Problem 4 - Algebra 2 and Trigonometry

Question

In \(3-10\) , the terminal side of \(\angle R O P\) in standard position intersects the unit circle at \(P .\) If \(\mathrm{m} \angle R O P\) is \(\theta,\) find: a. \(\sin \theta\) b. \(\cos \theta\) c. the quadrant of \(\angle R O P\). $$ P(0.6,-0.8) $$

Step-by-Step Solution

Answer

a. \(\sin \theta = -0.8\), b. \(\cos \theta = 0.6\), c. Fourth quadrant.

Step 1: Understand the unit circle and point P

The unit circle is a circle with a radius of 1 centered at the origin, (0,0). The point \(P(0.6, -0.8)\) lies on this circle, meaning the distance from the origin to \(P\) is 1.

Step 2: Calculate sin(θ) from point's coordinates

For a point \((x, y)\) on the unit circle, \(\sin \theta = y\). Since the point \(P\) is given as \(P(0.6, -0.8)\), we have \(\sin \theta = -0.8\).

Step 3: Calculate cos(θ) from point's coordinates

Similarly, for a point \((x, y)\) on the unit circle, \(\cos \theta = x\). From point \(P(0.6, -0.8)\), we find \(\cos \theta = 0.6\).

Step 4: Determine the quadrant of angle ROP

Since \(\sin \theta = -0.8\) is negative and \(\cos \theta = 0.6\) is positive, angle \(ROP\) is in the fourth quadrant of the unit circle.

Key Concepts

Unit Circle

The unit circle is a fundamental concept in trigonometry. It's essentially a circle with a radius of 1. The center of the unit circle is located at the origin, which is the point

• (0,0) on the coordinate plane.

This simple definition makes the unit circle crucial for understanding trigonometric functions. Why? Because every angle in trigonometry can be mapped to a point on this circle. The unit circle helps us simplify complex trigonometric problems by providing a visual and geometric representation of angles and their corresponding sine and cosine values. Imagine wrapping the angles around a circle with steps where the entire circle sums up to 360 degrees, or 2π radians. When dealing with the unit circle:

• The length of the radius always remains 1.
• Any point along the circle obeys the equation: \[ x^2 + y^2 = 1 \]This equation ensures that the point's distance from the origin remains 1.

By placing trigonometric problems onto the unit circle, calculations become much more intuitive and visually clear.

Sine and Cosine

Sine and cosine are two of the primary trigonometric functions. These functions are essential for determining the ratios of sides in right-angled triangles. When we move to the unit circle, determining these functions becomes much simpler. Each angle on the unit circle corresponds to a specific point

• (x, y), where
• \(\cos \theta = x \)
• \(\sin \theta = y \)

This allows us to easily extract the sine and cosine values directly from the coordinates of points on the unit circle.Let's say you have a point \(P(x, y)\) on the unit circle. To find the sine and cosine of an angle \(\theta\), which determines point \(P\), you simply observe:

• \(\sin \theta\) is the y-coordinate of point \(P\).
• \(\cos \theta\) is the x-coordinate of point \(P\).

This straightforward approach helps students quickly identify trigonometric ratios and comprehend their significance without the need for complex calculations.

Quadrants of the Unit Circle

The unit circle is divided into four sections known as quadrants. These quadrants help determine the sign of the sine and cosine functions. Each quadrant is a unique part of the circle, covering 90 degrees or \(\frac{\pi}{2} \) radians. Let's explore how these quadrants affect the trigonometric functions:

• **First Quadrant:** Both \(\sin\) and \(\cos\) are positive.
• **Second Quadrant:** \(\sin\) is positive, but \(\cos\) is negative.
• **Third Quadrant:** Both \(\sin\) and \(\cos\) are negative.
• **Fourth Quadrant:** \(\sin\) is negative, but \(\cos\) is positive.

In the problem with point \(P(0.6, -0.8)\), since the \(\sin \theta = -0.8\) is negative and \(\cos \theta = 0.6\) is positive, this tells us that the angle is in the fourth quadrant. This quadrant-based understanding is essential because it helps predict and confirm the signs of trigonometric functions based on the position of angles. By mastering the concept of quadrants, you can easily deduce the behavior of trigonometric functions without directly calculating their exact values.