Problem 34 - Precalculus Mathematics for Calculus

Question

Simplify the expression by using a Double-Angle Formula or a Half-Angle Formula. (a) \(\sqrt{\frac{1-\cos 30^{\circ}}{2}}\) (b) \(\sqrt{\frac{1-\cos 8 \theta}{2}}\)

Step-by-Step Solution

Answer

(a) \( \sin 15^{\circ} = \frac{\sqrt{2-\sqrt{3}}}{2} \); (b) \( \sin 4\theta \).

Step 1: Recognize the Half-Angle Formula

The expression \( \sqrt{\frac{1-\cos \alpha}{2}} \) corresponds to the half-angle formula used to find \( \sin \frac{\alpha}{2} \). This formula rearranges the trigonometric identity for cosine in terms of sine.

Step 2: Apply the Formula to Part (a)

For part (a), identify \( \alpha = 30^{\circ} \). The formula gives us \( \sin \frac{\alpha}{2} = \sqrt{\frac{1-\cos \alpha}{2}} \). Thus, we need to compute \( \sin 15^{\circ} \).

Step 3: Compute \( \cos 30^{\circ} \)

Use known trigonometric values: \( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \).

Step 4: Simplify the Expression for \( \sin 15^{\circ} \)

Substitute the value of \( \cos 30^{\circ} \) into the half-angle formula: \[ \sin 15^{\circ} = \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} = \sqrt{\frac{2-\sqrt{3}}{4}} = \sqrt{\frac{2-\sqrt{3}}{2^2}} \]Further simplification gives \( \sin 15^{\circ} = \frac{\sqrt{2-\sqrt{3}}}{2} \).

Step 5: Apply the Formula to Part (b)

For part (b), identify \( \alpha = 8\theta \). The expression \( \sqrt{\frac{1 - \cos 8\theta}{2}} \) simplifies to \( \sin 4\theta \). There are no further numerical simplifications without a specific value for \( \theta \).

Key Concepts

Trigonometric Identities

Trigonometric identities are fundamental equations involving trigonometric functions that hold true for all values of the included variables. They are incredibly useful in simplifying expressions

• They help identify relationships between different trigonometric functions.
• Common identities include Pythagorean identities, angle-sum formulas, double-angle formulas, and importantly for this task, half-angle formulas.

In this exercise, we focus on the half-angle formula: \[ \sin \left(\frac{\alpha}{2}\right) = \sqrt{\frac{1 - \cos \alpha}{2}} \]This particular identity allows us to determine the sine of half an angle, using only the cosine of the full angle. This property is vital when angles themselves are impractical to measure directly. Here, recognizing the half-angle formula helps us simplify the given expressions substantially.

Sine Function

The sine function is a fundamental trigonometric function that relates an angle of a right triangle to the ratio of the length of the opposite side over the hypotenuse. Here’s how it’s typically used:

• The sine of an angle measures the vertical component of a unit circle.
• The range of the sine function is from -1 to 1.
• In our context, the half-angle identity lets us calculate sine for angles such as 15° in part (a), derived from the cosine of 30°.

For example, by applying the half-angle formula, the exercise transforms the expression \( \sqrt{\frac{1 - \cos 30^{\circ}}{2}} \) into \( \sin 15^{\circ} \). Without knowing the direct sine values for some angles, identities allow us to deduce them indirectly, showcasing the sine function's versatility in connection to its geometric interpretation.

Simplifying Expressions

Simplifying expressions involves reducing them to their simplest form, which often makes equations easier to solve or more elegant to read. Here's how we can approach simplification:

• Use known identities to transform complex forms into simple, familiar ones.
• Replacing parts of the expression with equivalent terms helps streamline the process.

In our exercise:

For part (a):

We take \( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \) and substitute it into the half-angle formula. After that, simplifying further gives \( \sin 15^{\circ} = \frac{\sqrt{2-\sqrt{3}}}{2} \).

For part (b):

Although it seems more abstract without a specific value for \( \theta \), recognizing the formula tells us that the expression ultimately resolves to \( \sin 4\theta \), completing the simplification.The art of simplification often relies on recognizing when to use particular identities to transition expressions to their most useful form.