Problem 8
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) \(P(-t-\pi)\) $$ \left(\frac{7}{25},-\frac{24}{25}\right) $$
Step-by-Step Solution
Verified Answer
(a) \((-\frac{7}{25}, \frac{24}{25})\), (b) \((-\frac{7}{25}, \frac{24}{25})\), (c) \((\frac{7}{25}, \frac{24}{25})\), (d) \((-\frac{7}{25}, -\frac{24}{25})\).
1Step 1: Understanding the Problem
We need to find four different points on the unit circle related to the given point \(P(t) = \left(\frac{7}{25},-\frac{24}{25}\right)\). The points to be found are \(P(t+\pi)\), \(P(t-\pi)\), \(P(-t)\), and \(P(-t-\pi)\). These points are transformations based on certain angular shifts on the unit circle.
2Step 2: Find \(P(t+\pi)\) using symmetry properties
On the unit circle, adding \(\pi\) to the angle corresponds to finding the opposite point, as it rotates 180 degrees. Given \(P(t) = (x, y) = \left(\frac{7}{25}, -\frac{24}{25}\right)\), the opposite point is \((-x, -y)\). So, \(P(t+\pi) = \left(-\frac{7}{25}, \frac{24}{25}\right)\).
3Step 3: Calculate \(P(t-\pi)\) using symmetry properties
Similarly, subtracting \(\pi\) is the same as adding \(\pi\), due to the properties of the circle, thus \(P(t-\pi) = P(t+\pi) = \left(-\frac{7}{25}, \frac{24}{25}\right)\).
4Step 4: Determine \(P(-t)\) with reflection over the x-axis
The \(-t\) reflects the point across the x-axis. For \(P(t) = (x, y) = \left(\frac{7}{25}, -\frac{24}{25}\right)\), reflection leads to \(P(-t) = (x, -y) = \left(\frac{7}{25}, \frac{24}{25}\right)\).
5Step 5: Find \(P(-t-\pi)\) by another reflection and rotation
Start with \(P(-t) = \left(\frac{7}{25}, \frac{24}{25}\right)\). Adding (or subtracting) \(\pi\) to this reflects it again to the opposite, resulting in \(P(-t-\pi) = \left(-\frac{7}{25}, -\frac{24}{25}\right)\).
Key Concepts
Trigonometric TransformationsSymmetry PropertiesAngle ReflectionCoordinate Geometry
Trigonometric Transformations
Trigonometric transformations on the unit circle tie together angles, coordinates, and their transformations.
For instance, when transforming an angle by adding or subtracting \(\pi\), it corresponds to shifting the point to the opposite side of the circle.
This operation changes the sign of both \(x\) and \(y\) coordinates of the original point.
For instance, when transforming an angle by adding or subtracting \(\pi\), it corresponds to shifting the point to the opposite side of the circle.
This operation changes the sign of both \(x\) and \(y\) coordinates of the original point.
- Adding \(\pi\) or subtracting \(\pi\) to an angle, as seen in \(t+\pi\) and \(t-\pi\), results in a 180-degree rotation, which flips the signs of the coordinates.
- Reflecting with \(-t\) means altering the angle to its negative, swapping the sign of the \(y\)-coordinate, reflecting across the x-axis.
- The combination transformation \(-t-\pi\) involves reflecting and then rotating, transforming the coordinates to their opposite values both in \(x\) and \(y\).
Symmetry Properties
Symmetry properties are fundamental for comprehending movements on the unit circle.
When a point on the unit circle is transformed by adding \(\pi\), it shifts to the directly opposite side, making it symmetric.
When a point on the unit circle is transformed by adding \(\pi\), it shifts to the directly opposite side, making it symmetric.
- This symmetry means if a point \(P(t)\) is \((x, y)\), then its symmetric point \(P(t+\pi)\) is \((-x, -y)\).
- Similarly, subtracting \(\pi\), or \(t-\pi\), confers the same symmetric result as \(t+\pi\).
Angle Reflection
Angle reflection on the unit circle highlights the role of reflection over axes in transforming angles.
For a point \(P(t) = \left(\frac{7}{25}, -\frac{24}{25}\right)\), reflecting it over the x-axis (i.e., finding \(P(-t)\)) inverts the \(y\)-coordinate while keeping the \(x\)-coordinate constant.
For a point \(P(t) = \left(\frac{7}{25}, -\frac{24}{25}\right)\), reflecting it over the x-axis (i.e., finding \(P(-t)\)) inverts the \(y\)-coordinate while keeping the \(x\)-coordinate constant.
- The new point appears as \((x, -y)\).
- Furthermore, applying reflection followed by a half-turn \(-t-\pi\) brings the entire point to its opposite with both coordinates inverted \((-x, -y)\).
Coordinate Geometry
Coordinate geometry on the unit circle is a powerful tool to visualize the transformations of points.
The unit circle is a circle with a radius of 1 centered at the origin of a coordinate plane.
The unit circle is a circle with a radius of 1 centered at the origin of a coordinate plane.
- The coordinates of any point \(P(t)\) on this circle are \((x, y)\), where \(x = \cos(t)\) and \(y = \sin(t)\).
- Given a coordinate such as \(\left(\frac{7}{25}, -\frac{24}{25}\right)\), transformations are calculated directly through simple arithmetic involving \(x\) and \(y\).
Other exercises in this chapter
Problem 7
Find the angle that is supplementary to \(\theta\). (a) \(\theta=48^{\circ} 51^{\prime} 37^{\prime \prime}\) (b) \(\theta=136.42^{\circ}\)
View solution Problem 8
Exer. 1-8: Given the indicated parts of triangle \(A B C\) with \(\gamma=90^{\circ}\), find the exact values of the remaining parts. $$ b=7 \sqrt{2}, \quad c=14
View solution Problem 8
Find the period and sketch the graph of the equation. Show the asymptotes. $$ y=\frac{1}{4} \sec x $$
View solution Problem 8
Exer. 5-40: Find the amplitude, the period, and the phase shift and sketch the graph of the equation. $$ y=2 \sin \left(x-\frac{\pi}{3}\right) $$
View solution