\( \DeclareMathOperator{\abs}{abs} \newcommand{\ensuremath}[1]{\mbox{$#1$}} \)

Die Ausgangsgleichungen:

(%i1) gl1:k1·sin(t1/sqrt(L·C))=k4·sin((toff-t3)/sqrt(L·C));
\[\tag{gl1}\label{gl1}\mathit{k1}\,\sin{\left( \frac{\mathit{t1}}{\sqrt{CL}}\right) }=\mathit{k4}\,\sin{\left( \frac{\mathit{toff}-\mathit{t3}}{\sqrt{CL}}\right) }\]
(%i2) gl2:k3·sin(t3/sqrt(L·C))=k2·sin((ton-t1)/sqrt(L·C));
\[\tag{gl2}\label{gl2}\mathit{k3}\,\sin{\left( \frac{\mathit{t3}}{\sqrt{CL}}\right) }=\mathit{k2}\,\sin{\left( \frac{\mathit{ton}-\mathit{t1}}{\sqrt{CL}}\right) }\]

Der Übersicht k1/k4=u, k3/k2=v und sqrt(L·C)=r substituieren:

(%i4) gl1/k4, k1=k4·u$
gl1a: subst(r, sqrt(L·C), %);
\[\tag{gl1a}\label{gl1a}\sin{\left( \frac{\mathit{t1}}{r}\right) }u=\sin{\left( \frac{\mathit{toff}-\mathit{t3}}{r}\right) }\]
(%i6) gl2/k2, k3=k2·v$
gl2a: subst(r, sqrt(L·C), %);
\[\tag{gl2a}\label{gl2a}\sin{\left( \frac{\mathit{t3}}{r}\right) }v=\sin{\left( \frac{\mathit{ton}-\mathit{t1}}{r}\right) }\]

Anwendung des Additionstheorems:

(%i7) gl1b: trigexpand(expand(gl1a));
\[\tag{gl1b}\label{gl1b}\sin{\left( \frac{\mathit{t1}}{r}\right) }u=\cos{\left( \frac{\mathit{t3}}{r}\right) }\,\sin{\left( \frac{\mathit{toff}}{r}\right) }-\sin{\left( \frac{\mathit{t3}}{r}\right) }\,\cos{\left( \frac{\mathit{toff}}{r}\right) }\]
(%i8) gl2b: trigexpand(expand(gl2a));
\[\tag{gl2b}\label{gl2b}\sin{\left( \frac{\mathit{t3}}{r}\right) }v=\cos{\left( \frac{\mathit{t1}}{r}\right) }\,\sin{\left( \frac{\mathit{ton}}{r}\right) }-\sin{\left( \frac{\mathit{t1}}{r}\right) }\,\cos{\left( \frac{\mathit{ton}}{r}\right) }\]

Cos durch sin ausdrücken:

(%i9) gl1c: subst(sqrt(1-sin(t3/r)^2), cos(t3/r), gl1b);
\[\tag{gl1c}\label{gl1c}\sin{\left( \frac{\mathit{t1}}{r}\right) }u=\sqrt{1-{{\sin{\left( \frac{\mathit{t3}}{r}\right) }}^{2}}}\,\sin{\left( \frac{\mathit{toff}}{r}\right) }-\sin{\left( \frac{\mathit{t3}}{r}\right) }\,\cos{\left( \frac{\mathit{toff}}{r}\right) }\]
(%i10) gl2c: subst(sqrt(1-sin(t1/r)^2), cos(t1/r), gl2b);
\[\tag{gl2c}\label{gl2c}\sin{\left( \frac{\mathit{t3}}{r}\right) }v=\sqrt{1-{{\sin{\left( \frac{\mathit{t1}}{r}\right) }}^{2}}}\,\sin{\left( \frac{\mathit{ton}}{r}\right) }-\sin{\left( \frac{\mathit{t1}}{r}\right) }\,\cos{\left( \frac{\mathit{ton}}{r}\right) }\]

sin(t1/r)=x1 und sin(t3/r)=x3 substituieren:

(%i14) subst(x1, sin(t1/r), [gl1c, gl2c])$
[gl1d, gl2d]: subst(x3, sin(t3/r), %)$
gl1d;
gl2d;
\[\tag{\%{}o13}\label{o13} u\,\mathit{x1}=\sin{\left( \frac{\mathit{toff}}{r}\right) }\,\sqrt{1-{{\mathit{x3}}^{2}}}-\cos{\left( \frac{\mathit{toff}}{r}\right) }\,\mathit{x3}\] \[\tag{\%{}o14}\label{o14} v\,\mathit{x3}=\sin{\left( \frac{\mathit{ton}}{r}\right) }\,\sqrt{1-{{\mathit{x1}}^{2}}}-\cos{\left( \frac{\mathit{ton}}{r}\right) }\,\mathit{x1}\]

Gleichungen so umstellen, dass die Wurzeln nicht mehr in einer Summe stehen:

(%i16) gl1e: gl1d-part(gl1d,2,2);
gl2e: gl2d-part(gl2d,2,2);
\[\tag{gl1e}\label{gl1e}\cos{\left( \frac{\mathit{toff}}{r}\right) }\,\mathit{x3}+u\,\mathit{x1}=\sin{\left( \frac{\mathit{toff}}{r}\right) }\,\sqrt{1-{{\mathit{x3}}^{2}}}\] \[\tag{gl2e}\label{gl2e}v\,\mathit{x3}+\cos{\left( \frac{\mathit{ton}}{r}\right) }\,\mathit{x1}=\sin{\left( \frac{\mathit{ton}}{r}\right) }\,\sqrt{1-{{\mathit{x1}}^{2}}}\]

Quadrieren, um die Wurzeln wegzubekommen:

(%i18) gl1f: gl1e^2;
gl2f: gl2e^2;
\[\tag{gl1f}\label{gl1f}{{\left( \cos{\left( \frac{\mathit{toff}}{r}\right) }\,\mathit{x3}+u\,\mathit{x1}\right) }^{2}}={{\sin{\left( \frac{\mathit{toff}}{r}\right) }}^{2}}\,\left( 1-{{\mathit{x3}}^{2}}\right) \] \[\tag{gl2f}\label{gl2f}{{\left( v\,\mathit{x3}+\cos{\left( \frac{\mathit{ton}}{r}\right) }\,\mathit{x1}\right) }^{2}}={{\sin{\left( \frac{\mathit{ton}}{r}\right) }}^{2}}\,\left( 1-{{\mathit{x1}}^{2}}\right) \]

Alles auf die linke Seite bringen und die Beziehung sin²(x)+cos²(x)=1 anwenden:

(%i20) gl1g: trigsimp(gl1f-rhs(gl1f));
gl2g: trigsimp(gl2f-rhs(gl2f));
\[\tag{gl1g}\label{gl1g}{{\mathit{x3}}^{2}}+2\cos{\left( \frac{\mathit{toff}}{r}\right) }u\,\mathit{x1}\,\mathit{x3}+{{u}^{2}}\,{{\mathit{x1}}^{2}}+{{\cos{\left( \frac{\mathit{toff}}{r}\right) }}^{2}}-1=0\] \[\tag{gl2g}\label{gl2g}{{v}^{2}}\,{{\mathit{x3}}^{2}}+2\cos{\left( \frac{\mathit{ton}}{r}\right) }v\,\mathit{x1}\,\mathit{x3}+{{\mathit{x1}}^{2}}+{{\cos{\left( \frac{\mathit{ton}}{r}\right) }}^{2}}-1=0\]

cos(ton/r)=s und cos(toff/s)=t substituieren:

(%i22) gl1h: subst(t, cos(toff/r), gl1g);
gl2h: subst(s, cos(ton/r), gl2g);
\[\tag{gl1h}\label{gl1h}{{\mathit{x3}}^{2}}+2tu\,\mathit{x1}\,\mathit{x3}+{{u}^{2}}\,{{\mathit{x1}}^{2}}+{{t}^{2}}-1=0\] \[\tag{gl2h}\label{gl2h}{{v}^{2}}\,{{\mathit{x3}}^{2}}+2sv\,\mathit{x1}\,\mathit{x3}+{{\mathit{x1}}^{2}}+{{s}^{2}}-1=0\]

x3² bzw. x1² eliminieren, so dass jeweils eine lineare Gleichung in x3 und x1 entsteht:

(%i24) gl3: factor(v^2·gl1h-gl2h);
gl4: factor(gl1h-u^2·gl2h);
\[\tag{gl3}\label{gl3}2tu\,{{v}^{2}}\,\mathit{x1}\,\mathit{x3}-2sv\,\mathit{x1}\,\mathit{x3}+{{u}^{2}}\,{{v}^{2}}\,{{\mathit{x1}}^{2}}-{{\mathit{x1}}^{2}}+{{t}^{2}}\,{{v}^{2}}-{{v}^{2}}-{{s}^{2}}+1=0\] \[\tag{gl4}\label{gl4}-\left( {{u}^{2}}\,{{v}^{2}}\,{{\mathit{x3}}^{2}}-{{\mathit{x3}}^{2}}+2s\,{{u}^{2}}v\,\mathit{x1}\,\mathit{x3}-2tu\,\mathit{x1}\,\mathit{x3}+{{s}^{2}}\,{{u}^{2}}-{{u}^{2}}-{{t}^{2}}+1\right) =0\]

Diese Gleichungen nach x3 bzw. x1 auflösen

(%i26) gl3a: solve(gl3, x3);
gl4a: solve(gl4, x1);
\[\tag{gl3a}\label{gl3a}[\mathit{x3}=-\frac{\left( {{u}^{2}}\,{{v}^{2}}-1\right) \,{{\mathit{x1}}^{2}}+\left( {{t}^{2}}-1\right) \,{{v}^{2}}-{{s}^{2}}+1}{\left( 2tu\,{{v}^{2}}-2sv\right) \,\mathit{x1}}]\] \[\tag{gl4a}\label{gl4a}[\mathit{x1}=-\frac{\left( {{u}^{2}}\,{{v}^{2}}-1\right) \,{{\mathit{x3}}^{2}}+\left( {{s}^{2}}-1\right) \,{{u}^{2}}-{{t}^{2}}+1}{\left( 2s\,{{u}^{2}}v-2tu\right) \,\mathit{x3}}]\]

1. Gleichung in die 2. Gleichung und
2. Gleichung in die 1. Gleichung einsetzen:

(%i28) gl5: gl4a, gl3a;
gl6: gl3a, gl4a;
\[\tag{gl5}\label{gl5}[\mathit{x1}=\frac{\left( 2tu\,{{v}^{2}}-2sv\right) \,\mathit{x1}\,\left( \frac{\left( {{u}^{2}}\,{{v}^{2}}-1\right) \,{{\left( \left( {{u}^{2}}\,{{v}^{2}}-1\right) \,{{\mathit{x1}}^{2}}+\left( {{t}^{2}}-1\right) \,{{v}^{2}}-{{s}^{2}}+1\right) }^{2}}}{{{\left( 2tu\,{{v}^{2}}-2sv\right) }^{2}}\,{{\mathit{x1}}^{2}}}+\left( {{s}^{2}}-1\right) \,{{u}^{2}}-{{t}^{2}}+1\right) }{\left( 2s\,{{u}^{2}}v-2tu\right) \,\left( \left( {{u}^{2}}\,{{v}^{2}}-1\right) \,{{\mathit{x1}}^{2}}+\left( {{t}^{2}}-1\right) \,{{v}^{2}}-{{s}^{2}}+1\right) }]\] \[\tag{gl6}\label{gl6}[\mathit{x3}=\frac{\left( 2s\,{{u}^{2}}v-2tu\right) \,\mathit{x3}\,\left( \frac{\left( {{u}^{2}}\,{{v}^{2}}-1\right) \,{{\left( \left( {{u}^{2}}\,{{v}^{2}}-1\right) \,{{\mathit{x3}}^{2}}+\left( {{s}^{2}}-1\right) \,{{u}^{2}}-{{t}^{2}}+1\right) }^{2}}}{{{\left( 2s\,{{u}^{2}}v-2tu\right) }^{2}}\,{{\mathit{x3}}^{2}}}+\left( {{t}^{2}}-1\right) \,{{v}^{2}}-{{s}^{2}}+1\right) }{\left( 2tu\,{{v}^{2}}-2sv\right) \,\left( \left( {{u}^{2}}\,{{v}^{2}}-1\right) \,{{\mathit{x3}}^{2}}+\left( {{s}^{2}}-1\right) \,{{u}^{2}}-{{t}^{2}}+1\right) }]\]

Anmerkung: Wenn man diese Gleichungen durch x1 bzw. x3 dividiert, sieht man, dass sie biquadratisch sind.
Da Maxima aber auch allgemeine Gleichungen 4. Grades lösen kann, ist dieser Schritt hier nicht notwendig.

1. Gleichungen nach x1 und 2. Gleichung nach x3 auflösen, es gibt jeweils 4 Lösungen:

(%i30) gl5a: factor(solve(gl5, x1));
gl6a: factor(solve(gl6, x3));
\[\tag{gl5a}\label{gl5a}[\mathit{x1}=-\sqrt{\frac{\left( 2tu\,{{v}^{2}}-2sv\right) \,\sqrt{\left( {{s}^{2}}-1\right) \,\left( {{t}^{2}}-1\right) \,\left( \left( {{u}^{2}}+{{t}^{2}}-1\right) \,{{v}^{2}}-2stuv+\left( {{s}^{2}}-1\right) \,{{u}^{2}}+1\right) }+\left( 1-{{t}^{2}}\right) \,{{u}^{2}}\,{{v}^{4}}+\left( 2s\,{{t}^{3}}-2st\right) u\,{{v}^{3}}+\left( \left( \left( 2-2{{s}^{2}}\right) \,{{t}^{2}}+{{s}^{2}}-1\right) \,{{u}^{2}}+\left( 1-2{{s}^{2}}\right) \,{{t}^{2}}+2{{s}^{2}}-1\right) \,{{v}^{2}}+\left( 2{{s}^{3}}-2s\right) tuv-{{s}^{2}}+1}{{{u}^{4}}\,{{v}^{4}}-4st\,{{u}^{3}}\,{{v}^{3}}+\left( 4{{t}^{2}}+4{{s}^{2}}-2\right) \,{{u}^{2}}\,{{v}^{2}}-4stuv+1}},\mathit{x1}=\sqrt{\frac{\left( 2tu\,{{v}^{2}}-2sv\right) \,\sqrt{\left( {{s}^{2}}-1\right) \,\left( {{t}^{2}}-1\right) \,\left( \left( {{u}^{2}}+{{t}^{2}}-1\right) \,{{v}^{2}}-2stuv+\left( {{s}^{2}}-1\right) \,{{u}^{2}}+1\right) }+\left( 1-{{t}^{2}}\right) \,{{u}^{2}}\,{{v}^{4}}+\left( 2s\,{{t}^{3}}-2st\right) u\,{{v}^{3}}+\left( \left( \left( 2-2{{s}^{2}}\right) \,{{t}^{2}}+{{s}^{2}}-1\right) \,{{u}^{2}}+\left( 1-2{{s}^{2}}\right) \,{{t}^{2}}+2{{s}^{2}}-1\right) \,{{v}^{2}}+\left( 2{{s}^{3}}-2s\right) tuv-{{s}^{2}}+1}{{{u}^{4}}\,{{v}^{4}}-4st\,{{u}^{3}}\,{{v}^{3}}+\left( 4{{t}^{2}}+4{{s}^{2}}-2\right) \,{{u}^{2}}\,{{v}^{2}}-4stuv+1}},\mathit{x1}=-\sqrt{-\frac{\left( 2tu\,{{v}^{2}}-2sv\right) \,\sqrt{\left( {{s}^{2}}-1\right) \,\left( {{t}^{2}}-1\right) \,\left( \left( {{u}^{2}}+{{t}^{2}}-1\right) \,{{v}^{2}}-2stuv+\left( {{s}^{2}}-1\right) \,{{u}^{2}}+1\right) }+\left( {{t}^{2}}-1\right) \,{{u}^{2}}\,{{v}^{4}}+\left( 2st-2s\,{{t}^{3}}\right) u\,{{v}^{3}}+\left( \left( \left( 2{{s}^{2}}-2\right) \,{{t}^{2}}-{{s}^{2}}+1\right) \,{{u}^{2}}+\left( 2{{s}^{2}}-1\right) \,{{t}^{2}}-2{{s}^{2}}+1\right) \,{{v}^{2}}+\left( 2s-2{{s}^{3}}\right) tuv+{{s}^{2}}-1}{{{u}^{4}}\,{{v}^{4}}-4st\,{{u}^{3}}\,{{v}^{3}}+\left( 4{{t}^{2}}+4{{s}^{2}}-2\right) \,{{u}^{2}}\,{{v}^{2}}-4stuv+1}},\mathit{x1}=\sqrt{-\frac{\left( 2tu\,{{v}^{2}}-2sv\right) \,\sqrt{\left( {{s}^{2}}-1\right) \,\left( {{t}^{2}}-1\right) \,\left( \left( {{u}^{2}}+{{t}^{2}}-1\right) \,{{v}^{2}}-2stuv+\left( {{s}^{2}}-1\right) \,{{u}^{2}}+1\right) }+\left( {{t}^{2}}-1\right) \,{{u}^{2}}\,{{v}^{4}}+\left( 2st-2s\,{{t}^{3}}\right) u\,{{v}^{3}}+\left( \left( \left( 2{{s}^{2}}-2\right) \,{{t}^{2}}-{{s}^{2}}+1\right) \,{{u}^{2}}+\left( 2{{s}^{2}}-1\right) \,{{t}^{2}}-2{{s}^{2}}+1\right) \,{{v}^{2}}+\left( 2s-2{{s}^{3}}\right) tuv+{{s}^{2}}-1}{{{u}^{4}}\,{{v}^{4}}-4st\,{{u}^{3}}\,{{v}^{3}}+\left( 4{{t}^{2}}+4{{s}^{2}}-2\right) \,{{u}^{2}}\,{{v}^{2}}-4stuv+1}}]\] \[\tag{gl6a}\label{gl6a}[\mathit{x3}=-\sqrt{\frac{\left( 2s\,{{u}^{2}}v-2tu\right) \,\sqrt{\left( {{s}^{2}}-1\right) \,\left( {{t}^{2}}-1\right) \,\left( \left( {{u}^{2}}+{{t}^{2}}-1\right) \,{{v}^{2}}-2stuv+\left( {{s}^{2}}-1\right) \,{{u}^{2}}+1\right) }+\left( \left( 1-{{s}^{2}}\right) \,{{u}^{4}}+\left( \left( 1-2{{s}^{2}}\right) \,{{t}^{2}}+2{{s}^{2}}-1\right) \,{{u}^{2}}\right) \,{{v}^{2}}+\left( \left( 2{{s}^{3}}-2s\right) t\,{{u}^{3}}+\left( 2s\,{{t}^{3}}-2st\right) u\right) v+\left( \left( 2-2{{s}^{2}}\right) \,{{t}^{2}}+{{s}^{2}}-1\right) \,{{u}^{2}}-{{t}^{2}}+1}{{{u}^{4}}\,{{v}^{4}}-4st\,{{u}^{3}}\,{{v}^{3}}+\left( 4{{t}^{2}}+4{{s}^{2}}-2\right) \,{{u}^{2}}\,{{v}^{2}}-4stuv+1}},\mathit{x3}=\sqrt{\frac{\left( 2s\,{{u}^{2}}v-2tu\right) \,\sqrt{\left( {{s}^{2}}-1\right) \,\left( {{t}^{2}}-1\right) \,\left( \left( {{u}^{2}}+{{t}^{2}}-1\right) \,{{v}^{2}}-2stuv+\left( {{s}^{2}}-1\right) \,{{u}^{2}}+1\right) }+\left( \left( 1-{{s}^{2}}\right) \,{{u}^{4}}+\left( \left( 1-2{{s}^{2}}\right) \,{{t}^{2}}+2{{s}^{2}}-1\right) \,{{u}^{2}}\right) \,{{v}^{2}}+\left( \left( 2{{s}^{3}}-2s\right) t\,{{u}^{3}}+\left( 2s\,{{t}^{3}}-2st\right) u\right) v+\left( \left( 2-2{{s}^{2}}\right) \,{{t}^{2}}+{{s}^{2}}-1\right) \,{{u}^{2}}-{{t}^{2}}+1}{{{u}^{4}}\,{{v}^{4}}-4st\,{{u}^{3}}\,{{v}^{3}}+\left( 4{{t}^{2}}+4{{s}^{2}}-2\right) \,{{u}^{2}}\,{{v}^{2}}-4stuv+1}},\mathit{x3}=-\sqrt{-\frac{\left( 2s\,{{u}^{2}}v-2tu\right) \,\sqrt{\left( {{s}^{2}}-1\right) \,\left( {{t}^{2}}-1\right) \,\left( \left( {{u}^{2}}+{{t}^{2}}-1\right) \,{{v}^{2}}-2stuv+\left( {{s}^{2}}-1\right) \,{{u}^{2}}+1\right) }+\left( \left( {{s}^{2}}-1\right) \,{{u}^{4}}+\left( \left( 2{{s}^{2}}-1\right) \,{{t}^{2}}-2{{s}^{2}}+1\right) \,{{u}^{2}}\right) \,{{v}^{2}}+\left( \left( 2s-2{{s}^{3}}\right) t\,{{u}^{3}}+\left( 2st-2s\,{{t}^{3}}\right) u\right) v+\left( \left( 2{{s}^{2}}-2\right) \,{{t}^{2}}-{{s}^{2}}+1\right) \,{{u}^{2}}+{{t}^{2}}-1}{{{u}^{4}}\,{{v}^{4}}-4st\,{{u}^{3}}\,{{v}^{3}}+\left( 4{{t}^{2}}+4{{s}^{2}}-2\right) \,{{u}^{2}}\,{{v}^{2}}-4stuv+1}},\mathit{x3}=\sqrt{-\frac{\left( 2s\,{{u}^{2}}v-2tu\right) \,\sqrt{\left( {{s}^{2}}-1\right) \,\left( {{t}^{2}}-1\right) \,\left( \left( {{u}^{2}}+{{t}^{2}}-1\right) \,{{v}^{2}}-2stuv+\left( {{s}^{2}}-1\right) \,{{u}^{2}}+1\right) }+\left( \left( {{s}^{2}}-1\right) \,{{u}^{4}}+\left( \left( 2{{s}^{2}}-1\right) \,{{t}^{2}}-2{{s}^{2}}+1\right) \,{{u}^{2}}\right) \,{{v}^{2}}+\left( \left( 2s-2{{s}^{3}}\right) t\,{{u}^{3}}+\left( 2st-2s\,{{t}^{3}}\right) u\right) v+\left( \left( 2{{s}^{2}}-2\right) \,{{t}^{2}}-{{s}^{2}}+1\right) \,{{u}^{2}}+{{t}^{2}}-1}{{{u}^{4}}\,{{v}^{4}}-4st\,{{u}^{3}}\,{{v}^{3}}+\left( 4{{t}^{2}}+4{{s}^{2}}-2\right) \,{{u}^{2}}\,{{v}^{2}}-4stuv+1}}]\]

Anhand eines Beispiels ermitteln, welche der 4 Lösungen die gewünschte ist:

(%i32) sqrt(200e-6·100e-6)$
bsp:[r=%, u=2.5, v=3.2, s=cos(70e-6/%), t=cos(30e-6/%)];
\[\tag{bsp}\label{bsp}[r=1.414213562373095{{10}^{-4}},u=2.5,v=3.2,s=0.8799807056103829,t=0.9775842485391512]\]
(%i33) x1=r·asin(map(rhs, gl5a)), bsp;
\[\tag{\%{}o33}\label{o33} \mathit{x1}=[-2.227884439606365{{10}^{-5}},2.227884439606365{{10}^{-5}},-4.019093945220097{{10}^{-6}},4.019093945220097{{10}^{-6}}]\]
(%i34) x3=r·asin(map(rhs, gl6a)), bsp;
\[\tag{\%{}o34}\label{o34} \mathit{x3}=[-2.699761822406673{{10}^{-5}},2.699761822406673{{10}^{-5}},-1.99451485261793{{10}^{-5}},1.99451485261793{{10}^{-5}}]\]

Die jeweils 4. Lösung entspricht der des iterativen Verfahrens, also lauten die allgemeinen Lösungen

(%i36) t1=r·asin(x1);
t3=r·asin(x3);
\[\tag{\%{}o35}\label{o35} \mathit{t1}=r\,\operatorname{asin}\left( \mathit{x1}\right) \] \[\tag{\%{}o36}\label{o36} \mathit{t3}=r\,\operatorname{asin}\left( \mathit{x3}\right) \]

wobei

(%i43) gl5a[4];
gl6a[4];
s=cos(ton/r);
t=cos(toff/r);
u=k1/k4;
v=k3/k2;
r=sqrt(L·C);
\[\tag{\%{}o37}\label{o37} \mathit{x1}=\sqrt{-\frac{\left( 2tu\,{{v}^{2}}-2sv\right) \,\sqrt{\left( {{s}^{2}}-1\right) \,\left( {{t}^{2}}-1\right) \,\left( \left( {{u}^{2}}+{{t}^{2}}-1\right) \,{{v}^{2}}-2stuv+\left( {{s}^{2}}-1\right) \,{{u}^{2}}+1\right) }+\left( {{t}^{2}}-1\right) \,{{u}^{2}}\,{{v}^{4}}+\left( 2st-2s\,{{t}^{3}}\right) u\,{{v}^{3}}+\left( \left( \left( 2{{s}^{2}}-2\right) \,{{t}^{2}}-{{s}^{2}}+1\right) \,{{u}^{2}}+\left( 2{{s}^{2}}-1\right) \,{{t}^{2}}-2{{s}^{2}}+1\right) \,{{v}^{2}}+\left( 2s-2{{s}^{3}}\right) tuv+{{s}^{2}}-1}{{{u}^{4}}\,{{v}^{4}}-4st\,{{u}^{3}}\,{{v}^{3}}+\left( 4{{t}^{2}}+4{{s}^{2}}-2\right) \,{{u}^{2}}\,{{v}^{2}}-4stuv+1}}\] \[\tag{\%{}o38}\label{o38} \mathit{x3}=\sqrt{-\frac{\left( 2s\,{{u}^{2}}v-2tu\right) \,\sqrt{\left( {{s}^{2}}-1\right) \,\left( {{t}^{2}}-1\right) \,\left( \left( {{u}^{2}}+{{t}^{2}}-1\right) \,{{v}^{2}}-2stuv+\left( {{s}^{2}}-1\right) \,{{u}^{2}}+1\right) }+\left( \left( {{s}^{2}}-1\right) \,{{u}^{4}}+\left( \left( 2{{s}^{2}}-1\right) \,{{t}^{2}}-2{{s}^{2}}+1\right) \,{{u}^{2}}\right) \,{{v}^{2}}+\left( \left( 2s-2{{s}^{3}}\right) t\,{{u}^{3}}+\left( 2st-2s\,{{t}^{3}}\right) u\right) v+\left( \left( 2{{s}^{2}}-2\right) \,{{t}^{2}}-{{s}^{2}}+1\right) \,{{u}^{2}}+{{t}^{2}}-1}{{{u}^{4}}\,{{v}^{4}}-4st\,{{u}^{3}}\,{{v}^{3}}+\left( 4{{t}^{2}}+4{{s}^{2}}-2\right) \,{{u}^{2}}\,{{v}^{2}}-4stuv+1}}\] \[\tag{\%{}o39}\label{o39} s=\cos{\left( \frac{\mathit{ton}}{r}\right) }\] \[\tag{\%{}o40}\label{o40} t=\cos{\left( \frac{\mathit{toff}}{r}\right) }\] \[\tag{\%{}o41}\label{o41} u=\frac{\mathit{k1}}{\mathit{k4}}\] \[\tag{\%{}o42}\label{o42} v=\frac{\mathit{k3}}{\mathit{k2}}\] \[\tag{\%{}o43}\label{o43} r=\sqrt{CL}\]
Created with wxMaxima.