how can I show that this identity is a development of a fourier series? $$f(x)=\sin^3 x=\frac{3}4 \sin x-\frac{1}4 \sin 3x$$
I tried this:
obtain the Fourier coefficients whih $$b_n=\frac{2}\pi \int_0^\pi{\sin^3 x\;\sin nx}\;dx\qquad n=1,2,3,\cdots$$
where $$f(x)=\sin^3 x=\sum_{n=1}^\infty b_n\sin{nx},$$ as clearly $b_n=0\;\;\forall n\not=1,3. \quad b_1=3/4\wedge b_3=-1/4$. But this integral is too long from the way I have tried
Thanks in advance
By using the addition formulas for the sine function: $$\begin{eqnarray*}\sin(3x)&=&\sin x\cos(2x)+\cos x\sin(2x) = \sin x(1-2\sin^2 x)+2\cos^2 x\sin x\\ &=& \sin x -2\sin^3 x + 2(1-\sin^2 x)\sin x = -4\sin^3+3\sin x\end{eqnarray*}$$ hence: $$\sin^3 x = \frac{3\sin x-\sin(3x)}{4}$$ as wanted. As an alternative: $$\sin^3 x = \left(\frac{e^{ix}-e^{-ix}}{2}\right)^3 = \frac{e^{3ix}-e^{-3ix}-3e^{ix}+3e^{-ix}}{-8i} = \frac{3\sin x-\sin(3x)}{4}.$$ The Fourier series is unique, hence if such an identity holds, it is the Fourier series you are looking for.