Problem 58 - Precalculus Mathematics for Calculus

Question

Write the first expression in terms of the second if the terminal point determined by \(t\) is in the given quadrant. \(\tan t, \cos t ; \quad\) Quadrant III

Step-by-Step Solution

Answer

In Quadrant III, \( \tan(t) = -\sqrt{\frac{1}{\cos^2(t)} - 1} \).

Step 1: Understand the relationship between trigonometric functions in Quadrant III

In Quadrant III, the cosine of an angle is negative and the tangent is positive. Knowing the identities: \( \tan(t) = \frac{\sin(t)}{\cos(t)} \) and \( \cos(t) = x \), where \( \sin(t) = y \). Contemplate how these relate in Quadrant III.

Step 2: Use the Pythagorean identity for sinus and cosinus

Remember the identity \( \sin^2(t) + \cos^2(t) = 1 \). Solve this for \( \sin(t) \) to get \( \sin(t) = \pm \sqrt{1 - \cos^2(t)} \). In Quadrant III, \( \sin(t) \) is negative.

Step 3: Express tangent in terms of cosine

Substitute the expression for \( \sin(t) \) into the tangent identity: \( \tan(t) = \frac{\sin(t)}{\cos(t)} = \frac{-\sqrt{1 - \cos^2(t)}}{\cos(t)} \) which simplifies to \( \tan(t) = -\sqrt{\frac{1 - \cos^2(t)}{\cos^2(t)}} \).

Step 4: Simplify the expression

Further simplify the equation: \( \tan(t) = -\sqrt{\frac{1}{\cos^2(t)} - 1} = -\sqrt{\sec^2(t) - 1} \). Substitute back that \( \sec(t) = \frac{1}{\cos(t)} \) to achieve \( \sec^2(t) = \frac{1}{\cos^2(t)} \).

Step 5: Final expression for tan in terms of cos

Thus, \( \tan(t) = -\sqrt{\frac{1}{\cos^2(t)} - 1} \). Therefore, suppose \( \tan(t) = k \) then \( k = -\sqrt{\frac{1}{\cos^2(t)} - 1} \) holds true, concluding the relationship.

Key Concepts

Quadrant Analysis

When analyzing trigonometric functions, understanding the behavior of these functions in different quadrants is crucial. The coordinate plane is divided into four quadrants, each affecting the sign of trigonometric functions differently. In Quadrant III:

• The x-coordinate, or the cosine (\( ext{cos}(t)\), is negative.
• The y-coordinate, or the sine (\( ext{sin}(t)\), is also negative.
• The tangent (\( an(t)\)), being the ratio of \( ext{sin}(t)\) to \( ext{cos}(t)\), remains positive since a negative divided by a negative results in a positive value.

This understanding is essential for interpreting trigonometric functions based on the quadrant in which the terminal side of the angle lies. For instance, if given that \( an(t)\) is positive in Quadrant III, we can deduce that both sine and cosine must be negative, confirming consistency with the quadrant's characteristics.

Pythagorean Identity

The Pythagorean Identity is a keystone in trigonometry, exemplified by the equation: \[ \sin^2(t) + \cos^2(t) = 1 \] This equation derives from the Pythagorean Theorem, focusing on a unit circle with radius 1. Its practical application shines when isolating one component in terms of another. For instance, solving for sine gives the expression: \[ \sin(t) = \pm \sqrt{1 - \cos^2(t)} \] In Quadrant III, since sine is negative, it applies that: \[ \sin(t) = -\sqrt{1 - \cos^2(t)} \] This identity not only helps express one trigonometric function in terms of another, but it also facilitates deeper exploration of relation patterns and trigonometric proofs.

Trigonometric Functions

Trigonometric functions—sine, cosine, and tangent—are fundamental in describing relationships within right triangles and rotational angles. Here’s how they interplay when angles lie in different quadrants. In the main problem, expressing tangent in terms of cosine using known identities serves as a prime example. Initially, tangent is defined as: \[ \tan(t) = \frac{\sin(t)}{\cos(t)} \] Linking trigonometric identities to \( an(t)\), we remember: \[ \tan^2(t) + 1 = \sec^2(t) \] Stemming from this, secant is expressed as: \[ \sec(t) = \frac{1}{\cos(t)} \] Substituting secant in our problem becomes: \[ \tan(t) = -\sqrt{\sec^2(t) - 1} = -\sqrt{\frac{1}{\cos^2(t)} - 1} \] The art of relating different trigonometric functions to one another lies in comprehending these identities flawlessly, achieving seamless problem-solving and advancing to more complex trigonometric challenges.