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What is the identity for hyperbolic functions?

The identity for hyperbolic functions is given by

cosh2(x)sinh2(x)=1.\cosh^2(x) - \sinh^2(x) = 1.

Hyperbolic functions serve as an analogy to trigonometric functions and are defined using the exponential function exe^x and its inverse, ln(x)\ln(x). The two primary hyperbolic functions are the hyperbolic sine, denoted as sinh(x)\sinh(x), and the hyperbolic cosine, denoted as cosh(x)\cosh(x).

The identity

cosh2(x)sinh2(x)=1\cosh^2(x) - \sinh^2(x) = 1

is reminiscent of the Pythagorean identity for trigonometric functions, which states

sin2(x)+cos2(x)=1.\sin^2(x) + \cos^2(x) = 1.

To prove the hyperbolic identity, we start with the definitions of cosh(x)\cosh(x) and sinh(x)\sinh(x):

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

Substituting these definitions into the left-hand side of our identity yields:

cosh2(x)sinh2(x)=(ex+ex2)2(exex2)2.\cosh^2(x) - \sinh^2(x) = \left(\frac{e^x + e^{-x}}{2}\right)^2 - \left(\frac{e^x - e^{-x}}{2}\right)^2.

Next, we expand the squares and simplify:

cosh2(x)sinh2(x)=(ex+ex)24(exex)24=e2x+2+e2x4e2x2+e2x4=44=1.\begin{align*} \cosh^2(x) - \sinh^2(x) & = \frac{(e^x + e^{-x})^2}{4} - \frac{(e^x - e^{-x})^2}{4} \\ & = \frac{e^{2x} + 2 + e^{-2x}}{4} - \frac{e^{2x} - 2 + e^{-2x}}{4} \\ & = \frac{4}{4} \\ & = 1. \end{align*}

Thus, we conclude that the identity

cosh2(x)sinh2(x)=1\cosh^2(x) - \sinh^2(x) = 1

holds true for all values of xx. This identity is valuable for simplifying expressions that involve hyperbolic functions and has practical applications in fields such as physics and engineering.

Answered by: Dr. Michael Green
A-Level Maths Tutor
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