Problem 93
Question
The hyperbolic cosine and hyperbolic sine functions are defined by $$ \cosh x-\frac{e^{x}+e^{-x}}{2} \text { and } \sinh x-\frac{e^{x}-e^{-x}}{2} $$ a. Show that \(\cosh x\) is an even function. b. Show that \(\sinh x\) is an odd function. c. Prove that \((\cosh x)^{2}-(\sinh x)^{2}-1\)
Step-by-Step Solution
Verified Answer
\(\cosh x\) is an even function. \(\sinh x\) is an odd function. \((\cosh x)^{2}-(\sinh x)^{2} = 1\).
1Step 1: Prove that \(\cosh x\) is an even function
Given \(\cosh x = \frac{e^{x}+e^{-x}}{2}\), we want to show that \(\cosh (-x) = \cosh x\). Starting from \(\cosh (-x) = \frac{e^{-x}+e^{x}}{2}\), we see that this is the same as the original definition of \(\cosh x\), meaning that \(\cosh x\) is indeed an even function.
2Step 2: Prove that \(\sinh x\) is an odd function
From \(\sinh x = \frac{e^{x}-e^{-x}}{2}\), we want to show that \(\sinh (-x) = -\sinh x\). Substituting \(-x\) into the definition we get \(\sinh (-x) = \frac{e^{-x}-e^{x}}{2}\), which is indeed the same as \(-\sinh x\), thereby showing that \(\sinh x\) is an odd function.
3Step 3: Prove \((\cosh x)^{2}-(\sinh x)^{2} = 1\)
Starting with the left hand side (LHS), \((\cosh x)^{2}-(\sinh x)^{2} = \left(\frac{e^{x}+e^{-x}}{2}\right)^{2} - \left(\frac{e^{x}-e^{-x}}{2}\right)^{2} = \frac{e^{2x} + 2 + e^{-2x}}{4} - \frac{e^{2x} - 2 + e^{-2x}}{4} = 1\), matching with the right hand side (RHS) and completing the proof.
Key Concepts
Even FunctionsOdd FunctionsIdentity ProofCollege Algebra
Even Functions
An even function is a special type of mathematical function that exhibits symmetry. Imagine a graph where if you fold it along the y-axis, the two halves would match perfectly. This is a signature property of even functions. For an even function, the equation can be defined as: - For any input value \( x \), \( f(-x) = f(x) \). This means that if you substitute \(-x\) into the function and it returns the same result as when \( x \) is substituted, the function is even.
In the case of hyperbolic functions, the hyperbolic cosine function, \( \cosh x \), is even. - Defined by \( \cosh x = \frac{e^{x} + e^{-x}}{2} \), - If you replace \( x \) with \( -x \), the expression remains unchanged, reaffirming its even nature.
In the case of hyperbolic functions, the hyperbolic cosine function, \( \cosh x \), is even. - Defined by \( \cosh x = \frac{e^{x} + e^{-x}}{2} \), - If you replace \( x \) with \( -x \), the expression remains unchanged, reaffirming its even nature.
Odd Functions
Odd functions are another category of functions known for their symmetry, but in a different way compared to even functions. Think about a graph of an odd function. If you rotate it 180 degrees around the origin, it will look unchanged. The defining property for odd functions is: - For any input value \( x \), \( f(-x) = -f(x) \). This indicates that substituting \(-x\) into the function should give the negative of the original output.
In hyperbolic functions, the hyperbolic sine function, \( \sinh x \), is odd. - Given by \( \sinh x = \frac{e^{x} - e^{-x}}{2} \), - When replaced with \(-x\), the outcome is \( -\sinh x \), confirming its odd function nature.
In hyperbolic functions, the hyperbolic sine function, \( \sinh x \), is odd. - Given by \( \sinh x = \frac{e^{x} - e^{-x}}{2} \), - When replaced with \(-x\), the outcome is \( -\sinh x \), confirming its odd function nature.
Identity Proof
In mathematics, proofs are important for verifying the accuracy of equations and identities. An identity proof shows that an equation holds true for all values within its domain.
For hyperbolic functions, an important identity is: - \( (\cosh x)^2 - (\sinh x)^2 = 1 \). To prove this identity, you start with the definitions: - \( \cosh x = \frac{e^{x} + e^{-x}}{2} \) and \( \sinh x = \frac{e^{x} - e^{-x}}{2} \). When you square these functions and subtract \( (\sinh x)^2 \) from \( (\cosh x)^2 \), complex computation simplifies to 1. This shows that the identity is indeed valid and demonstrates consistency in hyperbolic functions.
For hyperbolic functions, an important identity is: - \( (\cosh x)^2 - (\sinh x)^2 = 1 \). To prove this identity, you start with the definitions: - \( \cosh x = \frac{e^{x} + e^{-x}}{2} \) and \( \sinh x = \frac{e^{x} - e^{-x}}{2} \). When you square these functions and subtract \( (\sinh x)^2 \) from \( (\cosh x)^2 \), complex computation simplifies to 1. This shows that the identity is indeed valid and demonstrates consistency in hyperbolic functions.
College Algebra
College algebra is an essential subject for building a solid foundation in advanced mathematics. It involves studying a wide range of mathematical concepts, including functions, polynomials, equations, and so forth.
Hyperbolic functions, such as \( \cosh x \) and \( \sinh x \), are often part of college algebra. They are analogous to trigonometric functions but based on hyperbolas rather than circles.
Understanding these functions can involve:
Hyperbolic functions, such as \( \cosh x \) and \( \sinh x \), are often part of college algebra. They are analogous to trigonometric functions but based on hyperbolas rather than circles.
Understanding these functions can involve:
- Recognizing their properties, like being even or odd
- Deriving and proving identities, such as the one we explored: \( (\cosh x)^2 - (\sinh x)^2 = 1 \)
- Applying them in calculus and in real-world problems where these functions describe certain types of growth and decay
Other exercises in this chapter
Problem 93
Evaluate or simplify each expression without using a calculator. $$ e^{\ln 125} $$
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Solve each equation. $$ 5^{2 x} \cdot 5^{4 x}=125 $$
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Evaluate or simplify each expression without using a calculator. $$ e^{\ln 300} $$
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In Exercises \(89-102,\) determine whether each equation is true or false. Where possible, show work to support your conclusion. If the statement is false, make
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