# Hyperbolic Functions Formula

Hyperbolic functions refer to the exponential functions that share similar properties to trigonometric functions. These functions are analogous trigonometric functions in that they are named the same as trigonometric functions with the letter ‘h’ appended to each name.  These have the same relationship to the hyperbola that trigonometric functions have to the circle. Thus, they are collectively known as hyperbolic functions and are individually called hyperbolic sine, hyperbolic cosine, and so on. In addition to modeling, they can be used as solutions to some types of partial differential equations. Let us start learning the Hyperbolic functions formula.

## Equation of Hyperbola

A hyperbola is a plane curve that is generated by a point so moving that the difference of the distances from two fixed points is constant. The two fixed points are the foci and the mid-point of the line segment joining the foci is the center of the hyperbola. Transverse axis is the line through the foci. The conjugate axis is the line through the center and perpendicular to the transverse axis.

The vertices of the hyperbola are the points at which the hyperbola intersects the transverse axis. 2c is the distance between the two foci. The distance between the two vertices is 2a. 2a s is also the length of the transverse axis. The length of the conjugate axis is 2b. The value of b is $$sqrt{ (c^2 – a^2)}$$

Source: en.wikipedia.org

## Definition of Hyperbolic Functions

• Hyperbolic sine of x: $$\sinh x = \frac{(e^x – e^{-x})}{2}$$
• Hyperbolic cosine of x: $$\cosh x = \frac{ (e^x – e^{-x})}{2}$$
• Hyperbolic tangent of x: $$\tanh x = \frac {(e^x – e^{-x})}{((e^x – e^{-x})}$$
• Hyperbolic cotangent of x: $$\coth x = \frac {(e^x – e^{-x})}{(e^x – e^{-x})}$$
• Hyperbolic secant of x: $$\sec h x = \frac {2}{(e^x – e^{-x})}$$
• Hyperbolic cosecant of x: $$csc h x = \frac {2}{(e^x – e^{-x})}$$

## Relationships Among Hyperbolic Functions

• $$\tanh x = \frac \sinh x/\cosh x$$
• $$\coth x = \frac{1}{\tanh x} = \frac{\cosh x}{\sinh x}$$
• $$\sec h x = \frac {1}{\cos h x}$$
• $$\csc h x = \frac {1}{\sinh x}$$
• $$\cosh^2x – \sinh^2x = 1$$
• $$\sec h^2x + \tan h^2x = 1$$
• $$\cot h^2x – \csc h^2x = 1$$

• $$\sinh (x ± y) = \sinh x \times \cosh y ± \cosh x\times \sinh y$$
• $$\cosh (x ± y) = \cosh x \times \cosh y ± \sinh x \times \sinh y$$
• $$\tanh(x ± y) = \frac {(\tanh x ± \tanh y)}{(1 ± \tanh x\times\tanh y)}$$
• $$\coth(x ± y) = \frac {(\coth x \times \coth y ± l)}{(\coth y ± \coth x)}$$

## Sum, Difference And Product Of Hyperbolic Functions

• $$\sinh x + \sinh y = 2 \sinh \frac{1}{2}(x + y) \cosh \frac{1}{2}(x – y)$$
• $$\sinh x – \sinh y = 2 \cosh \frac{1}{2}(x + y) \sinh \frac{1}{2}(x – y)$$
• $$\cosh x + \cosh y = 2 \cosh \frac{1}{2}(x + y) \cosh \frac{1}{2}(x – y)$$
• $$\cosh x – \cosh y = 2 \sinh \frac{1}{2}(x + y) \sinh \frac{1}{2}(x — y)$$
• $$\sinh x \sinh y = \frac{1}{2}(\cosh (x + y) – \cosh (x – y))$$
• $$\cosh x \cosh y = \frac{1}{2}(\cosh (x + y) + \cosh (x — y))$$
• $$\sinh x \cosh y = \frac{1}{2}(\sinh (x + y) + \sinh (x – y))$$

## Derivative for Hyperbolic Functions

• $$\frac{\mathrm{d}}{\mathrm{d} x}\sinh x = \cosh x$$
• $$\frac{\mathrm{d}}{\mathrm{d} x}\cosh x = \sinh x$$
• $$\frac{\mathrm{d}}{\mathrm{d} x}\tan h x = \sec h^2 x$$
• $$\frac{\mathrm{d}}{\mathrm{d} x}\sec h x = -\sec h x \times \tan h x)$$
• $$\frac{\mathrm{d}}{\mathrm{d} x}\coth x = -\csc h^2 x$$
• $$\frac{\mathrm{d}}{\mathrm{d} x}\csc h x = – \csc h x \times \cot h x$$

## Solved Examples for Hyperbolic Functions Formula

Q.1: Find addition identities for \tanh(x-y) using the hyperbolic functions formula

Solution: $$\tanh(x-y)=\frac{\sinh(x-y)}{cosh(x-y)}$$

As,  $$\sinh(x-y)=\sinh x\cosh y-\cosh x\sinh y and \cosh(x-y)=\cosh x\cosh y-\sinh x\sinh y$$

$$\tanh(x-y)=\frac{\sinh x\cosh y-\cosh x\sinh y}{\cosh x\cosh y-sinh x\sinh y}$$

Dividing numerator and denominator by $$\cosh x\cosh y$$

$$\tanh(x-y) = \frac{\frac{\sinh x\cosh y-\cosh x\sinh y}{\cosh x\cosh y}}{\frac{\cosh x\cosh y – sinh x\sinh y}{\cosh x\cosh y}}$$

$$\tan h(x-y) =\frac{\tan h x-\tan h y}{1-\tan h x \times \tan h y}$$

Q.2. Find $$\frac{\mathrm{d}}{\mathrm{d} t} coth e^{2tx}.$$

Solution: $$\frac{\mathrm{d} }{\mathrm{d} t}coth e^{2tx}$$

= $$(– \csc h^{2} e^{2tx})e^{2tx}(2x)$$

= $$\frac{– 2xe^{2tx}}{csch2 e^{2tx}}$$

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KUCKOO B

I get a different answer for first example.
I got Q1 as 20.5
median 23 and
Q3 26

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Yashitha

Hi
Same

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virat

yes