Converting Polar Coordinates to Cartesian Coordinates

From Polar Position to Cartesian Coordinates

A polar description locates a point by giving its distance from the origin and the direction of that distance. A Cartesian description instead gives the point’s signed horizontal and vertical coordinates. Converting between the two descriptions is useful whenever a point or curve is easier to understand in one coordinate system than the other.

Let a point have polar coordinates \((r,\theta)\), where \(r\ge 0\) is its distance from the origin and \(\theta\) is measured from the positive \(x\)-axis. Its Cartesian coordinates \((x,y)\) are determined by the angle and the two signed coordinate displacements.

Coordinate formulas: For a polar point \((r,\theta)\), its Cartesian coordinates are \(x=r\cos\theta\) and \(y=r\sin\theta\). Here \(x\) and \(y\) are signed coordinates: they may be positive, negative, or zero.

The formulas apply for angles in every quadrant. The signs of cosine and sine determine the signs of the horizontal and vertical coordinates. This matters because lengths in a right triangle are nonnegative, while Cartesian coordinates are signed.

Why the Formulas Work

Suppose first that the point is not the origin, so \(r>0\). Draw the vertical segment from the point to the \(x\)-axis. The resulting right triangle has hypotenuse \(r\). Its horizontal and vertical leg lengths are \(|x|\) and \(|y|\), respectively. If \(\alpha\) is the reference angle—the acute angle between the terminal side and the nearer part of the \(x\)-axis—then right-triangle trigonometry gives

\cos\alpha=\frac{|x|}{r} \qquad\text{and}\qquad \sin\alpha=\frac{|y|}{r}.

These equations determine the magnitudes of the coordinates, not their signs. To obtain the signed coordinates, use the quadrant of \(\theta\). In quadrant II, for example, the horizontal coordinate is negative and the vertical coordinate is positive; cosine is negative and sine is positive. Thus the signs on the trigonometric values match the signs on the coordinates.

The same reasoning works in each quadrant, and on the axes one coordinate is zero. It follows that \(x/r=\cos\theta\) and \(y/r=\sin\theta\), so multiplying by \(r\) gives the coordinate formulas. If \(r=0\), the point is the origin, \(x=y=0\), and both formulas still hold for any angle.

Find the leg magnitudes.
The reference-angle ratios give \(|x|=r|\cos\theta|\) and \(|y|=r|\sin\theta|\).

Apply the quadrant signs.
The signs of \(\cos\theta\) and \(\sin\theta\) give the signs of \(x\) and \(y\).

Write the signed coordinates.
The resulting formulas are \(x=r\cos\theta\) and \(y=r\sin\theta\).

A related identity follows immediately from the coordinate formulas and \(\cos^2\theta+\sin^2\theta=1\):

x^2+y^2 =r^2\cos^2\theta+r^2\sin^2\theta =r^2.

Because \(r\ge 0\), this also says \(r=\sqrt{x^2+y^2}\). The squared identity is particularly useful when converting equations that involve \(r\).

Converting Polar Points

To convert a point, substitute its radius and angle into the coordinate formulas. Exact trigonometric values are preferable when available. Before calculating, identify the signs expected from the quadrant; this provides a quick check on the result.

Worked Example: A Point in Quadrant II

Convert the polar point \((4,2\pi/3)\) to Cartesian coordinates.

The angle \(2\pi/3\) is in quadrant II, so \(x\) should be negative and \(y\) positive. The exact trigonometric values are \(\cos(2\pi/3)=-1/2\) and \(\sin(2\pi/3)=\sqrt{3}/2\). Therefore,

x=4\cos(2\pi/3)=4\left(-\frac12\right)=-2, \qquad y=4\sin(2\pi/3)=4\left(\frac{\sqrt3}{2}\right)=2\sqrt3.

The Cartesian point is \((-2,2\sqrt3)\). Check its distance from the origin:

\sqrt{x^2+y^2} =\sqrt{(-2)^2+(2\sqrt3)^2} =\sqrt{4+12} =\sqrt{16} =4.

The distance agrees with the given radius, and the signs agree with quadrant II.

Worked Example: A Point with a Negative Vertical Coordinate

Convert \((2,7\pi/4)\) to Cartesian coordinates.

The angle is in quadrant IV, where \(x\) is positive and \(y\) is negative. Since \(\cos(7\pi/4)=\sqrt2/2\) and \(\sin(7\pi/4)=-\sqrt2/2\),

x=2\cos(7\pi/4)=2\left(\frac{\sqrt2}{2}\right)=\sqrt2, \qquad y=2\sin(7\pi/4)=2\left(-\frac{\sqrt2}{2}\right)=-\sqrt2.

The Cartesian point is \((\sqrt2,-\sqrt2)\). Its distance is

\sqrt{(\sqrt2)^2+(-\sqrt2)^2} =\sqrt{2+2} =\sqrt4 =2.

This confirms the radius, and the coordinate signs match quadrant IV.

Converting Polar Equations

For an equation, use the substitutions \(r\cos\theta=x\), \(r\sin\theta=y\), and \(r^2=x^2+y^2\). Each substitution should be made in a way that preserves the points described by the equation. If an algebraic step involves multiplying by a variable, dividing by a variable, or squaring, check whether the reverse step is valid as well. Such operations can change which parameter values satisfy an equation, even when the resulting point sets agree.

Worked Example: Converting \(r=4\cos\theta\)

Convert the polar equation \(r=4\cos\theta\) to Cartesian form.

Multiply both sides by \(r\). Then use \(r^2=x^2+y^2\) and \(r\cos\theta=x\):

r=4\cos\theta \quad\Longrightarrow\quad r^2=4r\cos\theta \quad\Longrightarrow\quad x^2+y^2=4x.

Thus the Cartesian equation is \(x^2+y^2=4x\). To check a point, take \(r=2\) and \(\theta=\pi/3\). The polar equation holds because \(4\cos(\pi/3)=4(1/2)=2=r\). Its Cartesian coordinates are \(x=2(1/2)=1\) and \(y=2(\sqrt3/2)=\sqrt3\). Substitution gives

x^2+y^2=1^2+(\sqrt3)^2=1+3=4, \qquad 4x=4(1)=4.

The converted equation is satisfied. The algebra multiplied by \(r\), so it is worth checking the origin: it satisfies the Cartesian equation, and it is represented by the original polar equation when \(r=0\) and \(\theta=\pi/2\).

Worked Example: Converting \(r=6\sin\theta\)

Convert the polar equation \(r=6\sin\theta\) to Cartesian form.

Multiply both sides by \(r\), then substitute \(r^2=x^2+y^2\) and \(r\sin\theta=y\):

r=6\sin\theta \quad\Longrightarrow\quad r^2=6r\sin\theta \quad\Longrightarrow\quad x^2+y^2=6y.

For a check, choose \(r=3\) and \(\theta=\pi/6\). Since \(6\sin(\pi/6)=6(1/2)=3\), this point satisfies the polar equation. Its Cartesian coordinates are

x=3\cos(\pi/6)=\frac{3\sqrt3}{2}, \qquad y=3\sin(\pi/6)=\frac32.

Substitute them into the Cartesian equation:

x^2+y^2 =\left(\frac{3\sqrt3}{2}\right)^2+\left(\frac32\right)^2 =\frac{27}{4}+\frac94 =\frac{36}{4} =9, \qquad 6y=6\left(\frac32\right)=9.

Both sides agree, as required.

Worked Example: An Equation That Gives a Vertical Line

Convert \(r\cos\theta=-2\) to Cartesian form.

Here the left side is exactly the Cartesian coordinate \(x\), so the equation becomes

r\cos\theta=-2 \quad\Longrightarrow\quad x=-2.

For example, the polar point \((2,\pi)\) satisfies the original equation because \(2\cos\pi=2(-1)=-2\). Its Cartesian coordinates are \(x=2(-1)=-2\) and \(y=2(0)=0\), which satisfy \(x=-2\). This check confirms the sign as well as the conversion.

A Common Sign Error

A frequent mistake is to use the positive length of a triangle leg as if it were the signed Cartesian coordinate. For the point \((4,2\pi/3)\), the reference-angle calculation gives a horizontal leg length of \(4(1/2)=2\). It would be wrong to conclude that \(x=2\): the point is in quadrant II, so its horizontal coordinate is negative.

\text{Incorrect: }x=4\left|\cos(2\pi/3)\right|=2. \qquad \text{Correct: }x=4\cos(2\pi/3)=-2.

The absolute value gives the leg’s length, \(|x|=2\). The coordinate itself is \(x=-2\). Keeping this distinction clear is essential when converting points in quadrants II and III, and when converting equations that depend on the signs of \(x\) or \(y\).

What the Conversion Preserves

The coordinate formulas assign a Cartesian point to each polar description. Converting an equation means expressing the same geometric set of points in terms of \(x\) and \(y\). A transformed equation should therefore be checked for whether it describes exactly that set, rather than merely whether one sample point works.

For instance, multiplying an equation by \(r\) can make additional angle values satisfy the transformed relation when \(r=0\). Those values all represent the origin, but the original equation may allow the origin only for particular angles. The Cartesian equation describes points, not the full list of polar parameter pairs, so inspect the origin when it is involved. More generally, after an algebraic conversion, test the reverse implication wherever a step may not be reversible.

Key takeaway: Use \(x=r\cos\theta\) and \(y=r\sin\theta\) for signed coordinates, and \(x^2+y^2=r^2\) when an equation involves the radius squared. Track signs and verify that algebraic steps preserve the intended set of points.

Check Your Understanding

Answer each question by using the coordinate formulas and checking signs or substitutions where appropriate.

Convert \((5,5\pi/6)\) to Cartesian coordinates, and verify the distance from the origin.
What are the signs of \(x\) and \(y\) for an angle in quadrant III? How do the signs of cosine and sine produce them?
Convert \(r\sin\theta=3\) to Cartesian form.
Convert \(r=2\cos\theta\) to an equation in \(x\) and \(y\). State which identity you use after multiplying by \(r\).
Why is it incorrect to use the positive length of a triangle leg as the coordinate \(x\) when the point lies in quadrant II?