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Given that\[\cosh x=\dfrac{5}{4}\], determine the value of \[\cosh 2x\]. Use the formula for \[\cosh \left( 2x+x \right)\] to determine the value of \[\cosh 3x\].

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Last updated date: 16th Jul 2024
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Answer
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Hint: In order to find the value of \[\cosh 2x\] and \[\cosh 3x\], firstly, we have to find the value of \[\cosh 2x\] by evaluating it in terms of \[\cosh x=\dfrac{5}{4}\] and then we must find the exponential function values from \[\cosh x=\dfrac{5}{4}\]. Then we will be finding the value of \[\cosh 3x\] by splitting it into \[\cosh \left( 2x+x \right)\] by evaluating each term of the formula of \[\cosh \left( 2x+x \right)\].

Complete step by step answer:
Now let us learn about the hyperbolic functions. The hyperbolic functions are the analogues of trigonometric functions but we will be using hyperbola instead of circle. The domain of the various functions varies. We can easily obtain the derivative formula for the hyperbolic tangent. Inverse hyperbolic functions are also called area hyperbolic functions.
Now let us start solving our given problem.
Let us find the value of \[\cosh 2x\].
We are given that, \[\cosh x=\dfrac{5}{4}\]
We know that, \[\cosh 2x=2{{\left( \cosh x \right)}^{2}}-1\]
Upon substituting the value, we get
\[\begin{align}
  & \Rightarrow \cosh 2x=2{{\left( \cosh x \right)}^{2}}-1 \\
 & \Rightarrow \cosh 2x=2{{\left( \dfrac{5}{4} \right)}^{2}}-1 \\
 & \Rightarrow \cosh 2x=2\left( \dfrac{25}{16} \right)-1 \\
 & \Rightarrow \cosh 2x=\dfrac{25-8}{8}=\dfrac{17}{8} \\
\end{align}\]
Now let us find the exponential function of the given hyperbolic function \[\cosh x=\dfrac{5}{4}\].
We have, \[\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}=\dfrac{5}{4}\]
Upon solving it, we get
\[\begin{align}
  & \dfrac{{{e}^{x}}+{{e}^{-x}}}{2}=\dfrac{5}{4} \\
 & \Rightarrow 4{{e}^{x}}+4{{e}^{-x}}=10 \\
 & \Rightarrow 4{{e}^{2x}}-10ex+4=0 \\
 & \Rightarrow {{e}^{x}}=\dfrac{10\pm \sqrt{100-64}}{8} \\
 & \Rightarrow {{e}^{x}}=2or\dfrac{1}{2} \\
\end{align}\]
Now let us find the value of \[\cosh 3x\] by splitting it in the form of \[\cosh \left( 2x+x \right)\].
We know that, \[\sinh x=\dfrac{3}{4}or-\dfrac{3}{4}\]
Then, \[\sinh 2x=2\cosh x\sinh x\]
Upon solving it, we get
\[\begin{align}
  & \sinh 2x=2\cosh x\sinh x \\
 & \Rightarrow \sinh 2x=2\left( \dfrac{5}{4} \right)\left( \dfrac{3}{4} \right)=\dfrac{15}{8} \\
\end{align}\]
Now, we have
\[\begin{align}
  & \cosh 3x=\cosh \left( 2x+x \right)=\left( \cosh 2x \right)\left( \cosh x \right)+\left( \sinh 2x \right)\left( \sinh x \right) \\
 & \Rightarrow \cosh 3x=\dfrac{17}{8}\left( \dfrac{5}{4} \right)+\dfrac{15}{8}\left( \dfrac{3}{4} \right) \\
 & \Rightarrow \cosh 3x=\dfrac{5}{4} \\
\end{align}\]
\[\therefore \] \[\cosh 3x=\dfrac{5}{4}\]

Note: We can apply these hyperbolic functions in our everyday life. For example, hyperbolic cosine function can be used in describing the shape of the curve formed by the voltage line formed between two towers. These functions can also be used in defining a measure of some of the non-Euclidean geometry.