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Ch. 5 - Complex Numbers, Polar Coordinates and Parametric Equations
Blitzer - Trigonometry 3rd Edition
Blitzer3rd EditionTrigonometryISBN: 9780137316601Non è quello che usi tu?Cambia libro di testo
Capitolo 5, Problema 81

In Exercises 81–82, find the rectangular coordinates of each pair of points. Then find the distance, in simplified radical form, between the points. (2, 2π/3) and (4, π/6)

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Identify that the given points are in polar coordinates, where each point is given as \((r, \theta)\) with \(r\) being the radius (distance from the origin) and \(\theta\) the angle in radians.
Convert each polar coordinate to rectangular coordinates using the formulas: \(x = r \cdot \cos(\theta)\) and \(y = r \cdot \sin(\theta)\).
For the first point \((2, \frac{2\pi}{3})\), calculate \(x_1 = 2 \cdot \cos\left(\frac{2\pi}{3}\right)\) and \(y_1 = 2 \cdot \sin\left(\frac{2\pi}{3}\right)\).
For the second point \((4, \frac{\pi}{6})\), calculate \(x_2 = 4 \cdot \cos\left(\frac{\pi}{6}\right)\) and \(y_2 = 4 \cdot \sin\left(\frac{\pi}{6}\right)\).
Use the distance formula between two points in rectangular coordinates: \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\), and simplify the expression to get the distance in simplified radical form.

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Polar to Rectangular Coordinate Conversion

Polar coordinates (r, θ) represent points using a radius and an angle. To convert to rectangular coordinates (x, y), use x = r cos θ and y = r sin θ. This conversion is essential for comparing points or calculating distances in the Cartesian plane.
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Convert Points from Polar to Rectangular

Distance Formula in the Cartesian Plane

The distance between two points (x₁, y₁) and (x₂, y₂) in rectangular coordinates is given by the formula √[(x₂ - x₁)² + (y₂ - y₁)²]. This formula derives from the Pythagorean theorem and is used to find the straight-line distance between points.
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Quadratic Formula

Simplifying Radical Expressions

Simplifying radicals involves expressing square roots in their simplest form by factoring out perfect squares. This process makes the distance expression cleaner and easier to interpret, which is often required in final answers for trigonometry problems.
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