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Eliminating trigonometric parameters

Eliminate trigonometric parameters using double-angle, triple-angle, compound-angle and reciprocal identities. Select the correct radical branch and preserve every original domain restriction.

Before you startTrigonometric identities, sine/cosine signs, parametric elimination

01 / Choose an identity that connects the two rules

Elimination is often an identity problem before it is an algebra problem.

If x and y contain different trigonometric expressions, first rewrite them using a shared function. A double-angle identity may turn a curve into a polynomial; a square-root expression may need a sign chosen from the parameter interval.

Identify shared trig functions → substitute → retain the domain and sign

The manual model shows why squaring can combine two branches that came from different parameter intervals.

Choose the cosine signExplore

Parameter intervalPosition through the interval

x = sin θ; y = sin 2θ = 2 sin θ cos θ.

y = 2x√(1 − x²), −1 ≤ x ≤ 1.

θ = −π/2 gives (−1, 0).

cos θ = 0 at this endpoint; both branch expressions agree there.

Grey shows the full two-branch relation. Blue is the locus allowed by your interval. The sign of cos θ chooses the radical sign; it does not always equal the sign of y.

02 / Turn a double angle into a polynomial

Pick the identity written in the function you already know.

Eliminate θ from x = 2 sin θ and y = 3 cos 2θ for 0 ≤ θ ≤ 2π.Worked example

sin θ = x/2

Use the x-rule.

cos 2θ = 1 − 2 sin² θ

This identity uses sine only.

y = 3 − 3x²/2

Substitute x/2 and simplify.

−2 ≤ x ≤ 2

Sine’s range restricts the parabola. The y-range is −3 ≤ y ≤ 3.

The same Cartesian curve can be retraced during the interval. The final equation records the locus, not the number of visits to each point.

03 / Choose the positive cosine branch

A square root is non-negative; cosine need not be.

Let x = sin θ and y = sin 2θ, with −π/2 ≤ θ ≤ π/2.Worked example

y = 2 sin θ cos θ = 2x cos θ

Use sin 2θ = 2 sin θ cos θ.

cos² θ = 1 − x²

Use the Pythagorean identity.

cos θ = √(1 − x²)

Cosine is non-negative throughout this interval.

y = 2x√(1 − x²), −1 ≤ x ≤ 1

The sign of y still depends on x.

Do not claim cosine is strictly positive at θ = ±π/2. It is zero at both endpoints, and those points are included.

Watch the parameter interval select a branch

Pause, replay or seek freely. The notes explain the same idea and stay in view.

04 / Choose the negative cosine branch

The same x-range can correspond to a different curve.

Use x = sin θ, y = sin 2θ on π/2 ≤ θ ≤ 3π/2 instead.Worked example

cos θ ≤ 0

This interval lies on the left half of the unit circle.

cos θ = −√(1 − x²)

The negative root is required.

y = −2x√(1 − x²), −1 ≤ x ≤ 1

The x-range is unchanged; the y-values are reflected in the x-axis.

At θ = 5π/6, x = 1/2 and y = −√3/2

This distinguishes the branches at a shared x-coordinate.

Choosing a sign from the sign of x would be wrong here. It is the sign of the eliminated cosine that matters.

05 / Use an implicit relation when both signs occur

Keep all permitted branches, but no impossible x-values.

Over 0 ≤ θ ≤ 2π, the preceding rules reach both cosine signs. Squaring gives:

y² = 4x²(1 − x²), with −1 ≤ x ≤ 1.

Every point of both radical branches on this interval is reached. At x = 0 and at x = ±1, the two branch formulas agree at y = 0; do not count those as two distinct points.

Why is the squared relation too broad for −π/2 ≤ θ ≤ π/2?Worked example

It permits (1/2, −√3/2)

That point satisfies y² = 4x²(1 − x²).

The original restricted rules require y = +√3/2 at x = 1/2

Cosine is non-negative, so the negative-y point is extra.

A valid equation for every original point may still contain additional points. Check the converse or retain the unsquared branch condition.

06 / Use a triple-angle polynomial

A cubic identity can eliminate the parameter directly.

For x = cos θ and y = cos 3θ, with 0 ≤ θ ≤ π, find the Cartesian form.Worked example

cos 3θ = 4 cos³ θ − 3 cos θ

Choose the cosine triple-angle identity.

y = 4x³ − 3x

Replace cos θ by x.

−1 ≤ x ≤ 1

Cosine decreases from 1 to −1 on the interval.

The curve is traced from x = 1 towards x = −1

Elimination does not retain this direction by itself.

Similarly, x = sin θ and y = sin 3θ give y = 3x − 4x³, with the x-range selected by the given θ-domain.

07 / Expand a compound angle before substituting

A remaining sine or cosine may need a sign condition.

Eliminate θ from x = cos θ and y = cos(θ − π/3), where 0 ≤ θ ≤ π.Worked example

y = (1/2) cos θ + (√3/2) sin θ

Expand the cosine of a difference.

sin θ = √(1 − x²)

Sine is non-negative on 0 ≤ θ ≤ π.

y = [x + √3√(1 − x²)]/2, −1 ≤ x ≤ 1

Substitute the positive branch.

(2y − x)² = 3(1 − x²), with 2y − x ≥ 0

An equivalent implicit description retains the lost sign.

On −π ≤ θ ≤ 0 the sine sign is reversed, so use the negative radical instead. The squared equation alone cannot distinguish the two intervals.

08 / Normalise reciprocal functions before using their identity

Original poles are always excluded.

For x = 2 sec θ and y = 3 tan θ, with −π/2 < θ < π/2, eliminate θ.Worked example

sec θ = x/2; tan θ = y/3

Normalise each rule.

sec² θ − tan² θ = 1

Use the reciprocal identity.

x²/4 − y²/9 = 1

Substitute the normalised coordinates.

x ≥ 2

Cosine is positive on this interval, so sec θ ≥ 1. Only the positive-x branch is traced.

Every real y is reached, because tan θ ranges over all real values on this interval. The endpoints ±π/2 are excluded because both original rules are undefined there. They must not be drawn as finite included points.

09 / A reciprocal identity can also give a parabola

The parameter domain decides how much of it is reached.

Eliminate θ from x = tan θ and y = sec² θ on 0 ≤ θ < π/2.Worked example

sec² θ = 1 + tan² θ

Use the Pythagorean reciprocal identity.

y = 1 + x², x ≥ 0

Tangent runs from 0 upwards without bound.

y ≥ 1

The minimum is included at θ = 0.

There is no finite endpoint at θ = π/2

The original functions are undefined there; the coordinates diverge.

Using −π/2 < θ < π/2 gives the complete parabola y = 1 + x². If an interval crosses a pole, split it into valid pieces before describing its image.

10 / Check the final equation against the original rules

A quick substitution can expose a lost sign or domain.

List every original denominator, logarithm or root restriction.
Find the actual range of the function replaced by x or y.
When taking a square root, determine the sign on each part of the interval.
Check endpoints and at least one point that distinguishes possible branches.
Ask whether every point described by the final answer can be reached.

The common value must be zero.

Worked solution

At (−1, 0), (0, 0) and (1, 0). These are three distinct points, not six.

12 / Recap

The identity removes the parameter; its interval completes the answer.

Select an identity in the function you can replace.
Use the interval to choose each radical sign.
Keep both branches only when both are actually reached.
Exclude every original pole, even if a simplified equation looks defined.
Check coordinate ranges, endpoints and reachability.

Section 1 of 12 · Choose an identity that connects the two rules