Let P be the point have the polar coordinates (r, θ) and its rectangular coordinates will be (x, y).

Then,

x = r cosθ and y = r sin θ

r^{2} = x^{2} + y^{2}

tan θ = y/x

**Example 1 :**

Convert the given polar coordinates to rectangular coordinates.

(a) (-4, 2π/3) (b) (√3, π/6)

**Solution :**

From the point (-4, 2π/3), r is -4 and θ is 2π/3

x = r cos θ x = -4 cos 2π/3 x = -4(-1/2) x = 2 |
y = r sin θ y = -4 sin 2π/3 y = -4(√3/2) y = -2√3 |

So, the required rectangular co ordinate is (2, -2√3).

(b) (√3, π/6)

From the point (√3, π/6), r is √3 and θ is π/6

x = r cos θ x = √3 cos π/6 x = √3 (√3/2) x = 3/2 |
y = r sin θ y = √3 sin π/6 y = √3 (1/2) y = √3/2 |

So, the required rectangular co ordinate is (3/2, √3/2).

**Example 2 :**

Convert the given polar coordinates to rectangular coordinates.

(a) (2, π/4) (b) (-3, 5π/6)

(c) (5, 10π/3) (d) (47, 17π/2)

**Solution :**

(a) (2, π/4)

From the point (2, π/4), r is 2 and θ is π/4

x = r cos θ x = 2 cos π/4 x = 2(1/√2) x = 2/√2 |
y = r sin θ y = 2 sin π/4 y = 2(1/√2) y = 2/√2 |

So, the required rectangular co ordinate is (2/√2, 2/√2).

(b) (-3, 5π/6)

From the point (-3, 5π/6), r is -3 and θ is 5π/6

x = r cos θ x = -3 cos 5π/6 x = -3(√3/2) x = -3√3/2 |
y = r sin θ y = -3 cos 5π/6 y = -3(1/2) y = -3/2 |

So, the required rectangular co ordinate is

(-3√3/2, -3/2).

(c) (5, 10π/3)

From the point (5, 10π/3), r is 5 and θ is 10π/3

x = r cos θ x = 5 cos 10π/3 x = 5(-√3/2) x = -5√3/2 |
y = r sin θ y = 5 cos 10π/3 y = 5(-1/2) y = -5/2 |

So, the required rectangular co ordinate is

(-5√3/2, -5/2).

(d) (47, 17π/2)

From the point (47, 17π/2), r is 47 and θ is 17π/2

x = r cos θ x = 47 cos 17π/2 x = 47(0) x = 0 |
y = r sin θ x = 47 sin 17π/2 x = 47(1) x = 47 |

So, the required rectangular co ordinate is (0, 47).

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