One may also use ''[[dual numbers]]'', denoteddefined ε,as oftennumbers consideredin similarthe toform an<math>a infinitesimal+ amountb\varepsilon</math>, with <math>a,b\in\mathbb squareR</math> ofand <math>\varepsilon</math> satisfying by definition <math>\varepsilon^2 = 0</math> and <math>\varepsilon \ne 0</math>. By using the MaclaurinMacLaurin series of cosine and sine and substituting θ=θε, theone resultcan isshow that <math>\cos(θε\theta\varepsilon) = 1</math> and <math>\sin(θε\theta\varepsilon) =θε \theta\varepsilon</math>. TheseFurthermore, approximationsit satisfyis not hard to prove that the [[Pythagorean identity]] holds:<math display="block">\sin^2(\theta\varepsilon) + \cos^2(\theta\varepsilon) = (\theta\varepsilon)^2 + 1^2 = \theta^2\varepsilon^2 + 1 = \theta^2 \cdot 0 + 1 = 1</math>
: cos²(θε)+sin²(θε)=1²+(θε)²=1+θ²ε²=1+θ²0=1.
==Error of the approximations==
[[File:Small angle compare error.svg|thumb|upright=2|'''Figure 3.''' A graph of the [[relative error]]s for the small angle approximations.]]
