Problem 13 - 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. $$ \sin \theta=-\frac{3}{5} ; 180^{\circ}<\theta<270^{\circ} $$

Step-by-Step Solution

Answer

\( \sin 2\theta = \frac{24}{25} \), \( \cos 2\theta = \frac{7}{25} \), \( \sin \frac{\theta}{2} = \frac{3\sqrt{10}}{10} \), \( \cos \frac{\theta}{2} = -\frac{\sqrt{10}}{10} \).

Step 1: Determine the Quadrant

Since \( 180^{\circ} < \theta < 270^{\circ} \), \( \theta \) is in the third quadrant. In the third quadrant, both sine and cosine values are negative.

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

Use the identity \( \sin^2 \theta + \cos^2 \theta = 1 \) to find \( \cos\theta \).\[ \sin\theta = -\frac{3}{5} \Rightarrow \sin^2\theta = \left( -\frac{3}{5} \right)^2 = \frac{9}{25} \]\[ \cos^2\theta = 1 - \sin^2\theta = 1 - \frac{9}{25} = \frac{16}{25} \]Since cos is negative in the third quadrant, \( \cos\theta = -\frac{4}{5} \).

Step 3: Calculate \( \sin 2\theta \) Using Double Angle Formula

Use the formula \( \sin 2\theta = 2 \sin\theta \cos\theta \).\[ \sin 2\theta = 2 \times \left(-\frac{3}{5}\right) \times \left(-\frac{4}{5}\right) = 2 \times \frac{12}{25} = \frac{24}{25} \]

Step 4: Calculate \( \cos 2\theta \) Using Double Angle Formula

Use the formula \( \cos 2\theta = \cos^2\theta - \sin^2\theta \).\[ \cos 2\theta = \left( -\frac{4}{5} \right)^2 - \left( -\frac{3}{5} \right)^2 = \frac{16}{25} - \frac{9}{25} = \frac{7}{25} \]

Step 5: Calculate \( \sin \frac{\theta}{2} \) Using Half-Angle Formula

Use the formula \( \sin \frac{\theta}{2} = \pm \sqrt{\frac{1 - \cos\theta}{2}} \). Since the half-angle for the third quadrant \( \theta \) is in the second quadrant where sine is positive, choose the positive root.\[ \sin \frac{\theta}{2} = \sqrt{\frac{1 - (-\frac{4}{5})}{2}} = \sqrt{\frac{1 + \frac{4}{5}}{2}} = \sqrt{\frac{9/5}{2}} = \sqrt{\frac{9}{10}} = \frac{3}{\sqrt{10}} \] Rationalizing gives: \( \sin \frac{\theta}{2} = \frac{3\sqrt{10}}{10} \).

Step 6: Calculate \( \cos \frac{\theta}{2} \) Using Half-Angle Formula

Use the formula \( \cos \frac{\theta}{2} = \pm \sqrt{\frac{1 + \cos\theta}{2}} \). Since the half-angle for the third quadrant \( \theta \) is in the second quadrant where cosine is negative, choose the negative root.\[ \cos \frac{\theta}{2} = -\sqrt{\frac{1 - \frac{4}{5}}{2}} = -\sqrt{\frac{\frac{1}{5}}{2}} = -\sqrt{\frac{1}{10}} = -\frac{1}{\sqrt{10}} \] Rationalizing gives: \( \cos \frac{\theta}{2} = -\frac{\sqrt{10}}{10} \).

Key Concepts

Double Angle Formulas

Double angle formulas are essential in trigonometry, as they allow us to express trigonometric functions of double angles in terms of single angles. Specifically, for any angle \( \theta \), the double angle formulas are:

• \( \sin 2\theta = 2 \sin\theta \cos\theta \)
• \( \cos 2\theta = \cos^2\theta - \sin^2\theta \)

These formulas are vital when dealing with trigonometric equations, as they help simplify expressions and calculate angles. In our exercise, knowing that \( \sin\theta = -\frac{3}{5} \) and \( \cos\theta = -\frac{4}{5} \) because \( \theta \) is in the third quadrant, we can directly apply these double angle formulas. For \( \sin 2\theta \), we substitute the given values:\[ \sin 2\theta = 2 \times \left(-\frac{3}{5}\right) \times \left(-\frac{4}{5}\right) = \frac{24}{25} \] For \( \cos 2\theta \), substitute again:\[ \cos 2\theta = \left(-\frac{4}{5}\right)^2 - \left(-\frac{3}{5}\right)^2 = \frac{7}{25} \] These formulas are indispensable when tackling problems involving the doubling of angles, making solving trigonometric equations more straightforward.

Half Angle Formulas

Half angle formulas are another extremely useful set of identities in trigonometry. They allow us to find the trigonometric functions of half angles, which can simplify calculations in various problems. The formulas are:

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

The sign depends on the quadrant of the resulting angle after halving. In our exercise, the given angle \( \theta \) is in the third quadrant, and its half-angle lies in the second quadrant (since a third quadrant angle divided by two results in a second quadrant angle). This means we choose the positive sign for \( \sin \frac{\theta}{2} \) but the negative sign for \( \cos \frac{\theta}{2} \). Thus, for sine:\[ \sin \frac{\theta}{2} = \sqrt{\frac{1 - (-\frac{4}{5})}{2}} = \frac{3\sqrt{10}}{10} \] And for cosine:\[ \cos \frac{\theta}{2} = -\sqrt{\frac{1 + (-\frac{4}{5})}{2}} = -\frac{\sqrt{10}}{10} \] These formulas elegantly break down complex full angle problems into manageable halves, aiding immensely in precise trigonometric calculations.

Pythagorean Identity

The Pythagorean identity is foundational in trigonometry. It relates the square of sine and cosine of a particular angle \( \theta \), expressed as: \[ \sin^2\theta + \cos^2\theta = 1 \] This identity is derived from the Pythagorean theorem and is applicable for any angle \( \theta \). It is useful whenever either \( \sin \theta \) or \( \cos \theta \) is given, allowing us to solve for the other. In our exercise, with \( \sin\theta = -\frac{3}{5} \), we find \( \cos\theta \) by rearranging the identity:\[ \cos^2\theta = 1 - \sin^2\theta \] Plugging in the given sine value:\[ \cos^2\theta = 1 - \left(-\frac{3}{5}\right)^2 = \frac{16}{25} \] Because \( \theta \) is in the third quadrant, where cosine values are negative, we have:\[ \cos\theta = -\frac{4}{5} \] This identity is not only pivotal for solving equations but also provides a connection between sine, cosine, and the inherent geometry of the unit circle.

Third Quadrant Angles

Understanding the quadrant in which an angle lies is crucial for determining the sign of sine and cosine values. The unit circle helps us visualize this. In the third quadrant, both sine and cosine are negative. Angles between \( 180^{\circ} \) and \( 270^{\circ} \) fall into this category. Consequently, recognizing this can determine the sign in trigonometric identities:

• \( \sin \theta < 0 \)
• \( \cos \theta < 0 \)

For the original exercise, knowing \( 180^{\circ} < \theta < 270^{\circ} \) allows us to correctly identify that \( \sin\theta = -\frac{3}{5} \) and \( \cos\theta = -\frac{4}{5} \). Understanding this aspect allows for accurate calculation of double and half angles, as the signs directly influence which root to choose in identities. It plays a key role in solving trigonometric equations by providing contextual clues based on angle positioning within the unit circle.