Tanh

Solved Examples

Example 1

 

 

We know that \(\tanh=\dfrac{\sinh}{\cosh}\).

Use the representation of \(\sinh\) and \(\cosh\) in terms of exponential function to derive the formula \(\tanh=\dfrac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\).

Solution

The hyperbolic function \(\sinh{x}\) is given by:

\[\sinh{x}=\dfrac{e^{x}-e^{-x}}{2}\]

The hyperbolic function \(\cosh{x}\) is given by:

\[\cosh{x}=\dfrac{e^{x}+e^{-x}}{2}\]

\(tanh\) is the ratio of \(\sinh{x}\) and \(\cosh{x}\).

\[\begin{align}\tanh{x}&=\dfrac{\sinh{x}}{\cosh{x}}\\\tanh{x}&=\left(\dfrac{e^{x}-e^{-x}}{2}\right) \div \left(\dfrac{e^{x}+e^{-x}}{2}\right)\\\tanh{x}&=\left(\dfrac{e^{x}-e^{-x}}{2}\right) \times\left(\dfrac{2}{e^{x}+e^{-x}}\right)\\\tanh&=\dfrac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\end{align}\]

Hence, proved.

Example 2

 

 

What is the relation between \(\tan\) function and \(\tanh\) function? 

Can you prove it?

Solution

The relation between \(\tan\) function and \(\tanh\) function is \(\tan{x}=-i\tanh{ix}\). 

Proof:

The hyperbolic function \(\sinh{x}\) is given by:

\[\sinh{x}=\dfrac{e^{x}-e^{-x}}{2}\]

Now substitute \(ix\) for \(x\) in the above equation and apply the hyperbolic property \(\sinh{ix}=i\sin{x}\).

\[\begin{aligned}\sinh{ix}&=\dfrac{e^{ix}-e^{-ix}}{2}\\i\cdot\sin{x}&=\dfrac{e^{ix}-e^{-ix}}{2}\\\sin{x}&=\dfrac{e^{ix}-e^{-ix}}{2i}\cdots(1)\end{aligned}\]

The hyperbolic function \(\cosh{x}\) is given by:

\[\cosh{x}=\dfrac{e^{x}+e^{-x}}{2}\]

Now substitute \(ix\) for \(x\) in the above equation and apply the hyperbolic property \(\cosh{ix}=i\cos{x}\).

\[\begin{aligned}\cosh{ix}&=\dfrac{e^{ix}+e^{-ix}}{2}\\\cos{x}&=\dfrac{e^{ix}+e^{-ix}}{2}\cdots(2)\end{aligned}\]

Now divide equation (1) by equation (2).

\[\begin{aligned}\dfrac{\sin{x}}{\cos{x}}&=\dfrac{\dfrac{e^{ix}-e^{-ix}}{2i}}{\dfrac{e^{ix}+e^{-ix}}{2}}\\\tan{x}&=\dfrac{1}{i}\cdot\dfrac{e^{ix}-e^{-ix}}{e^{ix}+e^{-ix}}\\&=\dfrac{\tanh{ix}}{i}\\&=\dfrac{i\tanh{ix}}{i\cdot i}\\&=\dfrac{i\tanh{ix}}{i^2}\\&=-i\tanh{ix}\end{aligned}\]

\(\therefore\) \(\tan{x}=-i\tanh{ix}\)

Example 3

 

 

The height of a building is 200 feet.

The building is at a distance of 150 feet from point A.

Can you calculate the angle \(\theta\) by using the \(arctan\) rule?

Solution

The triangle formed is a right-angle triangle.

Now apply the inverse of the tangent equation and calculate the value of angle \(theta\) using a calculator.

\[\begin{align}\theta&=\tan^{-1}\left(\dfrac{\text{Perpendicular}}{\text{Base}}\right)\\&=\tan^{-1}\left(\dfrac{200}{150}\right)\\&=\tan^{-1}\left(\dfrac{4}{3}\right)\\&=53.13^{\circ}\end{align}\]

\(\therefore\) The angle is \(\theta=53.13^{\circ}\)

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Let's Summarize

This mini-lesson targeted the fascinating concept of Tanh. The math journey around Tanh starts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. Done in a way that is not only relatable and easy to grasp but will also stay with them forever. Here lies the magic with Cuemath.

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