ln(e^x) = x = |x|\)

Case 2: \(x < 0\)

Now let's consider the case when \(x < 0\). By using the identity \(\cosh(-x) = \cosh x\), we can rewrite the expression as: \(\cosh^{-1}(\cosh x) = \cosh^{-1}(\cosh (-x))\) Since \(-x > 0\), we can apply the result from Case 1 to this expression: \(\cosh^{-1}(\cosh (-x)) = |-x|\) Since \(x < 0\), we have \(|-x| = -x\), thus: \(\cosh^{-1}(\cosh x) = |-x| = -x = |x|\)

Conclusion

In both cases, we have shown that \(\cosh^{-1}(\cosh x) = |x|\). Therefore, the inverse identity holds for the hyperbolic cosine function.

Key concepts

Hyperbolic Cosine

The hyperbolic cosine function, denoted as \( \cosh x \), is one of the fundamental hyperbolic functions, similar in concept to the regular cosine function but applied to hyperbolic geometry. It is defined using the exponential function as:
\[ \cosh x = \frac{e^x + e^{-x}}{2} \]This formula reveals that \( \cosh x \) is an even function, meaning \( \cosh(-x) = \cosh x \). The hyperbolic cosine is always positive for all real numbers \( x \), and it grows exponentially as \( x \) increases.

• It is often used in calculations involving hyperbolic angles and can be visualized as forming the shape of a hanging cable, known as a catenary.
• It helps to understand the behavior of this function, as it lays the groundwork for grasping its inverse and related identities.

Understanding \( \cosh x \) will help you navigate the concept of its inverse, \( \cosh^{-1} x \), as explored in the problem.

Inverse Identity

The inverse identity for hyperbolic functions addresses finding the original input from the output of a hyperbolic function. In our exercise, we focus on proving:
\[ \cosh^{-1} (\cosh x) = |x| \]To find \( \cosh^{-1} \), we use:
\[ \cosh^{-1} t = \ln(t + \sqrt{t^2 - 1}) \]When \( x \geq 0 \), direct substitution gives us \( \cosh^{-1} \cosh x = x \). However, for \( x < 0 \), the symmetry of \( \cosh \) helps us see \( \cosh x = \cosh(-x) \). This results in \( \cosh^{-1} \cosh x = |x| \) for all real \( x \).
This illustrates that the inverse hyperbolic cosine retrieves the magnitude of the input. The logic here is not just about reversing the function but ensuring it fits the definition of the inverse.

• In practical terms, this identity checks the idea that the action of \( \cosh \) can be undone and leads us back to the magnitude of our starting point.

Logarithmic Function

The logarithmic function plays a crucial role in the definition of the inverse hyperbolic cosine, \( \cosh^{-1} \). It's defined using the natural logarithm \( \ln \), which offers the ability to solve for \( x \) in exponential equations.
The formula:
\[ \cosh^{-1} t = \ln(t + \sqrt{t^2 - 1}) \]shows how logarithms translate hyperbolic problems into a more manageable linear form. This conversion is vital for expressing inverse hyperbolic functions.

• Logarithms are critical because they help in linearizing the solution, which is otherwise difficult to interpret directly via exponential functions.
• By using \( \ln \), the inverse identity proofs for hyperbolic functions become clearer, transforming complex roots and expressions into simpler comparisons.

Effective use of logarithms thus allows for solutions that incorporate inverse identities straightforwardly.

Even and Odd Functions

Even and odd functions are essential in understanding the symmetry and behavior of functional outputs. An even function, such as \( \cosh x \), satisfies:
\[ f(-x) = f(x) \]This symmetric property around the y-axis means if you input \(-x\), it behaves the same as \( x \). This provides a simplified model for functions and enhances calculation efficiency.
Odd functions, in contrast, satisfy:
\[ f(-x) = -f(x) \]Though not directly involved in our current example, recognizing this symmetry helps when learning other hyperbolic functions, like \( \sinh x \), which is odd.

• The even nature of \( \cosh x \) was pivotal in solving \( \cosh^{-1} \cosh x = |x| \), as it allows simplification across positive and negative domains.
• It provides essential insights into how transformations and inverses behave under point reflection symmetry.

Overall, understanding even and odd properties solidifies foundational skills necessary for mastering hyperbolic function identities.