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Exponential functions play a key role in defining hyperbolic functions. These functions are characterized by expressions involving powers of the mathematical constant \(e\), approximately equal to 2.71828. In the context of the original exercise, the hyperbolic sine and cosine functions, \(\sinh x\) and \(\cosh x\), are expressed in terms of exponential functions.

• The hyperbolic sine, \(\sinh x\), is defined as \(\frac{e^x - e^{-x}}{2}\).
• The hyperbolic cosine, \(\cosh x\), is defined as \(\frac{e^x + e^{-x}}{2}\).

These definitions enable us to express complex relationships between hyperbolic functions using algebraic manipulation of exponential functions. In the original exercise, the identity \(\sinh 2x = 2 \sinh x \cosh x\) is verified using these exponential definitions.

Mathematical identities are equations that are true for all values of the variables involved. For hyperbolic functions, there exist several important identities, similar to those for trigonometric functions. The identity \(\sinh 2x = 2 \sinh x \cosh x\) is one such identity. It reflects the close relationship between the hyperbolic sine and cosine functions.

These identities are not just mathematical curiosities; they are essential tools in calculus and engineering.

• Identities help simplify expressions and solve complex equations.
• They are used in integration and differentiation of functions.

Verifying an identity, as done in the exercise, involves proving both sides of the equation are equal for all permissible values of \(x\). This can be achieved through substitution, similar to the step-by-step approach used in demonstrating \(\sinh 2x = 2 \sinh x \cosh x\).

Algebraic manipulations involve rewriting and rearranging expressions to achieve a desired form. This skill is vital in solving mathematical problems and proving identities. In the exercise, we used algebraic manipulations to show that \(\sinh 2x\) and \(2 \sinh x \cosh x\) are equivalent.

• First, we expanded \(\sinh 2x\) using its exponential definition, which was \(\frac{e^{2x} - e^{-2x}}{2}\).
• Next, we expanded \(2 \sinh x \cosh x\) by multiplying their exponential expressions which simplified to the same form.

By equating these forms, we demonstrated that the identity holds true. These steps highlight how understanding the underlying relationships and performing precise algebraic operations can lead to solving complex problems.