Questions: Simplify the exponents. = (e^(-x) - e^x) / 2 = -sinh x (cosh x)^2 - (sinh x)^2 = ((e^x + e^(-x)) / 2)^2 - ((e^x - e^(-x)) / 2)^2 = 1/4 (□^2 - 1/4(□)^2)

Solution Steps

1. For the first expression, \(\frac{e^{-x}-e^{x}}{2}\), recognize that it simplifies to \(-\sinh x\) using the definition of the hyperbolic sine function: \(\sinh x = \frac{e^x - e^{-x}}{2}\).

2. For the second expression, \((\cosh x)^{2}-(\sinh x)^{2}\), use the definitions of hyperbolic cosine and sine: \(\cosh x = \frac{e^x + e^{-x}}{2}\) and \(\sinh x = \frac{e^x - e^{-x}}{2}\). Then apply the identity \((\cosh x)^2 - (\sinh x)^2 = 1\).

We start with the expression

\[ \frac{e^{-x} - e^{x}}{2}. \]

Using the definition of the hyperbolic sine function, we know that

\[ \sinh x = \frac{e^{x} - e^{-x}}{2}. \]

Thus, we can rewrite the expression as

\[ \frac{e^{-x} - e^{x}}{2} = -\sinh x. \]

Next, we consider the expression

\[ (\cosh x)^{2} - (\sinh x)^{2}. \]

Using the definitions of hyperbolic functions, we have

\[ \cosh x = \frac{e^{x} + e^{-x}}{2} \quad \text{and} \quad \sinh x = \frac{e^{x} - e^{-x}}{2}. \]

Substituting these into the expression gives us

\[ \left(\frac{e^{x} + e^{-x}}{2}\right)^{2} - \left(\frac{e^{x} - e^{-x}}{2}\right)^{2}. \]

This simplifies to

\[ \frac{1}{4}\left((e^{x} + e^{-x})^{2} - (e^{x} - e^{-x})^{2}\right). \]

Using the identity \(a^2 - b^2 = (a-b)(a+b)\), we find that

\[ (e^{x} + e^{-x})^{2} - (e^{x} - e^{-x})^{2} = 4e^{x}e^{-x} = 4. \]

Thus, we have

\[ (\cosh x)^{2} - (\sinh x)^{2} = \frac{1}{4} \cdot 4 = 1. \]

Final Answer