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Marla Payton
2021-12-28
Answered

Eliminate $\theta \text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}\varphi$ between the following equations:

$\{\begin{array}{c}\mathrm{sin}\theta +\mathrm{sin}\varphi =x\\ \mathrm{cos}\theta +\mathrm{cos}\varphi =y\\ \mathrm{tan}\frac{\theta}{2}\mathrm{tan}\frac{\varphi}{2}=z\end{array}$

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Cassandra Ramirez

Answered 2021-12-29
Author has **30** answers

By sum to product and product to sum formulas we have

$\{\begin{array}{c}\mathrm{sin}\theta +\mathrm{sin}\varphi =2\mathrm{sin}\left(\frac{\theta +\varphi}{2}\right)\mathrm{cos}\left(\frac{\theta -\varphi}{2}\right)=x\\ \mathrm{cos}\theta +\mathrm{cos}\varphi =2\mathrm{cos}\left(\frac{\theta +\varphi}{2}\right)\mathrm{cos}\left(\frac{\theta -\varphi}{2}\right)=y\\ \mathrm{tan}\frac{\theta}{2}\mathrm{tan}\frac{\varphi}{2}=\frac{\mathrm{cos}\left(\frac{\theta -\varphi}{2}\right)-\mathrm{cos}\left(\frac{\theta +\varphi}{2}\right)}{\mathrm{cos}\left(\frac{\theta -\varphi}{2}\right)+\mathrm{cos}\left(\frac{\theta +\varphi}{2}\right)}=z\end{array}\Rightarrow \{\begin{array}{c}2ab=x\\ 2cb=y\\ \frac{b-c}{b+c}=z\end{array}$

and since${a}^{2}+{c}^{2}=1$ we obtain

$b=\pm \frac{12}{\sqrt{{x}^{2}+{y}^{2}}}$

$c=\pm \frac{y}{\sqrt{{x}^{2}+{y}^{2}}}$

and then

$z=\frac{\pm \frac{12}{\sqrt{{x}^{2}+{y}^{2}}}\mp \frac{y}{\sqrt{{x}^{2}+{y}^{2}}}}{\pm \frac{12}{\sqrt{{x}^{2}+{y}^{2}}}\pm \frac{y}{\sqrt{{x}^{2}+{y}^{2}}}}=\frac{{x}^{2}+{y}^{2}-2y}{{x}^{2}+{y}^{2}+2y}$

and since

and then

Joseph Lewis

Answered 2021-12-30
Author has **43** answers

Write $u=\mathrm{tan}\frac{\theta}{2},v=\mathrm{tan}\frac{\varphi}{2}$ . Then we get

$\{\begin{array}{c}2(u+v)(1+uv)=x(1+{u}^{2})(1+{v}^{2})\\ 2-2{u}^{2}{v}^{2}=y(1+{u}^{2})(1+{v}^{2})\\ uv=z\end{array}$

Now setting s=u+v and using the third relation, we obtain

$\{\begin{array}{c}2s(1+z)=x(1-2z+{z}^{2}+{s}^{2})\\ 2(1-{z}^{2})=y(1-2z+{z}^{2}+{s}^{2})\end{array}$

Dividing, we obtain the relation$s=\frac{x(1-z)}{y}$ , which youve

Now setting s=u+v and using the third relation, we obtain

Dividing, we obtain the relation

karton

Answered 2022-01-08
Author has **368** answers

Hints

Writing

We have

Using this one can find

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