Problem 25 - Algebra 2 and Trigonometry

Question

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

Step-by-Step Solution

Answer

\( \sin 4A = 2 \sin 2A \cos 2A \), \( \cos 4A = \cos^2 2A - \sin^2 2A \), \( \tan 4A = \frac{2 \tan 2A}{1 - \tan^2 2A} \).

Step 1: Understanding the Problem

We are tasked to find trigonometric identities for \( \sin 4A \), \( \cos 4A \), and \( \tan 4A \) in terms of trigonometric functions of \( 2A \). We can use the hint provided: let \( 4A = 2\theta \), thus \( \theta = 2A \). This means we need to express these functions using \( \theta \).

Step 2: Using Double Angle Formula for Sine

We use the double angle formula for sine, which is: \( \sin 2\theta = 2\sin \theta \cos \theta \). By substituting \( \theta = 2A \), we have \( \sin 4A = \sin (2 \cdot 2A) = 2\sin 2A \cos 2A \).

Step 3: Expressing \( \sin 2A \) and \( \cos 2A \)

Since \( \sin 2A = 2 \sin A \cos A \) and \( \cos 2A = \cos^2 A - \sin^2 A \), substitute these into the \( \sin 4A \) formula: \( \sin 4A = 2(2 \sin A \cos A)(\cos^2 A - \sin^2 A) \).

Step 4: Using Double Angle Formula for Cosine

For \( \cos 4A \), we use: \( \cos 2\theta = \cos^2 \theta - \sin^2 \theta \). Substituting \( \theta = 2A \) gives \( \cos 4A = \cos^2 2A - \sin^2 2A \).

Step 5: Expressing \( \cos 2A \) and \( \sin 2A \)

Substitute \( \cos 2A = \cos^2 A - \sin^2 A \) and \( \sin 2A = 2 \sin A \cos A \) back into \( \cos 4A \): \( \cos 4A = (\cos^2 A - \sin^2 A)^2 - (2 \sin A \cos A)^2 \).

Step 6: Using Double Angle Formula for Tangent

The double angle formula for tangent is: \( \tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta} \). Using \( \theta = 2A \), we find \( \tan 4A = \tan (2 \cdot 2A) = \frac{2 \tan 2A}{1 - \tan^2 2A} \).

Step 7: Expressing \( \tan 2A \)

Since \( \tan 2A = \frac{2 \tan A}{1 - \tan^2 A} \), substitute into \( \tan 4A \): \( \tan 4A = \frac{2 \left(\frac{2 \tan A}{1 - \tan^2 A}\right)}{1 - \left(\frac{2 \tan A}{1 - \tan^2 A}\right)^2} \). Simplify this expression to get the identity for \( \tan 4A \).

Step 8: Simplifying \( \sin 4A \), \( \cos 4A \), \( \tan 4A \)

Final simplified expressions are: \( \sin 4A = 8 \sin A \cos^3 A - 8 \sin^3 A \cos A \), \( \cos 4A = \cos^4 A - 6\cos^2 A \sin^2 A + \sin^4 A \), \( \tan 4A = \frac{4 \tan A (1 - \tan^2 A)}{1 - 6 \tan^2 A + \tan^4 A} \).

Key Concepts

Double Angle Formulas

The double angle formulas are key tools in trigonometry that allow us to express functions of angles, like sine, cosine, and tangent, in terms of twice an angle. These formulas are particularly helpful when simplifying expressions or solving trigonometric equations.
In the context of the exercise, the double angle formulas help us derive identities for overarching angles like \(4A\) using smaller angles like \(2A\). Here are the main double angle formulas:

• \(\sin 2\theta = 2\sin \theta \cos \theta\)
• \(\cos 2\theta = \cos^2 \theta - \sin^2 \theta\)
• \(\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}\)

Understanding these can make complex trigonometric problems much more approachable. In our exercise, these formulas are creatively applied to find the values of \(\sin 4A\), \(\cos 4A\), and \(\tan 4A\) by expressing them in terms of \(2A\).
Substituting \(\theta = 2A\) into these formulas allows us to derive identities for \(4A\) — a powerful technique to manage complex angles.

Sine Function

The sine function is one of the fundamental functions in trigonometry. It relates the angles of a right-angled triangle to the ratios of its sides. For any angle \(\theta\) in the unit circle, \(\sin \theta\) equals the y-coordinate of the point on the circle that forms angle \(\theta\) with the positive x-axis.
The double angle formula for sine, \(\sin 2\theta = 2\sin \theta \cos \theta\), helps us express \(\sin 4A\) in terms of \(2A\).

• First, understand that \(\sin 4A = \sin (2 \cdot 2A)\).
• Using the double angle formula, it becomes: \(\sin 4A = 2 \sin 2A \cos 2A\).

By expressing \(\sin 2A\) and \(\cos 2A\) with \(\sin A\) and \(\cos A\) using their respective double angle identities, we can further simplify \(\sin 4A\). Ultimately, this is given by:
\[\sin 4A = 8 \sin A \cos^3 A - 8 \sin^3 A \cos A\]
This identity is particularly useful in solving problems related to wave functions or oscillations where angles are doubled.

Cosine Function

The cosine function is another crucial function in trigonometry. It represents the adjacent side of a triangle over the hypotenuse in a right-angle triangle. In the unit circle, \(\cos \theta\) corresponds to the x-coordinate of a point making angle \(\theta\) with the x-axis.
The double angle formula for cosine, \(\cos 2\theta = \cos^2 \theta - \sin^2 \theta\), helps simplify expressions like \(\cos 4A\).
Here's how it works:

• We express \(\cos 4A\) as \(\cos 2(2A)\).
• Then, apply the double angle formula: \(\cos 4A = \cos^2 2A - \sin^2 2A\).

Further substitution using \(\cos 2A = \cos^2 A - \sin^2 A\) and \(\sin 2A = 2 \sin A \cos A\) provides a deeper simplification:
\[\cos 4A = \cos^4 A - 6 \cos^2 A \sin^2 A + \sin^4 A\]
This formula is particularly handy in physics and engineering, where precise angle measurements are necessary.

Tangent Function

The tangent function in trigonometry is the ratio of sine to cosine functions. In right triangles, it represents the opposite side's length over the adjacent's. When dealing with the tangent function, especially in the context of angle doubling, the double angle formula becomes very practical.
The double angle formula for tangent is given by:
\[\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}\]
By substituting \(\theta = 2A\), we can derive \(\tan 4A\) in terms of \(\tan 2A\). The detailed steps are as follows:

• We start with \(\tan 4A = \tan(2 \cdot 2A)\).
• Using the double angle identity, this becomes \(\tan 4A = \frac{2 \tan 2A}{1 - \tan^2 2A}\).
• Substitute \(\tan 2A = \frac{2 \tan A}{1 - \tan^2 A}\) to express \(\tan 4A\) using \(\tan A\):

After simplification, the identity emerges as:
\[\tan 4A = \frac{4 \tan A (1 - \tan^2 A)}{1 - 6 \tan^2 A + \tan^4 A}\]
This identity can help solve problems where the angles are repetitive or cyclical, ensuring simpler calculations in trigonometry.