Problem 89 - Precalculus: Functions and Graphs

Question

Use fundamental identities to find the values of the trigonometric functions for the given conditions. $$\cos \theta=-\frac{1}{3} \text { and } \sin \theta<0$$

Step-by-Step Solution

Answer

In the third quadrant, \( \sin \theta = -\frac{2\sqrt{2}}{3} \) and \( \tan \theta = 2\sqrt{2} \).

Step 1: Identify the Quadrant

Given that \( \cos \theta = -\frac{1}{3} \) and \( \sin \theta < 0 \), determine which quadrant \( \theta \) lies in. Since cosine is negative and sine is also negative, \( \theta \) is in the third quadrant.

Step 2: Apply Pythagorean Identity

Use the Pythagorean identity: \( \sin^2 \theta + \cos^2 \theta = 1 \). Substitute \( \cos \theta = -\frac{1}{3} \) into the equation:\[sin^2 \theta + \left(-\frac{1}{3}\right)^2 = 1.\]Simplify this to get:\[sin^2 \theta + \frac{1}{9} = 1.\]

Step 3: Solve for \( \sin^2 \theta \)

Rearrange the equation from Step 2 to solve for \( \sin^2 \theta \):\[sin^2 \theta = 1 - \frac{1}{9} = \frac{9}{9} - \frac{1}{9} = \frac{8}{9}.\]

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

Since \( \theta \) is in the third quadrant, both sine and cosine are negative. Thus, take the negative root:\[sin \theta = -\sqrt{\frac{8}{9}} = -\frac{\sqrt{8}}{3} = -\frac{2\sqrt{2}}{3}.\]

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

Use the identity \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). Substitute \( \sin \theta = -\frac{2\sqrt{2}}{3} \) and \( \cos \theta = -\frac{1}{3} \):\[\tan \theta = \frac{-\frac{2\sqrt{2}}{3}}{-\frac{1}{3}} = 2\sqrt{2}.\]

Key Concepts

Quadrants in Trigonometry

In trigonometry, understanding the concept of quadrants is essential. A quadrant is a part of the Cartesian plane, divided by the x-axis and y-axis. There are four quadrants in trigonometry, each representing a unique combination of signs for sine, cosine, and tangent.

• First Quadrant: Both sine (\( \ ext{sin} \theta \)) and cosine (\( \ ext{cos} \theta \)) are positive.
• Second Quadrant: Sine is positive while cosine is negative.
• Third Quadrant: Both sine and cosine are negative.
• Fourth Quadrant: Sine is negative while cosine is positive.

To determine the quadrant for a given angle, observe the signs of the trigonometric functions. In the exercise, \( \ ext{cos} \theta = -\frac{1}{3} \) and \( \ ext{sin} \theta < 0 \) indicate that \( \theta \) is in the third quadrant.

Here, both sine and cosine are negative, reinforcing that identifying the correct quadrant is crucial before proceeding with calculations.

Pythagorean Identity

The Pythagorean identity is a fundamental tool in trigonometry, allowing us to link sine and cosine. It states that for any angle \( \theta \), the following equation holds:\[ \sin^2 \theta + \cos^2 \theta = 1 \]This identity is derived from the Pythagorean theorem, applying to the unit circle where the radius is 1.

In our exercise, using the given \( \cos \theta = -\frac{1}{3} \), we substitute it into the identity:\[ \sin^2 \theta + \left(-\frac{1}{3}\right)^2 = 1 \]which simplifies to:\[ \sin^2 \theta + \frac{1}{9} = 1 \]

By solving, we find \( \sin^2 \theta = \frac{8}{9} \). This step illustrates how the Pythagorean identity serves as a bridge to finding one trigonometric function when another is known.

Sine and Cosine Functions

Sine and cosine are among the primary trigonometric functions. Understanding their roles and how they interact is vital for mastering trigonometry. Both functions relate to the unit circle and describe various properties of angles and triangles.

• Sine (\( \sin \theta \)) represents the y-coordinate of the point on the unit circle.
• Cosine (\( \cos \theta \)) represents the x-coordinate.

In problems where only one is given, the Pythagorean identity assists in determining the other.
In our step-by-step solution, we found \( \sin \theta \) using the identity and given \( \cos \theta \). Since \( \theta \) lies in the third quadrant, where both functions are negative, we chose the negative value for \( \sin \theta \). For trigonometric identities involving tangent, one can calculate it by taking \( \tan \theta = \frac{\sin \theta}{\cos \theta} \), highlighting the interplay between these functions.