17.2 Identities for Hyperbolic Functions
- 1.
- \(\cosh ^2x-\sinh ^2x=1\), equivalently \(\cosh ^{2}x=\sinh ^{2}x+1\)
- 2.
- \(\sinh 2x=2\sinh x\cosh x\)
- 3.
- \(\cosh 2x=\cosh ^2x+\sinh ^2x\)
- 4.
- \(\cosh ^2x=\frac {\cosh 2x+1}{2}\)
- 5.
- \(\sinh ^2x=\frac {\cosh 2x-1}{2}\)
- 6.
- \(\tanh ^2x=1-\text {sech}^2x\)
- 7.
- \(\coth ^2x=1+\text {csch}^2x\)
Note 17.3. The first identity is the hyperbolic counterpart of \(\cos ^{2}x+\sin ^{2}x=1\), and the change from a plus to a minus is the essential difference between the two families. It is also where the name comes from: the point \((\cosh t,\sinh t)\) satisfies \(X^{2}-Y^{2}=1\) and therefore traces a hyperbola, exactly as \((\cos t,\sin t)\) satisfies \(X^{2}+Y^{2}=1\) and traces a circle.
Verify it directly from the definitions, which takes one line: \[\cosh ^{2}x-\sinh ^{2}x =\frac {\left (e^{x}+e^{-x}\right )^{2}-\left (e^{x}-e^{-x}\right )^{2}}{4} =\frac {4e^{x}e^{-x}}{4}=1 .\]
Every other identity in the list matches the trigonometric one of the same name except for a sign, and the pattern behind those signs is Osborn’s rule, stated later in this chapter. It is why \(\cosh 2x=\cosh ^{2}x+\sinh ^{2}x\) carries a plus where \(\cos 2x=\cos ^{2}x-\sin ^{2}x\) carries a minus.
Solution.
\begin {align*} \sinh ^2x -4\cosh x +5 & = 0\\ \implies \sinh ^2x-4\cosh x & = -4\\ \implies (\cosh x-2)^2 & =-4+4\\ \implies \cosh x -2 & = 0\\\\ \therefore \quad \cosh x & =2 \end {align*}
Now, \(\cosh x =\frac {1}{2}(e^x+e^{-x})\)
\begin {align*} \displaystyle \implies \frac {1}{2}(e^x+e^{-x}) & =2\\ \implies e^x\{e^x+e^{-x} & = 4\}\\ \implies e^{2x}+1 & =4e^x\\\\ \therefore \quad e^{2x}-4e^x+1 &=0\\\\ \therefore \quad e^x=\frac {4\pm \sqrt {16-4}}{2}& =\frac {4\pm 2\sqrt {3}}{2}\\\\ \therefore \quad e^x &=2\pm \sqrt {3} \end {align*}
Thus, \(e^x=2+\sqrt {3}\) or \(e^x=2-\sqrt {3}\).
Introduce ’\(\ln \)’ to both sides, we have that
\(\ln e^x=\ln (2\pm \sqrt {3})\)
Hence, \(x=\ln (2\pm \sqrt {3})\).
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