Problem 31 - Precalculus

Question

Find the indicated trigonometric function values. If \(\sec \theta=-2,\) and the terminal side of \(\theta\) lies in quadrant III, find tan \(\theta\)

Step-by-Step Solution

Answer

\( \tan \theta = \sqrt{3} \)

Step 1: Understanding Given Information

We know that \( \sec \theta = \frac{1}{\cos \theta} = -2 \). This implies \( \cos \theta = -\frac{1}{2} \). Additionally, since \( \theta \) is in the third quadrant, both sine and tangent must be positive while cosine is negative.

Step 2: Using the Identity for Sine

Recall the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \). Substituting for \( \cos \theta \), we have: \[ \sin^2 \theta + \left(-\frac{1}{2}\right)^2 = 1 \]This simplifies to \( \sin^2 \theta + \frac{1}{4} = 1 \).

Step 3: Solving for \( \sin \theta \)

Subtract \( \frac{1}{4} \) from both sides: \[ \sin^2 \theta = 1 - \frac{1}{4} = \frac{3}{4} \]Taking the square root gives \( \sin \theta = \pm \frac{\sqrt{3}}{2} \). Since \( \theta \) is in the third quadrant where sine is negative, \( \sin \theta = -\frac{\sqrt{3}}{2} \).

Step 4: Using the Ratio Definition of Tangent

Tangent is defined as \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). Substituting the known values:\[ \tan \theta = \frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = \frac{\sqrt{3}}{1} = \sqrt{3} \]

Step 5: Confirming the Sign of \( \tan \theta \)

In the third quadrant, tangent is positive because both sine and cosine are negative. Our calculation agrees with this property.

Key Concepts

Quadrants in Trigonometry

When working with trigonometric functions, understanding the concept of quadrants is essential. The unit circle is divided into four quadrants. Each of these quadrants has specific characteristics that affect the signs of the trigonometric functions.

• Quadrant I: All trigonometric functions (sine, cosine, tangent, etc.) are positive.
• Quadrant II: Sine is positive; cosine and tangent are negative.
• Quadrant III: Tangent is positive; sine and cosine are negative.
• Quadrant IV: Cosine is positive; sine and tangent are negative.

Understanding which quadrant a particular angle \( \theta \) lies in helps determine the sign for trigonometric ratios. For example, in the third quadrant, as given in the exercise, both sine and cosine are negative, making tangent positive since it is the ratio of sine to cosine. Paying close attention to quadrants is crucial for solving problems correctly in trigonometry.

Pythagorean Identity

One of the most important identities in trigonometry is the Pythagorean identity. This identity states that: \[ \sin^2 \theta + \cos^2 \theta = 1 \]It arises from the equation of a circle with radius 1 on the coordinate system, representing the unit circle.
This identity is powerful for finding one function knowing another. In the given problem, knowing that \( \cos \theta = -\frac{1}{2} \), the identity allows us to find \( \sin \theta \). By substituting the known value, you isolate \( \sin^2 \theta \) and solve:\[ \sin^2 \theta = 1 - \left( -\frac{1}{2} \right)^2 = \frac{3}{4} \]Taking the square root yields two possible values for \( \sin \theta \), positive or negative. However, knowing the quadrant of \( \theta \) (Quadrant III) helps determine the sign of \( \sin \theta \), which is negative in this case.

Ratio Definitions of Trigonometric Functions

The trigonometric functions can be defined as ratios of the sides of a right-angled triangle. This perspective helps in visualizing the relationships between different functions.
The primary ratio definitions are:

• Sine: \( \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} \)
• Cosine: \( \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} \)
• Tangent: \( \tan \theta = \frac{\text{opposite}}{\text{adjacent}} \)

The problem utilizes these definitions to find \( \tan \theta \). From the ratio \( \tan \theta = \frac{\sin \theta}{\cos \theta} \), combining the calculated values for sine and cosine gives us:\[ \tan \theta = \frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = \sqrt{3} \]Understanding these ratios gives a solid grasp on how these functions interrelate, allowing students to solve trigonometric problems with confidence.