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

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)

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.

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.
• @Metastasis Just expand it; $3\cos\theta (1-\cos^2 \theta)$ is $3\cos\theta - 3\cos^3 \theta$. Nov 9, 2015 at 12:21
• @Metastasis I realize that you make a mistake in the calculation. You should get $\cos3\theta = \cos^3\theta - 3\cos\theta\sin^2\theta$, not +. Nov 9, 2015 at 12:21
• Thanks for this. I have derived the answer. Should it not be $3cos\theta + 3cos^3\theta$ then to get $4cos^3\theta$ from expansion? Nov 9, 2015 at 12:27