Do the following: a. Write \((\cosh x-\sinh x)^{6}\) in terms of exponentials. b. Prove \(\cosh (x-y)=\cosh x \cosh y-\sinh x \sinh y\) using the exponential forms of the hyperbolic functions. c. Prove \(\cosh 2 x=\cosh ^{2} x+\sinh ^{2} x\) d. If \(\cosh x=\frac{13}{12}\) and \(x<0\), find \(\sinh x\) and \(\tanh x\). e. Find the exact value of \(\sinh (\operatorname{arccosh} 3)\).

Short Answer

Expert verified
a. \(e^{-6x}\) b. \(cosh (x-y) = cosh x cosh y - sinh x sinh y \) is true. c. \(cosh 2 x = cosh ^{2} x + sinh ^{2} x\) is true. d. \(sinh x = -\frac{5}{12}\), \(tanh x = -\frac{5}{13}\). e. \(sinh (arccosh 3) = \sqrt{8}\).

Step by step solution

01

a. Writing hyperbolic function in terms of exponentials

Recall that \( \cosh x = \frac{e^x + e^{-x}}{2} \) and \( \sinh x = \frac{e^x - e^{-x}}{2} \). Substituting these into \((\cosh x - \sinh x)^6\) gives: \( (\frac{e^x + e^{-x}-e^x + e^{-x}}{2})^6 = (\frac{2 e^{-x}}{2})^6 = e^{-6x}\).
02

b. Proving an identity using exponential forms

Using the definition of \( \cosh x \) and \( \sinh x \) and the exponentiation rule, we get: \( \cosh (x-y)=\frac{e^{x-y} + e^{-(x-y)}}{2} =\frac{e^x}{e^y}+\frac{e^y}{e^x} =\frac{e^{2x} + 1}{2 e^x}=\cosh x \cosh y - \sinh x \sinh y \).
03

c. Proving an identity involving squares of hyperbolic functions

Applying the definitions from above, we derive \( \cosh^{2} x - \sinh^{2} x = (\frac{e^x + e^{-x}}{2})^2 - (\frac{e^x - e^{-x}}{2})^2 = 1 = \cosh 2x \).
04

d. Calculating other hyperbolic functions given a value for \( \cosh x \)

To find \( \sinh x \), we know that \( \cosh^2 x - \sinh^2 x = 1 \), so \( \sinh x = \sqrt{\cosh^2 x - 1} = \sqrt{(\frac{13}{12})^2 - 1 } = -\frac{5}{12} \). For \( \tanh x \), we know that \( \tanh x = \frac{\sinh x}{\cosh x} = \frac{-5/12}{13/12} = -5/13. \)
05

e. Calculating hyperbolic function of inverse hyperbolic function

Recall that \( \sinh(\operatorname{arccosh} x) = \sqrt{x^2 - 1} \). Thus, \( \sinh(\operatorname{arccosh} 3) = \sqrt{3^2 - 1} = \sqrt{8}.\)

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