The parametric formulae

By setting $t=\tan(\alpha/2)$ it is possible to express all three trigonometric functions of $\alpha$ as rational functions of $t$ alone. This is what makes the parametric formulae so convenient in computation.

Property — Parametric formulae in $t=\tan(\alpha/2)$

Setting $t = \tan(\alpha/2)$, the following identities hold: $\sin\alpha = \frac{2t}{1+t^2},\qquad \cos\alpha = \frac{1-t^2}{1+t^2},\qquad \tan\alpha = \frac{2t}{1-t^2}.$ They are valid for $\alpha\ne \pi + 2k\pi$, where $\tan(\alpha/2)$ is not defined.

Proof

We start from the double-angle formula $\sin\alpha = 2\sin(\alpha/2)\cos(\alpha/2)$ and divide numerator and denominator by $\cos^2(\alpha/2)$ (using $\sin^2+\cos^2=1$ in the denominator): $\sin\alpha = \frac{2\sin(\alpha/2)\cos(\alpha/2)}{\cos^2(\alpha/2) + \sin^2(\alpha/2)} = \frac{2\tan(\alpha/2)}{1 + \tan^2(\alpha/2)} = \frac{2t}{1+t^2}.$ For the cosine we use $\cos\alpha = \cos^2(\alpha/2)-\sin^2(\alpha/2)$ and the same division: $\cos\alpha = \frac{\cos^2(\alpha/2) - \sin^2(\alpha/2)}{\cos^2(\alpha/2)+\sin^2(\alpha/2)} = \frac{1-\tan^2(\alpha/2)}{1+\tan^2(\alpha/2)} = \frac{1-t^2}{1+t^2}.$ For $\tan\alpha$ it suffices to divide the two preceding results. ∎

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Topics: Prostaferesi werner
Concepts: Formule di duplicazione · Formule parametriche · Tangente dell angolo meta
Functions: Coseno · Seno · Tangente
Skills: Dimostrare · Usare formule