Problem 8 - Algebra 2 and Trigonometry

Question

In \(3-11, P\) is the point at which the terminal side of an angle in standard position intersects the unit circle. The measure of the angle is \(\theta .\) For each point \(P\) the \(x\) -coordinate and the quadrant is given. Find: a. the \(y\) -coordinate of \(P\) b. \(\cos \theta\) c. \(\sin \theta\) d. \(\tan \theta\) \(\left(\frac{\sqrt{5}}{3}, y\right),\) first quadrant

Step-by-Step Solution

Answer

The \( y \)-coordinate is \( \frac{2}{3} \), \( \cos \theta = \frac{\sqrt{5}}{3} \), \( \sin \theta = \frac{2}{3} \), \( \tan \theta = \frac{2\sqrt{5}}{5} \).

Step 1: Identify Characteristics of the Point

The point \( P \) is the intersection of the terminal side of angle \( \theta \) with the unit circle, positioned in the first quadrant with an \( x \)-coordinate of \( \frac{\sqrt{5}}{3} \). In the first quadrant, both \( x \) and \( y \) coordinates are positive.

Step 2: Find the y-coordinate

Since the point \( (\frac{\sqrt{5}}{3}, y) \) lies on the unit circle, it satisfies the equation: \( x^2 + y^2 = 1 \). Substituting the known \( x \)-coordinate gives \( \left(\frac{\sqrt{5}}{3}\right)^2 + y^2 = 1 \). Simplifying, \( \frac{5}{9} + y^2 = 1 \). Solving for \( y^2 \), we find \( y^2 = 1 - \frac{5}{9} = \frac{4}{9} \). Therefore, \( y = \pm \frac{2}{3} \). Since the point is in the first quadrant, \( y = \frac{2}{3} \).

Step 3: Calculate cos(θ)

The cosine of angle \( \theta \) corresponds to the \( x \)-coordinate of the point \( P \) on the unit circle. Thus, \( \cos \theta = \frac{\sqrt{5}}{3} \).

Step 4: Calculate sin(θ)

The sine of angle \( \theta \) corresponds to the \( y \)-coordinate of the point on the unit circle. Hence, \( \sin \theta = \frac{2}{3} \).

Step 5: Calculate tan(θ)

The tangent of angle \( \theta \) is given by the ratio \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). Substituting the known values, \( \tan \theta = \frac{\frac{2}{3}}{\frac{\sqrt{5}}{3}} = \frac{2}{\sqrt{5}} \). To rationalize the denominator, multiply by \( \frac{\sqrt{5}}{\sqrt{5}} \) to get \( \tan \theta = \frac{2\sqrt{5}}{5} \).

Key Concepts

Unit Circle

Imagine the coordinate plane with a circle centered at the origin (0,0) and having a radius of 1. That's the unit circle.
This circle is crucial for trigonometry because it helps visualize trigonometric functions. Points on the unit circle are described by coordinates \((x, y)\) that satisfy the equation: \(x^2 + y^2 = 1\). No matter the angle, if a line intersects the circle, the coordinates will meet this equation. This property allows us to easily find sine and cosine values because on the unit circle:

• Cosine is the x-coordinate of the point where the line intersects the circle.
• Sine is the y-coordinate.

If you know one of the coordinates and the quadrant, you can find the other using the equation above. For instance, if \(x = \frac{\sqrt{5}}{3}\)in the first quadrant, you can solve for \(y:\) \(y = \frac{2}{3}\).
This shows how the unit circle connects geometry and trigonometry visually.

Trigonometric Functions

Trigonometric functions relate angles and ratios in triangles and are represented on the unit circle. The primary functions are:

• Cosine (cos): the ratio of the adjacent side to the hypotenuse in a right triangle. For the unit circle, it’s the x-coordinate of a point.
• Sine (sin): the ratio of the opposite side to the hypotenuse. On the unit circle, this is the y-coordinate of a point.
• Tangent (tan): the ratio of sine to cosine. It’s useful for determining angles and is given by \(\tan \theta = \frac{\sin \theta}{\cos \theta}\).

Each function provides unique insights into angles and distances.
This insight is tremendously helpful in solving problems. Using our example, where \(\cos \theta = \frac{\sqrt{5}}{3}\)and \(\sin \theta = \frac{2}{3}\),tangent becomes \(\tan \theta = \frac{2 \sqrt{5}}{5}\)after rationalizing the denominator. Understanding these functions through the unit circle simplifies calculations and solidifies comprehension.

Quadrants in Coordinate Plane

The coordinate plane is divided into four quadrants by the x and y axes.These quadrants assist in determining the sign of trigonometric functions by the coordinates’ sign:

• First Quadrant: both x and y are positive.
• Second Quadrant: x is negative, y is positive.
• Third Quadrant: both x and y are negative.
• Fourth Quadrant: x is positive, y is negative.

Since each quadrant affects the signs of sine, cosine, and tangent, it’s essential to know where your angle lies.
In the given problem, the point \(\left(\frac{\sqrt{5}}{3}, \frac{2}{3}\right)\) is in the first quadrant. Here, all trigonometric functions are positive. Recognizing the quadrant helps ensure that calculated values make sense in context. Whether working with algebra or geometry, understanding quadrants aids in visualizing and verifying solution accuracy.