Rearranging a trigonometric expression \(\displaystyle{\tan{\beta}}={\frac{{\lambda-{\cos{\theta}}}}{{{\sin{\theta}}}}}\Leftrightarrow{\tan{\beta}}=\lambda\cdot{\csc{\theta}}-{\cot{\theta}}\) Where \(\displaystyle\lambda\) is

David Rodgers

David Rodgers

Answered question

2022-03-30

Rearranging a trigonometric expression
tanβ=λcosθsinθtanβ=λcscθcotθ
Where λ 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
θ=f(β)
Any help is appreciated.

Answer & Explanation

Alexzander Evans

Alexzander Evans

Beginner2022-03-31Added 9 answers

Note that
sinθsinβ+cosθcosβ=λcosβ
and the LHS equivalent to cos(θβ)
SofZookywookeoybd

SofZookywookeoybd

Beginner2022-04-01Added 15 answers

The answer to your question is that the inverse equation, θ=f(β) has three values:
θ=arccos(λcos(β))
θ=arcsin(λcos(β))+β+3π2
θ=βarccos(λcos(β))

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