Problem 50 - A Graphical Approach to Precalculus with Limits

Question

Use a half-number identity to find an expression for the exact value for each function, given the information about \(x\). $$\sin x, \text { given } \cos 2 x=\frac{2}{3} \text { and } \pi < x < \frac{3 \pi}{2}$$

Step-by-Step Solution

Answer

\( \sin x = -\frac{\sqrt{6}}{6} \)

Step 1: Understand the Given Information

We are given that \( \cos 2x = \frac{2}{3} \) and \( \pi < x < \frac{3\pi}{2} \). This means \( x \) is in the third quadrant.

Step 2: Apply Trigonometric Identity

Use the identity \( \cos 2x = 1 - 2\sin^2 x \) to find \( \sin^2 x \). Since \( \cos 2x = \frac{2}{3} \), we equate and solve: \( \frac{2}{3} = 1 - 2\sin^2 x \).

Step 3: Solve for \(\sin^2 x\)

Re-arrange the equation from Step 2: \( 2\sin^2 x = 1 - \frac{2}{3} = \frac{1}{3} \). Then divide both sides by 2: \( \sin^2 x = \frac{1}{6} \).

Step 4: Determine the Sign of \(\sin x\)

Since \( x \) is in the third quadrant, \( \sin x \) is negative. Thus, \( \sin x = -\sqrt{\frac{1}{6}} = -\frac{\sqrt{6}}{6} \).

Key Concepts

Cosine Double Angle Identity

The Cosine Double Angle Identity is a crucial tool in trigonometry. It relates the cosine of a double angle, such as \( 2x \), to expressions involving the sine or cosine of the original angle \( x \). The most common form of this identity is given by:

• \( \cos 2x = \cos^2 x - \sin^2 x \)
• \( \cos 2x = 2\cos^2 x - 1 \)
• \( \cos 2x = 1 - 2\sin^2 x \)

In the provided exercise, the identity version \( \cos 2x = 1 - 2\sin^2 x \) was used. This form is particularly useful when you have \( \cos 2x \) and need to solve for \( \sin x \). By equating the given \( \cos 2x \) value to \( 1 - 2\sin^2 x \), one can solve for \( \sin^2 x \) by simple algebraic manipulation. It's essential to choose the appropriate identity form depending on what's given in the problem for ease of calculation.

Sine Function

The sine function is one of the fundamental trigonometric functions. It describes the ratio of the length of the opposite side to the hypotenuse in a right-angled triangle. In the unit circle representation, sine indicates the y-coordinate of a point on the circle's circumference.
For angles between \( 0 \) and \( 2\pi \), the sine function can take positive or negative values depending on the quadrant. In trigonometry, determining the appropriate sign of \( \sin x \) is crucial.
In this exercise, the signs become essential as \( x \) is specifically located in the third quadrant \( (\pi < x < \frac{3\pi}{2}) \), where sine values are negative. Knowing this helps to determine that \( \sin x = -\sqrt{\frac{1}{6}} \) after calculation, because the question gives \( \cos 2x \) and requires finding \( \sin x \). Whenever you solve for a square root, remembering the appropriate sign based on the quadrant is key.

Quadrants of the Unit Circle

The unit circle is a circle with a radius of one centered at the origin in the coordinate plane. It plays a prominent role in trigonometry.
The circle is divided into four quadrants. These quadrants help us understand the sign of trigonometric functions like sine and cosine at various angles:

• First Quadrant \( (0, \frac{\pi}{2}) \): Both sine and cosine are positive.
• Second Quadrant \( (\frac{\pi}{2}, \pi) \): Sine is positive; cosine is negative.
• Third Quadrant \( (\pi, \frac{3\pi}{2}) \): Both sine and cosine are negative.
• Fourth Quadrant \( (\frac{3\pi}{2}, 2\pi) \): Sine is negative; cosine is positive.

In the context of the exercise, the angle \( x \) is in the third quadrant where both sine and cosine are negative. Thus, the sign of \( \sin x \) had to be negative. This quadrant-specific knowledge helps when using trigonometric identities to solve for unknown values.