# Eliminate \theta \ and \ \phi between the following eq

Eliminate between the following equations:
$\left\{\begin{array}{c}\mathrm{sin}\theta +\mathrm{sin}\varphi =x\\ \mathrm{cos}\theta +\mathrm{cos}\varphi =y\\ \mathrm{tan}\frac{\theta }{2}\mathrm{tan}\frac{\varphi }{2}=z\end{array}$
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Cassandra Ramirez
By sum to product and product to sum formulas we have
$\left\{\begin{array}{c}\mathrm{sin}\theta +\mathrm{sin}\varphi =2\mathrm{sin}\left(\frac{\theta +\varphi }{2}\right)\mathrm{cos}\left(\frac{\theta -\varphi }{2}\right)=x\\ \mathrm{cos}\theta +\mathrm{cos}\varphi =2\mathrm{cos}\left(\frac{\theta +\varphi }{2}\right)\mathrm{cos}\left(\frac{\theta -\varphi }{2}\right)=y\\ \mathrm{tan}\frac{\theta }{2}\mathrm{tan}\frac{\varphi }{2}=\frac{\mathrm{cos}\left(\frac{\theta -\varphi }{2}\right)-\mathrm{cos}\left(\frac{\theta +\varphi }{2}\right)}{\mathrm{cos}\left(\frac{\theta -\varphi }{2}\right)+\mathrm{cos}\left(\frac{\theta +\varphi }{2}\right)}=z\end{array}⇒\left\{\begin{array}{c}2ab=x\\ 2cb=y\\ \frac{b-c}{b+c}=z\end{array}$
and since ${a}^{2}+{c}^{2}=1$ we obtain
$b=±\frac{12}{\sqrt{{x}^{2}+{y}^{2}}}$
$c=±\frac{y}{\sqrt{{x}^{2}+{y}^{2}}}$
and then
$z=\frac{±\frac{12}{\sqrt{{x}^{2}+{y}^{2}}}\mp \frac{y}{\sqrt{{x}^{2}+{y}^{2}}}}{±\frac{12}{\sqrt{{x}^{2}+{y}^{2}}}±\frac{y}{\sqrt{{x}^{2}+{y}^{2}}}}=\frac{{x}^{2}+{y}^{2}-2y}{{x}^{2}+{y}^{2}+2y}$
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Joseph Lewis
Write $u=\mathrm{tan}\frac{\theta }{2},v=\mathrm{tan}\frac{\varphi }{2}$. Then we get
$\left\{\begin{array}{c}2\left(u+v\right)\left(1+uv\right)=x\left(1+{u}^{2}\right)\left(1+{v}^{2}\right)\\ 2-2{u}^{2}{v}^{2}=y\left(1+{u}^{2}\right)\left(1+{v}^{2}\right)\\ uv=z\end{array}$
Now setting s=u+v and using the third relation, we obtain
$\left\{\begin{array}{c}2s\left(1+z\right)=x\left(1-2z+{z}^{2}+{s}^{2}\right)\\ 2\left(1-{z}^{2}\right)=y\left(1-2z+{z}^{2}+{s}^{2}\right)\end{array}$
Dividing, we obtain the relation $s=\frac{x\left(1-z\right)}{y}$, which youve
###### Not exactly what you’re looking for?
karton

Hints
Writing $\theta =2p,\varphi =2q$
We have
$\frac{\mathrm{sin}\left(p+q\right)}{x}=\frac{\mathrm{cos}\left(p+q\right)}{y}=±\frac{1}{\sqrt{{x}^{2}+{y}^{2}}}$
Using this one can find $\mathrm{cos}\left(p-q\right)$
Now use
$\frac{z-1}{z+1}=\dots =\frac{\mathrm{cos}\left(p+q\right)}{\mathrm{cos}\left(p-q\right)}$