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I must use de Moivre's Theorem to derive the following relation although I'm not exactly sure where to start:

$$\sin(3 \theta) = -4 \sin^3(\theta) + 3\sin(\theta)$$

Thanks in advance.

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2 Answers 2

Well, DeMoivre's theorem tells us that $$\cos(n\theta)+i\sin(n\theta)=(\cos\theta+i\sin\theta)^n$$ for all integers $n.$ Consider in particular when $n=3,$ and expand the perfect cube to see what happens. Don't forget your trig identities....

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Thanks. So once I have it expanded, I can just take the real and imaginary parts and separate them? –  Simon M Oct 16 '13 at 20:05
    
That's the idea! –  Cameron Buie Oct 16 '13 at 20:05

This answer goes through most of the steps, but I've gone through pretty fast, so it will take some going over. It's correct, but do the working out :-)

De Moivre's theorem:

$$ (\cos{(x)} + i\sin{(x)})^{n} = \cos{(nx)} + i\sin{(nx) } $$

Let $n = 3$, and use a bit of the binomial theorem:

$$\cos^{3}{(x)} + 3i\cos^{2}{(x)}\sin{(x)} - 3\cos{(x)}\sin^2{(x)} - i\sin^{3}{(x)} = \cos{(3x)} + i\sin{(3x)}$$

Now we have an identity of some description, let's try expanding that $\cos{(3x)}$, and see what happens:

$$ \cos{(2x + x)} = \cos{(2x)}\cos{(x)} - \sin{(2x)}\sin{(x)} \\ = \cos^3{(x)} - \sin^2{(x)}\cos{(x)} - 2\sin^2{(x)}\cos{(x)} $$

That seems pretty promising. So let's put that in and leave the identity we want on its own on the other side.

$$ i\sin{(3x)} = 3i\cos^2{(x)}\sin{(x)} - i\sin^3{(x)} $$

Now note that $\cos^2{(x)} = 1 - \sin^2{(x)}$, and the answer is obtained.

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Using angle sum formulas here is a rather roundabout approach, and misses a rather important fact: the cosine and sine functions are real-valued on the reals. Hence, since $$\cos(3x)+i\sin(3x)=\cos^3x-3\cos x\sin^2x+i(3\cos^2x\sin x-\sin^3x)$$ for all real $x,$ then $$\cos(3x)=\cos^3x-3\cos x\sin^2x$$ and $$\sin(3x)=3\cos^2x\sin x-\sin^3x$$ for all real $x$, as well. The latter, together with Pythagorean identity, gets us what we need. –  Cameron Buie Oct 16 '13 at 20:18
    
I'm fully aware of that method. I could say that I was demonstrating a second method, which is technically true but untrue in spirit. It simply never occurred to me to separate the real and imaginary parts. I'm not sure why. –  Max Bucknell Oct 19 '13 at 15:42
    
We all miss the obvious (in hindsight) sometimes. –  Cameron Buie Oct 19 '13 at 16:05

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