Problem 3 - Calculus Early Transcendentals

Question

What is the fundamental identity for hyperbolic functions?

Step-by-Step Solution

Answer

Answer: The fundamental identity for hyperbolic functions is \((\sinh(x))^2 + (\cosh(x))^2 = \cosh(2x)\).

Step 1: Define hyperbolic sine and cosine functions

Hyperbolic sine and cosine functions are defined as follows, where x is a real number: Hyperbolic sine: \(\sinh(x) = \frac{e^x - e^{-x}}{2}\) Hyperbolic cosine: \(\cosh(x) = \frac{e^x + e^{-x}}{2}\)

Step 2: Find the square of hyperbolic sine and cosine

We need to find the square of both hyperbolic sine and cosine functions: \((\sinh(x))^2 = \left(\frac{e^x - e^{-x}}{2}\right)^2 = \frac{(e^x - e^{-x})^2}{4}\) \((\cosh(x))^2 = \left(\frac{e^x + e^{-x}}{2}\right)^2 = \frac{(e^x + e^{-x})^2}{4}\)

Step 3: Add both squares and simplify

The next step is to add both squares and simplify the expression: \((\sinh(x))^2 + (\cosh(x))^2 = \frac{(e^x - e^{-x})^2}{4} + \frac{(e^x + e^{-x})^2}{4}\) Now, let's combine the fractions and expand: \(= \frac{(e^x - e^{-x})^2 + (e^x + e^{-x})^2}{4}\) \(= \frac{(e^{2x} - 2e^xe^{-x} + e^{-2x}) + (e^{2x} + 2e^xe^{-x} + e^{-2x})}{4}\)

Step 4: Identify the fundamental identity for hyperbolic functions

We notice that some terms cancel out when we add: \(= \frac{2e^{2x} + 2e^{-2x}}{4}\) Simplify by dividing by 2: \(= \frac{e^{2x} + e^{-2x}}{2}\) Finally, we observe that the expression on the right is equal to \(\cosh(2x)\), so the fundamental identity for hyperbolic functions is: \((\sinh(x))^2 + (\cosh(x))^2 = \cosh(2x)\) This identity relates the squares of the hyperbolic sine and cosine functions to the hyperbolic cosine of double the angle.