Problem 30 - Algebra and Trigonometry

Question

find the exact value of each of the remaining trigonometric functions of \(\theta\) $$ \tan \theta=-\frac {1}{3}, \quad \sin \theta>0 $$

Step-by-Step Solution

Answer

Therefore, the exact values of the six trigonometric functions of \( \theta \) are \( \sin \theta = \frac {3}{\sqrt{10}} \), \( \cos \theta = - \frac {9}{\sqrt{10}} \), \( \tan \theta = - \frac {1}{3} \), \( \csc \theta = \frac {\sqrt{10}}{3} \), \( \sec \theta = - \frac {\sqrt{10}}{9} \), and \( \cot \theta = -3 \).

Step 1: Identify Quadrant

The problem states that \( \tan \theta \) is negative and \( \sin \theta \) is positive. The tangent function is negative either in the second or fourth quadrant, and the sine function is positive in the first or second quadrant. So the angle \( \theta \) lies in the second quadrant.

Step 2: Use Pythagorean Identity

For any right triangle, the Pythagorean identity states that \( \sin^2 \theta + \cos^2 \theta = 1 \). Now, write the given \( \tan \theta \) as a fraction. In a right triangle, \( \tan \theta = \frac {\sin \theta}{\cos \theta} \). Thus, \( \tan \theta = -\frac {1}{3} \) implies \( \sin \theta = \frac {k}{\sqrt{k^2+9}} \) and \( \cos \theta = -\frac {3k}{\sqrt{k^2+9}} \) for some positive constant \( k \). Choose \( k = 3 \) so that \( \sin \theta > 0 \), which gives \( \sin \theta = \frac {3}{\sqrt{10}} \) and \( \cos \theta = -\frac {9}{\sqrt{10}} \).

Step 3: Compute Remaining Functions

With \( \sin \theta \) and \( \cos \theta \) determined, we can compute the remaining trigonometric functions. Recall that \( \csc \theta = \frac {1}{\sin \theta} \), \( \sec \theta = \frac {1}{\cos \theta} \), and \( \cot \theta = \frac {1}{\tan \theta} \). Thus, we find \( \csc \theta = \frac {\sqrt{10}}{3} \), \( \sec \theta = - \frac {\sqrt{10}}{9} \), and \( \cot \theta = -3 \).

Key Concepts

Pythagorean Identity

Understanding the Pythagorean identity is crucial for solving many trigonometric problems. At its core, the identity is an equation expressing a fundamental relationship between the sine and cosine of an angle. The Pythagorean identity is written as \( \sin^2 \theta + \cos^2 \theta = 1 \). This means that for any angle \( \theta \) in a right-angled triangle, the square of the sine of \( \theta \) plus the square of the cosine of \( \theta \) will always equal one.

This identity is derived from the Pythagorean theorem, which states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. In trigonometric terms, when considering a unit circle (a circle with a radius of 1), the sine and cosine values represent the lengths of the sides opposite and adjacent to \( \theta \), with the hypotenuse being 1. Hence, the Pythagorean identity is always true for any angle, and therefore it serves as a powerful tool to solve for unknown trigonometric functions when one is known.

Tangent Function

The tangent function is another fundamental trigonometric function often symbolized as \( \tan \). It is defined as the ratio of the sine of an angle to the cosine of that same angle, expressed as \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). This function is particularly unique because it can be both positive and negative depending on the angle \( \theta \) and which quadrant it lies in.

In the context of the exercise, \( \tan \theta \) is given as \( -\frac{1}{3} \) suggesting that the \( \sin \theta \) and \( \cos \theta \) have opposite signs. The information \( \sin \theta > 0 \) helps to determine that \( \cos \theta \) must be negative for \( \tan \theta \) to be negative. Additionally, knowing that the tangent function has a period of \( \pi \) radians helps in understanding that it repeats its values after each \( \pi \) interval, which is vital for analyzing the function over the different quadrants of the unit circle.

Unit Circle Quadrant Analysis

In trigonometry, quadrant analysis of the unit circle is a method for determining the signs of trigonometric functions based on the angle's location. The unit circle is divided into four quadrants, and the sign of \( \sin \theta \) and \( \cos \theta \) will vary depending on which quadrant \( \theta \) falls into.

To simplify, the first quadrant (0 to \( \frac{\pi}{2} \)) has both \( \sin \theta \) and \( \cos \theta \) as positive. In the second quadrant (\( \frac{\pi}{2} \) to \( \pi \) ), \( \sin \theta \) is positive while \( \cos \theta \) is negative. In the third quadrant (\( \pi \) to \( \frac{3\pi}{2} \) ), both are negative, and in the fourth quadrant (\( \frac{3\pi}{2} \) to \( 2\pi \) ), \( \sin \theta \) is negative and \( \cos \theta \) is positive.

As stated in the provided solution, since \( \sin \theta > 0 \) and \( \tan \theta \) is negative, the angle \( \theta \) must be in the second quadrant. This piece of information is the key to determining the exact values of the trigonometric functions for \( \theta \) in this scenario. Utilizing quadrant analysis in conjunction with the Pythagorean identity and our understanding of tangent can greatly simplify complex trigonometric problems.