Problem 1 - Algebra 2

Question

If the given point \(P\) is located on the unit circle, find \(\sin \theta\) and \(\cos \theta\) $$ P\left(\frac{5}{13},-\frac{12}{13}\right) $$

Step-by-Step Solution

Answer

\(\sin \theta = -\frac{12}{13}\), \(\cos \theta = \frac{5}{13}\).

Step 1: Understanding the Unit Circle

A point \(P(x, y)\) is located on the unit circle if it satisfies the equation \(x^2 + y^2 = 1\). The unit circle has a radius of 1.

Step 2: Check If Point Satisfies the Unit Circle Equation

For the point \(P\left(\frac{5}{13}, -\frac{12}{13}\right)\), verify if it satisfies the unit circle equation by calculating: \(\left(\frac{5}{13}\right)^2 + \left(-\frac{12}{13}\right)^2\).

Step 3: Calculate Components

Calculate separately: \(\left(\frac{5}{13}\right)^2 = \frac{25}{169}\) and \(\left(-\frac{12}{13}\right)^2 = \frac{144}{169}\).

Step 4: Sum Components

Add both components: \(\frac{25}{169} + \frac{144}{169} = \frac{169}{169} = 1\). The point lies on the unit circle.

Step 5: Identify Sine and Cosine

On the unit circle, the x-coordinate represents \(\cos \theta\) and the y-coordinate represents \(\sin \theta\). Thus, \(\cos \theta = \frac{5}{13}\) and \(\sin \theta = -\frac{12}{13}\).

Key Concepts

Understanding the Unit Circle

The unit circle is a fundamental concept in trigonometry. It is a circle centered at the origin of a coordinate plane with a radius of 1. This simple geometry makes it a powerful tool in understanding angles and trigonometric functions.
The equation of the unit circle is \[ x^2 + y^2 = 1. \] This means that any point \( P(x, y) \) on the circle satisfies this equation. In the context of trigonometry, angles are measured from the positive x-axis. From this reference, an angle \( \theta \) corresponds to a point on the unit circle with coordinates \( (\cos \theta, \sin \theta) \). This effectively links trigonometric functions to the geometry of the circle.

• The radius of the unit circle is always equal to 1.
• X-coordinates represent the cosine of the angle \(\theta\).
• Y-coordinates represent the sine of the angle \(\theta\).

This connection makes the unit circle a handy reference for both evaluating and visualizing trigonometric functions.

Exploring the Sine Function

The sine function traces the y-coordinate of a point on the unit circle. Given an angle \( \theta \), \( \sin \theta \) represents how far up or down the point is from the x-axis.
For example, when angle \( \theta \) is at 90 degrees or \( \frac{\pi}{2} \) radians, this point sits at the top of the unit circle with coordinates \( (0, 1) \). Here, \( \sin \theta = 1 \), which is the maximum value for a sine function. At 0 and \( \pi \) radians (or 180 degrees), the point is on the x-axis, making \( \sin \theta = 0 \).

• Values of \( \sin \theta \) range from -1 to 1.
• \( \sin(0) = 0 \), since the point is at \((1, 0)\).
• \( \sin(\pi/2) = 1 \), representing the topmost point of the circle.
• \( \sin(\pi) = 0 \), as the point returns to the x-axis.

Understanding the sine function in terms of the unit circle helps to visualize and solve trigonometric problems easily.

Understanding the Cosine Function

Similar to the sine function, the cosine function relates to points on the unit circle as well. Specifically, it represents the x-coordinate of a point corresponding to an angle \( \theta \).
Consider when \( \theta \) is 0 degrees or 0 radians; the x-coordinate of the point is at its maximum, making \( \cos \theta = 1 \). At \( \theta = \pi \) radians (or 180 degrees), the point is at the far left of the unit circle, meaning \( \cos \theta = -1 \). Thus, 1 and -1 are the extrema for the cosine function.

• The cosine function values span from -1 to 1.
• \( \cos(0) = 1 \), as the point is at \((1, 0)\).
• \( \cos(\pi/2) = 0 \), since the point reaches the top of the circle.
• \( \cos(\pi) = -1 \), reflecting the point's position on the left-most side.

Understanding cosine through the unit circle provides a clear and detailed visualization of how angles translate into coordinates, facilitating easier computation and problem-solving in trigonometry.