Problem 85 - Precalculus: Functions and Graphs

Question

Use fundamental identities to find the values of the trigonometric functions for the given conditions. $$\tan \theta=-\frac{3}{4} \text { and } \sin \theta>0$$

Step-by-Step Solution

Answer

\( \sin \theta = \frac{3}{5} \) and \( \cos \theta = -\frac{4}{5} \). \( \tan \theta = -\frac{3}{4} \), consistent with calculations.

Step 1: Understanding the Condition on Quadrants

Given that \( \tan \theta = -\frac{3}{4} \) and \( \sin \theta > 0 \), we need to determine the quadrant. Since the tangent of \( \theta \) is negative and sine is positive, \( \theta \) must be in the second quadrant where sine is positive and tangent is negative.

Step 2: Use Tangent Identity to Find Sine and Cosine

We have \( \tan \theta = -\frac{3}{4} \), which means \( \frac{\sin \theta}{\cos \theta} = -\frac{3}{4} \). Assume \( \sin \theta = 3k \) and \( \cos \theta = -4k \) (as cosine is negative in the second quadrant). Solving for \( k \), use \( \tan \theta = \frac{3k}{-4k} = -\frac{3}{4} \), which is consistent.

Step 3: Apply Pythagorean Identity

Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), substitute \( \sin \theta = 3k \) and \( \cos \theta = -4k \). So, \( (3k)^2 + (-4k)^2 = 1 \) becomes \( 9k^2 + 16k^2 = 1 \), leading to \( 25k^2 = 1 \). Solve for \( k^2 \) to get \( k^2 = \frac{1}{25} \), hence \( k = \pm \frac{1}{5} \). Since \( \sin \theta > 0 \), choose \( k = \frac{1}{5} \).

Step 4: Determine Specific Values for Sine and Cosine

Substitute \( k = \frac{1}{5} \) back into the expressions: \( \sin \theta = 3k = \frac{3}{5} \) and \( \cos \theta = -4k = -\frac{4}{5} \).

Step 5: Verify the Solution Using Identity

Verify \( \sin^2 \theta + \cos^2 \theta = 1 \). Calculate \( \sin^2 \theta + \cos^2 \theta = \left(\frac{3}{5}\right)^2 + \left(-\frac{4}{5}\right)^2 = \frac{9}{25} + \frac{16}{25} = 1 \), confirming the calculations.

Key Concepts

Tangent and Sine Relationship

When exploring trigonometric identities, understanding the relationship between tangent and sine is important. The tangent of an angle, \( \theta \), can be expressed as the ratio of its sine and cosine. Thus, the fundamental identity for tangent is:

• \( \tan \theta = \frac{\sin \theta}{\cos \theta} \)

Given that \( \tan \theta = -\frac{3}{4} \), we interpret this as \( \frac{\sin \theta}{\cos \theta} = -\frac{3}{4} \). Here, the numerator \( 3 \) indicates the proportional value of \( \sin \theta \), and the denominator \( 4 \) shows the proportional value of \( \cos \theta \). The negative sign implies that sine and cosine must have opposite signs.
Recognizing how these values interact through their signs assists in locating the angle in its respective quadrant. The relationship demonstrates how changes in one function, like sine, inform the values and signs of others, such as tangent. This interconnectedness is a key element in solving trigonometric problems.

Pythagorean Identity

The Pythagorean identity is one of the core identities in trigonometry, essential for solving equations involving sine and cosine. It can be written as:

• \( \sin^2 \theta + \cos^2 \theta = 1 \)

Once you have expressions for \( \sin \theta \) and \( \cos \theta \) using the information from the tangent identity, you can substitute them into the Pythagorean identity to solve for any missing pieces. In our example, we use:

• \( (3k)^2 + (-4k)^2 = 1 \)
• \( 9k^2 + 16k^2 = 1 \)
• \( 25k^2 = 1 \)

Solving this, we find that \( k^2 = \frac{1}{25} \), so \( k = \pm \frac{1}{5} \). This result then allows us to find the specific values of sine and cosine, \( \sin \theta = \frac{3}{5} \) and \( \cos \theta = -\frac{4}{5} \) under given conditions. The identity acts as a robust verification step, ensuring our solutions remain consistent.

Quadrant Determination

Determining the correct quadrant for an angle \( \theta \) based on its trigonometric properties is crucial. Quadrants separate a plane into four sections, influencing the signs of trigonometric functions:

• First quadrant: \( \sin > 0 \), \( \cos > 0 \), \( \tan > 0 \)
• Second quadrant: \( \sin > 0 \), \( \cos < 0 \), \( \tan < 0 \)
• Third quadrant: \( \sin < 0 \), \( \cos < 0 \), \( \tan > 0 \)
• Fourth quadrant: \( \sin < 0 \), \( \cos > 0 \), \( \tan < 0 \)

In the original problem, \( \tan \theta = -\frac{3}{4} \) and \( \sin \theta > 0 \) direct us to the second quadrant.
Here, sine is positive while tangent is negative, matching the conditions. Recognizing these properties helps isolate correct function values and ensures each trigonometric function maintains the expected sign and relationship to each other. Quadrant determination simplifies and solidifies the problem-solving process.