Problem 119 - Precalculus

Question

For the following exercises, find the exact values of a ) \(\sin \left(\frac{x}{2}\right),\) b) \(\sin \left(\frac{x}{2}\right),\) and \(c )\) without solving for \(x\) If \(\sin x=-\frac{12}{13}, \quad\) and \(x\) is in quadrant III.

Step-by-Step Solution

Answer

\(\sin \left(\frac{x}{2}\right) = \frac{3\sqrt{13}}{13}\).

Step 1: Determine the Quadrant of x/2

Since \(x\) is in quadrant III, \(\frac{x}{2}\) will be in quadrant II, because dividing by 2 moves the angle to an earlier quadrant. Sine is positive in quadrant II.

Step 2: Use Sine Identity

The half-angle identity for sine is \( \sin \left(\frac{x}{2}\right) = \pm \sqrt{\frac{1 - \cos x}{2}}.\) Since \(\frac{x}{2}\) is in quadrant II, choose the positive root.

Step 3: Find Cosine of x

Using the Pythagorean identity \(sin^2 x + cos^2 x = 1,\) substitute \(\sin x = -\frac{12}{13}\):\[cos^2 x = 1 - \left(-\frac{12}{13}\right)^2 = \frac{25}{169}.\]Thus, \(cos x = -\frac{5}{13}\) because cosine is negative in quadrant III.

Step 4: Calculate the Exact Value of \(\sin \left(\frac{x}{2}\right)\)

Substitute \(cos x = -\frac{5}{13}\) into the half-angle identity:\[\sin \left(\frac{x}{2}\right) = \sqrt{\frac{1 + \frac{5}{13}}{2}} = \sqrt{\frac{18}{26}} = \sqrt{\frac{9}{13}} = \frac{3\sqrt{13}}{13}.\]

Key Concepts

Trigonometric Identities

Trigonometric identities are equations that involve trigonometric functions and are true for every value in the domain of the variables. These identities play a fundamental role in simplifying complex mathematical expressions and solving trigonometric equations.
There are several types of trigonometric identities, such as:

• Pythagorean Identities: These are derived from the Pythagorean theorem and include equations like \( \sin^2 x + \cos^2 x = 1 \).
• Reciprocal Identities: These define the reciprocals of trigonometric functions, such as \( \csc x = \frac{1}{\sin x} \).
• Half-Angle Identities: These help find the sine, cosine, or tangent of half-angles, like \( \sin \left( \frac{x}{2} \right) = \pm \sqrt{\frac{1 - \cos x}{2}} \).

In our specific exercise, we utilize the half-angle identity to find the value of \( \sin \left( \frac{x}{2} \right) \) when given \( \sin x = -\frac{12}{13} \). It's important to remember which quadrant the angle resides in when employing these identities, as this affects the sign of the resulting value.

Quadrants in Trigonometry

Understanding quadrants in trigonometry is essential for determining the signs of trigonometric functions and their values. The coordinate plane is divided into four quadrants. Each quadrant affects the sign of sine, cosine, and tangent differently, based on positive and negative axes.
The quadrants are:

• Quadrant I: Both sine and cosine are positive.
• Quadrant II: Sine is positive, but cosine is negative.
• Quadrant III: Both sine and cosine are negative.
• Quadrant IV: Sine is negative, but cosine is positive.

In this exercise, \( x \) resides in Quadrant III where sine is calculated as negative, which directly affects the function values. When finding \( \frac{x}{2} \), it shifts the angle into Quadrant II, where sine becomes positive, guiding us to choose the positive root using the half-angle identity.

Sine and Cosine Functions

Sine and cosine functions are key components in trigonometry, representing periodic oscillations. They are defined based on a right triangle or the unit circle.Here’s what to know about these functions:

• Sine Function: Given by \( \sin x = \frac{\text{opposite}}{\text{hypotenuse}} \), it varies between -1 and 1 as you move along the unit circle. It starts at 0, reaches 1 at \( \frac{\pi}{2} \), and goes back to 0 at \( \pi \).
• Cosine Function: Known as \( \cos x = \frac{\text{adjacent}}{\text{hypotenuse}} \), it also ranges from -1 to 1. It starts at 1, goes to 0 at \( \frac{\pi}{2} \), and reaches -1 at \( \pi \).

Using the given piece of the equation \( \sin x = -\frac{12}{13} \,\) we apply the Pythagorean identity to find \( \cos x \). This helps us calculate half-angle expressions. Understanding how sine and cosine shift through quadrants is crucial, as demonstrated in determining the half-angle value in this exercise.