Problem 29 - Algebra 2

Question

Find the exact value of each function. $$ 12\left(\sin 150^{\circ}\right)\left(\cos 150^{\circ}\right) $$

Step-by-Step Solution

Answer

The exact value is \(-3\sqrt{3}\).

Step 1: Identify the Angle Properties

It's important to first recognize that the angle given, \(150^{\circ}\), can be related to known angles, such as those in the unit circle. \(150^{\circ}\) is in the second quadrant where sine is positive and cosine is negative.

Step 2: Use Reference Angles

The reference angle for \(150^{\circ}\) in the unit circle is \(180^{\circ} - 150^{\circ} = 30^{\circ}\). Hence, \(\sin 150^{\circ} = \sin 30^{\circ}\) and \(\cos 150^{\circ} = -\cos 30^{\circ}\).

Step 3: Apply Trigonometric Values

From the fundamental trigonometric ratios, we know that \(\sin 30^{\circ} = \frac{1}{2}\) and \(\cos 30^{\circ} = \frac{\sqrt{3}}{2}\). Therefore, \(\cos 150^{\circ} = -\frac{\sqrt{3}}{2}\).

Step 4: Substitute Values into the Expression

Substitute \(\sin 150^{\circ} = \frac{1}{2}\) and \(\cos 150^{\circ} = -\frac{\sqrt{3}}{2}\) into the expression: \[12 \left(\frac{1}{2}\right) \left(-\frac{\sqrt{3}}{2}\right)\]

Step 5: Calculate the Expression

Perform the multiplication: \(12 \times \frac{1}{2} \times -\frac{\sqrt{3}}{2} = 6 \times -\frac{\sqrt{3}}{2} = -3\sqrt{3}\).

Key Concepts

Unit Circle

The unit circle is a fundamental concept in trigonometry. It is a circle with a radius of one, centered at the origin of a coordinate plane. This circle helps us understand the relationships between angles and the corresponding sine and cosine values.

Points on the unit circle correspond to specific angles measured from the positive x-axis. Each point \((x, y)\) represents \(\cos(\theta)\) and \(\sin(\theta)\), respectively, where \(\theta\) is the angle. The use of the unit circle allows us to determine trigonometric identities and relationships for any angle.

For example, an angle of \(150^{\circ}\) is located in the second quadrant of the unit circle. The quadrant's properties help us know that sine values are positive and cosine values are negative there. Understanding the placement of an angle in relation to the quadrants is crucial for solving trigonometric functions correctly.

Reference Angles

Reference angles are a useful tool in trigonometry, allowing us to simplify and solve problems with angles beyond the first quadrant. A reference angle is the acute angle formed by the terminal side of the given angle and the x-axis.

To find a reference angle for an angle in the unit circle, utilize the following:

• Angles in the second quadrant: Subtract the angle from \(180^{\circ}\).
• Angles in the third quadrant: Subtract \(180^{\circ}\) from the angle.
• Angles in the fourth quadrant: Subtract the angle from \(360^{\circ}\).

In our example, the angle \(150^{\circ}\) falls in the second quadrant. Therefore, the reference angle is \(180^{\circ} - 150^{\circ} = 30^{\circ}\).

Understanding reference angles allows us to use the known sine and cosine values of these common angles, regardless of their quadrant.

Sine and Cosine Values

Sine and cosine values of standard angles are fundamental to solving trigonometric equations. These are derived from the unit circle, where each angle corresponds to a specific point \((x, y)\).

For example, consider the angle \(30^{\circ}\), which in the unit circle corresponds to the coordinates \(\left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right)\). Hence, we have:

• \(\sin 30^{\circ} = \frac{1}{2}\)
• \(\cos 30^{\circ} = \frac{\sqrt{3}}{2}\)

These values are essential when determining the sine and cosine for angles like \(150^{\circ}\), which has the same reference angle of \(30^{\circ}\):

• \(\sin 150^{\circ} = \sin 30^{\circ} = \frac{1}{2}\)
• \(\cos 150^{\circ} = -\cos 30^{\circ} = -\frac{\sqrt{3}}{2}\)

Knowing these values allows for express calculation without having to derive each value from scratch every time.