Problem 113 - Precalculus with Limits

Question

In Exercises 111 - 124, verify the identity. \( \sin \dfrac{\alpha}{3} \cos \dfrac{\alpha}{3} = \dfrac{1}{2} \sin \dfrac{2\alpha}{3} \)

Step-by-Step Solution

Answer

By using the double angle formula for sine, the identity is proven. The left-hand side of the equation \( \sin \dfrac{\alpha}{3} \cos \dfrac{\alpha}{3} \) matches the right-hand side \( \dfrac{1}{2}\sin \dfrac{2\alpha}{3} \). So the identity is verified.

Step 1: Introduce the double angle formula

Commence by applying the sine double-angle formula on the right side of the identity. The double angle formula for sine is \( \sin2x = 2\sin x \cos x \). Hence, \( \sin \dfrac{2\alpha}{3} = 2\sin \dfrac{\alpha}{3} \cos \dfrac{\alpha}{3} \).

Step 2: Adjust the right side identity

To match the structures of both sides of the equation, divide both sides of the newly defined equation by 2. Consequently, the equation becomes \( \dfrac{1}{2}\sin \dfrac{2\alpha}{3} = \sin \dfrac{\alpha}{3} \cos \dfrac{\alpha}{3} \). This matches the structure of the given identity.

Step 3: Compare the two sides

On comparing the two sides the left-hand side (LHS) \( \sin \dfrac{\alpha}{3} \cos \dfrac{\alpha}{3} \) is equal to right-hand side (RHS) \( \dfrac{1}{2}\sin \dfrac{2\alpha}{3} \).

Key Concepts

Double Angle Formula

Understanding the double angle formula is key in trigonometry, especially when it comes to solving identities. The double angle formulas allow us to express trigonometric functions of double angles in terms of single angles.
One of the most common double angle formulas is for the sine function, written as \( \sin 2x = 2 \sin x \cos x \). This formula is extremely useful for simplifying expressions and proving identities involving angles that are multiples of one another.
In the given exercise, we applied the double angle formula by recognizing \( \sin \frac{2\alpha}{3} \) as equivalent to \( 2 \sin \frac{\alpha}{3} \cos \frac{\alpha}{3} \). By doing so, we simplified the expression, showing that the identity holds true. Remembering this formula can aid in transforming more complex trigonometric expressions into simpler forms.

Sine Function

The sine function is a fundamental component of trigonometry. It is used to determine the y-coordinate of a point on the unit circle corresponding to a given angle. The sine of an angle \( \theta \) is the ratio of the opposite side to the hypotenuse in a right-angled triangle.
To better understand this function, let's reflect on its key properties:

• The sine function forms a wave that starts at the origin (0,0), indicating that \( \sin 0 = 0 \).
• It reaches its maximum value of 1 at \( \frac{\pi}{2} \) (90 degrees), and the minimum value of -1 at \( \frac{3\pi}{2} \) (270 degrees).
• Sine is an odd function, meaning \( \sin(-x) = -\sin(x) \), which contributes to its symmetric nature about the origin.

When verifying identities, understanding the sine function helps us predict how it behaves and how it might simplify expressions.

Cosine Function

The cosine function is another cornerstone of trigonometry, closely related to the sine function. In a right triangle, it determines the ratio of the adjacent side to the hypotenuse for a given angle.
Let's explore some primary characteristics of the cosine function:

• Cosine of an angle \( x \) is associated with the x-coordinate on the unit circle.
• The cosine wave begins at its maximum point, \( \,\cos(0) = 1 \,\), dropping to its minimum, \( \,\cos(\pi) = -1 \,\), as the angle progresses around the circle.
• Unlike sine, cosine is an even function, so \( \cos(-x) = \cos(x) \), exhibiting symmetry about the y-axis.

In our exercise, both sine and cosine appear. Utilizing the properties of cosine in conjunction with the double angle identity lets us bridge the identities effectively. Comprehending these trigonometric functions allows us to deconstruct equations into manageable segments.