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How to derive the formulas for hyperbolic cotangent?

The formula for the hyperbolic cotangent is derived from the definitions of hyperbolic functions.

The hyperbolic cotangent, denoted as coth(x)\text{coth}(x), is defined as the ratio of the hyperbolic cosine and hyperbolic sine functions:

coth(x)=cosh(x)sinh(x)\text{coth}(x) = \frac{\cosh(x)}{\sinh(x)}

The hyperbolic cosine and sine functions are defined as follows:

cosh(x)=ex+ex2\cosh(x) = \frac{e^x + e^{-x}}{2} sinh(x)=exex2\sinh(x) = \frac{e^x - e^{-x}}{2}

By substituting these definitions into the formula for coth(x)\text{coth}(x), we get:

coth(x)=ex+ex2exex2=ex+exexex\text{coth}(x) = \frac{\frac{e^x + e^{-x}}{2}}{\frac{e^x - e^{-x}}{2}} = \frac{e^x + e^{-x}}{e^x - e^{-x}}

To simplify this expression, we can multiply both the numerator and the denominator by exe^x:

coth(x)=e2x+1e2x1\text{coth}(x) = \frac{e^{2x} + 1}{e^{2x} - 1}

This provides us with the formula for the hyperbolic cotangent. Additionally, it can also be expressed in terms of exponential functions as follows:

coth(x)=ex+exexex=e2x+1e2x1=1+e2xe2x1\text{coth}(x) = \frac{e^x + e^{-x}}{e^x - e^{-x}} = \frac{e^{2x} + 1}{e^{2x} - 1} = \frac{1 + e^{-2x}}{e^{-2x} - 1}

These various forms highlight the relationships between the hyperbolic cotangent and exponential functions.

Answered by: Dr. Sarah Wilson
A-Level Physics Tutor
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