Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Using the formula:

$$e^{i\omega t} = \cos {\omega t} + i\sin{\omega t}$$

I would like to prove that:

$$\sin^3\;x = -\frac{\sin{3x} - 3\sin{x}}{4} $$

However I haven't found any approach to this question. Just converting the first formula to $\sin^3$ doesn't seem to help as I'm still getting $\cos^3$ on the other side. Can anyone help me to guide me on the right way?

share|improve this question
2  
Try cubing the first formula as it is and taking the imaginary part on each side. You may also need to use $sin^2(x)+cos^2(x)=1$ Taking the real part gives the corresponding formula for cos. –  Mark Bennet Jun 29 '11 at 15:06
2  
@Mark: you should make that an answer! –  The Chaz 2.0 Jun 29 '11 at 15:08
1  
I was trying to give a hint rather than an answer, because I reckon you get more value working through the calculation at least once in your life. –  Mark Bennet Jun 29 '11 at 15:33
add comment

3 Answers 3

$$ \sin^3 x = \left(\frac{e^{ix}-e^{-ix}}{2i}\right)^3 = \frac{e^{i3x}-3e^{ix}+3e^{-ix}-e^{-i3x}}{-8i} $$

$$ = -\frac{1}{4}\left(\frac{e^{i3x}-e^{-i3x}}{2i}\right) +\frac{3}{4}\left(\frac{e^{ix}-e^{-ix}}{2i}\right) = -\frac{\sin 3x - 3\sin x}{4} $$

share|improve this answer
3  
+1 for showing that no ingenunity is required; the equation $e^{ix} = \cos x + i\sin x$ is really a "proof machine" for most trigonometric identities that can be carried out mechanically. –  ShreevatsaR Jun 29 '11 at 16:07
1  
@ShreevatsaR : Agreed, I think this way is the most straightforward. –  Joel Cohen Jun 29 '11 at 17:07
    
another answer gets bumped, and yours gets upvoted :-) –  robjohn Nov 11 '11 at 17:38
    
But it seems CuStud never came back to look at the answers... –  GEdgar Nov 11 '11 at 17:56
add comment

You don't need to know de Moivre's Theorem, in fact it is implicit in the first formula.

Try $(e^{i\omega t})^3 = (\cos {\omega t} + i\sin{\omega t})^3$.

But you also have

$$ (e^{i\omega t})^3 = e^{i3\omega t} = \cos {3\omega t} + i\sin{3\omega t}$$

using the first formula with $3\omega t$ instead of $\omega t$.

Equate the two, take real and imaginary parts, and use $1 - \cos^2 x = \sin^2 x$ to get everything in terms of the $\sin$ function.

share|improve this answer
add comment

Here's how you do it:

By de Moivre's theorem, you should know that

$(\cos 3x + i \sin 3x) = (\cos x + i \sin x)^3$.

Expand the term on the right hand side using the binomial theorem, viz $(\cos 3x + i \sin 3x) = \cos^3 x +3i\cos^2 x \sin x - 3\sin^2 x \cos x - i \sin^3 x.$

So equating imaginary parts and you should get

$\sin^3 x = 3\cos^2 x \sin x - \sin 3x$.

Then you can use the fact that $1 - \cos^2 x = \sin^2 x$ which gives:

$\sin^3 x = (3 - 3\sin^2 x)\sin x - \sin3x$, hence

$4\sin^3 x = 3\sin x - \sin 3x$

which is what you want.

share|improve this answer
    
But you ignored the "Using" clause in the question! See my answer... –  GEdgar Jun 29 '11 at 15:28
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.