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Question Number 30000 by ajfour last updated on 14/Feb/18

If cos α = sin β sin φ=sin γ cos ψ cos β = sin γ sin ψ =sin α cos θ cos γ = sin α sin θ =sin β cos φ then find cos α, cos β , cos γ briefly and if possible linearly in terms of only sin θ, cos θ, sin φ, cos φ, sin ψ, cos ψ .

$${If}\:\mathrm{cos}\:\alpha\:=\:\mathrm{sin}\:\beta\:\mathrm{sin}\:\phi=\mathrm{sin}\:\gamma\:\mathrm{cos}\:\psi \\ $$$$\:\:\:\:\:\:\mathrm{cos}\:\beta\:=\:\mathrm{sin}\:\gamma\:\mathrm{sin}\:\psi\:=\mathrm{sin}\:\alpha\:\mathrm{cos}\:\theta \\ $$$$\:\:\:\:\:\:\mathrm{cos}\:\gamma\:=\:\mathrm{sin}\:\alpha\:\mathrm{sin}\:\theta\:=\mathrm{sin}\:\beta\:\mathrm{cos}\:\phi \\ $$$${then}\:{find}\:\:\mathrm{cos}\:\alpha,\:\mathrm{cos}\:\beta\:,\:\mathrm{cos}\:\gamma\:\:\: \\ $$$${briefly}\:{and}\:{if}\:{possible}\:{linearly} \\ $$$${in}\:{terms}\:{of}\:{only}\:\mathrm{sin}\:\theta,\:\mathrm{cos}\:\theta, \\ $$$$\mathrm{sin}\:\phi,\:\mathrm{cos}\:\phi,\:\mathrm{sin}\:\psi,\:\mathrm{cos}\:\psi\:. \\ $$

Commented by ajfour last updated on 15/Feb/18

entangled ! please help ..

$$\:{entangled}\:!\:{please}\:{help}\:.. \\ $$

Answered by mrW2 last updated on 16/Feb/18

cos α = sin β sin φ=sin γ cos ψ=a cos β = sin γ sin ψ =sin α cos θ=b cos γ = sin α sin θ =sin β cos φ=c tan ψ=(b/a) tan θ=(c/b) tan φ=(a/c) sin^2 β (sin^2 φ+cos^2 φ)=a^2 +c^2 =1−b^2 ⇒a^2 +b^2 +c^2 =1 tan^2 ψ=(b^2 /a^2 ) tan^2 φ=(a^2 /c^2 )⇒(1/(tan^2 φ))=(c^2 /a^2 ) ⇒((b^2 +c^2 )/a^2 )=tan^2 ψ+(1/(tan^2 φ))=((tan^2 φ+tan^2 ψ)/(tan^2 φ)) ⇒((1−a^2 )/a^2 )=((tan^2 φ+tan^2 ψ)/(tan^2 φ)) ⇒(1/a^2 )−1=((tan^2 φ+tan^2 ψ)/(tan^2 φ)) ⇒(1/a^2 )=((2tan^2 φ+tan^2 ψ)/(tan^2 φ)) ⇒a^2 =((tan^2 φ)/(2tan^2 φ+tan^2 ψ))=(((sin^2 φ)/(cos^2 φ))/(((2sin^2 φ)/(cos^2 φ))+((sin^2 ψ)/(cos^2 ψ)))) ⇒a^2 =((sin^2 φ cos^2 ψ)/(2sin^2 φcos^2 ψ+cos^2 φsin^2 ψ)) ⇒a=cos α=±((sin φ cos ψ)/(√(2sin^2 φcos^2 ψ+cos^2 φsin^2 ψ))) or cos α=±(1/(√(2+(((sin ψ cos φ)/(sin φ cos ψ)))^2 ))) cos β, cos γ similarly.

$$\mathrm{cos}\:\alpha\:=\:\mathrm{sin}\:\beta\:\mathrm{sin}\:\phi=\mathrm{sin}\:\gamma\:\mathrm{cos}\:\psi={a} \\ $$$$\mathrm{cos}\:\beta\:=\:\mathrm{sin}\:\gamma\:\mathrm{sin}\:\psi\:=\mathrm{sin}\:\alpha\:\mathrm{cos}\:\theta={b} \\ $$$$\mathrm{cos}\:\gamma\:=\:\mathrm{sin}\:\alpha\:\mathrm{sin}\:\theta\:=\mathrm{sin}\:\beta\:\mathrm{cos}\:\phi={c} \\ $$$$ \\ $$$$\mathrm{tan}\:\psi=\frac{{b}}{{a}} \\ $$$$\mathrm{tan}\:\theta=\frac{{c}}{{b}} \\ $$$$\mathrm{tan}\:\phi=\frac{{a}}{{c}} \\ $$$$ \\ $$$$\mathrm{sin}^{\mathrm{2}} \:\beta\:\left(\mathrm{sin}^{\mathrm{2}} \:\phi+\mathrm{cos}^{\mathrm{2}} \:\phi\right)={a}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{1}−{b}^{\mathrm{2}} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{1} \\ $$$$ \\ $$$$\mathrm{tan}^{\mathrm{2}} \:\psi=\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} } \\ $$$$\mathrm{tan}^{\mathrm{2}} \:\phi=\frac{{a}^{\mathrm{2}} }{{c}^{\mathrm{2}} }\Rightarrow\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\phi}=\frac{{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} } \\ $$$$ \\ $$$$\Rightarrow\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\mathrm{tan}^{\mathrm{2}} \:\psi+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\phi}=\frac{\mathrm{tan}^{\mathrm{2}} \:\phi+\mathrm{tan}^{\mathrm{2}} \:\psi}{\mathrm{tan}^{\mathrm{2}} \:\phi} \\ $$$$\Rightarrow\frac{\mathrm{1}−{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\frac{\mathrm{tan}^{\mathrm{2}} \:\phi+\mathrm{tan}^{\mathrm{2}} \:\psi}{\mathrm{tan}^{\mathrm{2}} \:\phi} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{a}^{\mathrm{2}} }−\mathrm{1}=\frac{\mathrm{tan}^{\mathrm{2}} \:\phi+\mathrm{tan}^{\mathrm{2}} \:\psi}{\mathrm{tan}^{\mathrm{2}} \:\phi} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{a}^{\mathrm{2}} }=\frac{\mathrm{2tan}^{\mathrm{2}} \:\phi+\mathrm{tan}^{\mathrm{2}} \:\psi}{\mathrm{tan}^{\mathrm{2}} \:\phi} \\ $$$$\Rightarrow{a}^{\mathrm{2}} =\frac{\mathrm{tan}^{\mathrm{2}} \:\phi}{\mathrm{2tan}^{\mathrm{2}} \:\phi+\mathrm{tan}^{\mathrm{2}} \:\psi}=\frac{\frac{\mathrm{sin}^{\mathrm{2}} \:\phi}{\mathrm{cos}^{\mathrm{2}} \:\phi}}{\frac{\mathrm{2sin}^{\mathrm{2}} \:\phi}{\mathrm{cos}^{\mathrm{2}} \:\phi}+\frac{\mathrm{sin}^{\mathrm{2}} \:\psi}{\mathrm{cos}^{\mathrm{2}} \:\psi}} \\ $$$$\Rightarrow{a}^{\mathrm{2}} =\frac{\mathrm{sin}^{\mathrm{2}} \:\phi\:\mathrm{cos}^{\mathrm{2}} \:\psi}{\mathrm{2sin}^{\mathrm{2}} \:\phi\mathrm{cos}^{\mathrm{2}} \:\psi+\mathrm{cos}^{\mathrm{2}} \:\phi\mathrm{sin}^{\mathrm{2}} \:\psi} \\ $$$$\Rightarrow{a}=\mathrm{cos}\:\alpha=\pm\frac{\mathrm{sin}\:\phi\:\mathrm{cos}\:\psi}{\sqrt{\mathrm{2sin}^{\mathrm{2}} \:\phi\mathrm{cos}^{\mathrm{2}} \:\psi+\mathrm{cos}^{\mathrm{2}} \:\phi\mathrm{sin}^{\mathrm{2}} \:\psi}} \\ $$$${or}\:\mathrm{cos}\:\alpha=\pm\frac{\mathrm{1}}{\sqrt{\mathrm{2}+\left(\frac{\mathrm{sin}\:\psi\:\mathrm{cos}\:\phi}{\mathrm{sin}\:\phi\:\mathrm{cos}\:\psi}\right)^{\mathrm{2}} }} \\ $$$$\mathrm{cos}\:\beta,\:\mathrm{cos}\:\gamma\:{similarly}. \\ $$

Commented by ajfour last updated on 16/Feb/18

Thanks Sir, but cannot a, b, c be uniquely obtained, i merely have a notion that it can be so.

$${Thanks}\:{Sir},\:{but}\:{cannot}\:\:{a},\:{b},\:{c}\:\:{be} \\ $$$${uniquely}\:{obtained},\:{i}\:{merely} \\ $$$${have}\:{a}\:{notion}\:{that}\:{it}\:{can}\:{be}\:{so}. \\ $$

Commented by mrW2 last updated on 16/Feb/18

do you mean that a,b,c have fixed values, independently from θ,φ,ψ ?

$${do}\:{you}\:{mean}\:{that}\:{a},{b},{c}\:{have}\:{fixed} \\ $$$${values},\:{independently}\:{from}\:\theta,\phi,\psi\:? \\ $$

Commented by ajfour last updated on 16/Feb/18

no , i mean no ±sign .

$${no}\:,\:{i}\:{mean}\:{no}\:\pm{sign}\:. \\ $$

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