How to Prove the Fundamental Hyperbolic Identity

The phrase “hyperbolic function identity” does not specify one particular formula. The most common foundational identity is $\cosh^2 x-\sinh^2 x=1.$ This article proves it from the definitions and shows how the same method helps verify related identities. If you have a different formula in mind, the proof may use a different starting point.

The identity and its definitions

For real $x$, the hyperbolic sine and cosine are defined by

$$\sinh x=\frac{e^x-e^{-x}}{2},\qquad \cosh x=\frac{e^x+e^{-x}}{2}.$$

These definitions resemble the exponential formulas for sine and cosine, but the signs differ. In particular, the fundamental identity has a minus sign between the squares, unlike the circular-trigonometric identity $\cos^2 x+\sin^2 x=1$.

Proof using exponentials

Substitute the definitions directly, then expand the squares:

$$\begin{aligned} \cosh^2 x-\sinh^2 x &=\left(\frac{e^x+e^{-x}}{2}\right)^2-\left(\frac{e^x-e^{-x}}{2}\right)^2\\ &=\frac{(e^{2x}+2+e^{-2x})-(e^{2x}-2+e^{-2x})}{4}\\ &=\frac{4}{4}\\ &=1. \end{aligned}$$

The terms $e^{2x}$ and $e^{-2x}$ cancel. The middle terms add to $4$ because the second squared expression is being subtracted. This proves the identity for every real $x$; the same algebra also works for complex $x$.

Worked example

Take $x=\ln 2$. Since $e^x=2$ and $e^{-x}=1/2$,

$$\cosh(\ln 2)=\frac{2+1/2}{2}=\frac54, \qquad \sinh(\ln 2)=\frac{2-1/2}{2}=\frac34.$$

Then

$$\cosh^2(\ln 2)-\sinh^2(\ln 2) =\left(\frac54\right)^2-\left(\frac34\right)^2 =\frac{25-9}{16}=1.$$

This numerical check illustrates the identity, while the exponential derivation proves it for all $x$.

Deriving other hyperbolic identities

Since $\cosh x$ is positive for real $x$, you can divide by it. Dividing the fundamental identity by $\cosh^2 x$ gives

$$1-\tanh^2 x=\operatorname{sech}^2 x,$$

where $\tanh x=\sinh x/\cosh x$ and $\operatorname{sech} x=1/\cosh x$. Rearranging the original identity also gives $\cosh^2 x=1+\sinh^2 x$.

A different common task is proving an addition identity, such as $\cosh(a+b)=\cosh a\cosh b+\sinh a\sinh b$. This follows by substituting the exponential definitions and expanding. The terms involving opposite exponents cancel, leaving $(e^{a+b}+e^{-(a+b)})/2$, which is $\cosh(a+b)$. Likewise, expansion gives $\sinh(a+b)=\sinh a\cosh b+\cosh a\sinh b$. These formulas are not consequences of the fundamental identity alone; they are most directly proved from the same definitions.

Common mistakes

• Using the wrong sign. The identity is $\cosh^2 x-\sinh^2 x=1$, not $\cosh^2 x+\sinh^2 x=1$.
• Forgetting to square the denominator. Each function has denominator $2$, so its square has denominator $4$.
• Losing the subtraction when expanding. The second bracket is subtracted as a whole. Thus $-(e^{2x}-2+e^{-2x})=-e^{2x}+2-e^{-2x}$.
• Confusing hyperbolic and circular functions. Their identities have different signs. Start from the relevant definitions instead of transferring a formula from ordinary trigonometry.

If the requested identity is not the fundamental one, the exact formula matters: an addition formula, a double-angle formula, or an identity involving inverse hyperbolic functions may require a different proof.