Using De Moivre's theorem to prove Trigonemtric Identity

Question

How can I use De Moivre's Theorem to prove,

sin 3θ = 3 cos² θ sin θ - sin³ θ = 3 sin θ - 4 sin³ θ ?

Answer 1

By DeMoivre's Theorem, we have

(cosθ + i·sinθ)³ = cos(3θ) + i·sin(3θ).

To find sin(3θ), we can distribute and simplify the left side of the equation to determine that

(cosθ + i·sinθ)³ = cos³θ + 3i·sinθ·cos²θ - 3·sin²θ·cosθ - i·sin³θ.

If you get stuck here, start by finding (cosθ + i·sinθ)² first, and remember i² = -1.

Now, by DeMoivre's Theorem we have

cos(3θ) + i·sin(3θ) = cos³θ + 3i·sinθ·cos²θ - 3·sin²θ·cosθ - i·sin³θ.

Separating the real and imaginary parts of this complex equation (i.e., grouping the terms with and without i as a factor), we have

cos(3θ) + i·sin(3θ) = (cos³θ - 3·sin²θ·cosθ) + i·(3·sinθ·cos²θ - sin³θ).

The imaginary parts (i.e., the parts with i) on each side of the equation must be equal, so

sin(3θ) = 3·sinθ·cos²θ - sin³θ.

With this, the first part of the problem is solved! To solve the second part, it looks like we want to get everything in terms of just sinθ, not cosθ. Recalling the identity sin²θ + cos²θ = 1, we can replace cos²θ with 1 - sin²θ in the previous equation, which gives us

sin(3θ) = 3·sinθ·cos²θ - sin³θ

= 3sinθ·(1 - sin²θ) - sin³θ

= (3sinθ - 3sin³θ) - sin³θ

= 3sinθ - 4sin³θ.

Answer 2

the theorem implies (cosθ+isinθ)³=cos3θ+isin3θ

using just algebra (cosθ+isinθ)³=cos³θ+3icos²θ(sinθ)-3cosθ(sin²θ)-isin³θ

i²=-1

the i terms in both calculations are equal

so sin3θ=3cos²θ(sinθ)-sin³θ

now use cos²θ=1-sin²θ