Problem 59 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

\( \sec t = -\sqrt{1 + \tan^2 t} \)

Step 1: Recall Trigonometric Relationships

In the context of trigonometry, we have the identities \( \sec t = \frac{1}{\cos t} \) and \( \tan t = \frac{\sin t}{\cos t} \). Our goal is to express \( \sec t \) in terms of \( \tan t \).

Step 2: Consider the Signs of Functions in Quadrant II

In Quadrant II, the cosine function is negative, and the sine function is positive. Therefore, \( \sec t = \frac{1}{\cos t} \) will be negative, and \( \tan t = \frac{\sin t}{\cos t} \) will also be negative since \( \sin t \) is positive and \( \cos t \) is negative in this quadrant.

Step 3: Express Secant in Terms of Tangent

Using the identity \( 1 + \tan^2 t = \sec^2 t \), solve for \( \sec t \). We have \( \sec t = \pm \sqrt{1 + \tan^2 t} \). Since \( \sec t \) is negative in Quadrant II, we use \( \sec t = -\sqrt{1 + \tan^2 t} \).

Key Concepts

Quadrant Analysis

Quadrant analysis helps us understand the behavior and signs of trigonometric functions based on the angle's terminal position in the unit circle. The unit circle is divided into four quadrants, with each having its own unique sign properties for trigonometric functions.

In Quadrant II, where angles range from 90° to 180°, the sine function (\(\sin t\)) is positive because it represents the vertical component above the x-axis, while the cosine function (\(\cos t\)) is negative, illustrating the horizontal distance to the left of the origin.

Understanding these sign conventions is crucial because they determine the sign of derived trigonometric functions. For example:

• Secant (\(\sec t = \frac{1}{\cos t}\)) will be negative since cosine is negative.
• Tangent (\(\tan t = \frac{\sin t}{\cos t}\)) will also be negative as it involves division of positive sine by negative cosine.

This insight helps in simplifying expressions and solving trigonometric equations, indicating the importance of quadrant analysis in trigonometry.

Trigonometric Functions

Trigonometric functions are fundamental in the study of angles and their relationships. Most commonly used functions include sine, cosine, and tangent, each defined using right-angled triangles or the unit circle.

For any angle \(t\):

• Secant (\(\sec t\)) is defined as the reciprocal of cosine (\(\sec t = \frac{1}{\cos t}\)).
• Tangent (\(\tan t\)) is defined as the ratio of sine to cosine (\(\tan t = \frac{\sin t}{\cos t}\)).

Understanding these definitions helps simplify complex expressions and solve equations by expressing one trigonometric function in terms of another. Recognizing these functions on the unit circle enhances our capabilities of evaluating angles and understanding their properties in different quadrants.

Trigonometric Relationships

Trigonometric relationships connect different functions through identities that hold true for any angle. These relationships are pivotal in solving problems that need transformation or simplification of expressions.

A key identity used frequently is the Pythagorean identity, given by:\[1 + \tan^2 t = \sec^2 t\]This identity is incredibly useful when expressing one function in terms of another. For example, if we need to express secant in terms of tangent, we rearrange it to get:\[\sec t = \pm \sqrt{1 + \tan^2 t}\]In Quadrant II, since \(\sec t\) is negative, we select the negative root:\[\sec t = -\sqrt{1 + \tan^2 t}\]These relationships are essential tools that provide flexibility in handling various trigonometric scenarios, assisting in reaching a solution efficiently.