Problem 79 - 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. III; parallel to the line \(2 y-7 x+2=0\)

Step-by-Step Solution

Answer

In quadrant III, \(\sin(\theta) = -\frac{7}{\sqrt{53}}, \cos(\theta) = -\frac{2}{\sqrt{53}}, \tan(\theta) = \frac{7}{2}, \csc(\theta) = -\frac{\sqrt{53}}{7}, \sec(\theta) = -\frac{\sqrt{53}}{2}, \cot(\theta) = \frac{2}{7}.\)

Step 1: Identify the direction of the line

The given line is expressed as \(2y - 7x + 2 = 0\). A line parallel to this one will have the same slope. Rewrite the equation in slope-intercept form \(y = mx + b\), where \(m\) is the slope. From \(2y = 7x - 2\), we get \(y = \frac{7}{2}x - 1\). Therefore, the slope \(m\) is \(\frac{7}{2}\).

Step 2: Determine the angle's reference

In quadrant III, both \(x\) and \(y\) are negative. A line with a positive slope (\(\frac{7}{2}\)) in this quadrant would extend in such a way that both coordinates (\(x\), \(y\)) of points on the line are negative.

Step 3: Calculate exact values for sine and cosine

We can use the slope ratio to determine these trigonometric values. If the slope is \(\frac{7}{2}\), this implies \(\tan(\theta) = \frac{y}{x} = \frac{7}{2}\). For the unit circle, use \(y = 7\), \(x = 2\) and get the hypotenuse \(r\) using \(r = \sqrt{x^2 + y^2} = \sqrt{2^2 + 7^2} = \sqrt{53}\). Thus, \(\sin(\theta) = \frac{y}{r} = -\frac{7}{\sqrt{53}}\) and \(\cos(\theta) = \frac{x}{r} = -\frac{2}{\sqrt{53}}\), where the angles are negative in quadrant III.

Step 4: Determine the remaining trigonometric functions

Using the values of \(\sin(\theta)\) and \(\cos(\theta)\), find \(\tan(\theta)\), \(\csc(\theta)\), \(\sec(\theta)\), and \(\cot(\theta)\): 1. \(\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{-\frac{7}{\sqrt{53}}}{-\frac{2}{\sqrt{53}}} = \frac{7}{2}\) 2. \(\csc(\theta) = \frac{1}{\sin(\theta)} = -\frac{\sqrt{53}}{7}\) 3. \(\sec(\theta) = \frac{1}{\cos(\theta)} = -\frac{\sqrt{53}}{2}\) 4. \(\cot(\theta) = \frac{1}{\tan(\theta)} = \frac{2}{7}\).

Key Concepts

Trigonometric Identities

Trigonometric identities are fundamental tools in understanding and solving problems involving angles. These identities help relate different trigonometric functions to one another. In the context of this exercise, the most crucial identities are those that connect sine, cosine, and tangent, such as:

• Pythagorean Identity: \( ext{sin}^2 \theta + \text{cos}^2 \theta = 1\)
• Tangent Identity: \(\tan \theta = \frac{\sin \theta}{\cos \theta}\)

By using these identities, we can derive expressions for other trigonometric functions like cosecant, secant, and cotangent. For instance, the reciprocal identities define \(\csc \theta = \frac{1}{\sin \theta}\) and \(\sec \theta = \frac{1}{\cos \theta}\). Similarly, \(\cot \theta = \frac{1}{\tan \theta}\). Recognizing these identities allows us to switch between different forms and functions based on available information.

Reference Angles

Reference angles are an essential concept when working with trigonometric functions in different quadrants. The reference angle is the positive acute angle that a given angle makes with the x-axis. This angle is used to find trigonometric values easily by knowing their simplest form in the first quadrant.
To find the reference angle for an angle \(\theta\) located in any quadrant:

• In Quadrant III: The reference angle is \(\theta\) - 180° or in radians, \(\theta - \pi\).

In our exercise, understanding the reference angle helps in determining the correct signs of the trigonometric functions since sine and cosine are negative in Quadrant III. The reference angle essentially mirrors the angle in Quadrant I, so it gives a foundational understanding of potential values the functions might take.

Unit Circle

The unit circle is a central concept in trigonometry, and it helps to visualize trigonometric functions concerning their geometrical representation. The unit circle is defined as the circle with a radius of one unit centered at the origin of a coordinate plane. Comprehending this circle is crucial as it simplifies the calculation of sine, cosine, and tangent values.
On the unit circle:

• Sine: Represents the y-coordinate of the point on the circle.
• Cosine: Corresponds to the x-coordinate of the point.

Therefore, for any angle \(\theta\), coordinates (\( ext{cos} \theta, \text{sin} \theta\)) describe a point on the unit circle. In our task, to calculate sine and cosine, we use the reference from an extended line intersecting the unit circle. These values are subsequently utilized to find other trigonometric functions. The negative signs in Quadrant III reflect the position of points in this quadrant.

Slope of a Line

The slope of a line is an important concept when relating trigonometry with algebraic expressions of lines. The slope is a measure of the steepness or incline of a line and is calculated as the ratio of the change in vertical distance to the change in horizontal distance between two points on the line. Often, it is denoted by the symbol \(m\).

• Slope-Intercept Form: The equation of a line in slope-intercept form is \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept.

In our exercise, the line is parallel in the context of a given slope. Since the angle \(\theta\) mentioned has its terminal side aligning with this line \(m = \frac{7}{2}\), understanding the slope helps find \( an(\theta)\) as the ratio of its components. Using the Pythagorean Theorem, which relates all sides and angles, we can calculate the hypotenuse and derive the sine and cosine, among other trigonometric functions.