Problem 72 - Precalculus Mathematics for Calculus

Question

Find the values of the trigonometric functions of \(t\) from the given information. \(\tan t=-4, \quad \csc t > 0\)

Step-by-Step Solution

Answer

\( \sin t = \frac{4}{\sqrt{17}}, \cos t = -\frac{1}{\sqrt{17}}, \tan t = -4, \sec t = -\sqrt{17}, \csc t = \frac{\sqrt{17}}{4}, \cot t = -\frac{1}{4} \).

Step 1: Understanding the Given Information

We are given that \( \tan t = -4 \) and \( \csc t > 0 \). The tangent being negative means that \( t \) lies in either the second or fourth quadrant. The cosecant being positive implies that the sine is positive, which occurs in the first and second quadrants. Combining these, we conclude \( t \) is in the second quadrant.

Step 2: Determine the Sine and Cosine Values

In the second quadrant, the sine is positive, and the cosine is negative. Since \( \tan t = \frac{\sin t}{\cos t} = -4\), let's assume \( \sin t = a \) and \( \cos t = -b \). Then we have \( \frac{a}{-b} = -4 \). This implies \( \frac{a}{b} = 4 \), or \( a = 4b \).

Step 3: Use the Pythagorean Identity

The identity \( \sin^2 t + \cos^2 t = 1 \) applies. Substituting \( a = 4b \), we have \( (4b)^2 + b^2 = 1 \rightarrow 16b^2 + b^2 = 1 \rightarrow 17b^2 = 1 \). Solving for \( b^2 \), we get \( b^2 = \frac{1}{17} \) and so \( b = \frac{1}{\sqrt{17}} \). Therefore, \( a = 4b = \frac{4}{\sqrt{17}} \).

Step 4: Calculate the Trigonometric Functions

The sine is \( \sin t = \frac{4}{\sqrt{17}} \), the cosine is \( \cos t = -\frac{1}{\sqrt{17}} \), and the given \( \tan t = \frac{4}{-1} = -4 \). The secant is the reciprocal of the cosine, so \( \sec t = -\sqrt{17} \). The cosecant is the reciprocal of the sine, so \( \csc t = \frac{\sqrt{17}}{4} \). Lastly, the cotangent is the reciprocal of the tangent, \( \cot t = -\frac{1}{4} \).

Key Concepts

Tangent Function

The tangent function is one of the primary trigonometric functions related to a right-angle triangle. It expresses the relationship between the opposite side and the adjacent side. In terms of angle \( t \), the tangent function is given by:

• \( \tan t = \frac{\text{opposite}}{\text{adjacent}} \)
• In terms of sine and cosine: \( \tan t = \frac{\sin t}{\cos t} \)

In our exercise, \( \tan t = -4 \), indicates that the angle \( t \) is either in the second or fourth quadrant.
This is because the tangent of an angle is negative in these quadrants.
The given information that \( \csc t \) is positive helps narrow it down to the second quadrant where \( \sin t \) is positive, aligning with our conditions.

Sine Function

The sine function is a fundamental trigonometric function that represents the ratio of the length of the opposite side to the hypotenuse in a right-angle triangle. It is expressed as:

• \( \sin t = \frac{\text{opposite}}{\text{hypotenuse}} \)

In this exercise, we needed to determine \( \sin t \) given that \( \tan t = -4 \) and \( t \) is in the second quadrant.
Knowing that the sine is positive in this quadrant, we can use the relation \( a = 4b \) derived from \( \tan t \) and solve for the sine value using the Pythagorean identity:

• \( \sin^2 t + \cos^2 t = 1 \)

This helps us find \( \sin t = \frac{4}{\sqrt{17}} \), as the second quadrant has positive sine values.

Cosine Function

The cosine function is another primary trigonometric function related to a right-angle triangle. It is defined as the ratio of the length of the adjacent side to the hypotenuse, expressed by:

• \( \cos t = \frac{\text{adjacent}}{\text{hypotenuse}} \)

In the context of our problem, we know \( \cos t \) is negative since \( t \) is in the second quadrant.
Using the relationship from the tangent function \( \frac{a}{-b} = -4 \) and the Pythagorean identity, we solved:

• \( \sin^2 t + \cos^2 t = 1 \)

This helped us find \( \cos t = -\frac{1}{\sqrt{17}} \).
The negative sign is due to the fact that the cosine of an angle is negative in the second quadrant, making sure all calculations align accurately with trigonometric identities.