Chapter 6: Problem 2
Approximate the expression to four decimal places. $$ \begin{array}{llll}{\text { (a) } \operatorname{csch}(-1)} & {\text { (b) } \operatorname{sech}(\ln 2)} & {\text { (c) } \operatorname{coth} 1} & {\text { . }}\end{array} $$ $$ \text { (d) } \operatorname{sech}^{-1} \frac{1}{2} \quad \text { (e) } \operatorname{coth}^{-1} 3 \quad \text { (f) } \operatorname{csch}^{-1}(-\sqrt{3}) $$
Short Answer
Step by step solution
Calculate \( \operatorname{csch}(-1) \)
Calculate \( \operatorname{sech}(\ln 2) \)
Calculate \( \operatorname{coth}(1) \)
Calculate \( \operatorname{sech}^{-1}\left(\frac{1}{2}\right) \)
Calculate \( \operatorname{coth}^{-1}(3) \)
Calculate \( \operatorname{csch}^{-1}(-\sqrt{3}) \)
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Inverse Hyperbolic Functions
- The inverse hyperbolic secant, written as \( \operatorname{sech}^{-1}(x) \), finds \( y \) such that \( \operatorname{sech}(y) = x \).
- The inverse hyperbolic cotangent, \( \operatorname{coth}^{-1}(x) \), is used to find the value such that \( \operatorname{coth}(y) = x \).
- The inverse hyperbolic cosecant, \( \operatorname{csch}^{-1}(x) \), gives the variable value where \( \operatorname{csch}(y) = x \).
Hyperbolic Trigonometric Identities
- \( \sinh(-x) = -\sinh(x) \) and \( \cosh(-x) = \cosh(x) \), reflecting their hyperbolic nature.
- \( \cosh^2(x) - \sinh^2(x) = 1 \), analogous to the Pythagorean identity in trigonometry.
- \( \tanh(x) = \frac{\sinh(x)}{\cosh(x)} \), similar to \( \tan(x) = \frac{\sin(x)}{\cos(x)} \).
Approximation Methods
- The Taylor series expansion can approximate expressions like \( \,\sinh(x) \approx x + \frac{x^3}{3!} + \frac{x^5}{5!} + \ldots \).
- When calculating \( \cosh(\ln 2) \), one may compute or use known identities to get an approximate value, as precise calculation may require iterative computation.
- For \( \operatorname{csch}(-1) \), using \( \sinh(1) = \frac{e^1 - e^{-1}}{2} \approx 1.1752 \) allows finding \( \operatorname{csch}(-1) \approx -0.8509 \).