Problem 32 - Algebra 2

Question

Find the exact values of \(\sin 2 \theta, \cos 2 \theta, \sin \frac{\theta}{2},\) and \(\cos \frac{\theta}{2}\) for each of the following. $$ \sin \theta=-\frac{1}{3} ; 270^{\circ}<\theta<360^{\circ} $$

Step-by-Step Solution

Answer

\( \sin 2\theta = -\frac{4\sqrt{2}}{9}, \cos 2\theta = \frac{7}{9}, \sin \frac{\theta}{2} = \sqrt{\frac{3 - 2\sqrt{2}}{6}}, \cos \frac{\theta}{2} = -\sqrt{\frac{3 + 2\sqrt{2}}{6}} \).

Step 1: Identify the Quadrant

Since \( 270^\circ < \theta < 360^\circ \), \( \theta \) is in the fourth quadrant. In the fourth quadrant, sine is negative and cosine is positive.

Step 2: Calculate \( \cos \theta \) Using Pythagorean Identity

The Pythagorean identity is \( \sin^2 \theta + \cos^2 \theta = 1 \). Substitute \( \sin \theta = -\frac{1}{3} \) into the equation: \[ (-\frac{1}{3})^2 + \cos^2 \theta = 1 \]\[ \frac{1}{9} + \cos^2 \theta = 1 \]\[ \cos^2 \theta = \frac{8}{9} \]\[ \cos \theta = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3} \]Since \( \theta \) is in the fourth quadrant, \( \cos \theta \) is positive.

Step 3: Determine \( \sin 2\theta \) and \( \cos 2\theta \)

Use the double angle formulas \( \sin 2\theta = 2 \sin \theta \cos \theta \) and \( \cos 2\theta = \cos^2 \theta - \sin^2 \theta \).Substitute the known values:\[ \sin 2\theta = 2 \left(-\frac{1}{3}\right) \left(\frac{2\sqrt{2}}{3}\right) = -\frac{4\sqrt{2}}{9} \]\[ \cos 2\theta = \left(\frac{2\sqrt{2}}{3}\right)^2 - \left(-\frac{1}{3}\right)^2 = \frac{8}{9} - \frac{1}{9} = \frac{7}{9} \]

Step 4: Determine \( \sin \frac{\theta}{2} \) and \( \cos \frac{\theta}{2} \)

Use the half-angle formulas \( \sin \frac{\theta}{2} = \pm \sqrt{\frac{1 - \cos \theta}{2}} \) and \( \cos \frac{\theta}{2} = \pm \sqrt{\frac{1 + \cos \theta}{2}} \). Since \( \theta \) is in the fourth quadrant, \( \frac{\theta}{2} \) will be in the second quadrant, where sine is positive and cosine is negative.Calculate:\[ \sin \frac{\theta}{2} = \sqrt{\frac{1 - \frac{2\sqrt{2}}{3}}{2}} = \sqrt{\frac{3 - 2\sqrt{2}}{6}} \]\[ \cos \frac{\theta}{2} = -\sqrt{\frac{1 + \frac{2\sqrt{2}}{3}}{2}} = -\sqrt{\frac{3 + 2\sqrt{2}}{6}} \]

Key Concepts

Understanding \( \sin 2\theta \)

The double angle formula for sine, \( \sin 2\theta = 2 \sin \theta \cos \theta \), helps calculate the sine of twice an angle. To apply it, both the values of \( \sin \theta \) and \( \cos \theta \) must be known. In this exercise, \( \sin \theta = -\frac{1}{3} \) because \( \theta \) is located in the fourth quadrant where sine is negative.

• Using the identity, we plug in
  
• \( \sin 2\theta = 2 \times \left(-\frac{1}{3}\right) \times \frac{2\sqrt{2}}{3}\)
• This results in \( \sin 2\theta = -\frac{4\sqrt{2}}{9} \).

Thus, \( \sin 2\theta \) is calculated as negative, aligning with its fourth quadrant position.

Calculating \( \cos 2\theta \)

The formula for \( \cos 2\theta \) is \( \cos 2\theta = \cos^2 \theta - \sin^2 \theta \). This lets you find the cosine of double an angle. Beginning with known values \( \cos \theta = \frac{2\sqrt{2}}{3} \) and \( \sin \theta = -\frac{1}{3} \), we plug these into the formula:

• First square both terms: \( \cos^2 \theta = \left(\frac{2\sqrt{2}}{3}\right)^2 = \frac{8}{9} \) and \( \sin^2 \theta = \left(-\frac{1}{3}\right)^2 = \frac{1}{9} \)
• Next, subtract: \( \cos 2\theta = \frac{8}{9} - \frac{1}{9} = \frac{7}{9} \)

The result shows \( \cos 2\theta \) is positive, consistent with behavior in the fourth quadrant.

Exploring \( \sin \frac{\theta}{2} \)

For half-angle identities, \( \sin \frac{\theta}{2} = \pm \sqrt{\frac{1 - \cos \theta}{2}} \) comes in handy. In our setting, \( \cos \theta = \frac{2\sqrt{2}}{3} \). Since \( \frac{\theta}{2} \) lies in the second quadrant, where sine is positive, choose the positive root:

• Subtract within the formula: \( 1 - \cos \theta = 1 - \frac{2\sqrt{2}}{3} = \frac{3 - 2\sqrt{2}}{3} \)
• Plug this into the half-angle formula: \( \sin \frac{\theta}{2} = \sqrt{\frac{3 - 2\sqrt{2}}{6}} \)

This shows \( \sin \frac{\theta}{2} \) in the second quadrant remains positive.

Determining \( \cos \frac{\theta}{2} \)

Using the half-angle formula \( \cos \frac{\theta}{2} = \pm \sqrt{\frac{1 + \cos \theta}{2}} \), let's determine the cosine of half the angle. Remember, \( \frac{\theta}{2} \) lies in the second quadrant, making cosine negative.

• Figure under the square root: \( 1 + \cos \theta = 1 + \frac{2\sqrt{2}}{3} = \frac{3 + 2\sqrt{2}}{3} \)
• Input into the formula: \( \cos \frac{\theta}{2} = -\sqrt{\frac{3 + 2\sqrt{2}}{6}} \)

The negative result matches the expected sign of cosine for the second quadrant.