Savanah Boone

2022-07-10

Solving a set of 3 Nonlinear Equations

In the following 3 equations:

${k}_{1}{\mathrm{cos}}^{2}(\theta )+{k}_{2}{\mathrm{sin}}^{2}(\theta )={c}_{1}$

$2({k}_{2}-{k}_{1})\mathrm{cos}(\theta )\mathrm{sin}(\theta )={c}_{2}$

${k}_{1}{\mathrm{sin}}^{2}(\theta )+{k}_{2}{\mathrm{cos}}^{2}(\theta )={c}_{3}$

${c}_{1}$, ${c}_{2}$ and ${c}_{3}$ are given, and ${k}_{1}$, ${k}_{2}$ and $\theta $ are the unknowns. What is the best way to solve for the unknowns?

In the following 3 equations:

${k}_{1}{\mathrm{cos}}^{2}(\theta )+{k}_{2}{\mathrm{sin}}^{2}(\theta )={c}_{1}$

$2({k}_{2}-{k}_{1})\mathrm{cos}(\theta )\mathrm{sin}(\theta )={c}_{2}$

${k}_{1}{\mathrm{sin}}^{2}(\theta )+{k}_{2}{\mathrm{cos}}^{2}(\theta )={c}_{3}$

${c}_{1}$, ${c}_{2}$ and ${c}_{3}$ are given, and ${k}_{1}$, ${k}_{2}$ and $\theta $ are the unknowns. What is the best way to solve for the unknowns?

isscacabby17

Beginner2022-07-11Added 13 answers

Re-write in terms of double-angle expressions, and define values $A$, , $C$:

$\begin{array}{rlr}{k}_{1}{\mathrm{cos}}^{2}\theta +{k}_{2}{\mathrm{sin}}^{2}\theta ={c}_{1}& \phantom{\rule{1em}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{1em}{0ex}}\mathrm{cos}2\theta =\frac{2{c}_{1}-{k}_{1}-{k}_{2}}{{k}_{1}-{k}_{2}}& =:A\\ 2({k}_{1}-{k}_{2})\mathrm{sin}\theta \mathrm{cos}\theta ={c}_{2}& \phantom{\rule{1em}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{1em}{0ex}}\mathrm{sin}2\theta =\frac{{c}_{2}}{{k}_{1}-{k}_{2}}& =:B\\ {k}_{1}{\mathrm{sin}}^{2}\theta +{k}_{2}{\mathrm{cos}}^{2}\theta ={c}_{3}& \phantom{\rule{1em}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{1em}{0ex}}\mathrm{cos}2\theta =\frac{{k}_{1}+{k}_{2}-2{c}_{3}}{{k}_{1}-{k}_{2}}& =:C\end{array}$

Now, $A=C$ implies ${k}_{1}+{k}_{2}={c}_{1}+{c}_{3}$ (but we'd know that simply by adding your first and third equations together), so that ${k}_{2}={c}_{1}+{c}_{3}-{k}_{1}$ and

$A=\frac{{c}_{1}-{c}_{3}}{2{k}_{1}-{c}_{1}-{c}_{3}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}B=\frac{{c}_{2}}{2{k}_{1}-{c}_{1}-{c}_{3}}$

Observe that you can already write

$\mathrm{tan}2\theta =\frac{{c}_{2}}{{c}_{1}-{c}_{3}}$

which gives you $\theta $. For ${k}_{1}$ and ${k}_{2}$, note that ${A}^{2}+{B}^{2}=1$ implies

$({c}_{1}-{c}_{3}{)}^{2}+{c}_{2}^{2}={(2{k}_{1}-{c}_{1}-{c}_{3})}^{2}$

so that

$\begin{array}{rl}{k}_{1}& =\frac{1}{2}({c}_{1}+{c}_{3}\pm \sqrt{({c}_{1}-{c}_{3}{)}^{2}+{c}_{2}^{2}})\\ {k}_{2}& =\frac{1}{2}({c}_{1}+{c}_{3}\mp \sqrt{({c}_{1}-{c}_{3}{)}^{2}+{c}_{2}^{2}})\end{array}$

$\begin{array}{rlr}{k}_{1}{\mathrm{cos}}^{2}\theta +{k}_{2}{\mathrm{sin}}^{2}\theta ={c}_{1}& \phantom{\rule{1em}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{1em}{0ex}}\mathrm{cos}2\theta =\frac{2{c}_{1}-{k}_{1}-{k}_{2}}{{k}_{1}-{k}_{2}}& =:A\\ 2({k}_{1}-{k}_{2})\mathrm{sin}\theta \mathrm{cos}\theta ={c}_{2}& \phantom{\rule{1em}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{1em}{0ex}}\mathrm{sin}2\theta =\frac{{c}_{2}}{{k}_{1}-{k}_{2}}& =:B\\ {k}_{1}{\mathrm{sin}}^{2}\theta +{k}_{2}{\mathrm{cos}}^{2}\theta ={c}_{3}& \phantom{\rule{1em}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{1em}{0ex}}\mathrm{cos}2\theta =\frac{{k}_{1}+{k}_{2}-2{c}_{3}}{{k}_{1}-{k}_{2}}& =:C\end{array}$

Now, $A=C$ implies ${k}_{1}+{k}_{2}={c}_{1}+{c}_{3}$ (but we'd know that simply by adding your first and third equations together), so that ${k}_{2}={c}_{1}+{c}_{3}-{k}_{1}$ and

$A=\frac{{c}_{1}-{c}_{3}}{2{k}_{1}-{c}_{1}-{c}_{3}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}B=\frac{{c}_{2}}{2{k}_{1}-{c}_{1}-{c}_{3}}$

Observe that you can already write

$\mathrm{tan}2\theta =\frac{{c}_{2}}{{c}_{1}-{c}_{3}}$

which gives you $\theta $. For ${k}_{1}$ and ${k}_{2}$, note that ${A}^{2}+{B}^{2}=1$ implies

$({c}_{1}-{c}_{3}{)}^{2}+{c}_{2}^{2}={(2{k}_{1}-{c}_{1}-{c}_{3})}^{2}$

so that

$\begin{array}{rl}{k}_{1}& =\frac{1}{2}({c}_{1}+{c}_{3}\pm \sqrt{({c}_{1}-{c}_{3}{)}^{2}+{c}_{2}^{2}})\\ {k}_{2}& =\frac{1}{2}({c}_{1}+{c}_{3}\mp \sqrt{({c}_{1}-{c}_{3}{)}^{2}+{c}_{2}^{2}})\end{array}$

Ciara Mcdaniel

Beginner2022-07-12Added 4 answers

Hint:

${k}_{1}{\mathrm{cos}}^{2}(\theta )+{k}_{2}{\mathrm{sin}}^{2}(\theta )={c}_{1}$

${k}_{1}{\mathrm{sin}}^{2}(\theta )+{k}_{2}{\mathrm{cos}}^{2}(\theta )={c}_{3}$

${c}_{1}+{c}_{3}={k}_{1}+{k}_{2}$

$2({k}_{2}-{k}_{1})\mathrm{cos}(\theta )\mathrm{sin}(\theta )={c}_{2}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}({k}_{2}-{k}_{1})\mathrm{sin}2\theta ={c}_{2}$

$({k}_{2}-{k}_{1})\mathrm{sin}2\theta +{k}_{1}+{k}_{2}={c}_{1}+{c}_{2}+{c}_{3}$

${k}_{2}(\mathrm{sin}2\theta +1)+{k}_{1}(1-\mathrm{sin}2\theta )={c}_{1}+{c}_{2}+{c}_{3}$

${k}_{1}{\mathrm{cos}}^{2}(\theta )+{k}_{2}{\mathrm{sin}}^{2}(\theta )={c}_{1}$

${k}_{1}{\mathrm{sin}}^{2}(\theta )+{k}_{2}{\mathrm{cos}}^{2}(\theta )={c}_{3}$

${c}_{1}+{c}_{3}={k}_{1}+{k}_{2}$

$2({k}_{2}-{k}_{1})\mathrm{cos}(\theta )\mathrm{sin}(\theta )={c}_{2}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}({k}_{2}-{k}_{1})\mathrm{sin}2\theta ={c}_{2}$

$({k}_{2}-{k}_{1})\mathrm{sin}2\theta +{k}_{1}+{k}_{2}={c}_{1}+{c}_{2}+{c}_{3}$

${k}_{2}(\mathrm{sin}2\theta +1)+{k}_{1}(1-\mathrm{sin}2\theta )={c}_{1}+{c}_{2}+{c}_{3}$

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