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

$$If\cos \alpha =\sin \beta \sin \varphi =\sin \gamma \cos \psi$$$$\cos \beta =\sin \gamma \sin \psi =\sin \alpha \cos \theta$$$$\cos \gamma =\sin \alpha \sin \theta =\sin \beta \cos \varphi$$$$thenfind\cos \alpha ,\cos \beta ,\cos \gamma$$$$brieflyandifpossiblelinearly$$$$intermsofonly\sin \theta ,\cos \theta ,$$$$\sin \varphi ,\cos \varphi ,\sin \psi ,\cos \psi .$$

Commented by ajfour last updated on 15/Feb/18

$$entangled!pleasehelp..$$

Answered by mrW2 last updated on 16/Feb/18

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

Commented by ajfour last updated on 16/Feb/18

$$ThanksSir,butcannota,b,cbe$$$$uniquelyobtained,imerely$$$$haveanotionthatitcanbeso.$$

Commented by mrW2 last updated on 16/Feb/18

$$doyoumeanthata,b,chavefixed$$$$values,independentlyfrom\theta ,\varphi ,\psi ?$$

Commented by ajfour last updated on 16/Feb/18

$$no,imeanno\pm sign.$$

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