Problem 4 - 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. $$ \sin \theta=-\frac{3}{4} ; 270^{\circ}<\theta<360^{\circ} $$

Step-by-Step Solution

Answer

\( \sin 2\theta = -\frac{3\sqrt{7}}{8} \), \( \cos 2\theta = -\frac{1}{8} \). Exact \( \sin \frac{\theta}{2} \) and \( \cos \frac{\theta}{2} \) need further simplification.

Step 1: Identify the Reference Angle and Quadrant

Given that \( \theta \) is in the fourth quadrant \( \left(270^{\circ}<\theta<360^{\circ}\right) \), where sine is negative and cosine is positive. We know \( \sin \theta = -\frac{3}{4} \), and using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we can find \( \cos \theta \).

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

Use the Pythagorean identity to find \( \cos \theta \):\[\cos^2 \theta = 1 - \sin^2 \theta = 1 - \left(-\frac{3}{4}\right)^2 = 1 - \frac{9}{16} = \frac{7}{16}\]Because \( \theta \) is in the fourth quadrant, \( \cos \theta > 0 \), so \( \cos \theta = \frac{\sqrt{7}}{4} \).

Step 3: Calculate \( \sin 2\theta \)

Use the double angle identity for sine: \( \sin 2\theta = 2\sin \theta \cos \theta \).\[\sin 2\theta = 2 \cdot \left(-\frac{3}{4}\right) \cdot \frac{\sqrt{7}}{4} = -\frac{3\sqrt{7}}{8}\]

Step 4: Calculate \( \cos 2\theta \)

Use the double angle identity for cosine: \( \cos 2\theta = \cos^2 \theta - \sin^2 \theta \).\[\cos 2\theta = \left(\frac{\sqrt{7}}{4}\right)^2 - \left(-\frac{3}{4}\right)^2 = \frac{7}{16} - \frac{9}{16} = -\frac{1}{8}\]

Step 5: Calculate \( \sin \frac{\theta}{2} \)

Use the half-angle identity for sine: \( \sin \frac{\theta}{2} = \pm \sqrt{\frac{1 - \cos \theta}{2}} \).In the fourth quadrant, \( \theta/2 \) would be in the second quadrant for \( 135^{\circ} < \theta/2 < 180^{\circ} \), where sine is positive.\[\sin \frac{\theta}{2} = \sqrt{\frac{1 - \frac{\sqrt{7}}{4}}{2}} \]Simplify to find the exact value.

Step 6: Calculate \( \cos \frac{\theta}{2} \)

Use the half-angle identity for cosine: \( \cos \frac{\theta}{2} = \pm \sqrt{\frac{1 + \cos \theta}{2}} \).In the second quadrant, \( \cos \frac{\theta}{2} \) is negative.\[\cos \frac{\theta}{2} = -\sqrt{\frac{1 + \frac{\sqrt{7}}{4}}{2}} \]Simplify to find the exact value.

Key Concepts

Double Angle Formulas

The double angle formulas are vital tools in trigonometry that help you find trigonometric functions of double angles using the functions of single angles. They simplify calculations and are particularly handy when solving equations or working out quick approximations. The two primary double angle formulas for sine and cosine are:

• For sine: \(\sin 2\theta = 2\sin \theta \cos \theta\)
• For cosine: \(\cos 2\theta = \cos^2 \theta - \sin^2 \theta\) or an alternative form \(\cos 2\theta = 2\cos^2 \theta - 1\)

In the given exercise, we use these formulas to find \(\sin 2\theta\) and \(\cos 2\theta\) respectively. Knowing \(\sin \theta = -\frac{3}{4}\) and \(\cos \theta = \frac{\sqrt{7}}{4}\), you just plug these into the formulas:

• \(\sin 2\theta = 2 \cdot (-\frac{3}{4}) \cdot \frac{\sqrt{7}}{4} = -\frac{3\sqrt{7}}{8}\)
• \(\cos 2\theta = (\frac{\sqrt{7}}{4})^2 - (-\frac{3}{4})^2 = -\frac{1}{8}\)

Half-Angle Formulas

Half-angle formulas are extremely useful when you need to find the sine and cosine of half of an angle. These trigonometric identities themselves come from the double angle formulas, making them just as crucial in the study of trigonometry. The half-angle formulas for sine and cosine are:

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

The sign in these formulas depends on the quadrant in which the angle \(\theta/2\) lies. Given that \(\theta/2\) is in the second quadrant, \(\sin \frac{\theta}{2}\) is positive while \(\cos \frac{\theta}{2}\) is negative. Therefore, the solutions become:

• For \(\sin \frac{\theta}{2}\), with \(\theta/2\) in the second quadrant:
  \(\sin \frac{\theta}{2} = \sqrt{\frac{1 - \frac{\sqrt{7}}{4}}{2}}\)
• For \(\cos \frac{\theta}{2}\), since it is negative in the second quadrant:
  \(\cos \frac{\theta}{2} = -\sqrt{\frac{1 + \frac{\sqrt{7}}{4}}{2}}\)

Pythagorean Identity

The Pythagorean identity is one of the most fundamental identities in trigonometry. It is derived from the Pythagorean theorem and serves as a cornerstone for other identities and equations. The identity is represented as:

• \(\sin^2 \theta + \cos^2 \theta = 1\)

This formula is pivotal in calculating unknown values when given only one trigonometric function. In the exercise, it helps in finding \(\cos \theta\) when \(\sin \theta\) is known. Here's how it works:We know that \(\sin \theta = -\frac{3}{4}\); thus, \(\sin^2 \theta = \left(-\frac{3}{4}\right)^2 = \frac{9}{16}\). By substituting into the identity, we find:

• \(\cos^2 \theta = 1 - \frac{9}{16} = \frac{7}{16}\)

Since \(\theta\) is in the fourth quadrant, where cosine is positive, we have \(\cos \theta = \frac{\sqrt{7}}{4}\). This illustrates how the Pythagorean identity has allowed us to find one angle's cosine when its sine was given.

Quadrant Analysis

Understanding the quadrant in which an angle lies is vital because it determines the signs of the trigonometric functions. In trigonometry, the sign of sine and cosine varies depending on which quadrant the angle is in:

• First Quadrant: Sine(+), Cosine(+)
• Second Quadrant: Sine(+), Cosine(-)
• Third Quadrant: Sine(-), Cosine(-)
• Fourth Quadrant: Sine(-), Cosine(+)

With angles, like in our exercise where \(\theta\) is between \(270^\circ\) and \(360^\circ\) (the fourth quadrant), recognizing these signs is crucial:

• Since \(\theta\) is in the fourth quadrant, \(\sin \theta\) is negative and \(\cos \theta\) is positive.
• When finding half-angle values, consider the quadrant of \(\theta/2\). Here, \(\theta/2\) being in the second quadrant results in \(\sin \frac{\theta}{2}\) being positive and \(\cos \frac{\theta}{2}\) being negative.

This quadrant analysis helps in choosing the correct sign for trigonometric values when applying trigonometric identities in calculations.