geg.: $\overrightarrow{O{P}_n}(\phi )=\left(\begin{array}{c}5\cdot \sin \phi \\ 5\cdot \cos \phi \end{array}\right)\backslash overrightarrow\; \{OP\_n\}(\backslash varphi)=\; \backslash begin\{pmatrix\}\; 5\backslash cdot\; \backslash sin\backslash varphi\; \backslash \backslash \; 5\backslash cdot\; \backslash cos\backslash varphi\; \backslash end\{pmatrix\}$; $\overrightarrow{O{Q}_n}(\phi )=\left(\begin{array}{c}-2\cdot {\sin }^{\mathbf{2}}\mathbf{\phi }\\ {\displaystyle \frac{4}{\sin \phi }}\end{array}\right)\backslash overrightarrow\; \{OQ\_n\}(\backslash varphi)=\; \backslash begin\{pmatrix\}\; -2\backslash cdot\; \backslash mathbf\{\backslash sin^2\backslash varphi\}\; \backslash \backslash \; \backslash dfrac\{4\}\{\backslash sin\backslash varphi\}\; \backslash end\{pmatrix\}$; $\phi =80\mathrm{°}\backslash varphi=80°$

ges.: Koordinaten $\overrightarrow{O{P}_1}\backslash overrightarrow\; \{OP\_1\}$ und $\overrightarrow{O{Q}_1}\backslash overrightarrow\; \{OQ\_1\}$ für $\phi =80\mathrm{°}\backslash varphi=80°$; Zeichnung Dreieck $O{P}_1{Q}_{1}OP\_1Q\_1$

Berechnung der Koordinaten für $\overrightarrow{O{P}_1}(\phi =80\mathrm{°})\backslash overrightarrow\; \{OP\_1\}(\backslash varphi=80°)$:

$\overrightarrow{O{P}_1}(\phi =80\mathrm{°})=\left(\begin{array}{c}5\cdot \sin 80\mathrm{°}\\ 5\cdot \cos 80\mathrm{°}\end{array}\right)=\left(\begin{array}{c}5\cdot \mathrm{0,98481}\\ 5\cdot \mathrm{0,17365}\end{array}\right)=\left(\begin{array}{c}\mathbf{4,92}\\ \mathbf{0,87}\end{array}\right)\backslash overrightarrow\; \{OP\_1\}(\backslash varphi=80°)=\; \backslash begin\{pmatrix\}\; 5\backslash cdot\; \{\backslash sin80°\}\; \backslash \backslash \; 5\backslash cdot\; \{\backslash cos\; 80°\}\; \backslash end\{pmatrix\}=\backslash begin\{pmatrix\}\; 5\backslash cdot\; \{0\{,\}98481\}\; \backslash \backslash \; 5\backslash cdot\; \{0\{,\}17365\}\; \backslash end\{pmatrix\}=\backslash mathbf\{\backslash begin\{pmatrix\}\; 4\{,\}92\; \backslash \backslash \; 0\{,\}87\; \backslash end\{pmatrix\}\}$

Berechnung der Koordinaten für $\overrightarrow{O{Q}_1}(\phi =80\mathrm{°})\backslash overrightarrow\; \{OQ\_1\}(\backslash varphi=80°)$

$\Rightarrow \overrightarrow{O{Q}_1}(\phi =80\mathrm{°})=\left(\begin{array}{c}-2\cdot {\sin }^2(80\mathrm{°})\\ {\displaystyle \frac{4}{\sin 80\mathrm{°}}}\end{array}\right)\backslash Rightarrow\; \backslash overrightarrow\; \{OQ\_1\}(\backslash varphi=80°)=\; \backslash begin\{pmatrix\}\; -2\backslash cdot\; \backslash sin^2(80°)\; \backslash \backslash \; \backslash dfrac\{4\}\{\backslash sin\; 80°\}\; \backslash end\{pmatrix\}$

$\phantom{\Rightarrow }=\left(\begin{array}{c}-2\cdot {\mathrm{0,98481}}^2\\ {\displaystyle \frac{4}{\mathrm{0,98481}}}\end{array}\right)=\left(\begin{array}{c}-\mathbf{1,94}\\ \mathbf{4,06}\end{array}\right)\backslash hphantom\{\backslash Rightarrow\}=\backslash begin\{pmatrix\}\; \{-2\backslash cdot\; 0\{,\}98481^2\}\; \backslash \backslash \; \backslash dfrac\{4\}\{0\{,\}98481\}\; \backslash end\{pmatrix\}=\backslash mathbf\; \{\backslash begin\{pmatrix\}\; \{-1\{,\}94\}\; \backslash \backslash \; 4\{,\}06\; \backslash end\{pmatrix\}\}$

Einzeichnen des Dreiecks $O{P}_1{Q}_{1}OP\_1Q\_1$ in das KOSY

geg.: Behauptung : Strecke $[O{P}_n]=const\mathrm{.}[OP\_n]=const.$

ges.: Begründung durch rechnerischen Nachweis

Berechnung der Strecken aus den Koordinaten

Mit dem Satz des Pythagoras gilt:

$\overrightarrow{O{P}_n}(\phi )=\left(\begin{array}{c}5\cdot \sin \phi \\ 5\cdot \cos \phi \end{array}\right)\backslash overrightarrow\; \{OP\_n\}(\backslash varphi)=\; \backslash begin\{pmatrix\}\; 5\backslash cdot\; \backslash sin\backslash varphi\; \backslash \backslash \; 5\backslash cdot\; \backslash cos\backslash varphi\; \backslash end\{pmatrix\}\backslash \backslash$$\overline{\mathbf{O}{\mathbf{P}}_{\mathbf{n}}}(\mathbf{\phi }{)}^{\mathbf{2}}=(\mathbf{5}\cdot \mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\phi }{)}^{\mathbf{2}}+(\mathbf{5}\cdot \mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\phi }{)}^{\mathbf{2}}\backslash mathbf\{\backslash overline\{OP\_n\}(\phi )^2=(5\backslash cdot\; sin\backslash varphi)^2+(5\backslash cdot\; cos\backslash varphi)^2\}$

${\overline{O{P}_n}}^2(\phi )=25\cdot (\sin \phi {)}^2+25\cdot (\cos \phi {)}^2\{\backslash overline\{OP\_n\}^2(\phi )=\{25\backslash cdot\; (\backslash sin\backslash varphi)^2+25\backslash cdot\; (\backslash cos\backslash varphi)^2\}\}$

$\phantom{{\overline{O{P}_n}}^2(\phi )}=25\cdot ({\sin }^2\phi +{\cos }^2\phi )\backslash hphantom\{\{\backslash overline\{OP\_n\}^2(\phi )\}\}=25\; \backslash cdot(\backslash sin^2\; \backslash varphi+\backslash cos^2\; \backslash varphi)$

$\phantom{{\overline{O{P}_n}}^2(\phi )}=25\backslash hphantom\{\{\backslash overline\{OP\_n\}^2(\phi )\}\}=25$ (trigonometrischer Pythagoras)

Daher ist:

$\overline{\mathbf{O}{\mathbf{P}}_{\mathbf{n}}}(\mathbf{\phi })=\sqrt{\mathbf{25}}\mathbf{L}\mathbf{E}=\mathbf{5}\mathbf{L}\mathbf{E}\backslash mathbf\{\; \backslash overline\{OP\_n\}(\phi )=\backslash sqrt\{25\}\backslash \; LE\; =5\backslash \; LE\}$ für alle $\phi \in ]0\mathrm{°};90\mathrm{°}]\backslash varphi\; \backslash in]0°;90°]\backslash qquad$q.e.d.

geg.: Vektoren $\overline{O{P}_1}(80\mathrm{°})=\left(\begin{array}{c}\mathrm{4,92}\\ \mathrm{0,87}\end{array}\right)\backslash overline\{OP\_1\}(80°)=\backslash begin\{pmatrix\}\; 4\{,\}92\; \backslash \backslash \; 0\{,\}87\; \backslash end\{pmatrix\}$ und $\overline{O{Q}_1}(80\mathrm{°})=\left(\begin{array}{c}-\mathrm{1,94}\\ \mathrm{4,06}\end{array}\right)\backslash overline\{OQ\_1\}(80°)=\backslash begin\{pmatrix\}\; -1\{,\}94\; \backslash \backslash \; 4\{,\}06\; \backslash end\{pmatrix\}$ gem. Aufgabe 3a

ges.: $\mathrm{\measuredangle }{P}_{1}O{Q}_1\measuredangle \; P\_1OQ\_1$

Berechnung mit Kosinussatz aufgelöst nach Kosinus $\gamma \gamma$

$c^2=a^2+b^2-2ab\cdot \cos (\gamma )c^2=a^2+b^2-2ab\backslash cdot\; \backslash cos(\gamma )\backslash qquad\backslash qquad$|$-c^2-c^2$

$0=c^2+a^2+b^2-2ab\cdot \cos (\gamma )0=c^2+a^2+b^2-2ab\backslash cdot\; \backslash cos(\gamma )\backslash qquad\backslash :\backslash :$|$+2ab\cdot \cos (\gamma )+\; 2ab\backslash cdot\; \backslash cos(\gamma )$

$2ab\cdot \cos (\gamma )=-c^2+a^2+b^{2}2ab\backslash cdot\; \backslash cos(\gamma )=-c^2+a^2+b^2\backslash qquad\backslash quad$ | $:2ab:\; 2ab$

${\displaystyle \cos (\gamma )=\frac{-c^2+a^2+b^2}{2ab}}\backslash displaystyle\backslash cos(\gamma )=\backslash frac\{-c^2+a^2+b^2\}\{2ab\}$

Berechnung von $a,b,ca,\; b,\; c$ und $cos(\gamma )cos(\gamma )$

$\Rightarrow a:\mathrm{\stackrel{\frown}{=}}\mathrm{\mid }\overline{O{P}_1}\mathrm{\mid }=5\backslash Rightarrow\; a:\; \stackrel{\wedge}{=}\; |\backslash overline\; \{OP\_1\}|=5\backslash :\backslash qquad$für alle $\phi \backslash varphi$ nach Teil b)

$\Rightarrow b:\mathrm{\stackrel{\frown}{=}}\mathrm{\mid }\overline{O{Q}_1}\mathrm{\mid }=\sqrt{-{\mathrm{1,94}}^2+{\mathrm{4,06}}^2}=\sqrt{\mathrm{3,76}+\mathrm{16,48}}=\sqrt{\mathrm{20,24}}=\mathrm{4,50}\backslash Rightarrow\; b:\; \stackrel{\wedge}{=}\; |\backslash overline\; \{OQ\_1\}|=\backslash sqrt\{-1\{,\}94^2+4\{,\}06^2\}=\backslash sqrt\{3\{,\}76+16\{,\}48\}=\backslash sqrt\{20\{,\}24\}=4\{,\}50$

Berechnung Vektor $\overrightarrow{P_1{Q}_1}\backslash overrightarrow\; \{P\_1Q\_1\}$ mit Vektoraddition $\overrightarrow{O{P}_1}\oplus \overrightarrow{P_1{Q}_1}=\overrightarrow{O{Q}_1}\backslash overrightarrow\{OP\_1\}\backslash oplus\backslash overrightarrow\{P\_1Q\_1\}=\backslash overrightarrow\; \{OQ\_1\}$

umgestellt nach $\overrightarrow{P_1{Q}_1}=\overrightarrow{O{Q}_1}-\overrightarrow{O{P}_1}=\left(\begin{array}{c}-\mathrm{1,94}\\ \mathrm{4,06}\end{array}\right)-\left(\begin{array}{c}\mathrm{4,92}\\ \mathrm{0,87}\end{array}\right)=\left(\begin{array}{c}-\mathrm{6,86}\\ \mathrm{3,19}\end{array}\right)\backslash overrightarrow\{P\_1Q\_1\}=\backslash overrightarrow\; \{OQ\_1\}-\backslash overrightarrow\{OP\_1\}=\backslash begin\{pmatrix\}\; -1\{,\}94\; \backslash \backslash \; 4\{,\}06\; \backslash end\{pmatrix\}-\backslash begin\{pmatrix\}\; 4\{,\}92\; \backslash \backslash \; 0\{,\}87\; \backslash end\{pmatrix\}=\backslash begin\{pmatrix\}\; -6\{,\}86\; \backslash \backslash \; 3\{,\}19\; \backslash end\{pmatrix\}$

$c^2=\overrightarrow{\mathbf{\mid }{\mathbf{P}}_{\mathbf{1}}{\mathbf{Q}}_{\mathbf{1}}{\mathbf{\mid }}^{\mathbf{2}}}=(-\mathrm{6,86}{)}^2+{\mathrm{3,22}}^2=\mathrm{47,06}+\mathrm{10,18}=\mathrm{57,24}c^2=\; \backslash mathbf\{\backslash overrightarrow\{|P\_1Q\_1|^2\}\}=\{(-6\{,\}86)^2+3\{,\}22^2\}=\{47\{,\}06+10\{,\}18\}=\{57\{,\}24\}$

Die Werte von $a^{2}a^2$ und $b^{2}b^2$ sind schon oben berechnet worden.

${\displaystyle \Rightarrow \cos (\mathbf{\measuredangle }{\mathbf{P}}_{\mathbf{1}}\mathbf{O}{\mathbf{Q}}_{\mathbf{1}})=\frac{-c^2+a^2+b^2}{2ab}}\backslash displaystyle\backslash Rightarrow\; \backslash mathbf\{\backslash cos(\measuredangle P\_1\; OQ\_1\; )\}=\backslash frac\{-c^2+a^2+b^2\}\{2ab\}$

${\displaystyle =\frac{-\mathrm{57,46}+25+\mathrm{20,24}}{2\cdot 5\cdot \mathrm{4,50}}=\frac{-\mathrm{12,22}}{45}=-\mathbf{0,27}}\backslash displaystyle=\backslash frac\{-57\{,\}46+25+20\{,\}24\}\{2\backslash cdot5\backslash cdot4\{,\}50\}\; =\backslash frac\{-12\{,\}22\}\{45\}=\backslash mathbf\{-0\{,\}27\}$

Falls $\cos (\mathrm{\measuredangle }{P}_{1}O{Q}_1)\backslash cos(\measuredangle P\_1OQ\_1\; )$ negativ, so ist $\mathrm{\measuredangle }{P}_{1}O{Q}_1\measuredangle P\_1OQ\_1$ zwischen $90\mathrm{°}90°$ und $180\mathrm{°}180°$ groß.

Kosinuswert $\mathbf{\measuredangle }{\mathbf{P}}_{\mathbf{1}}\mathbf{O}{\mathbf{Q}}_{\mathbf{1}}=-\mathbf{0,27}\Rightarrow \mathbf{\measuredangle }{\mathbf{P}}_{\mathbf{1}}\mathbf{O}{\mathbf{Q}}_{\mathbf{1}}={\mathbf{105,76}}^{\circ }\backslash mathbf\{\measuredangle P\_1OQ\_1=-0\{,\}27\; \backslash qquad\; \backslash Rightarrow\; \measuredangle P\_1\; OQ\_1=105\{,\}76^\backslash circ\}$