\(\sin 3x = 3\sin x - 4\sin^3 x\)
\(\cos 3x = -3\cos x + 4\cos^3 x\)
\[
\begin{align}
\sin 3x &= \sin x\cos2x +\cos x\sin2x\\
&= \sin x(1-2\sin^2x) +2\cos^2x\sin x\\
&= \sin x(1-2\sin^2x) +2(1-\sin^2x)\sin x\\
&= \sin x -2\sin^3x +2\sin x-2\sin^3x)\\
&= 3\sin x - 4\sin^3 x\\
\end{align}
\]
\[
\begin{align}
\cos 3x &= \cos x\cos2x - \sin x\sin2x\\
&= \cos x(2\cos^2x-1) - 2\sin^2x\cos x\\
&= \cos x(2\cos^2x-1) - 2(1-\cos^2x)\cos x\\
&= 2\cos^3x - \cos x - 2\cos x +2\cos^3x\\
&= -3\cos x + 4\cos^3 x\\
\end{align}
\]
\[
\begin{align}
\sin 3x &= 3\sin x - 4\sin^3 x\\
\cos 3x &= -3\cos x + 4\cos^3 x\\
\end{align}
\]
\(\sin 4x = 4\sin x\cos x\left(\cos^2 x - \sin^2 x\right)\)
\(\cos 4x = 8\cos^4 x - 8\cos^2 x + 1\)
\[
\begin{align}
\sin 4x &= 4\sin x\cos x\left(\cos^2 x - \sin^2 x\right)\\
\cos 4x &= 8\cos^4 x - 8\cos^2 x + 1\\
\end{align}
\]
\(\sin 5 x = 5\sin(x) - 20\sin^3(x) + 16\sin^5(x)\)
\[ \begin{align} \sin 5x &= \sin(x+4x)\\ &= \sin(x)\cos(4x) + \cos(x)\sin(4x)\\ &= \sin(x)\left(1-2\sin^2(2x)\right) + \cos(x)2\sin(2x)\cos(2x)\\ &= \sin(x)\left(1-8\sin^2(x)\cos^2(x)\right) + 4\sin(x)\cos^2(x)\left(1-2\sin^2(x)\right)\\ &= \sin(x) - 8\sin^3(x)\cos^2(x) + 4\sin(x)\cos^2(x) - 8\sin^3(x)\cos^2(x)\\ &= \sin(x) - 8\sin^3(x) + 8\sin^5(x) + 4\sin(x) - 4\sin^3(x) - 8\sin^3(x) + 8\sin^5(x)\\ &= 5\sin(x) - 20\sin^3(x) + 16\sin^5(x)\\ \end{align} \] Si \(\sin 5x =0\) llavors \(5x=0\) o \(5x=\pi\) o \(5x=2\pi\) o ... \[x=0, x=\frac{\pi}{5},x=\frac{2\pi}{5},...\] Llavors \(\sin(x)\left(16\sin^4(x) - 20\sin^2(x) + 5=0\right)\) i per tant \[\Delta=400-4\cdot16\cdot5=400-320=80\] \[ \sin^2x = \frac{20\pm\sqrt{\Delta}}{2\cdot16} = \frac{20\pm\sqrt{80}}{32} = \frac{5\pm\sqrt{5}}{8} = \left\{\matrix{ \sin^2x = \frac{5+\sqrt{5}}{8} \\ \sin^2x = \frac{5-\sqrt{5}}{8} }\right\} \rightarrow \left\{\matrix{\sin x = \pm\sqrt{\frac{5+\sqrt{5}}{8}}\\ \sin x = \pm\sqrt{\frac{5-\sqrt{5}}{8}}}\right. \] \[\sin 5x =0 \rightarrow \left\{ \begin{array}{ccl} x=0 & x=0^{\circ} & \sin 0^{\circ} = \sin 0 = 0 \\ x=\frac{\pi}{5} & x=72^{\circ} & \sin 72^{\circ} = \sin \frac{\pi}{5} = \sqrt{\frac{5+\sqrt{5}}{8}} = \frac{\sqrt{10+2\sqrt{5}}}{4} \\ x=\frac{2\pi}{5}& x=144^{\circ} & \sin 144^{\circ} = \sin \frac{2\pi}{5} = \sqrt{\frac{5-\sqrt{5}}{8}} = \frac{\sqrt{10-2\sqrt{5}}}{4} \\ x=\frac{3\pi}{5}& x=216^{\circ} & \sin 216^{\circ} = \sin \frac{3\pi}{5} = -\sqrt{\frac{5-\sqrt{5}}{8}} = -\frac{\sqrt{10-2\sqrt{5}}}{4} \\ x=\frac{4\pi}{5}& x=288^{\circ} & \sin 288^{\circ} = \sin \frac{4\pi}{5} = -\sqrt{\frac{5+\sqrt{5}}{8}} = -\frac{\sqrt{10+2\sqrt{5}}}{4} \end{array} \right.\]\(\cos 5 x = 5\cos(x) - 20\cos^3(x) + 16\cos^5(x)\)
\[ \begin{align} \sin 5x &= \sin(x+4x)\\ &= \sin(x)\cos(4x) + \cos(x)\sin(4x)\\ &= \sin(x)\left(1-2\sin^2(2x)\right) + \cos(x)2\sin(2x)\cos(2x)\\ &= \sin(x)\left(1-8\sin^2(x)\cos^2(x)\right) + 4\sin(x)\cos^2(x)\left(1-2\sin^2(x)\right)\\ &= \sin(x) - 8\sin^3(x)\cos^2(x) + 4\sin(x)\cos^2(x) - 8\sin^3(x)\cos^2(x)\\ &= \sin(x) - 8\sin^3(x) + 8\sin^5(x) + 4\sin(x) - 4\sin^3(x) - 8\sin^3(x) + 8\sin^5(x)\\ &= 5\sin(x) - 20\sin^3(x) + 16\sin^5(x)\\ \end{align} \] Si \(\sin 5x =0\) llavors \(5x=0\) o \(5x=\pi\) o \(5x=2\pi\) o ... \[x=0, x=\frac{\pi}{5},x=\frac{2\pi}{5},...\] Llavors \(\sin(x)\left(16\sin^4(x) - 20\sin^2(x) + 5=0\right)\) i per tant \[ \sin^2x = \frac{20\pm\sqrt{400-4\cdot16\cdot5}}{2\cdot16} = \frac{20\pm\sqrt{80}}{32} = \frac{5\pm\sqrt{5}}{8} = \left\{\matrix{ \sin^2x = \frac{5+\sqrt{5}}{8} \\ \sin^2x = \frac{5-\sqrt{5}}{8} }\right\} \rightarrow \left\{\matrix{\sin x = \pm\sqrt{\frac{5+\sqrt{5}}{8}}\\ \sin x = \pm\sqrt{\frac{5-\sqrt{5}}{8}}}\right. \] \[\sin 5x =0 \rightarrow \left\{ \begin{array}{ccRrl} x=0 & x=0^{\circ} & \sin 0^{\circ} = \sin 0 &= 0 \\ x=\frac{\pi}{5} & x=72^{\circ} & \sin 72^{\circ} = \sin \frac{\pi}{5} &= \sqrt{\frac{5+\sqrt{5}}{8}} = \frac{\sqrt{10+2\sqrt{5}}}{4} \\ x=\frac{2\pi}{5}& x=144^{\circ} & \sin 144^{\circ} = \sin \frac{2\pi}{5} &= \sqrt{\frac{5-\sqrt{5}}{8}} = \frac{\sqrt{10-2\sqrt{5}}}{4} \\ x=\frac{3\pi}{5}& x=216^{\circ} & \sin 216^{\circ} = \sin \frac{3\pi}{5} &= -\sqrt{\frac{5-\sqrt{5}}{8}} = -\frac{\sqrt{10-2\sqrt{5}}}{4} \\ x=\frac{4\pi}{5}& x=288^{\circ} & \sin 288^{\circ} = \sin \frac{4\pi}{5} &= -\sqrt{\frac{5+\sqrt{5}}{8}} = -\frac{\sqrt{10+2\sqrt{5}}}{4} \end{array} \right.\]
\(\sin 6 x = 2\sin(x)\left(16\cos^5(x)-16\cos^3(x)+3\cos(x)\right)\)
\(\cos 6 x\)
\[
\begin{align}
\sin 6x &= \sin(2x+4x)\\
&= \sin(2x)\cos(4x) + \cos(2x)\sin(4x)\\
&= \sin(2x)\cos(4x) + \cos(2x)2\sin(2x)\cos(2x)\\
&= \sin(2x)\left(\cos(4x) + 2\cos^2(2x)\right)\\
&= \sin(2x)\left(2\cos^2(2x) - 1 + 2\cos^2(2x)\right)\\
&= \sin(2x)\left(4\cos^2(2x) - 1\right)\\
&= \sin(2x)\left(4\left(2\cos^2(x)-1\right)^2 - 1\right)\\
&= \sin(2x)\left(4\left(4\cos^4(x)-4\cos^2(x)+1\right) - 1\right)\\
&= \sin(2x)\left(16\cos^4(x)-16\cos^2(x)+4 - 1\right)\\
&= \sin(2x)\left(16\cos^4(x)-16\cos^2(x)+3\right)\\
&= 2\sin(x)\cos(x)\left(16\cos^4(x)-16\cos^2(x)+3\right)\\
\end{align}
\]
Si \(\sin 6x =0\) llavors \(6x=0\) o \(6x=\pi\) o \(6x=2\pi\) o ...
\[x=0, x=\frac{\pi}{6},x=\frac{\pi}{3},...\]
Llavors \(16\cos^4(x)-16\cos^2(x)+3=0\) i per tant
\[
\Delta = 256-4\cdot16\cdot3 = 256 - 192 = 64
\]
\[
\cos^2x = \frac{16\pm\sqrt{\Delta}}{2\cdot16}
= \frac{16\pm8}{32}
= \frac{4\pm1}{4}
= \left\{\matrix{\cos^2x = \frac{3}{4}\\ \cos^2x = \frac{1}{4}}\right\}
\rightarrow
\left\{\matrix{\cos x = \pm\frac{\sqrt{3}}{2}\\ \cos x = \pm\frac{1}{2}}\right.
\]
\[
\begin{align}
\cos 6x &= \cos(2x+4x)\\
&= \cos(2x)\cos(4x) - \sin(2x)\sin(4x)\\
&= \cos(2x)\cos(4x) - \sin(2x)2\sin(2x)\cos(2x)\\
&= \cos(2x)\left(\cos(4x) - 2\sin^2(2x)\right)\\
&= \cos(2x)\left(2\cos^2(2x) - 1 - 2 + 2\cos^2(2x)\right)\\
&= \cos(2x)\left(4\cos^2(2x) - 3\right)\\
\end{align}
\]
Solució general per \(\sin n x\)
\[ \begin{align} \sin nx &= \Im\left[e^{i\cdot n\cdot x}\right]\\ &= \Im\left[\left(e^{i\cdot x}\right)^n\right]\\ &= \Im\left[\left(\cos x + i\cdot \sin x\right)^n\right]\\ &= \Im\left[\sum_{k=0}^n \left(\matrix{n\\k}\right)i^k\cdot(\sin x)^k\cdot(\cos x)^{n-k}\right]\\ &= \sum_{k=0}^n \left(\matrix{n\\k}\right)\frac{i^k-(-i)^k}{2i}\cdot(\sin x)^k\cdot(\cos x)^{n-k}\\ &= \sum_{k=0}^n \left(\matrix{n\\k}\right)\frac{1-(-1)^k}{2}i^{k-1}\cdot(\sin x)^k\cdot(\cos x)^{n-k}\\ &= \sum_{l=0}^{2l+1\leq n} \left(\matrix{n\\2l+1}\right)(-1)^l\cdot(\sin x)^{2l+1}\cdot(\cos x)^{n-2l-1}\\ &= \sin x\sum_{l=0}^{2l+1\leq n} \left(\matrix{n\\2l+1}\right)(-1)^l\cdot(\sin^2 x)^{l}\cdot(\cos x)^{n-2l-1} \end{align} \]Solució general per \(\cos n x\)
\[ \begin{align} \cos nx &= \Re\left[e^{i\cdot n\cdot x}\right]\\ &= \Re\left[\left(e^{i\cdot x}\right)^n\right]\\ &= \Re\left[\left(\cos x + i\cdot \sin x\right)^n\right]\\ &= \Re\left[\sum_{k=0}^n \left(\matrix{n\\k}\right)i^k\cdot(\sin x)^k\cdot(\cos x)^{n-k}\right]\\ &= \sum_{k=0}^n \left(\matrix{n\\k}\right)\frac{i^k+(-i)^k}{2}\cdot(\sin x)^k\cdot(\cos x)^{n-k}\\ &= \sum_{k=0}^n \left(\matrix{n\\k}\right)\frac{1+(-1)^k}{2}i^{k}\cdot(\sin x)^k\cdot(\cos x)^{n-k}\\ &= \sum_{l=0}^{2l\leq n} \left(\matrix{n\\2l}\right)(-1)^l\cdot(\sin x)^{2l}\cdot(\cos x)^{n-2l}\\ &= \sum_{l=0}^{2l\leq n} \left(\matrix{n\\2l}\right)(-1)^l\cdot(1-\cos^2x)^{l}\cdot(\cos x)^{n-2l} \end{align} \]\[ \begin{align} \cos nx &= \Re\left[\sum_{k=0}^n \left(\matrix{n\\k}\right)i^k\cdot(\sin x)^k\cdot(\cos x)^{n-k}\right]\\ \end{align} \] \[ \begin{align} \sin nx &= \Im\left[\sum_{k=0}^n \left(\matrix{n\\k}\right)i^k\cdot(\sin x)^k\cdot(\cos x)^{n-k}\right]\\ \cos nx &= \Re\left[\sum_{k=0}^n \left(\matrix{n\\k}\right)i^k\cdot(\sin x)^k\cdot(\cos x)^{n-k}\right]\\ \end{align} \] \[ \begin{align} a_n&=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(n\cdot x) dx\\ &=\frac{1}{\pi}\int_{-\pi}^{\pi}x\cdot\cos(n\cdot x) dx\\ &=\frac{1}{\pi n}\left.\left[x\cdot\sin(n\cdot x) + \sin(n\cdot x)\right]\right|_{-\pi}^{\pi}\\ &=\frac{1}{\pi n}\left[\pi\cdot\sin(n \pi) + \sin(n \pi)-\pi\cdot \sin(n \pi) + \sin(n \pi) \right]\\ &=0\\ &\\ b_n&=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin(n\cdot x) dx\\ &=\frac{1}{\pi n}\left[(1 - x)\cos(n\cdot x)\right]|_{-\pi}^{\pi}\\ &=\frac{1}{\pi n}\left[(1 - \pi)\cos(n\cdot \pi)-(1 + \pi)\cos(n\cdot \pi)\right]\\ &=\frac{1}{\pi n}\left[(1 - \pi -1 - \pi)\cos(n\cdot \pi)\right]\\ &=-\frac{2}{n}\cos(n\cdot \pi)\\ &=\frac{2}{n}\frac{1+(-1)^{n+1}}{2}\\ \end{align} \]
Definició
Una funció periòdica \(f(x)\) que repeteix el esquema de l'interval \([-\pi,+\pi]\) és pot desenvolupar en funcions periòdiques d'aquest mateix interval:
\[f(x)=a_0+\sum_{n=1}^{\infty}\left(a_n\cdot\cos(n\cdot x)+b_n\cdot\sin(n\cdot x)\right)\]on
\[ \begin{align} a_0&=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)dx&\\ a_n&=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(n\cdot x) dx &\hbox{ per }n>0\\ b_n&=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin(n\cdot x) dx &\hbox{ per }n>0 \end{align} \]Sèrie de Fourier especials
| Sèrie de Fourier de la funció esglaonada | Demostració |
| \[ f(x)=\left\{ \begin{align} -1&\;\;\;-\pi\le x<0\\ +1&\;\;\;0\le x<\pi \end{align} \right. \] | \[ f(x)=\sum_{n=1}^{\infty}\frac{4}{(2n-1)\pi}\sin((2n-1)\cdot x) \] |
| \[ f(x)=\frac{4}{\pi}\sum_{n=1}^{\infty}\frac{\sin((2n-1)\cdot x)}{2n-1} \] | |
| \[\frac{4}{\pi}\left(\frac{\sin(x)}{1} + \frac{\sin(3x)}{3} + \frac{\sin(5x)}{5} + ...\right)\] |