Problem 23 - Algebra 2 and Trigonometry

Question

a. Derive an identity for \(\sin \frac{1}{4} A\) in terms of \(\cos \frac{1}{2} A .\) b. Derive an identity for \(\cos \frac{1}{4} A\) in terms of \(\cos \frac{1}{2} A\) c. Derive an identity for \(\tan \frac{1}{4} A\) in terms of \(\cos \frac{1}{2} A\) Hint: Let \(\frac{1}{4} A=\frac{1}{2} \theta\)

Step-by-Step Solution

Answer

a. \( \sin\frac{1}{4}A = \pm \sqrt{\frac{1-\cos\frac{1}{2}A}{2}} \); b. \( \cos\frac{1}{4}A = \pm \sqrt{\frac{1+\cos\frac{1}{2}A}{2}} \); c. \( \tan\frac{1}{4}A = \pm \sqrt{\frac{1-\cos\frac{1}{2}A}{1+\cos\frac{1}{2}A}} \).

Step 1: Define the Variables

Let \( \frac{1}{4}A = \frac{1}{2}\theta \). Thus, \( A = 2\theta \) and \( \sin\frac{1}{4}A = \sin\frac{1}{2}\theta \). We need to express \( \sin\frac{1}{2}\theta \), \( \cos\frac{1}{2}\theta \), and \( \tan\frac{1}{2}\theta \) in terms of \( \cos\theta \).

Step 2: Derive Identity for \( \sin\frac{1}{4}A \)

Using the half-angle identity, \( \sin\frac{1}{2}\theta = \pm \sqrt{\frac{1 - \cos\theta}{2}} \). Therefore, \( \sin\frac{1}{4}A = \pm \sqrt{\frac{1 - \cos\frac{1}{2}A}{2}} \).

Step 3: Derive Identity for \( \cos\frac{1}{4}A \)

Again using the half-angle formula, \( \cos\frac{1}{2}\theta = \pm \sqrt{\frac{1 + \cos\theta}{2}} \). Therefore, \( \cos\frac{1}{4}A = \pm \sqrt{\frac{1 + \cos\frac{1}{2}A}{2}} \).

Step 4: Derive Identity for \( \tan\frac{1}{4}A \)

Using \( \tan\frac{1}{2}\theta = \frac{\sin\frac{1}{2} \theta}{\cos\frac{1}{2}\theta} = \pm \sqrt{\frac{1-\cos\theta}{1+\cos\theta}} \), we get \( \tan\frac{1}{4}A = \pm \sqrt{\frac{1-\cos\frac{1}{2}A}{1+\cos\frac{1}{2}A}} \).

Key Concepts

Understanding Half-Angle Identities

Half-angle identities are a vital part of trigonometry that allow us to express trigonometric functions of half angles in terms of full angles. They give us a direct link between the angle and its half. In our context, we are looking to express functions of \( \frac{1}{4}A \), which is half of \( \frac{1}{2}A \).
The formula for the sine half-angle identity is given by:

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

Similarly, the cosine half-angle identity is:

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

The tangent half-angle identity is derived from these two and simplifies to:

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

These identities help significantly in simplifying complex expressions and solving trigonometric equations.

Exploring the Sine Function

The sine function is a fundamental trigonometric function that represents the ratio of the opposite side to the hypotenuse in a right triangle. For angles measured from the unit circle, it corresponds to the y-coordinate of a point moving around the circle.
In half-angle formulas, the sine function allows us to take a given angle and determine the sine of its half efficiently. Consider the derived identity:

• \( \sin\frac{1}{4}A = \pm \sqrt{\frac{1 - \cos\frac{1}{2}A}{2}} \)

This formula enables one to find the sine of \(\frac{1}{4}A\) directly from \(\cos\frac{1}{2}A\), a crucial link in simplifying expressions where angles are halved.

Connecting with the Cosine Function

The cosine function is another core component of trigonometry, representing the ratio of the adjacent side to the hypotenuse in a right triangle. On the unit circle, it corresponds to the x-coordinate.
Using the half-angle identity for cosine allows us to transition from a full angle to its half effectively. The relationship is established as follows:

• \( \cos\frac{1}{4}A = \pm \sqrt{\frac{1 + \cos\frac{1}{2}A}{2}} \)

This identity makes use of \(\cos\frac{1}{2}A\) to express \(\cos\frac{1}{4}A\), providing a straightforward way to manage problems involving divided angles. This can be particularly helpful in simplifying complex trigonometric identities or resolving analytical problems.

Decoding the Tangent Function

Tangent is a dynamic trigonometric function, defined as the ratio of sine to cosine, or equivalently the y-coordinate divided by the x-coordinate on the unit circle. It's particularly useful in various trigonometric applications due to its versatility.
To derive the half-angle identity for the tangent function, one utilizes both the sine and cosine half-angle identities:

• \( \tan\frac{1}{4}A = \pm \sqrt{\frac{1-\cos\frac{1}{2}A}{1+\cos\frac{1}{2}A}} \)

This identity allows us to express \(\tan\frac{1}{4}A\) using the cosine of half that angle, setting a foundation to simplify calculations involving multiple angle divisions while accommodating different mathematical tasks. It's an elegant solution that ties together our understanding of trigonometric relationships.