Problem 6 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 5-8: Let \(P(t)\) be the point on the unit circle \(U\) that corresponds to \(t\). If \(P(t)\) has the given rectangular coordinates, find (a) \(P(t+\pi)\) (b) \(P(t-\pi)\) (c) \(P(-t)\) (d) \(P(-t-\pi)\) $$ \left(-\frac{8}{17}, \frac{15}{17}\right) $$

Step-by-Step Solution

Answer

(a) \( \left(\frac{8}{17}, -\frac{15}{17}\right) \), (b) \( \left(\frac{8}{17}, -\frac{15}{17}\right) \), (c) \( \left(-\frac{8}{17}, -\frac{15}{17}\right) \), (d) \( \left(\frac{8}{17}, \frac{15}{17}\right) \).

Step 1: Understanding the Unit Circle and Reference Points

The point \( P(t) = \left(-\frac{8}{17}, \frac{15}{17}\right) \) is on the unit circle, so it corresponds to some angle \( t \) on this circle. The unit circle has a radius of 1, and any point \( (x, y) \) on it follows the equation \( x^2 + y^2 = 1 \). Indeed, \( \left(-\frac{8}{17}\right)^2 + \left(\frac{15}{17}\right)^2 = 1 \) confirms that the point is on the unit circle.

Step 2: Calculating P(t+π)

The point \( P(t+\pi) \) is directly opposite \( P(t) \) on the unit circle, making its coordinates the negatives of \( P(t) \). Thus, \( P(t+\pi) = \left(-(-\frac{8}{17}), -\frac{15}{17}\right) = \left(\frac{8}{17}, -\frac{15}{17}\right) \).

Step 3: Calculating P(t-π)

Similarly, \( P(t-\pi) \) is also opposite \( P(t) \) on the circle. Hence, the coordinates are the same as for \( P(t+\pi) \) due to symmetry. Thus, \( P(t-\pi) = \left(\frac{8}{17}, -\frac{15}{17}\right) \).

Step 4: Calculating P(-t)

Point \( P(-t) \) is the reflection of \( P(t) \) about the x-axis. Consequently, its coordinates are \( P(-t) = \left(-\frac{8}{17}, -\frac{15}{17}\right) \).

Step 5: Calculating P(-t-π)

To find \( P(-t-\pi) \), we take \( P(-t) \) and find its opposite point on the circle. The coordinates are \( P(-t-\pi) = \left(\frac{8}{17}, \frac{15}{17}\right) \).

Key Concepts

Trigonometric Functions

Trigonometric functions are crucial when working with the unit circle, mainly because they help us connect angles with their corresponding point positions on the circle. Each point on the unit circle corresponds to an angle \( t \) that we measure from the positive x-axis.
These functions, including sine and cosine, help express these points' coordinates. Specifically, for a given angle \( t \), the coordinates of the point \( P(t) \) on the unit circle are given by \((\cos(t), \sin(t))\).

In the exercise, starting with the point \(-\frac{8}{17}, \frac{15}{17}\), we see it aligns with this principle, where \( x = \cos(t) \) and \( y = \sin(t) \). This means the trigonometric functions not only define the positions but also enable transformations when angles change, such as when they are increased or decreased by \( \pi \), which flips the point to the opposite side of the circle.

Reference Angles

Reference angles are a powerful concept that helps simplify the understanding of points on the unit circle. They allow us to track symmetry and relational positions within different quadrants.
A reference angle is the acute angle formed by the terminal side of the angle \( t \) and the x-axis. This provides a standard way to evaluate trigonometric function values, especially when dealing with reflections and opposites.

In our unit circle context, calculating \( P(t+\pi) \) and \( P(t-\pi) \) leverages the idea that adding or subtracting \( \pi \) (which is half a circle, or 180 degrees) flips the point across to the opposite side. As such, while absolute values of coordinates remain the same, signs change due to this symmetry; using reference angles makes these transformations intuitive.

Coordinate Geometry

Coordinate geometry involves understanding how coordinates and geometric concepts interrelate. On the unit circle, this means connecting algebraic expressions—like point coordinates—to geometric transformations.
Unit circles use a radius of 1, providing simplicity, as each coordinate point \((x, y)\) satisfying \(x^2 + y^2 = 1\) lies on the circle.

For example, when you find the point \( P(-t) \), this transformation reflects the original point \(P(t)\) across the x-axis. Here, we are using the principles of coordinate geometry to understand that the x-coordinate remains the same while the y-coordinate changes its sign.
Understanding such reflection and rotation within coordinate systems can precisely determine points' locations and relationships as seen with transformations like \( P(-t) \) and \( P(-t-\pi) \).