Problem 52 - Precalculus Mathematics for Calculus

Question

Evaluate each expression under the given conditions. \(\sin (\theta / 2) ; \tan \theta=-\frac{5}{12}, \theta\) in Quadrant IV

Step-by-Step Solution

Answer

\( \sin (\theta / 2) = \frac{1}{\sqrt{26}} \)

Step 1: Identify the formula

To find \( \sin (\theta / 2) \), we can use the half-angle identity for sine: \( \sin(\theta/2) = \pm \sqrt{\frac{1 - \cos \theta}{2}} \). The sign depends on the quadrant where \( \theta/2 \) lies.

Step 2: Determine \(\cos \theta\)

Since \(\tan \theta = -\frac{5}{12}\), we can use the identity \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) to find \( \sin \theta\) and \( \cos \theta\). In a right triangle, if opposite/adjacent sides for tangent \(\theta\) are 5 and 12, respectively, the hypotenuse \(h\) is \( \sqrt{5^2 + 12^2} = 13 \). \( \cos \theta = \frac{12}{13} \) in Quadrant IV, since \(\cos > 0\) there.

Step 3: Solve for \(\cos \theta\) considering Quadrant IV

Since \( \tan \theta = -\frac{5}{12} \) and \( \theta \) is in Quadrant IV, where cosine is positive and sine is negative. Therefore, \( \cos \theta = \frac{12}{13} \) and \(\sin \theta = -\frac{5}{13} \).

Step 4: Apply the half-angle identity

Using \( \cos \theta = \frac{12}{13} \), substitute into the half-angle formula: \( \sin(\theta/2) = \pm \sqrt{\frac{1 - \cos \theta}{2}} = \pm \sqrt{\frac{1 - \frac{12}{13}}{2}} \).

Step 5: Simplify the expression

Simplify the expression \( \frac{1 - \frac{12}{13}}{2} = \frac{\frac{1}{13}}{2} = \frac{1}{26} \). Thus, \( \sin(\theta/2) = \pm \sqrt{\frac{1}{26}} = \pm \frac{1}{\sqrt{26}} \).

Step 6: Determine the Sign of \(\sin(\theta/2)\)

Since \(\theta\) is in Quadrant IV, \(\theta/2\) is in Quadrant I, where all trigonometric functions are positive. Therefore, \(\sin(\theta/2) = \frac{1}{\sqrt{26}}\).

Key Concepts

Half-Angle Identities

In trigonometry, half-angle identities are crucial for simplifying expressions and solving equations involving trigonometric functions. One important half-angle identity is for sine:

• \(\sin(\theta/2) = \pm \sqrt{ \frac{1 - \cos \theta}{2} } \)

This formula helps to find the sine value for half of an angle \(\theta\). The sign of \(\sin(\theta/2)\) depends on the quadrant in which \(\theta/2\) is located. In the exercise, we first find the cosine of \(\theta\) to use this formula. Understanding the half-angle identities allows you to break down complex trigonometric problems into simpler tasks. Consider practicing using these identities with different angles to become more comfortable with their application.

Quadrant Analysis

Quadrant analysis is essential in understanding the signs of trigonometric functions. The coordinate system is divided into four quadrants, each indicating specific behaviors for sine, cosine, and tangent:

• Quadrant I: All trigonometric functions are positive.
• Quadrant II: Sine is positive, cosine and tangent are negative.
• Quadrant III: Tangent is positive, sine and cosine are negative.
• Quadrant IV: Cosine is positive, sine and tangent are negative.

In this exercise, \(\theta\) is in Quadrant IV, which means \(\cos \theta\) is positive and \(\sin \theta\) is negative. By understanding the quadrant location of \(\theta\), we can also infer that \(\theta/2\) will be in a different quadrant, influencing the sign of \(\sin(\theta/2)\). Here, \(\theta/2\) lies in Quadrant I, where all trigonometric values are positive.

Trigonometric Functions

Trigonometric functions are the foundation of studying angles and their relationships. The primary trigonometric functions include sine, cosine, and tangent. Each function captures a different relationship:

• Sine (\(\sin\)): Opposite side over hypotenuse in a right triangle.
• Cosine (\(\cos\)): Adjacent side over hypotenuse.
• Tangent (\(\tan\)): Opposite side over adjacent side.

In this problem, the given value is \(\tan \theta = -\frac{5}{12}\). You can use these relationships to solve the triangle and find other trigonometric values like \(\sin \theta\) and \(\cos \theta\). Understanding how these functions intertwine helps solve complex angle problems efficiently, especially when working with identities and angle formulas.

Triangle Relationships

Triangle relationships arise naturally when dealing with trigonometric problems. Specifically, knowing how the sides and angles of a right triangle relate informs the value of trigonometric functions. Typically, you'll see a problem expressed as a fraction of side lengths in a right triangle:

• Use the Pythagorean theorem: \(c^2 = a^2 + b^2\) to find missing side lengths.
• Use these side lengths to find \(\sin\), \(\cos\), and \(\tan\).

In the exercise, \(\tan \theta = -\frac{5}{12}\) suggests a triangle with opposite side 5, adjacent side 12, and hypotenuse \(\sqrt{5^2 + 12^2} = 13\). From this, you find \(\cos \theta = \frac{12}{13}\) and \(\sin \theta = -\frac{5}{13}\). Mastering these relationships enables you to break down complex trigonometric tasks into manageable steps.