Problem 51 - Algebra and Trigonometry

Question

If \(\theta=\pi / 3,\) find the value of each expression. (a) \(\sin 2 \theta, \quad 2 \sin \theta \quad\) (b) \(\sin \frac{1}{2} \theta, \quad \frac{1}{2} \sin \theta\) (c) \(\sin ^{2} \theta, \quad \sin \left(\theta^{2}\right)\)

Step-by-Step Solution

Answer

(a) \( \sin 2\theta = \frac{\sqrt{3}}{2}, 2\sin\theta = \sqrt{3} \) (b) \( \sin \frac{1}{2}\theta = \frac{1}{2}, \frac{1}{2}\sin\theta = \frac{\sqrt{3}}{4} \) (c) \( \sin^2\theta = \frac{3}{4}, \sin(\theta^2) = \sin\left(\frac{\pi^2}{9}\right) \)

Step 1: Evaluate \( \sin 2\theta \)

The formula for sine of double angle is \( \sin 2\theta = 2 \sin \theta \cos \theta \). Substituting \( \theta = \frac{\pi}{3} \), we have:\[\sin 2\theta = 2 \sin\left(\frac{\pi}{3}\right) \cos\left(\frac{\pi}{3}\right)\]Since \( \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \) and \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \), substitute these values:\[\sin 2\theta = 2 \times \frac{\sqrt{3}}{2} \times \frac{1}{2} = \frac{\sqrt{3}}{2}\]

Step 2: Evaluate \( 2 \sin \theta \)

Substituting \( \theta = \frac{\pi}{3} \), we calculate:\[2 \sin \theta = 2 \sin\left(\frac{\pi}{3}\right) = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3}\]

Step 3: Evaluate \( \sin \frac{1}{2}\theta \)

The value for the angle \( \frac{1}{2}\theta \) is \( \frac{1}{2}\times\frac{\pi}{3} = \frac{\pi}{6} \). Using the sine function:\[\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}\]

Step 4: Evaluate \( \frac{1}{2} \sin \theta \)

Substituting \( \theta = \frac{\pi}{3} \), we calculate:\[\frac{1}{2} \sin \theta = \frac{1}{2} \times \sin\left(\frac{\pi}{3}\right) = \frac{1}{2} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4}\]

Step 5: Evaluate \( \sin^{2} \theta \)

Using the value \( \sin \theta = \frac{\sqrt{3}}{2} \), the square is:\[\sin^{2} \theta = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}\]

Step 6: Evaluate \( \sin\left(\theta^{2}\right) \)

Since \( \theta = \frac{\pi}{3} \) and \( \theta^2 = \left(\frac{\pi}{3}\right)^2 = \frac{\pi^2}{9} \), then:\[\sin\left(\theta^{2}\right) = \sin\left(\frac{\pi^2}{9}\right)\]This expression cannot be simplified further without numerical approximation as \( \frac{\pi^2}{9} \) is not a standard angle.

Key Concepts

Sine function

The sine function is one of the fundamental trigonometric functions, often represented as \( \sin \theta \), where \( \theta \) is an angle measured in radians or degrees. The sine of an angle in a right triangle gives the ratio of the length of the side opposite the angle to the hypotenuse. It is periodic with a period of \( 2\pi \), meaning its values repeat every \( 2\pi \) radians.

To calculate \( \sin \theta \) for specific angles, certain standard values are commonly used, such as \( \sin\left( \frac{\pi}{6} \right) = \frac{1}{2} \), \( \sin\left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} \), and \( \sin\left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2} \).

When working with expressions like \( 2 \sin \theta \), you are simply multiplying the sine of the angle by 2. This operation can be useful when evaluating or transforming trigonometric expressions in problems or computations.

Double angle formula

The double angle formula is a critical trigonometric identity that allows the calculation of the sine, cosine, or tangent of twice an angle in terms of the sine, cosine, and tangent of the original angle. For sine, the formula is:

\[ \sin 2\theta = 2 \sin \theta \cos \theta \]

This formula can simplify complex trigonometric expressions and is especially useful in calculus and higher-level algebra. When \( \theta = \frac{\pi}{3} \), the double angle formula gives \( \sin(2\theta) = \frac{\sqrt{3}}{2} \), as derived by substituting known sine and cosine values. Knowing the double angle identities helps in solving equations and can reduce computational complexity in various trigonometric problems.

Half angle formula

The half angle formulas are another set of identities useful for finding the sine, cosine, or tangent of half an angle. These formulas are derived from the double angle formulas and are crucial in situations where precise value calculation or trigonometric simplification is required.

For sine, the half angle formula is given by:

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

In our problem, we are tasked with finding \( \sin\left( \frac{1}{2} \theta \right) \) for \( \theta = \frac{\pi}{3} \). Here, the calculated quarter angle, \( \theta /2 \), is \( \frac{\pi}{6} \), and using standard sine values, the result is \( \frac{1}{2} \). This illustrates the half angle formula's effectiveness in simplifying complex trigonometric calculations.

Trigonometric expressions evaluation

Evaluating trigonometric expressions involves using known values and identities to find exact numerical outcomes for given variables. These evaluations commonly require standard angle values, special identities, or both.

For instance, in solving \( \sin^2 \theta \), you take the square of the sine value, here \( \sin \theta = \frac{\sqrt{3}}{2} \), leading to \( \sin^2 \theta = \frac{3}{4} \).

However, situations like \( \sin(\theta^2) \) when \( \theta \) is not a standard angle, might necessitate approximation techniques or numerical computation, as exact values cannot easily be found. Understanding how to manipulate and simplify these expressions requires a deep grasp of trigonometric properties and can provide solutions for both theoretical and practical mathematical problems.