Parametric Equations Using Trigonometric Identities
When x and y are both built from sin t and cos t, one identity is usually all it takes to reveal the shape underneath.
9 min read3 worked examples
Eliminating t with sin²t + cos²t ≡ 1
Whenever $x$ and $y$ are both given in terms of $\sin t$ and $\cos t$, the identity $\sin^2t+\cos^2t\equiv1$ is the natural tool for eliminating the parameter: isolate $\cos t$ and $\sin t$ separately (usually by rearranging and dividing), square each, and add.
A standard circle
The pair $x = 3\cos t$, $y = 3\sin t$ traces a circle of radius 3, centred at the origin. This works because $x^2+y^2 = 9\cos^2t + 9\sin^2t = 9(\cos^2t+\sin^2t) = 9$, using $\sin^2t+\cos^2t \equiv 1$ — every point $(x,y)$ on the curve satisfies $x^2+y^2=9$, the Cartesian equation of that circle.
More generally, $x = a+r\cos t$, $y = b+r\sin t$ traces a circle of radius $r$ centred at $(a,b)$: the constants added to $\cos t$ and $\sin t$ shift the centre, and the coefficient multiplying them sets the radius. If the coefficients of $\cos t$ and $\sin t$ differ instead of matching, the same method still eliminates $t$ — but produces an ellipse rather than a circle, as the second worked example below shows.
Table of values — x = 3cos t, y = 3sin t
$t=0$: $(3,0)$. $\quad t=\dfrac{\pi}{2}$: $(0,3)$. $\quad t=\pi$: $(-3,0)$. $\quad t=\dfrac{3\pi}{2}$: $(0,-3)$. $\quad t=2\pi$: back to $(3,0)$. Plotting these four points and joining them smoothly traces out the full circle of radius 3, matching the Cartesian equation found by elimination.
Worked example 1
A shifted, scaled circle
Describe the curve given by $x=2+5\cos t$, $y=-4+5\sin t$ for $0\le t\le2\pi$, and state its centre and radius.
MethodCompare to the standard shifted-circle form x = a + r cos t, y = b + r sin t
1
The standard form for a circle of radius $r$ centred at $(a,b)$ is $x=a+r\cos t$, $y=b+r\sin t$.
General parametric form for any circle.
2
Comparing, $a=2$, $b=-4$, $r=5$.
Match coefficients directly against the standard form.
3
So the curve is a circle centred at $(2,-4)$ with radius $5$.
Read off centre and radius.
4
Check: $(x-2)^2+(y+4)^2 = 25\cos^2t+25\sin^2t = 25(\cos^2t+\sin^2t)=25$, matching a circle of radius $5$.
Verify by eliminating t using sin²t + cos²t ≡ 1.
5
Since $0\le t\le2\pi$ covers a full revolution, the complete circle is traced (not just an arc).
Always check the domain of t to confirm the full curve is described.
▪ solved
Examiner's eye
The constants added to $\cos t$ and $\sin t$ shift the centre; the coefficient multiplying them sets the radius. Confusing which number plays which role is the most common error when reading off a shifted circle from its parametric form.
Worked example 2
Isolate, square, add — even when the coefficients differ
Find the Cartesian equation of $x=4\cos t$, $y=2\sin t$ for $0\le t\le2\pi$, and identify the curve.
MethodIsolate cos t and sin t separately, then apply sin²t + cos²t ≡ 1
1
From $x=4\cos t$: $\cos t = \dfrac{x}{4}$. From $y=2\sin t$: $\sin t=\dfrac{y}{2}$.
Isolate the trig functions separately, ready to use the Pythagorean identity.
2
Substitute into $\sin^2t+\cos^2t\equiv1$: $\left(\dfrac{y}{2}\right)^2+\left(\dfrac{x}{4}\right)^2=1$.
This identity eliminates t entirely, since it holds for every value of t.
3
$\dfrac{x^2}{16}+\dfrac{y^2}{4}=1$.
Simplify each squared fraction.
4
This is the standard Cartesian equation of an ellipse centred at the origin, with semi-axis $4$ along $x$ and semi-axis $2$ along $y$.
Recognise the standard form x²/a² + y²/b² = 1.
▪ solved
Look deeper
Unlike Example 1's circle (where the coefficients of $\cos t$ and $\sin t$ were equal), here they differ ($4$ and $2$), which stretches the circle into an ellipse. The same elimination method — isolate, square, add — works regardless of whether the coefficients match.
Worked example 3
Effect of a restricted parameter domain · Challenge
Sketch the curve $x=3\cos t$, $y=3\sin t$ for $\dfrac{\pi}{2}\le t\le\pi$, and describe the part of the circle it represents.
MethodFind the start and end points, then determine which quadrant(s) t sweeps through
1
At $t=\dfrac{\pi}{2}$: $x=3\cos\dfrac{\pi}{2}=0$, $y=3\sin\dfrac{\pi}{2}=3$, giving start point $(0,3)$.
Evaluate at the lower bound of the domain.
2
At $t=\pi$: $x=3\cos\pi=-3$, $y=3\sin\pi=0$, giving end point $(-3,0)$.
Evaluate at the upper bound of the domain.
3
For $\dfrac{\pi}{2}\le t\le\pi$, $\cos t\le0$ and $\sin t\ge0$, so $x\le0$ and $y\ge0$ throughout.
Check the sign of each coordinate across the whole domain, not just the endpoints.
4
So the curve is the quarter-circle arc lying entirely in the second quadrant, from $(0,3)$ to $(-3,0)$.
Combine the endpoint and sign information to describe the arc precisely.
▪ solved
The key insight
Checking only the two endpoints is not enough to describe a restricted parametric curve — a curve could leave and re-enter a region between the endpoints. Here, because $\cos t$ and $\sin t$ are monotonic on $\left[\dfrac{\pi}{2},\pi\right]$, the endpoints do bound the whole arc, but this should always be verified rather than assumed.
Common mistake
Applying $\sin^2t+\cos^2t\equiv1$ before correctly isolating $\sin t$ and $\cos t$ — for instance, forgetting to divide by the radius first — which leads to an equation that doesn't actually eliminate $t$. Also common: sketching the full circle or ellipse implied by elimination while ignoring a restricted domain that only traces part of it.
Check yourself
A curve has parametric equations $x=4\cos t$, $y=4\sin t$ for $\pi\le t\le\dfrac{3\pi}{2}$. Find the coordinates of the start and end points, and describe the part of the circle traced. (Hint: evaluate x and y at each end of the domain, then check the sign of cos t and sin t throughout.)
Method
1
At $t=\pi$: $x=4\cos\pi=-4$, $y=4\sin\pi=0$, giving $(-4,0)$.
Evaluate at the lower bound.
2
At $t=\dfrac{3\pi}{2}$: $x=4\cos\dfrac{3\pi}{2}=0$, $y=4\sin\dfrac{3\pi}{2}=-4$, giving $(0,-4)$.
Evaluate at the upper bound.
3
For $\pi\le t\le\dfrac{3\pi}{2}$, $\cos t\le0$ and $\sin t\le0$, so $x\le0$ and $y\le0$ throughout — the arc lies entirely in the third quadrant.
The curve is the quarter-circle from (−4, 0) to (0, −4), radius 4, centred at the origin.