miniaufgabe.js ==== 27. November 2017 bis 1. Dezember 2017 ==== === Dienstag 28. November 2017 und Donnerstag 30. November 2017 === Berechnen Sie folgende Teilsumme einer geometrischen Reihe. Der TR darf eingesetzt werden. Achtung: Wenn Sie mit dem TR nach $n$ auflösen wollen, lassen Sie die Vorzeichen weg! miniAufgabe("#exogeomsum","#solgeomsum", [["$s=\\frac{3}{5}-1+\\frac{5}{3}-\\frac{25}{9}+ \\ldots +\\frac{5^{7}}{3^{7}}-\\frac{5^{8}}{3^{8}}+\\frac{5^{9}}{3^{9}}$", "$g_1=\\frac{3}{5}=\\frac{3}{5}$, $q=\\frac{g_2}{g_1} = -\\frac{5}{3}$, \n$\\frac{g_n}{g_1} = q^{n-1} = \\frac{5^{10}}{3^{10}}$ daraus folgt $n=11$, Resultat: $s_n=g_1 \\cdot \\frac{1-q^n}{1-q} = \\frac{6125659}{98415}$"], ["$s=\\frac{27}{50}+\\frac{9}{10}+\\frac{3}{2}+\\frac{5}{2}+ \\ldots +\\frac{5^{3}}{2\\cdot3^{2}}+\\frac{5^{4}}{2\\cdot3^{3}}+\\frac{5^{5}}{2\\cdot3^{4}}$", "$g_1=\\frac{27}{50}=\\frac{3^{3}}{2\\cdot5^{2}}$, $q=\\frac{g_2}{g_1} = \\frac{5}{3}$, \n$\\frac{g_n}{g_1} = q^{n-1} = \\frac{5^{7}}{3^{7}}$ daraus folgt $n=8$, Resultat: $s_n=g_1 \\cdot \\frac{1-q^n}{1-q} = \\frac{96016}{2025}$"], ["$s=\\frac{24}{25}+\\frac{6}{5}+\\frac{3}{2}+\\frac{15}{8}+ \\ldots +\\frac{3\\cdot5^{4}}{2^{9}}+\\frac{3\\cdot5^{5}}{2^{11}}+\\frac{3\\cdot5^{6}}{2^{13}}$", "$g_1=\\frac{24}{25}=\\frac{2^{3}\\cdot3}{5^{2}}$, $q=\\frac{g_2}{g_1} = \\frac{5}{4}$, \n$\\frac{g_n}{g_1} = q^{n-1} = \\frac{5^{8}}{2^{16}}$ daraus folgt $n=9$, Resultat: $s_n=g_1 \\cdot \\frac{1-q^n}{1-q} = \\frac{5072943}{204800}$"], ["$s=-\\frac{16}{27}+\\frac{8}{9}-\\frac{4}{3}+2- \\ldots +\\frac{3^{6}}{2^{5}}-\\frac{3^{7}}{2^{6}}+\\frac{3^{8}}{2^{7}}$", "$g_1=-\\frac{16}{27}=-\\frac{2^{4}}{3^{3}}$, $q=\\frac{g_2}{g_1} = -\\frac{3}{2}$, \n$\\frac{g_n}{g_1} = q^{n-1} = -\\frac{3^{11}}{2^{11}}$ daraus folgt $n=12$, Resultat: $s_n=g_1 \\cdot \\frac{1-q^n}{1-q} = \\frac{105469}{3456}$"], ["$s=\\frac{27}{50}+\\frac{9}{10}+\\frac{3}{2}+\\frac{5}{2}+ \\ldots +\\frac{5^{6}}{2\\cdot3^{5}}+\\frac{5^{7}}{2\\cdot3^{6}}+\\frac{5^{8}}{2\\cdot3^{7}}$", "$g_1=\\frac{27}{50}=\\frac{3^{3}}{2\\cdot5^{2}}$, $q=\\frac{g_2}{g_1} = \\frac{5}{3}$, \n$\\frac{g_n}{g_1} = q^{n-1} = \\frac{5^{10}}{3^{10}}$ daraus folgt $n=11$, Resultat: $s_n=g_1 \\cdot \\frac{1-q^n}{1-q} = \\frac{24325489}{109350}$"], ["$s=-\\frac{32}{625}+\\frac{16}{125}-\\frac{8}{25}+\\frac{4}{5}- \\ldots +\\frac{5^{3}}{2^{2}}-\\frac{5^{4}}{2^{3}}+\\frac{5^{5}}{2^{4}}$", "$g_1=-\\frac{32}{625}=-\\frac{2^{5}}{5^{4}}$, $q=\\frac{g_2}{g_1} = -\\frac{5}{2}$, \n$\\frac{g_n}{g_1} = q^{n-1} = -\\frac{5^{9}}{2^{9}}$ daraus folgt $n=10$, Resultat: $s_n=g_1 \\cdot \\frac{1-q^n}{1-q} = \\frac{1394943}{10000}$"], ["$s=\\frac{128}{27}+\\frac{32}{9}+\\frac{8}{3}+2+ \\ldots +\\frac{3^{3}}{2^{5}}+\\frac{3^{4}}{2^{7}}+\\frac{3^{5}}{2^{9}}$", "$g_1=\\frac{128}{27}=\\frac{2^{7}}{3^{3}}$, $q=\\frac{g_2}{g_1} = \\frac{3}{4}$, \n$\\frac{g_n}{g_1} = q^{n-1} = \\frac{3^{8}}{2^{16}}$ daraus folgt $n=9$, Resultat: $s_n=g_1 \\cdot \\frac{1-q^n}{1-q} = \\frac{242461}{13824}$"], ["$s=\\frac{9}{5}+\\frac{6}{5}+\\frac{4}{5}+\\frac{8}{15}+ \\ldots +\\frac{2^{6}}{3^{4}\\cdot5}+\\frac{2^{7}}{3^{5}\\cdot5}+\\frac{2^{8}}{3^{6}\\cdot5}$", "$g_1=\\frac{9}{5}=\\frac{3^{2}}{5}$, $q=\\frac{g_2}{g_1} = \\frac{2}{3}$, \n$\\frac{g_n}{g_1} = q^{n-1} = \\frac{2^{8}}{3^{8}}$ daraus folgt $n=9$, Resultat: $s_n=g_1 \\cdot \\frac{1-q^n}{1-q} = \\frac{19171}{3645}$"], ["$s=-\\frac{125}{12}+\\frac{25}{6}-\\frac{5}{3}+\\frac{2}{3}- \\ldots +\\frac{2^{3}}{3\\cdot5^{2}}-\\frac{2^{4}}{3\\cdot5^{3}}+\\frac{2^{5}}{3\\cdot5^{4}}$", "$g_1=-\\frac{125}{12}=-\\frac{5^{3}}{2^{2}\\cdot3}$, $q=\\frac{g_2}{g_1} = -\\frac{2}{5}$, \n$\\frac{g_n}{g_1} = q^{n-1} = -\\frac{2^{7}}{5^{7}}$ daraus folgt $n=8$, Resultat: $s_n=g_1 \\cdot \\frac{1-q^n}{1-q} = -\\frac{18589}{2500}$"], ["$s=\\frac{27}{125}+\\frac{9}{25}+\\frac{3}{5}+1+ \\ldots +\\frac{5^{6}}{3^{6}}+\\frac{5^{7}}{3^{7}}+\\frac{5^{8}}{3^{8}}$", "$g_1=\\frac{27}{125}=\\frac{3^{3}}{5^{3}}$, $q=\\frac{g_2}{g_1} = \\frac{5}{3}$, \n$\\frac{g_n}{g_1} = q^{n-1} = \\frac{5^{11}}{3^{11}}$ daraus folgt $n=12$, Resultat: $s_n=g_1 \\cdot \\frac{1-q^n}{1-q} = \\frac{121804592}{820125}$"]] );});

=== Freitag 1. Dezember 2017 === //Unterlagen zur {{lehrkraefte:blc:math:unterlagen:trigonometrie-sv.pdf|Trigonometrie}} (insbesondere Definitionen 26,27 und A222)// Machen Sie eine Skizze des Einheitskreises mit dem entsprechenden Winkel $\alpha$ und dem entsprechenden speziellen rechtwinkligen Dreieck, mit dem Sie die Werte von $\sin(\alpha)$, $\cos(\alpha)$ und $\tan(\alpha)$ berechnen: miniAufgabe("#exotrigo","#soltrigo", [["$\\alpha = -330^\\circ$", "$\\sin(-330^\\circ) = \\frac{1}{2}$, $\\cos(-330^\\circ) = \\frac{\\sqrt{3}}{2}$, $\\tan(-330^\\circ) = \\frac{\\sqrt{3}}{3}$"], ["$\\alpha = -300^\\circ$", "$\\sin(-300^\\circ) = \\frac{\\sqrt{3}}{2}$, $\\cos(-300^\\circ) = \\frac{1}{2}$, $\\tan(-300^\\circ) = \\sqrt{3}$"], ["$\\alpha = -240^\\circ$", "$\\sin(-240^\\circ) = \\frac{\\sqrt{3}}{2}$, $\\cos(-240^\\circ) = -\\frac{1}{2}$, $\\tan(-240^\\circ) = -\\sqrt{3}$"], ["$\\alpha = -210^\\circ$", "$\\sin(-210^\\circ) = \\frac{1}{2}$, $\\cos(-210^\\circ) = -\\frac{\\sqrt{3}}{2}$, $\\tan(-210^\\circ) = -\\frac{\\sqrt{3}}{3}$"], ["$\\alpha = -150^\\circ$", "$\\sin(-150^\\circ) = -\\frac{1}{2}$, $\\cos(-150^\\circ) = -\\frac{\\sqrt{3}}{2}$, $\\tan(-150^\\circ) = \\frac{\\sqrt{3}}{3}$"], ["$\\alpha = -120^\\circ$", "$\\sin(-120^\\circ) = -\\frac{\\sqrt{3}}{2}$, $\\cos(-120^\\circ) = -\\frac{1}{2}$, $\\tan(-120^\\circ) = \\sqrt{3}$"], ["$\\alpha = -60^\\circ$", "$\\sin(-60^\\circ) = -\\frac{\\sqrt{3}}{2}$, $\\cos(-60^\\circ) = \\frac{1}{2}$, $\\tan(-60^\\circ) = -\\sqrt{3}$"], ["$\\alpha = -30^\\circ$", "$\\sin(-30^\\circ) = -\\frac{1}{2}$, $\\cos(-30^\\circ) = \\frac{\\sqrt{3}}{2}$, $\\tan(-30^\\circ) = -\\frac{\\sqrt{3}}{3}$"], ["$\\alpha = 30^\\circ$", "$\\sin(30^\\circ) = \\frac{1}{2}$, $\\cos(30^\\circ) = \\frac{\\sqrt{3}}{2}$, $\\tan(30^\\circ) = \\frac{\\sqrt{3}}{3}$"], ["$\\alpha = 60^\\circ$", "$\\sin(60^\\circ) = \\frac{\\sqrt{3}}{2}$, $\\cos(60^\\circ) = \\frac{1}{2}$, $\\tan(60^\\circ) = \\sqrt{3}$"], ["$\\alpha = 120^\\circ$", "$\\sin(120^\\circ) = \\frac{\\sqrt{3}}{2}$, $\\cos(120^\\circ) = -\\frac{1}{2}$, $\\tan(120^\\circ) = -\\sqrt{3}$"], ["$\\alpha = 150^\\circ$", "$\\sin(150^\\circ) = \\frac{1}{2}$, $\\cos(150^\\circ) = -\\frac{\\sqrt{3}}{2}$, $\\tan(150^\\circ) = -\\frac{\\sqrt{3}}{3}$"], ["$\\alpha = 210^\\circ$", "$\\sin(210^\\circ) = -\\frac{1}{2}$, $\\cos(210^\\circ) = -\\frac{\\sqrt{3}}{2}$, $\\tan(210^\\circ) = \\frac{\\sqrt{3}}{3}$"], ["$\\alpha = 240^\\circ$", "$\\sin(240^\\circ) = -\\frac{\\sqrt{3}}{2}$, $\\cos(240^\\circ) = -\\frac{1}{2}$, $\\tan(240^\\circ) = \\sqrt{3}$"], ["$\\alpha = 300^\\circ$", "$\\sin(300^\\circ) = -\\frac{\\sqrt{3}}{2}$, $\\cos(300^\\circ) = \\frac{1}{2}$, $\\tan(300^\\circ) = -\\sqrt{3}$"], ["$\\alpha = 330^\\circ$", "$\\sin(330^\\circ) = -\\frac{1}{2}$, $\\cos(330^\\circ) = \\frac{\\sqrt{3}}{2}$, $\\tan(330^\\circ) = -\\frac{\\sqrt{3}}{3}$"], ["$\\alpha = -315^\\circ$", "$\\sin(-315^\\circ) = \\frac{\\sqrt{2}}{2}$, $\\cos(-315^\\circ) = \\frac{\\sqrt{2}}{2}$, $\\tan(-315^\\circ) = 1$"], ["$\\alpha = -225^\\circ$", "$\\sin(-225^\\circ) = \\frac{\\sqrt{2}}{2}$, $\\cos(-225^\\circ) = -\\frac{\\sqrt{2}}{2}$, $\\tan(-225^\\circ) = -1$"], ["$\\alpha = -135^\\circ$", "$\\sin(-135^\\circ) = -\\frac{\\sqrt{2}}{2}$, $\\cos(-135^\\circ) = -\\frac{\\sqrt{2}}{2}$, $\\tan(-135^\\circ) = 1$"], ["$\\alpha = -45^\\circ$", "$\\sin(-45^\\circ) = -\\frac{\\sqrt{2}}{2}$, $\\cos(-45^\\circ) = \\frac{\\sqrt{2}}{2}$, $\\tan(-45^\\circ) = -1$"], ["$\\alpha = 45^\\circ$", "$\\sin(45^\\circ) = \\frac{\\sqrt{2}}{2}$, $\\cos(45^\\circ) = \\frac{\\sqrt{2}}{2}$, $\\tan(45^\\circ) = 1$"], ["$\\alpha = 135^\\circ$", "$\\sin(135^\\circ) = \\frac{\\sqrt{2}}{2}$, $\\cos(135^\\circ) = -\\frac{\\sqrt{2}}{2}$, $\\tan(135^\\circ) = -1$"], ["$\\alpha = 225^\\circ$", "$\\sin(225^\\circ) = -\\frac{\\sqrt{2}}{2}$, $\\cos(225^\\circ) = -\\frac{\\sqrt{2}}{2}$, $\\tan(225^\\circ) = 1$"], ["$\\alpha = 315^\\circ$", "$\\sin(315^\\circ) = -\\frac{\\sqrt{2}}{2}$, $\\cos(315^\\circ) = \\frac{\\sqrt{2}}{2}$, $\\tan(315^\\circ) = -1$"]] );});