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Pure · Coordinate geometry

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Parametric curves and trig identities

Converting trigonometric parametric equations to Cartesian form using identities — sin²+cos²≡1 for circles and ellipses, sec²−tan²≡1, and the double-angle formulae — and recognising the curve that results.

Pure · Coordinate geometry

Parametric curves and trig identities

Converting trigonometric parametric equations to Cartesian form using identities — sin²+cos²≡1 for circles and ellipses, sec²−tan²≡1, and the double-angle formulae — and recognising the curve that results.

Why it works

When the parametric equations involve sin⁡t\sin t, cos⁡t\cos t (or sec⁡t\sec t, tan⁡t\tan t), you usually cannot isolate tt tidily. Instead you eliminate it with a trigonometric identity: arrange each equation so that a trig function is the subject, then feed those into an identity that the functions must satisfy.

Circles and ellipses come from sin⁡2t+cos⁡2t≡1\sin^2 t + \cos^2 t \equiv 1. This is the workhorse. For x=acos⁡t,y=bsin⁡t,x = a\cos t, \qquad y = b\sin t, write cos⁡t=xa\cos t = \dfrac{x}{a} and sin⁡t=yb\sin t = \dfrac{y}{b}, then substitute into cos⁡2t+sin⁡2t=1\cos^2 t + \sin^2 t = 1: x2a2+y2b2=1.\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1. If a=ba = b this is a circle of radius aa; if a≠ba \ne b it is an ellipse. A shift like x=p+acos⁡t, y=q+bsin⁡tx = p + a\cos t,\ y = q + b\sin t just moves the centre to (p,q)(p, q): (x−p)2a2+(y−q)2b2=1\dfrac{(x-p)^2}{a^2} + \dfrac{(y-q)^2}{b^2} = 1.

The curve below is x=3cos⁡t, y=2sin⁡tx = 3\cos t,\ y = 2\sin t — an ellipse, x29+y24=1\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1, reaching ±3\pm 3 along xx and ±2\pm 2 along yy.-3-2-1123-2-112xyOther identities work the same way. If xx and yy involve sec⁡t\sec t and tan⁡t\tan t, use 1+tan⁡2t≡sec⁡2t1 + \tan^2 t \equiv \sec^2 t, i.e. sec⁡2t−tan⁡2t≡1\sec^2 t - \tan^2 t \equiv 1, to get x2−y2=1x^2 - y^2 = 1. And when one coordinate is a double angle, expand it so both coordinates share the same single-angle function: x=2sin⁡t,y=cos⁡2t=1−2sin⁡2t ⇒ y=1−2(x2)2=1−x22.x = 2\sin t,\quad y = \cos 2t = 1 - 2\sin^2 t \ \Rightarrow\ y = 1 - 2\left(\tfrac{x}{2}\right)^2 = 1 - \tfrac{x^2}{2}.

Don't forget the range. Because sin⁡t\sin t and cos⁡t\cos t stay between −1-1 and 11, x=acos⁡tx = a\cos t is trapped in −a≤x≤a-a \le x \le a. Quoting that restriction is part of a full answer, exactly as in the previous concept.