Problem 12
Question
Use the definitions of cosh x and sinh x to show that $$\cosh ^{2} x-\sinh ^{2} x=1$$
Step-by-Step Solution
Verified Answer
The identity \( \cosh^2 x - \sinh^2 x = 1 \) is proven using the definitions of \( \cosh x \) and \( \sinh x \).
1Step 1: Recall Definitions
Recall the definitions of hyperbolic cosine and hyperbolic sine: \( \cosh x = \frac{e^x + e^{-x}}{2} \) and \( \sinh x = \frac{e^x - e^{-x}}{2} \). We will use these definitions in the next steps.
2Step 2: Square Each Function
Calculate \( \cosh^2 x \) and \( \sinh^2 x \): \( \cosh^2 x = \left( \frac{e^x + e^{-x}}{2} \right)^2 = \frac{(e^x + e^{-x})^2}{4} \) and \( \sinh^2 x = \left( \frac{e^x - e^{-x}}{2} \right)^2 = \frac{(e^x - e^{-x})^2}{4} \).
3Step 3: Expand Squares
Expand \((e^x + e^{-x})^2\) and \((e^x - e^{-x})^2\): \( (e^x + e^{-x})^2 = e^{2x} + 2 + e^{-2x} \) and \( (e^x - e^{-x})^2 = e^{2x} - 2 + e^{-2x} \).
4Step 4: Substitute Results
Substitute the results from Step 3 back into the expressions for \( \cosh^2 x \) and \( \sinh^2 x \): \( \cosh^2 x = \frac{e^{2x} + 2 + e^{-2x}}{4} \) and \( \sinh^2 x = \frac{e^{2x} - 2 + e^{-2x}}{4} \).
5Step 5: Calculate \( \cosh^2 x - \sinh^2 x \)
Subtract \( \sinh^2 x \) from \( \cosh^2 x \): \[ \cosh^2 x - \sinh^2 x = \frac{(e^{2x} + 2 + e^{-2x}) - (e^{2x} - 2 + e^{-2x})}{4} = \frac{4}{4} = 1 \].
6Step 6: Conclude the Identity
Thus, by simplifying the expression, we have shown that \( \cosh^2 x - \sinh^2 x = 1 \). This confirms the identity.
Key Concepts
cosh (hyperbolic cosine)sinh (hyperbolic sine)hyperbolic identities
cosh (hyperbolic cosine)
The hyperbolic cosine function, represented as \( \cosh x \), is a fundamental building block of hyperbolic functions. It's similar to the cosine function in trigonometry but adapted for hyperbolic geometry. The formula for \( \cosh x \) is given by:
\( \cosh x \) describes the shape of a hanging cable or chain, known as a catenary. In its properties, it shows exponential growth as \( |x| \) increases.
In terms of application, \( \cosh x \) frequently occurs in calculations involving the angles or slopes of hyperbolic relationships in real-world physics and engineering problems.
- \( \cosh x = \frac{e^x + e^{-x}}{2} \)
\( \cosh x \) describes the shape of a hanging cable or chain, known as a catenary. In its properties, it shows exponential growth as \( |x| \) increases.
In terms of application, \( \cosh x \) frequently occurs in calculations involving the angles or slopes of hyperbolic relationships in real-world physics and engineering problems.
sinh (hyperbolic sine)
The hyperbolic sine function, denoted by \( \sinh x \), pairs with the hyperbolic cosine to describe hyperbolic functions. It mirrors the sine function from trigonometry but for hyperbolic spaces.
In practice, \( \sinh x \) is used to model phenomena such as the shape of a flexible beam subject to linear distributed loads.
Unlike the hyperbolic cosine, which has a minimum value of one, \( \sinh x \) can take any real number value, reflecting both positive and negative growth as \( x \) changes. It grows exponentially like \( \cosh x \) but without a horizontal baseline shift.
- \( \sinh x = \frac{e^x - e^{-x}}{2} \)
In practice, \( \sinh x \) is used to model phenomena such as the shape of a flexible beam subject to linear distributed loads.
Unlike the hyperbolic cosine, which has a minimum value of one, \( \sinh x \) can take any real number value, reflecting both positive and negative growth as \( x \) changes. It grows exponentially like \( \cosh x \) but without a horizontal baseline shift.
hyperbolic identities
Hyperbolic identities are various equations that involve hyperbolic functions, much like trigonometric identities involve sine and cosine. One central identity is:
The proof of this hyperbolic identity leverages the definitions of \( \cosh x \) and \( \sinh x \):
- \( \cosh^2 x - \sinh^2 x = 1 \)
The proof of this hyperbolic identity leverages the definitions of \( \cosh x \) and \( \sinh x \):
- Square \( \cosh x \) and \( \sinh x \)
- Plug in: \( \cosh^2 x = \frac{(e^x + e^{-x})^2}{4} \) and \( \sinh^2 x = \frac{(e^x - e^{-x})^2}{4} \)
- Simplify to show the identity
Other exercises in this chapter
Problem 11
In Exercises \(7-38,\) find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$ y=\ln \frac{3}{x} $$
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When is a polynomial \(f(x)\) of smaller order than a polynomial \(g(x)\) as \(x \rightarrow \infty ?\) Give reasons for your answer.
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Solve the differential equations in Exercises \(9-22\) $$\frac{d y}{d x}=3 x^{2} e^{-y}$$
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Use l'Hopital's rule to find the limits in Exercises \(7-50\) . $$ \lim _{x \rightarrow \infty} \frac{x-8 x^{2}}{12 x^{2}+5 x} $$
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