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In each part. find the exact numerical value of the expression. (a) \(\sinh (\ln 3)\) (b) \(\cosh (-\ln 2)\) (c) \(\tanh (2 \ln 5)\) (d) \(\sinh (-3 \ln 2)\)

Short Answer

Expert verified
(a) \(\frac{4}{3}\), (b) \(\frac{5}{4}\), (c) \(\frac{312}{313}\), (d) \(-\frac{63}{16}\).

Step by step solution

01

Understand Hyperbolic Functions

Hyperbolic sine, cosine, and tangent functions are defined in terms of exponential functions. Specifically, they are:- \( \sinh(x) = \frac{e^x - e^{-x}}{2} \)- \( \cosh(x) = \frac{e^x + e^{-x}}{2} \)- \( \tanh(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}} \). These will be used to find the numerical values in parts (a), (b), (c), and (d).
02

Calculate \( \sinh(\ln 3) \)

Using the formula for \( \sinh(x) \):\[ \sinh(\ln 3) = \frac{e^{\ln 3} - e^{-\ln 3}}{2} \]Since \( e^{\ln 3} = 3 \) and \( e^{-\ln 3} = \frac{1}{3} \), it becomes:\[ \sinh(\ln 3) = \frac{3 - \frac{1}{3}}{2} = \frac{\frac{9}{3} - \frac{1}{3}}{2} = \frac{\frac{8}{3}}{2} = \frac{8}{6} = \frac{4}{3} \]
03

Calculate \( \cosh(-\ln 2) \)

Using the formula for \( \cosh(x) \):\[ \cosh(-\ln 2) = \frac{e^{-\ln 2} + e^{\ln 2}}{2} \]Since \( e^{\ln 2} = 2 \) and \( e^{-\ln 2} = \frac{1}{2} \), it becomes:\[ \cosh(-\ln 2) = \frac{\frac{1}{2} + 2}{2} = \frac{\frac{1}{2} + \frac{4}{2}}{2} = \frac{\frac{5}{2}}{2} = \frac{5}{4} \]
04

Calculate \( \tanh(2 \ln 5) \)

Using the formula for \( \tanh(x) \):\[ \tanh(2 \ln 5) = \frac{e^{2 \ln 5} - e^{-2 \ln 5}}{e^{2 \ln 5} + e^{-2 \ln 5}} \]Calculate \( e^{2 \ln 5} = (e^{\ln 5})^2 = 5^2 = 25 \) and \( e^{-2 \ln 5} = \frac{1}{25} \). Thus, it becomes:\[ \tanh(2 \ln 5) = \frac{25 - \frac{1}{25}}{25 + \frac{1}{25}} = \frac{\frac{625}{25} - \frac{1}{25}}{\frac{625}{25} + \frac{1}{25}} = \frac{\frac{624}{25}}{\frac{626}{25}} = \frac{624}{626} = \frac{312}{313} \]
05

Calculate \( \sinh(-3 \ln 2) \)

Using the formula for \( \sinh(x) \):\[ \sinh(-3 \ln 2) = \frac{e^{-3 \ln 2} - e^{3 \ln 2}}{2} \]Calculate \( e^{3 \ln 2} = (e^{\ln 2})^3 = 2^3 = 8 \) and \( e^{-3 \ln 2} = \frac{1}{8} \). Therefore:\[ \sinh(-3 \ln 2) = \frac{\frac{1}{8} - 8}{2} = \frac{\frac{1}{8} - \frac{64}{8}}{2} = \frac{-\frac{63}{8}}{2} = -\frac{63}{16} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

hyperbolic sine
The hyperbolic sine function, represented as \( \sinh(x) \), is a fundamental hyperbolic function. It is similar to the regular sine function in trigonometry but defined using exponential functions. The formula for the hyperbolic sine function is given by:
  • \( \sinh(x) = \frac{e^x - e^{-x}}{2} \)
This formula involves two exponential terms \( e^x \) and \( e^{-x} \). The difference between these terms, divided by two, gives us the value of \( \sinh(x) \).

The hyperbolic sine has applications in describing hanging cables and even in the theory of special relativity. Understanding \( \sinh(x) \) requires grasping how exponential growth and decay contribute to its output.

For instance, to evaluate \( \sinh(\ln 3) \), you substitute \( \ln 3 \) into the equation, finding that \( e^{\ln 3} = 3 \) and \( e^{-\ln 3} = \frac{1}{3} \). Therefore, \( \sinh(\ln 3) \) simplifies to \( \frac{3 - \frac{1}{3}}{2} = \frac{4}{3} \). This demonstrates how the hyperbolic sine function can express numbers in terms of exponential growth and contraction.
hyperbolic cosine
The hyperbolic cosine function, or \( \cosh(x) \), is closely related to the hyperbolic sine, and it resembles the cosine function from trigonometry. Its formula incorporates exponential functions, designed to always yield non-negative values. It is expressed as:
  • \( \cosh(x) = \frac{e^x + e^{-x}}{2} \)
Notice how both exponential terms are positive and summed up, divided by two. This results in \( \cosh(x) \) being always non-negative.

\( \cosh(x) \) finds utility in physics and engineering, for example, in modeling hyperbolic structures. To calculate \( \cosh(-\ln 2) \), you plug \( \ln 2 \) into the formula. By realizing that \( e^{\ln 2} = 2 \) and \( e^{-\ln 2} = \frac{1}{2} \), the calculation simplifies to \( \frac{\frac{1}{2} + 2}{2} = \frac{5}{4} \). This simplification illustrates how exponential transformations help reveal fundamental properties within these hyperbolic functions.
hyperbolic tangent
The hyperbolic tangent, denoted as \( \tanh(x) \), connects to hyperbolic sine and cosine much like tangent in trigonometric terms. It provides the ratio of \( \sinh(x) \) to \( \cosh(x) \), derived using exponential functions:
  • \( \tanh(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}} \)
This creates a measure that ranges between -1 and 1, useful in various hyperbolic geometry and hyperbolic models for phenomena in nature.

When calculating \( \tanh(2 \ln 5) \), first evaluate \( e^{2 \ln 5} = 25 \) and \( e^{-2 \ln 5} = \frac{1}{25} \). Inserting these into the equation yields \( \frac{25 - \frac{1}{25}}{25 + \frac{1}{25}} \), simplifying further to \( \frac{312}{313} \). This simplification shows the balance between hyperbolic growth and attenuation captured by \( \tanh(x) \).

The hyperbolic tangent has a wide array of applications in fields such as neural networks, where it helps in modeling activation functions.
exponential functions
Exponential functions, central to calculus and many scientific fields, provide the backbone for all hyperbolic functions. Defined by equations of the form \( e^x \), where \( e \) represents Euler's number (approximately 2.71828), these functions display continuous growth or decay patterns.
  • Exponential Growth: In \( e^x \), the value grows rapidly towards infinity as \( x \) increases.
  • Exponential Decay: In \( e^{-x} \), the value diminishes towards zero as \( x \) increases.
Exponential functions are the essence of natural growth processes, how populations multiply, and how radioactive elements decay.

In hyperbolic functions, exponential functions create relationships mimicking natural growth and decay. For instance, \( \sinh(x) \) and \( \cosh(x) \) transform exponential differences and sums into expressions that describe hyperbolic geometry.
When solving problems like our example exercise, recognizing how logarithmic counterparts (\( \ln x \)) reverse exponential transformations ((\( e^{\ln x} = x \))) is vital. Understanding these connections is fundamental in math and science disciplines, enriching your comprehension of natural and engineered systems.

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