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Question Number 211075 by ajfour last updated on 27/Aug/24
sin (θ+φ)=θ  sin θ=φ  find θ and φ.
$$\mathrm{sin}\:\left(\theta+\phi\right)=\theta \\ $$$$\mathrm{sin}\:\theta=\phi \\ $$$${find}\:\theta\:{and}\:\phi. \\ $$
Answered by Ghisom last updated on 27/Aug/24
we can only approximate  φ=sin θ  sin (θ+sin θ) =θ  1. θ=φ=0  2. θ≈±.973696169714∧φ≈±.826969516935
$$\mathrm{we}\:\mathrm{can}\:\mathrm{only}\:\mathrm{approximate} \\ $$$$\phi=\mathrm{sin}\:\theta \\ $$$$\mathrm{sin}\:\left(\theta+\mathrm{sin}\:\theta\right)\:=\theta \\ $$$$\mathrm{1}.\:\theta=\phi=\mathrm{0} \\ $$$$\mathrm{2}.\:\theta\approx\pm.\mathrm{973696169714}\wedge\phi\approx\pm.\mathrm{826969516935} \\ $$
Answered by ajfour last updated on 27/Aug/24
cos (θ+φ)=(√(1−θ^2 ))  tan (θ+φ)=(θ/( (√(1−θ^2 ))))  ((tan θ+tan φ)/(1−tan θtan φ))=(θ/( (√(1−θ^2 ))))  tan φ=((θ−(√(1−θ^2 ))tan θ)/( (√(1−θ^2 ))+θtan θ))  ...yes so b it
$$\mathrm{cos}\:\left(\theta+\phi\right)=\sqrt{\mathrm{1}−\theta^{\mathrm{2}} } \\ $$$$\mathrm{tan}\:\left(\theta+\phi\right)=\frac{\theta}{\:\sqrt{\mathrm{1}−\theta^{\mathrm{2}} }} \\ $$$$\frac{\mathrm{tan}\:\theta+\mathrm{tan}\:\phi}{\mathrm{1}−\mathrm{tan}\:\theta\mathrm{tan}\:\phi}=\frac{\theta}{\:\sqrt{\mathrm{1}−\theta^{\mathrm{2}} }} \\ $$$$\mathrm{tan}\:\phi=\frac{\theta−\sqrt{\mathrm{1}−\theta^{\mathrm{2}} }\mathrm{tan}\:\theta}{\:\sqrt{\mathrm{1}−\theta^{\mathrm{2}} }+\theta\mathrm{tan}\:\theta} \\ $$$$…{yes}\:{so}\:{b}\:{it} \\ $$

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