Problem 46 - Algebra and Trigonometry

Question

Find the values of the trigonometric functions of \(\theta\) from the information given. $$ \cos \theta=-\frac{7}{12}, \quad \theta \text { in Quadrant III } $$

Step-by-Step Solution

Answer

\( \sin \theta = \frac{\sqrt{95}}{12}, \tan \theta = -\frac{\sqrt{95}}{7} \).

Step 1: Identify Given Information

We are given that \( \cos \theta = -\frac{7}{12} \) and that \( \theta \) is in Quadrant III.

Step 2: Recall Properties of Trigonometric Functions in Quadrant III

In Quadrant III, the cosine of an angle is negative, which matches our given information. Additionally, both the sine and tangent functions are positive.

Step 3: Use Pythagorean Identity to Find Sine

The Pythagorean identity is \( \sin^2 \theta + \cos^2 \theta = 1 \). Given \( \cos \theta = -\frac{7}{12} \), first calculate \( \cos^2 \theta = \left( -\frac{7}{12} \right)^2 = \frac{49}{144} \).

Step 4: Calculate \( \sin^2 \theta \)

Substitute the value of \( \cos^2 \theta \) into the Pythagorean identity: \( \sin^2 \theta = 1 - \frac{49}{144} = \frac{95}{144} \).

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

Since \( \sin^2 \theta = \frac{95}{144} \), take the square root to find \( \sin \theta = \pm \sqrt{\frac{95}{144}} = \pm \frac{\sqrt{95}}{12} \). In Quadrant III, \( \sin \theta \) is positive, so \( \sin \theta = \frac{\sqrt{95}}{12} \).

Step 6: Use Sine and Cosine to Find Tangent

The tangent function is the ratio \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). Substitute the values: \( \tan \theta = \frac{\frac{\sqrt{95}}{12}}{-\frac{7}{12}} = -\frac{\sqrt{95}}{7} \).

Key Concepts

Cosine

The cosine of an angle, often shown as \( \cos \theta \), is one of the foundational trigonometric functions. Essentially, for a right-angled triangle, cosine is the ratio of the length of the adjacent side to the hypotenuse. In mathematical terms, that is: \[ \cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} \] In this exercise, we're given that \( \cos \theta = -\frac{7}{12} \). This means the adjacent side in reference to our angle \( \theta \) is 7, while the hypotenuse is 12. However, note that \( \theta \) is in the third quadrant, which means the cosine, or x-coordinate, is negative. So our final representation is negative, demonstrating the characteristic of cosine in Quadrant III. Understanding the cosine function and its behavior in different quadrants is crucial for evaluating angles and solving trigonometric problems.

Sine

The sine function, represented as \( \sin \theta \), is another fundamental aspect of trigonometry. Like cosine, it's also a ratio from a right-angle triangle, but it involves the opposite side and the hypotenuse. It's written as: \[ \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} \] In Quadrant III, the sine function is actually positive, even though cosine is negative. This is because, in this quadrant, both x (cosine) and y (sine) coordinates in the unit circle become negative, but the ratio involved in the \( \sin \) definition results positive when squared (as in the identity used in calculation). Here, we found \( \sin \theta \) as \( \frac{\sqrt{95}}{12} \), after applying the Pythagorean identity. Always remember, the sine remains positive in Quadrant III.

Tangent

The tangent function \( \tan \theta \) combines both sine and cosine, expressing the relationship flipping sides: \[ \tan \theta = \frac{\sin \theta}{\cos \theta} \] In terms of the triangle, this translates to the ratio of the opposite side over the adjacent side. Remembering that \( \tan \theta = \sin \theta \) divided by \( \cos \theta \), we see this formula gives us \( \tan \theta = -\frac{\sqrt{95}}{7} \). Note how the outcome is negative, because in Quadrant III the sine is positive but the cosine is negative, which results in a negative tangent.

Pythagorean Identity

The Pythagorean identity is an essential conjecture in trigonometry. It ties together sine and cosine through this simple, yet powerful equation: \[ \sin^2 \theta + \cos^2 \theta = 1 \] This identity is derived from the Pythagorean theorem and serves as a fundamental building block for working with trigonometric functions. If you know one function's value, you can find the other using this equation, as shown in the given solution. By starting with the known \( \cos \theta \), you square it to find \( \cos^2 \theta \). Subtract from 1 to retrieve \( \sin^2 \theta \). Then, use the positive or negative root to determine \( \sin \theta \), keeping in mind the specific quadrant's properties.

Quadrant III

Understanding the coordinate plane's quadrants is vitally important in trigonometry. Quadrant III is where both x and y values are negative, which affects trigonometric functions. Key attributes of Quadrant III:

• \( \cos \theta \) (adjacent over hypotenuse) is negative, as it represents the negative x-values.
• \( \sin \theta \) (opposite over hypotenuse) is also negative in a raw form, but as a ratio with squares or specific properties, it affects overall positivity.
• \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) ends up being positive because both sine and cosine are negative.

Understanding how each trigonometric function behaves in each quadrant allows you to solve angles and problems in any trigonometric exercise, just like the current example.