# The Fourier series of $\sin^3 t$ in trigonometric form

I'm trying to calculate the Fourier series of $\sin^3t$ in trigonometric form. In previous excercises I have been able to use trigonometric identities to be able to calculate the coefficents, but here I can rewrite the function, but I cannot get what period the function has, and moreover, I can't solve for for the $b_n$ coefficents.

Am I missing something here? Because rewriting into complex form first and than raise that to 3 and calculate seems unnecessary complicated.

Any ideas?

• ..Why not use De Moivre's theorem? – FireGarden Sep 9 '13 at 15:01
• FireGarden: How will De Moivre's theorem help me here? It was a long time since I did excercises on the theorem, so I might miss something obvious. – Curtain Sep 9 '13 at 15:10
• OP: How is life in Ecuador's embassy? – Did Sep 9 '13 at 16:50

$$8(\sin t)^3=\mathrm i(\mathrm e^{\mathrm it}-\mathrm e^{-\mathrm it})^3=\mathrm i\mathrm e^{3\mathrm it}-3\mathrm i\mathrm e^{\mathrm it}+3\mathrm i\mathrm e^{-\mathrm it}-\mathrm i\mathrm e^{-3\mathrm it}=2\,(3\sin t-\sin 3t)$$

• Thanks, Did. However, how should I proceed calculating the period when the $\omega$ is different in both sin terms? – Curtain Sep 9 '13 at 15:05
• A period of $t\mapsto\sin t$ is $2\pi$. Hence a period of $t\mapsto(\sin t)^3$ is $2\pi$. // Alternatively, a period of $t\mapsto\sin t$ is $2\pi$ and a period of $t\mapsto\sin 3t$ is $2\pi/3$. Hence a period of their linear combination $t\mapsto(\sin t)^3$ is $\max(2\pi,2\pi/3)=2\pi$. – Did Sep 9 '13 at 15:09
• Hm, how do we know that the maximum of one of the two periods actually are the linear combination's period? I've checked it in Wolfram Alpha, just want to know. – Curtain Sep 9 '13 at 16:51
• Maximum was stupid, I should have said: least common multiple. Thanks for the remark. – Did Sep 9 '13 at 17:42

I would write it in complex form because trying to guess your way through rearranging terms from known trigonometric identities is a little silly. Why spend time guessing when you can quickly get the answer?

$$\sin t = \frac{1}{2i}(e^{it}-e^{-it})$$

Thus we have $\sin^3t = \frac{i}{8}(e^{it}-e^{-it})^3 = \frac{i}{8}(e^{3it}-3e^{it}+3e^{-it}-e^{-3it})$. Can you take it from here?

hint: use power reduction identities.

Seeing as $\sin 3x=3\sin x-4\sin^3x$ it follows $\sin^3x=\dfrac34\sin x-\dfrac14\sin 3x$ hence $a_i=0$ and $b_1=3/4,b_3=-1/4$ and other $b_j=0$. The period is essentially the 'least common multiple' i.e. $\operatorname{lcm}(1,3)\cdot\dfrac{2\pi}3=2\pi$.

• This is what I was about to write (+1) – robjohn Sep 9 '13 at 15:55
• Thanks for the correction, @robjohn. I posted this from my phone and it's difficult to use the small touchscreen keyboard effectively. – oldrinb Sep 9 '13 at 16:38
• @oldrinb: Why is that the period is the least common multiple? – Curtain Sep 9 '13 at 17:09

A related problem. First, evaluate the integral

$$b_n = \frac{1}{\pi}\int_{-\pi}^{\pi} \sin^3(t)\sin(nt)dt = 12\,{\frac {\sin \left( \pi \,n \right) }{{n}^{4}-10\,{n}^{2}+9}}.$$

Now, you can see that, the only non zero values for $b_n$ comes from $n=1$ and $n=3$ which they give the values

$$b_1= \frac{3}{4}\quad \rm and \quad b_3 = -\frac{1}{4}.$$

Note that, $a_n=0$. So, your final result is the well known identity

$$\sin^3(t)= \frac{3}{4}\sin(t) - \frac{1}{4}\sin(3t).$$ .....

Making my comment into a full answer;

Given $\sin^3(x)$ is a simple case, De Moivre's is much more efficient.

De Moivre's says $(\cos(x)+i\sin(x))^n = (\cos(nx)+i\sin(nx))$

Now, $(\cos(x)+i\sin(x))^3 = -4i\sin^3(x) + 3i\sin(x) + \cos(x) -4\sin(x)^2\cos(x)$.

By De Moivre's and comparing real and imaginary parts, we get $$\sin(3x)=-4sin(x)^3+3\sin(x)$$

Which you can rearrange for the fourier series.