Problem 135 - Calculus Volume 1

Question

For the following exercises, \(P\) is a point on the unit circle. a. Find the (exact) missing coordinate value of each point and b. find the values of the six trigonometric functions for the angle \(\theta\) with a terminal side that passes through point \(P .\) Rationalize denominators. $$P\left(\frac{7}{25}, y\right), y>0$$

Step-by-Step Solution

Answer

The missing coordinate is \( \frac{24}{25} \); the six trigonometric functions are: \(\sin(\theta) = \frac{24}{25}, \cos(\theta) = \frac{7}{25}, \tan(\theta) = \frac{24}{7}, \csc(\theta) = \frac{25}{24}, \sec(\theta) = \frac{25}{7}, \cot(\theta) = \frac{7}{24}\).

Step 1: Identify Point on Unit Circle

Given the point \( P \left( \frac{7}{25}, y \right) \) lies on the unit circle, we know that any point \((x, y)\) on the unit circle satisfies the equation \( x^2 + y^2 = 1 \). In this case, \( x \) is given as \( \frac{7}{25} \).

Step 2: Solve for Missing Coordinate

Substitute \( x = \frac{7}{25} \) into the unit circle equation: \( \left(\frac{7}{25}\right)^2 + y^2 = 1 \). Calculate \( \left(\frac{7}{25}\right)^2 = \frac{49}{625} \). Simplify the equation to find \( y^2 = 1 - \frac{49}{625} = \frac{576}{625} \). Solving for \( y \), we take the positive square root since \( y > 0 \). Therefore, \( y = \frac{24}{25} \).

Step 3: Calculate Trigonometric Functions

Given that \( \cos(\theta) = \frac{7}{25} \) and \( \sin(\theta) = \frac{24}{25} \), we can find the other trigonometric functions. 1. \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{\frac{24}{25}}{\frac{7}{25}} = \frac{24}{7}\).2. \( \csc(\theta) = \frac{1}{\sin(\theta)} = \frac{25}{24}\). 3. \( \sec(\theta) = \frac{1}{\cos(\theta)} = \frac{25}{7}\). 4. \( \cot(\theta) = \frac{1}{\tan(\theta)} = \frac{7}{24}\).

Key Concepts

Trigonometric Functions

The unit circle is a fundamental tool in trigonometry. It is a circle with a radius of one, centered at the origin of a coordinate plane. Each point on the circle corresponds to an angle \( \theta \) and the coordinates of these points can give us trigonometric function values. In this exercise, we are working with a point \( P \left( \frac{7}{25}, y \right) \) on the unit circle. The x-coordinate, \( \frac{7}{25} \), represents the cosine of the angle, while the y-coordinate represents the sine. This helps us directly derive two of the six primary trigonometric functions:- \( \cos(\theta) = \frac{7}{25} \)- \( \sin(\theta) = \frac{24}{25} \)Other trigonometric functions can be determined using these values. To find tangent, which is the ratio of sine to cosine, we calculate:\( \tan(\theta) = \frac{\frac{24}{25}}{\frac{7}{25}} = \frac{24}{7} \).Reciprocal functions like cosecant (\( \csc(\theta) \)), secant (\( \sec(\theta) \)), and cotangent (\( \cot(\theta) \)) can be found through:- \( \csc(\theta) = \frac{1}{\sin(\theta)} = \frac{25}{24} \)- \( \sec(\theta) = \frac{1}{\cos(\theta)} = \frac{25}{7} \)- \( \cot(\theta) = \frac{1}{\tan(\theta)} = \frac{7}{24} \)

Solving Equations

When dealing with the unit circle, the equation \( x^2 + y^2 = 1 \) always holds true. This equation stems from the Pythagorean Theorem, since the radius is equal to one. In the problem, we have a given x-coordinate, \( x = \frac{7}{25} \), and need to solve for the unknown y-coordinate. First, substitute the known x-value into the circle equation:- \( \left(\frac{7}{25}\right)^2 + y^2 = 1 \)By computing \( \left(\frac{7}{25}\right)^2 \), we get \( \frac{49}{625} \).This transforms the equation into:- \( \frac{49}{625} + y^2 = 1 \)To isolate \( y^2 \), subtract the x-term's square from 1:- \( y^2 = 1 - \frac{49}{625} = \frac{576}{625} \)Finally, since we're asked to consider \( y > 0 \), we take the positive square root:- \( y = \frac{24}{25} \). Thus, the value of \( y \) completes the coordinate for the point on the unit circle.

Coordinate Geometry

Coordinate geometry is used to visualize and solve problems involving shapes, sizes, and their relative positions in space, such as points on the unit circle. The circle resides in a 2D plane where any point \( (x, y) \) must adhere to the equation \( x^2 + y^2 = 1 \) for a unit circle.Each point on the circle can be decomposed into its x-component, representing \( \cos(\theta) \), and its y-component, representing \( \sin(\theta) \). This decomposition helps in locating the point \( P \) on the circle graph.Coordinate geometry not only aids in understanding where each point lands on the circle but also in geometrically interpreting the properties of trigonometric functions. These functions can be understood as ratios of the lengths of sides in a right triangle formed within the unit circle. Additionally, when solving problems, coordinate geometry allows verification of calculations through graphical representation. It forms a bridge between algebraic equations and their physical shape on the graph. Understanding these principles makes it easier to approach problems related to trigonometric functions and unit circles.