Use De Moivre's Theorem to express $\sin(3\theta)$ in terms of the powers of $\sin (\theta)$ and $\cos(\theta)$

Question

There are quite a few resources on this question but I seem to be at a point where I cannot find a resource to match my scenario

I tackle questions like this in 3 steps:

1) Apply De Moivre's Theorem
2) Use Pascals Triangle (Proves quicker for me than the method of Binomial Expansion)
3) Know your Trig Identities because this is where you're headed

My working so far:

$(\cos\theta + i \sin\theta)^3$ = $(\cos3\theta + i \sin3\theta)$ By De Moivre's Theorem

For this specific problem I am using the 1 3 3 1 tier of Pascal's Triangle

Now I apply this with my powers incremented by 1 for cos and decremented by 1 for sin. (Hope this makes sense?)

$1\cos^3\theta i^0 \sin^0\theta + 3\cos^2\theta i^1 \sin^1\theta + 3\cos^1\theta i^2 \sin^2\theta + \cos^0\theta i^3 \sin^3\theta$

Now I know $i^2 = (-1)$ so I proceed as follows:

$\cos^3\theta + 3\cos^2\theta i \sin\theta - 3\cos\theta \sin^2\theta - i\sin^3\theta$

Now I would equate the real and imaginary parts as follows:

$\cos3\theta = \cos^3\theta + 3\cos\theta \sin^2\theta $

$\sin3\theta = 3\cos^2\theta i \sin\theta + \sin^3\theta$

I know that I am supposed to derive the cube trig identities here namely:

$\cos3\theta = 4\cos^3\theta - 3\cos\theta$
$\sin3\theta = 3\sin\theta - 4\sin^3\theta$

Perhaps I've missed something here. My workings of $\cos4\theta$ and $\cos2\theta$ are spot on though.

Thanks for taking the time.

Answer 1

You are almost done. I only evaluate the case $\cos3\theta$. You can easily check the remainder case.

You get $\cos3\theta = \cos^3\theta - 3\cos\theta \sin^2\theta$. We can apply the identity $\sin^2\theta = 1-\cos^2\theta$. Substitute it gives the formula you want.