etudiante9c2

2022-04-30

Rearranging a trigonometric expression

$\mathrm{tan}\beta =\frac{\lambda -\mathrm{cos}\theta}{\mathrm{sin}\theta}\Rightarrow \mathrm{tan}\beta =\lambda \cdot \mathrm{csc}\theta -\mathrm{cot}\theta $

Where $\lambda$ is a constant.

Is it possible to simplify the right hand side term into a single trig function? My goal is to have an equation of the form

$\theta =f\left(\beta \right)$

Any help is appreciated.

Brennen Davies

Beginner2022-05-01Added 25 answers

Note that

$\mathrm{sin}\theta \mathrm{sin}\beta +\mathrm{cos}\theta \mathrm{cos}\beta =\lambda \mathrm{cos}\beta$

and the LHS equivalent to$\mathrm{cos}(\theta -\beta )$

and the LHS equivalent to

Pedro Taylor

Beginner2022-05-02Added 19 answers

The answer to your question is that the inverse equation, $\theta =f\left(\beta \right)$ has three values:

$\theta =\mathrm{arccos}(\lambda \cdot \mathrm{cos}\left(\beta \right))$

$\theta =\mathrm{arcsin}(\lambda \cdot \mathrm{cos}\left(\beta \right))+\beta +\frac{3\pi}{2}$

$\theta =\beta -\mathrm{arccos}(\lambda \cdot \mathrm{cos}\left(\beta \right))$

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