Problem 116 - A Graphical Approach to Precalculus with Limits

Question

Find all trigonometric function values for each angle \(\boldsymbol{\theta}\). \(\cos \theta=-\frac{3}{5},\) given that \(\theta\) is in quadrant III

Step-by-Step Solution

Answer

\( \sin \theta = -\frac{4}{5}, \tan \theta = \frac{4}{3}, \csc \theta = -\frac{5}{4}, \sec \theta = -\frac{5}{3}, \cot \theta = \frac{3}{4} \).

Step 1: Understand the Problem

We are given that \( \cos \theta = -\frac{3}{5} \) and \( \theta \) is in quadrant III. We need to find \( \sin \theta, \tan \theta, \csc \theta, \sec \theta, \) and \( \cot \theta.\)

Step 2: Determine the Reference Triangle

Since \( \cos \theta = \frac{x}{r} = -\frac{3}{5} \), we can interpret this as \( x = -3 \) and \( r = 5 \). In quadrant III, both \( x \) and \( y \) are negative.

Step 3: Find \( y \) Using the Pythagorean Theorem

Utilize the Pythagorean identity: \( x^2 + y^2 = r^2 \). Substitute the known values:\[ (-3)^2 + y^2 = 5^2 \]\[ 9 + y^2 = 25 \]\[ y^2 = 16 \]\[ y = -4 \] because in quadrant III, \( y \) is negative.

Step 4: Calculate \( \sin \theta \)

We have \( \sin \theta = \frac{y}{r} = \frac{-4}{5} \).

Step 5: Calculate \( \tan \theta \)

\( \tan \theta = \frac{y}{x} = \frac{-4}{-3} = \frac{4}{3} \).

Step 6: Calculate Reciprocal Trigonometric Functions

- \( \csc \theta = \frac{1}{\sin \theta} = -\frac{5}{4} \)- \( \sec \theta = \frac{1}{\cos \theta} = -\frac{5}{3} \)- \( \cot \theta = \frac{1}{\tan \theta} = \frac{3}{4} \)

Key Concepts

Quadrant III and Its Characteristics

In trigonometry, understanding the quadrant where an angle lies is crucial since it influences the sign of the trigonometric functions. Quadrant III is unique because it is located in the bottom left of the Cartesian coordinate system. In this quadrant, both the x-coordinate and y-coordinate are negative.

• This impacts the sine (\(\sin\)) and cosine (\(\cos\)) functions, as their ratios change sign depending on the quadrant.
• Sine is negative because it corresponds to the y-value, and cosine is negative due to the x-value.
• The tangent (\(\tan\)) function, which is the ratio of sine to cosine, is positive in Quadrant III since dividing two negatives results in a positive.

Knowing these rules helps in determining the sign of the trigonometric functions, even before calculating their numeric values.

Understanding the Pythagorean Identity

The Pythagorean identity is a fundamental concept in trigonometry that expresses the relation between sine and cosine functions. It is derived from the Pythagorean theorem in geometry. The identity states: \[ \sin^2 \theta + \cos^2 \theta = 1 \] This identity helps in finding missing trigonometric function values when one of them is given. For instance, when \( \cos \theta = -\frac{3}{5} \) is known, you can use:

• First calculate the square of cosine: \(\left( -\frac{3}{5} \right)^2 = \frac{9}{25}\).
• Then subtract this from 1 to find the square of sine: \(1 - \frac{9}{25} = \frac{16}{25}\).
• Taking the square root gives \(\sin \theta = -\frac{4}{5}\), fitting the negative sign convention in Quadrant III.

This identity is powerful for solving trigonometric problems, especially when constructing reference triangles or verifying calculated values.

Reciprocal Trigonometric Functions

Trigonometric functions have reciprocal relationships which are very useful in various calculations.

• Reciprocal functions are defined as follows: \( \csc \theta = \frac{1}{\sin \theta}, \; \sec \theta = \frac{1}{\cos \theta}, \; \text{and} \; \cot \theta = \frac{1}{\tan \theta} \).
• They help in deriving values when the basic sine, cosine, and tangent values are known.
• For example, given \( \sin \theta = -\frac{4}{5} \), it leads to \( \csc \theta = -\frac{5}{4} \).
• Similarly, \( \sec \theta \) can be found by taking the reciprocal of \( \cos \theta = -\frac{3}{5} \), resulting in \( \sec \theta = -\frac{5}{3} \).

This approach aids in solving complex trigonometric problems efficiently by providing alternative means to express trigonometric values.