Problem 17 - Precalculus with Limits

Question

In Exercises 11 - 24, use the given values to evaluate (if possible)all six trigonometric functions. \( \sec \phi = \dfrac{3}{2} \), \( \csc \phi = -\dfrac{3\sqrt{5}}{5} \)

Step-by-Step Solution

Answer

The six trigonometric functions of the angle \( \phi \) are: \( \cos \phi = \dfrac{2}{3} \), \( \sin \phi = -\dfrac{\sqrt{5}}{3} \), \( \tan \phi = -\dfrac{\sqrt{5}}{2} \), \( \sec \phi = \dfrac{3}{2} \), \( \csc \phi = -\dfrac{3\sqrt{5}}{5} \), \( \cot \phi = -\dfrac{2}{\sqrt{5}} \)

Step 1: Videoscribe the Reciprocal Relations

Note that the secant is the reciprocal of the cosine function and the cosecant is the reciprocal of the sine function. Hence, \( \cos \phi = \dfrac{1}{\sec \phi} = \dfrac{2}{3} \) and \( \sin \phi = \dfrac{1}{\csc \phi} = -\dfrac{5}{3\sqrt{5}} = -\dfrac{\sqrt{5}}{3} \).

Step 2: Calculate the Remaining Functions

To find the remaining functions, we use the Pythagorean identities. The tangent of an angle is the sine divided by the cosine, the cotangent is the reciprocal of the tangent. Therefore, \( \tan \phi = \dfrac{\sin \phi}{\cos \phi} = -\dfrac{\sqrt{5}}{2} \) and \( \cot \phi = \dfrac{1}{\tan \phi} = -\dfrac{2}{\sqrt{5}} \).

Key Concepts

Reciprocal Trigonometric Functions

Understanding reciprocal trigonometric functions is essential when delving into trigonometric problems. These special functions are pairs of trigonometric functions that are related through the concept of 'reciprocity' — meaning one is the reciprocal of the other. The three main pairs are sine (sin) and cosecant (csc), cosine (cos) and secant (sec), and tangent (tan) and cotangent (cot).

For example, the secant of an angle is simply the reciprocal of the cosine of that angle. This can be represented mathematically as \( \text{sec}(\theta) = \frac{1}{\text{cos}(\theta)} \). Conversely, \( \text{cos}(\theta) = \frac{1}{\text{sec}(\theta)} \). Similarly, \( \text{csc}(\theta) = \frac{1}{\text{sin}(\theta)} \) and \( \text{sin}(\theta) = \frac{1}{\text{csc}(\theta)} \).

In the given exercise, this reciprocal relationship is used to find the values of cosine and sine from given secant and cosecant values. The reciprocal nature of these pairs implies that if you have one, you can easily find the other, making this concept a powerful tool in evaluating trigonometric functions.

Pythagorean Identities

The Pythagorean identities are a set of equations that provide a fundamental relationship among the trigonometric functions. They come from the Pythagorean Theorem, which you may recall is \( a^2 + b^2 = c^2 \), relating the sides of a right triangle. These trigonometric identities extend this relationship to the unit circle, where the radius (r) is always 1, and therefore the Pythagorean theorem leads to these identities:

• \( \text{sin}^2(\theta) + \text{cos}^2(\theta) = 1 \)
• \( 1 + \text{tan}^2(\theta) = \text{sec}^2(\theta) \)
• \( 1 + \text{cot}^2(\theta) = \text{csc}^2(\theta) \)

In practice, these identities allow us to find the value of one trigonometric function using another, which is exactly what is performed in the exercise. By using the already determined sine and cosine values, other functions like tangent and cotangent can be found with the assistance of the Pythagorean identities. These identities are very helpful to check the correctness of solutions and to solve trigonometric equations.

Tangent and Cotangent

Exploring the relationship between tangent and cotangent is crucial for mastering trigonometry. The tangent of an angle in a right-angled triangle is defined as the ratio of the opposite side to the adjacent side, while the cotangent is the reciprocal of the tangent.

This implies \( \text{tan}(\theta) = \frac{\text{sin}(\theta)}{\text{cos}(\theta)} \) and \( \text{cot}(\theta) = \frac{1}{\text{tan}(\theta)} = \frac{\text{cos}(\theta)}{\text{sin}(\theta)} \). Tan and cot are particularly important because they can represent the slope of a line, rate of change, and angles of elevation or depression. Knowing how to switch between tan and cot is a valuable skill, as seen in the example where after calculating the sine and cosine, the tangent was found directly and the cotangent was then derived from the tangent.

Moreover, understanding the sign of tangent and cotangent functions gives insight into the angle's quadrant, as in the exercise, where both came out as negative, which corresponds to angles in either the second or fourth quadrant where either sine or cosine (and thus tangent or cotangent) is negative.