Problem 80 - Precalculus: Functions and Graphs

Question

Find the exact values of the six trigonometric functions of \(\boldsymbol{\theta}\) if \(\boldsymbol{\theta}\) is in standard position and the terminal side of \(\boldsymbol{\theta}\) is in the specified quadrant and satisfles the given condition. II; parallel to the line through \(A(1,4)\) and \(B(3,-2)\)

Step-by-Step Solution

Answer

The trigonometric values are: \(\sin \theta = \frac{3}{\sqrt{10}}\), \(\cos \theta = \frac{-1}{\sqrt{10}}\), \(\tan \theta = -3\), \(\csc \theta = \frac{\sqrt{10}}{3}\), \(\sec \theta = -\sqrt{10}\), \(\cot \theta = \frac{-1}{3}\).

Step 1: Find the Slope of the Given Line

To find the slope of the line through points \(A(1,4)\) and \(B(3,-2)\), use the slope formula: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \]Substitute \((x_1, y_1) = (1, 4)\) and \((x_2, y_2) = (3, -2)\): \[ m = \frac{-2 - 4}{3 - 1} = \frac{-6}{2} = -3 \]Therefore, the slope is \(-3\).

Step 2: Determine the Slope of the Terminal Side

Since the terminal side of \(\theta\) is parallel to the line, it will have the same slope of \(-3\). This means the direction of the line related to \(\theta\) in the coordinate plane follows the slope of \(-3\).

Step 3: Create a Right Triangle

To create a right triangle that matches the slope, assume a change in \(x\) of 1 (run = 1). Hence, from the slope \(-3\), the change in \(y\) (rise) must be \(-3\). Since the question specifies the terminal side is in Quadrant II, adjust to: \(x = -1\) and \(y = 3\). So, the point is \((-1, 3)\).

Step 4: Compute the Hypotenuse (r) Using Pythagorean Theorem

The hypotenuse \(r\) forms from the origin \((0,0)\) to point \((-1, 3)\). Calculate \(r\) using:\[ r = \sqrt{x^2 + y^2} = \sqrt{(-1)^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10} \]

Step 5: Calculate Trigonometric Functions

Now, use the definitions of the trigonometric functions in terms of \(x, y,\) and \(r\):- \(\sin \theta = \frac{y}{r} = \frac{3}{\sqrt{10}}\)- \(\cos \theta = \frac{x}{r} = \frac{-1}{\sqrt{10}}\)- \(\tan \theta = \frac{y}{x} = \frac{3}{-1} = -3\)- \(\csc \theta = \frac{r}{y} = \frac{\sqrt{10}}{3}\)- \(\sec \theta = \frac{r}{x} = \frac{\sqrt{10}}{-1} = -\sqrt{10}\)- \(\cot \theta = \frac{x}{y} = \frac{-1}{3}\)

Key Concepts

Understanding Slope

The concept of slope is fundamental in understanding how steep a line is on a graph. The slope of a line can be determined by the change in vertical distance divided by the change in horizontal distance between two points on the line.

• It is often described as "rise over run," representing how much the line moves up or down as we move from left to right.
• In the formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \), \( m \) stands for the slope, where 'rise' is \( y_2 - y_1 \) and 'run' is \( x_2 - x_1 \).

Calculating the slope between points \(A(1,4)\) and \(B(3,-2)\) yielded a slope of \(-3\).
A negative slope indicates that the line is moving downward from left to right. This understanding helps in visualizing the line's behavior with respect to the coordinate plane.

Significance of Right Triangle

A right triangle is a triangle where one angle measures 90 degrees. These triangles are crucial to trigonometry because they allow us to define the trigonometric functions based on the triangle's sides.
For a right triangle formed from a given slope, we use the change in \(x\) (run) and change in \(y\) (rise) to determine side lengths. Keeping in mind that our slope is \(-3\) and we're in Quadrant II:

• We took \(x = -1\), making the 'run' negative since we move left towards the origin.
• We took \(y = 3\), making the 'rise' positive as we move upwards.

This configuration helps in setting up the legs of the triangle, \(-1\) (adjacent to angle \(\theta\)) and \(3\) (opposite to angle \(\theta\)).
Because the right triangle is in Quadrant II, it ensures the horizontal component remains negative, while the vertical stays positive.

Exploring Quadrants

In mathematics, the coordinate plane is divided into four quadrants, each representing a different combination of positive and negative values for \(x\) and \(y\). Quadrant II is specifically the region where \(x\) is negative and \(y\) is positive.
The problem specified our angle \(\theta\) had its terminal side in Quadrant II:

• This quadrant affects the signs of trigonometric functions. For example, sine, involving \(y\), remains positive, while cosine, involving \(x\), turns negative.
• Understanding the rules for the signs of these functions across different quadrants is crucial for applying trigonometry correctly.

This understanding ensures that calculations for \(\sin \theta, \cos \theta, \tan \theta\), etc., are correct based on their algebraic signs in Quadrant II.

Applying the Pythagorean Theorem

The Pythagorean Theorem is invaluable in trigonometry for determining the hypotenuse of a right triangle, which is crucial for calculating the trigonometric functions.
It is given by the formula \( r = \sqrt{x^2 + y^2} \), where \(x\) and \(y\) are the legs of the triangle. In our scenario:

• Selecting \((-1, 3)\) from changes in \(x\) and \(y\) enabled the determination of the hypotenuse with \( r = \sqrt{10} \).

The hypotenuse \( r \) forms the essential ratio in functions such as sine, cosine, and tangent, relating the triangle's side lengths.
Using \( r \), the trigonometric function values can be calculated accurately, ensuring their correct application in solving problems.