Question

Derive \(\cosh ^{2}(x)+\sinh ^{2}(x)=\cosh (2 x)\) from the definition.

Short answer

The identity holds: \( \cosh^2(x) + \sinh^2(x) = \cosh(2x) \).

Step-by-step solution

Recall the Definitions

The hyperbolic sine and cosine functions are defined as follows: \( \sinh(x) = \frac{e^x - e^{-x}}{2} \) and \( \cosh(x) = \frac{e^x + e^{-x}}{2} \). These definitions will be used to solve the problem.

Express \( \cosh^2(x) \) and \( \sinh^2(x) \)

Using the identities \( \cosh(x) = \frac{e^x + e^{-x}}{2} \) and \( \sinh(x) = \frac{e^x - e^{-x}}{2} \), compute the squares: \[ \cosh^2(x) = \left(\frac{e^x + e^{-x}}{2}\right)^2 = \frac{(e^{2x} + 2 + e^{-2x})}{4} \] and \[ \sinh^2(x) = \left(\frac{e^x - e^{-x}}{2}\right)^2 = \frac{(e^{2x} - 2 + e^{-2x})}{4}. \]

Add \( \cosh^2(x) \) and \( \sinh^2(x) \)

Add the results from Step 2: \[ \cosh^2(x) + \sinh^2(x) = \frac{(e^{2x} + 2 + e^{-2x})}{4} + \frac{(e^{2x} - 2 + e^{-2x})}{4} = \frac{(2e^{2x} + 2e^{-2x})}{4}. \]

Simplify the Expression

Factor out a 2 from the expression: \[ \cosh^2(x) + \sinh^2(x) = \frac{2(e^{2x} + e^{-2x})}{4} = \frac{e^{2x} + e^{-2x}}{2}. \] This is the definition of \( \cosh(2x) \).

Conclusion

Thus, we have derived that \( \cosh^2(x) + \sinh^2(x) = \cosh(2x) \), confirming the identity with the provided definitions.

Key concepts

Hyperbolic Sine

The hyperbolic sine, often denoted as \( \sinh(x) \), is a fundamental concept in hyperbolic functions. It's analogous to the sine function in trigonometry but tailored for hyperbolic angles, which arise in certain geometric contexts. It's defined as:

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

This equation represents the hyperbolic sine function using the exponential function \( e^x \).
Why use exponentials? Because they are conveniently used in many branches of mathematics, including calculus and complex analysis.
Understanding \( \sinh(x) \) requires familiarization with this exponential expression. It helps describe growth, decay, and wave-like phenomena in math and physics.By grasping this fundamental definition, you'll better apprehend hyperbolic identities and their derivations, as they form the building blocks for more complex expressions.

Hyperbolic Cosine

The hyperbolic cosine, denoted as \( \cosh(x) \), is the counterpart to the hyperbolic sine. It is defined as:

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

This function takes an average of \( e^x \) and its reciprocal \( e^{-x} \). The hyperbolic cosine reflects balanced growth and decay.
Much like \( \sinh(x) \), \( \cosh(x) \) often appears in solutions to differential equations and models of suspended cables and arcs.

Relation to the Hyperbolic Identity

By calculating \( \cosh^2(x) \), we can derive particular identities involving hyperbolic functions, just as the problem does.
We start with the square: \[ \cosh^2(x) = \left(\frac{e^x + e^{-x}}{2}\right)^2 = \frac{e^{2x} + 2 + e^{-2x}}{4} \]Our understanding of this identity contributes to simplifying expressions that feature hyperbolic functions.

Mathematical Identity Derivation

Deriving mathematical identities involves establishing relationships between functions or expressions. For hyperbolic functions, these identities emerge from foundational definitions, like \( \sinh(x) \) and \( \cosh(x) \).
The identity in question is derived from adding and simplifying \( \cosh^2(x) \) and \( \sinh^2(x) \):

• We have \( \cosh^2(x) = \frac{e^{2x} + 2 + e^{-2x}}{4} \) and \( \sinh^2(x) = \frac{e^{2x} - 2 + e^{-2x}}{4} \).
• Adding them results in \( \cosh^2(x) + \sinh^2(x) = \frac{2(e^{2x} + e^{-2x})}{4} \).

Simplifying this expression gives \( \frac{e^{2x} + e^{-2x}}{2} \), which corresponds to the definition of \( \cosh(2x) \). This identity shows how hyperbolic functions relate to each other and demonstrates the power of mathematical derivations in revealing these relations.