Problem 46 - Precalculus Mathematics for Calculus

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: Understanding the Given Information

We are given the value of \( \cos \theta = -\frac{7}{12} \) and that \( \theta \) is in Quadrant III. In Quadrant III, both sine and cosine are negative, while tangent is positive. We need to find the values of \( \sin \theta \), \( \tan \theta \), and other trigonometric functions.

Step 2: Using the Pythagorean Identity

The Pythagorean identity states that \( \sin^2 \theta + \cos^2 \theta = 1 \). We know \( \cos \theta = -\frac{7}{12} \), so we substitute this into the identity to find \( \sin \theta \).\[ \sin^2 \theta + \left(-\frac{7}{12}\right)^2 = 1 \]\[ \sin^2 \theta + \frac{49}{144} = 1 \]\[ \sin^2 \theta = 1 - \frac{49}{144} \]\[ \sin^2 \theta = \frac{95}{144} \]\[ \sin \theta = -\sqrt{\frac{95}{144}} \]\[ \sin \theta = -\frac{\sqrt{95}}{12} \]We take the negative square root because sine is negative in Quadrant III.

Step 3: Finding Tangent using Sine and Cosine

Use the relationship \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) to find \( \tan \theta \).\[ \tan \theta = \frac{-\frac{\sqrt{95}}{12}}{-\frac{7}{12}} \]\[ \tan \theta = \frac{\sqrt{95}}{7} \]Tangent is positive in Quadrant III, which matches our result.

Step 4: Calculating Other Trigonometric Functions

We can now find the remaining trigonometric functions. We already have \( \cos \theta = -\frac{7}{12} \), \( \sin \theta = -\frac{\sqrt{95}}{12} \), and \( \tan \theta = \frac{\sqrt{95}}{7} \).Calculate the reciprocals:- \( \csc \theta = \frac{1}{\sin \theta} = -\frac{12}{\sqrt{95}} = -\frac{12\sqrt{95}}{95} \)- \( \sec \theta = \frac{1}{\cos \theta} = -\frac{12}{7} \)- \( \cot \theta = \frac{1}{\tan \theta} = \frac{7}{\sqrt{95}} = \frac{7\sqrt{95}}{95} \)

Key Concepts

Pythagorean Identity

The Pythagorean Identity is a fundamental concept in trigonometry that connects the squares of the sine and cosine of an angle. It is an equation that states:

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

This identity is derived from the Pythagorean theorem when considering the unit circle (a circle with radius 1). The idea is that any point on the unit circle represents the sine and cosine of an angle where the x-coordinate is \( \cos\theta \) and the y-coordinate is \( \sin\theta \).
By rearranging and solving this equation, you can find one trigonometric function if the other is known, along with the quadrant information to determine the sign.
In our problem, we knew \( \cos\theta = -\frac{7}{12} \). Using the Pythagorean Identity:

• \( \sin^2\theta = 1 - \left(-\frac{7}{12}\right)^2 = \frac{95}{144} \)

To find \( \sin\theta \), we take the square root on both sides, considering that in Quadrant III, sine is negative:

• \( \sin\theta = -\frac{\sqrt{95}}{12} \)

Sine and Cosine Relationship

Sine and cosine are the primary trigonometric functions that describe the relationship between the sides of a right triangle and its angles. They are also fundamental in describing the movement on a unit circle.
The relationship between sine and cosine is most elegantly captured by the Pythagorean Identity discussed earlier. However, to find other functions like tangent, we utilize another relationship:

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

This equation shows that tangent is the ratio of sine to cosine, making it easy to calculate when both are known.
In our example, after finding \( \sin\theta = -\frac{\sqrt{95}}{12} \) and given \( \cos\theta = -\frac{7}{12} \), we find:

• \( \tan\theta = \frac{-\frac{\sqrt{95}}{12}}{-\frac{7}{12}} = \frac{\sqrt{95}}{7} \)

Here, the negatives cancel each other out, and we see tangent is positive in Quadrant III.

Quadrant III Trigonometry

Understanding the quadrant in which an angle resides is crucial for determining the signs of trigonometric functions. In trigonometry, a circle is divided into four quadrants based on the x and y axes.
Quadrant III is the bottom left section of this circle:

• Here, cosine and sine values are both negative because the x and y coordinates in this quadrant are negative.
• However, tangent, being the division of sine by cosine, becomes positive since a negative divided by a negative is a positive.

Knowing this helps us determine that in our problem, when \( \cos\theta = -\frac{7}{12} \), \( \sin\theta \) should be negative. We calculated \( \sin\theta = -\frac{\sqrt{95}}{12} \).
For tangent, \( \tan\theta = \frac{\sqrt{95}}{7} \), verifying positivity, which fits the characteristic of Quadrant III. Этот анализ также облегчает расчет обратных функций, таких как касательная, секанс и ко-секанс, на основе их основных функций.