How to use de moivre’s theorem to express ⅙ sin 6θ / sin θ as a polynomial in cos θ?

To express \( \frac{1}{6} \sin 6\theta / \sin \theta \) as a polynomial in \( \cos \theta \) using De Moivre's theorem, follow these steps:

Understanding De Moivre's Theorem

De Moivre's theorem states that for any real number \( \theta \) and integer \( n \):

(cos θ + i sin θ)^n = cos(nθ) + i sin(nθ)

Applying the Theorem

For \( n = 6 \):

cos(6θ) + i sin(6θ) = (cos θ + i sin θ)^6

From this, we can extract \( \sin 6\theta \) as:

sin(6θ) = Im((cos θ + i sin θ)^6)

Finding \( \sin 6\theta \)

Using the binomial expansion:

• Expand \( (cos θ + i sin θ)^6 \).
• Extract the imaginary part to find \( \sin 6\theta \).

Using the Sine Identity

We can relate \( \sin 6\theta \) to \( \sin \theta \) using the identity:

sin 6θ = 6 sin θ cos^5 θ - 20 sin^3 θ cos^3 θ + 6 sin^5 θ cos θ

Final Steps

Now, substitute this back into the expression:

\( \frac{1}{6} \sin 6\theta / \sin \theta = \frac{1}{6} (6 cos^5 θ - 20 sin^2 θ cos^3 θ + 6 sin^4 θ) \)

Replace \( sin^2 θ \) with \( 1 - cos^2 θ \) to express everything in terms of \( cos θ \).

Resulting Polynomial

After simplification, you will obtain a polynomial in \( cos θ \). This method effectively uses De Moivre's theorem to transition from trigonometric functions to polynomial form.