Problem 15 - Algebra and Trigonometry

Question

15–26 Use an appropriate half-angle formula to find the exact value of the expression. $$\sin 15^{\circ}$$

Step-by-Step Solution

Answer

The exact value of \(\sin 15^\circ\) is \(\frac{\sqrt{2 - \sqrt{3}}}{2}\).

Step 1: Recognize the Half-Angle Formula

We recognize that to solve \( \sin 15^\circ \), we can use the half-angle formula for sine. The half-angle formula is given by: \( \sin \frac{\theta}{2} = \pm \sqrt{\frac{1 - \cos \theta}{2}} \).

Step 2: Find the Related Angle

Notice that \( 15^\circ \) is half of \(30^\circ\). Therefore, set \( \theta = 30^\circ \). This will help us find \( \sin 15^\circ = \sin \frac{30^\circ}{2} \).

Step 3: Apply the Half-Angle Formula

Substitute \( \theta = 30^\circ \) into the half-angle formula: \( \sin 15^\circ = \sin \frac{30^\circ}{2} = \sqrt{\frac{1 - \cos 30^\circ}{2}} \).

Step 4: Find \(\cos 30^\circ\)

We know \( \cos 30^\circ = \frac{\sqrt{3}}{2} \). Substitute this value into the expression: \( \sin 15^\circ = \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} \).

Step 5: Simplify the Expression

Continue by simplifying inside the square root: \( 1 - \frac{\sqrt{3}}{2} = \frac{2}{2} - \frac{\sqrt{3}}{2} = \frac{2 - \sqrt{3}}{2} \). Substitute this back: \( \sin 15^\circ = \sqrt{\frac{2 - \sqrt{3}}{4}} \).

Step 6: Simplify Further

Simplify further to get \( \sin 15^\circ = \frac{\sqrt{2 - \sqrt{3}}}{2} \). This is the exact value.

Key Concepts

Half-Angle Formula

Understanding the half-angle formula is vital for solving certain trigonometric problems. In trigonometry, the half-angle formulas allow us to find the sine, cosine, or tangent of half an angle, if we know the cosine of the full angle.
To calculate the sine of half an angle, we use the following half-angle formula:

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

This formula is particularly useful because it helps us find values for angles that are not typically found on the unit circle. That means we can get exact values for angles like \( 15^{\circ} \) or \( 7.5^{\circ} \), which can be tricky to handle otherwise.
The positive or negative sign in front of the square root depends on the quadrant where the angle \( \frac{\theta}{2} \) lies. For angles between 0 and 90 degrees, like \( 15^{\circ} \), we use the positive sign.

Sine Function

The sine function is one of the fundamental trigonometric functions that helps relate the angles to the ratios of the sides in a right triangle. For a given angle \( \theta \), \( \sin \theta \) is defined in a right-angled triangle as the ratio of the length of the opposite side to the length of the hypotenuse.
The sine function is periodic with a period of \( 360^{\circ} \) or \( 2\pi \) radians, meaning it repeats its values every full circle. It's also important in defining the waveform of periodic phenomena and is thus a cornerstone in fields like physics and engineering.The function reaches its maximum value of 1 and its minimum value of -1. At \( 0^{\circ}, \) \( 180^{\circ}, \) and \( 360^{\circ} \) or \( \pi \), \( 2\pi \) radians, the sine function returns a value of 0. Understanding these properties aids in visualizing and solving trigonometric problems.

Exact Value Calculation

Exact value calculation in trigonometry refers to finding the specific value of a trigonometric expression without resorting to a calculator, instead using known identities and angles.
To calculate the exact value of \( \sin 15^{\circ} \) using the half-angle formula, follow these steps:

• Recognize the angle: \( 15^{\circ} \) is half of \( 30^{\circ} \).
• Apply the formula: \( \sin 15^{\circ} = \sin \frac{30^{\circ}}{2} \).
• Substitute \( \theta = 30^{\circ} \): \( \sin \frac{30^{\circ}}{2} = \sqrt{\frac{1 - \cos 30^{\circ}}{2}} \).
• Find and use \( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \): \( \sin 15^{\circ} = \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} \).
• Simplify further: \( \sqrt{\frac{2 - \sqrt{3}}{4}} = \frac{\sqrt{2 - \sqrt{3}}}{2} \).

The resulting value, \( \frac{\sqrt{2 - \sqrt{3}}}{2} \), is the precise and exact value of \( \sin 15^{\circ} \) in its simplest form. Mastering these steps for exact value calculation sharpens your skills in deriving values from known trigonometric properties, bypassing an over-reliance on technological aids.