SveinR skrev:gamer32 skrev:(I) [tex]Y+Z*\cos\left ( \frac{\pi}{4}\left ( 1982-t_0 \right ) \right )=1.5[/tex]
(II) [tex]Y+Z*\cos\left ( \frac{\pi}{4}\left ( 1984-t_0 \right ) \right )=1[/tex]
(III) [tex]Y+Z*\cos\left ( \frac{\pi}{4}\left ( 1985-t_0 \right ) \right )=0.6464[/tex]
Det kan nok lønne seg å forenkle [tex]\cos{}[/tex]-uttrykkene først, før man gjør noe annet. F.eks. den midterste blir:
[tex]\cos{\left( \frac{\pi}{4}(1984-t_0)\right)} = \cos{\left(\frac{\pi}{4}\cdot 1984\right)}\cdot \cos{\left(\frac{\pi}{4}\cdot t_0\right)} + \sin{\left(\frac{\pi}{4}\cdot 1984\right)}\cdot \sin{\left(\frac{\pi}{4}\cdot t_0\right)}[/tex]
Dette ser kanskje ikke så mye enklere ut, men observer at [tex]\cos{\left(\frac{\pi}{4}\cdot 1984\right)} = \cos{\left(\pi \cdot 496 \right)} = \cos{\left(2\pi \cdot 248 \right)} = 1[/tex] (siden et helt antall [tex]2\pi[/tex] har cosinus-verdi 1).
Tilsvarende blir [tex]\sin{\left(\frac{\pi}{4}\cdot 1984\right)} = \sin{\left(\pi \cdot 496 \right)} = \sin{\left(2\pi \cdot 248 \right)} = 0[/tex]
Dermed ender vi opp med:
[tex]\cos{\left( \frac{\pi}{4}(1984-t_0)\right)} = 1\cdot \cos{\left(\frac{\pi}{4}\cdot t_0\right)} + 0 \cdot \sin{\left(\frac{\pi}{4}\cdot t_0\right)} = \cos{\left(\frac{\pi}{4}\cdot t_0\right)}[/tex]
De andre [tex]\cos{}[/tex]-uttrykkene kan også forenkles på nogenlunde tilsvarende måte.
takk for innspill
jeg ender opp med :
[tex]I \Rightarrow \, Y-Z \sin \left ( \frac{\pi}{4} t_0 \right )=1.5[/tex]
[tex]II \Rightarrow Y+Z \cos\left ( \frac{\pi}{4}t_0 \right )=1[/tex]
[tex]III \Rightarrow \, Y+Z\left ( \frac{\sqrt{2}}{2} \cos\left ( \frac{\pi}{4}t_0 \right )+\frac{\sqrt{2}}{2} \sin\left ( \frac{\pi}{4} t_0 \right ) \right )=0.6464[/tex]
substitusjon
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[tex]Y=1.5+Z \sin\left ( \frac{\pi}{4}t_0 \right )[/tex] fra [tex]I[/tex] i [tex]II[/tex] gir;
[tex]\left (1.5+Z \sin\left ( \frac{\pi}{4}t_0 \right ) \right )+Z \cos \left ( \frac{\pi}{4}t_0 \right )=1[/tex]
[tex]Z \sin\left ( \frac{\pi}{4}t_0 \right )+Z \cos \left ( \frac{\pi}{4}t_0 \right )=-0.5[/tex]
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[tex]\sqrt{2}Z \sin\left ( \frac{\pi}{4}t_0+\tan^{-1}\left ( \frac{Z}{Z} \right ) \right )=-0.5\Rightarrow Z \sin\left ( \frac{\pi}{4}t_0+\frac{\pi}{4} \right )=-\frac{\sqrt{2}}{4}[/tex]
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[tex]I[/tex] i [tex]III[/tex] gir:
[tex]\left (1.5+Z \sin \left ( \frac{\pi}{4}t_0 \right ) \right )+Z\left ( \frac{\sqrt{2}}{2} \cos\left ( \frac{{\pi}}{4} t_0 \right )+\frac{\sqrt{2}}{2} \sin\left ( \frac{\pi}{4} t_0 \right ) \right ) = 0.6464[/tex]
[tex]Z\left ( \sin\left ( \frac{\pi}{4}t_0 \right )+\frac{\sqrt{2}}{2} \cos\left ( \frac{\pi}{4}t_0 \right )+ \frac{\sqrt{2}}{2} \sin \left ( \frac{\pi}{4} t_0 \right )\right )= -0.8536\Rightarrow Z\left (\frac{2+\sqrt{2}}{2} \sin\left ( \frac{\pi}{4}t_0 \right )+\frac{\sqrt{2}}{2} \cos\left ( \frac{\pi}{4}t_0 \right )\right )=-0.8536[/tex]
Har dermed uttrykkene:
[tex]y_1: Z \sin\left ( \frac{\pi}{4}t_0+\frac{\pi}{4} \right )=-\frac{\sqrt{2}}{4}[/tex]
[tex]y_2: Z\left (\frac{2+\sqrt{2}}{2} \sin\left ( \frac{\pi}{4}t_0 \right )+\frac{\sqrt{2}}{2} \cos\left ( \frac{\pi}{4}t_0 \right )\right )=-0.8536[/tex]
fra [tex]y_1[/tex]
[tex]Z=\frac{\frac{-\sqrt{2}}{4}}{\sin\left ( \frac{\pi}{4}t_0+\frac{\pi}{4} \right )}[/tex]
innsatt i [tex]y_2[/tex] gir dette:
[tex]\left ( \frac{\frac{-\sqrt{2}}{4}}{\sin\left ( \frac{\pi}{4}t_0+\frac{\pi}{4} \right )} \right )*\left (\frac{2+\sqrt{2}}{2} \sin\left ( \frac{\pi}{4}t_0 \right )+\frac{\sqrt{2}}{2} \cos\left ( \frac{\pi}{4}t_0 \right )\right )=-0.8536[/tex]
[tex]\frac{\frac{-\sqrt{2}}{4}*\frac{2+\sqrt{2}}{4} \sin\left ( \frac{\pi}{4} t_0 \right ) }{\sin\left ( \frac{\pi}{4} t_0 \right )}+\frac{\frac{-\sqrt{2}}{4}*\frac{\sqrt{2}}{2} \cos\left ( \frac{\pi}{4}t_0 \right )}{\sin\left ( \frac{\pi}{4}t_0+\frac{\pi}{4} \right )}=-0.8536[/tex]
Jeg ser virkelig ikke hvordan dette skal gå videre??