Problem 129 - A Graphical Approach to Precalculus with Limits

Question

Use fundamental identities to find each expression. Write \(\tan \theta\) in terms of \(\sin \theta\) if \(\theta\) is in quadrant I or IV.

Step-by-Step Solution

Answer

\(\tan \theta = \frac{\sin \theta}{\sqrt{1 - \sin^2 \theta}}\) in quadrant I, \(\tan \theta = -\frac{\sin \theta}{\sqrt{1 - \sin^2 \theta}}\) in quadrant IV.

Step 1: Recall the basic trigonometric identity

The tangent of an angle, \(\theta\), can be expressed in terms of sine and cosine using the identity \(\tan \theta = \frac{\sin \theta}{\cos \theta}\). We will use this identity to express \(\tan \theta\) in terms of \(\sin \theta\).

Step 2: Identify the Pythagorean identity

To write \(\tan \theta\) solely in terms of \(\sin \theta\), we need to express \(\cos \theta\) in terms of \(\sin \theta\). Recall the Pythagorean identity: \(\sin^2 \theta + \cos^2 \theta = 1\).

Step 3: Solve for \(\cos \theta\) in terms of \(\sin \theta\)

Rearrange the Pythagorean identity to find \(\cos^2 \theta = 1 - \sin^2 \theta\). Thus, \(\cos \theta = \sqrt{1 - \sin^2 \theta}\) for \(\theta\) in quadrant I, and \(\cos \theta = -\sqrt{1 - \sin^2 \theta}\) for \(\theta\) in quadrant IV, since cosine is positive in quadrant I and negative in quadrant IV.

Step 4: Express \(\tan \theta\) in terms of \(\sin \theta\) for quadrants I and IV

Using the relationship \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), substitute \(\cos \theta\) from Step 3: \(\tan \theta = \frac{\sin \theta}{\sqrt{1 - \sin^2 \theta}}\) in quadrant I, or \(\tan \theta = -\frac{\sin \theta}{\sqrt{1 - \sin^2 \theta}}\) in quadrant IV.

Key Concepts

Pythagorean Identity

The Pythagorean Identity is a cornerstone of trigonometry and comes from the Pythagorean Theorem applied to the unit circle. Its most common form is written as \( \sin^2 \theta + \cos^2 \theta = 1 \). This identity shows the intrinsic connection between the sine and cosine functions. It states that, for any angle \( \theta \), the sum of the square of its sine and cosine will always equal one.

In trigonometric problems, the Pythagorean Identity is often used to derive other relationships. For example, if you know the sine value for an angle, you can easily find the cosine value by rearranging the identity to \( \cos^2 \theta = 1 - \sin^2 \theta \).

Depending on the quadrant where the angle is located, the signs of sine and cosine change:

• In Quadrant I, both sine and cosine are positive.
• In Quadrant IV, sine is negative, and cosine is positive.

Understanding this identity is crucial for simplifying and solving trigonometric expressions.

Sine Function

The sine function, often abbreviated as \( \sin \), is a fundamental component in trigonometry. It defines the vertical distance from the center of a unit circle to the point on the circle, corresponding to a given angle \( \theta \).

For angles in different quadrants, the sine function has specific properties:

• In Quadrant I, sine values are positive.
• In Quadrant IV, sine values turn negative.

This sign change plays a vital role when converting between different trigonometric forms and identities.

An essential property of the sine function is that it is always between -1 and 1 for any angle. This is a direct consequence of its definition related to the unit circle, ensuring that \( \sin^2 \theta \) and \( \cos^2 \theta \) sum to one. Hence, sine plays a key role in the Pythagorean identity, making it a vital tool in solving trigonometric equations.

Tangent Function

The tangent function, denoted as \( \tan \), is another basic trigonometric function, which can be expressed in terms of sine and cosine as \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). This ratio has significant applications in solving trigonometric problems, as it relates two fundamental trigonometric functions.

In Quadrant I, where both sine and cosine are positive, \( \tan \theta \) will also be positive. In contrast, in Quadrant IV, where sine is negative and cosine is positive, \( \tan \theta \) becomes negative.

To express tangent solely in terms of sine, especially when solving trigonometric identities, you can utilize the Pythagorean identity. By rearranging it to find \( \cos \theta \) and substituting into the tangent expression, you derive:

• \( \tan \theta = \frac{\sin \theta}{\sqrt{1 - \sin^2 \theta}} \) in Quadrant I.
• \( \tan \theta = -\frac{\sin \theta}{\sqrt{1 - \sin^2 \theta}} \) in Quadrant IV.

This manipulation highlights how interlinked trigonometric functions are and how identities facilitate expressions and problem-solving in trigonometry.