Question Number 61461 by aliesam last updated on 02/Jun/19
Answered by MJS last updated on 03/Jun/19
![sin x +sin y =a cos x +cos y =b u=tan (x/2) ∧ v=tan (y/2) [sin 2arctan t =((2t)/(1+t^2 )); cos 2arctan t =((1−t^2 )/(1+t^2 ))] ((2(u+v)(uv+1))/((u^2 +1)(v^2 +1)))=a −(((uv−1)(uv+1))/((u^2 +1)(v^2 +1)))=b ⇒ u=((2a±(√((a^2 +b^2 )(4−a^2 −b^2 ))))/(a^2 +b^2 +2b)); v=((2a∓(√((a^2 +b^2 )(4−a^2 −b^2 ))))/(a^2 +b^2 +2b)) ⇒ tan (x/2) +tan (y/2) =u+v=((4a)/(a^2 +b^2 +2b)) in our case tan (x/2) +tan (y/2) =((8(√3)(7(√5)−10(√2)))/9)≈2.325321](Q61471.png)
$$\mathrm{sin}\:{x}\:+\mathrm{sin}\:{y}\:={a} \\ $$$$\mathrm{cos}\:{x}\:+\mathrm{cos}\:{y}\:={b} \\ $$$${u}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:\wedge\:{v}=\mathrm{tan}\:\frac{{y}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\left[\mathrm{sin}\:\mathrm{2arctan}\:{t}\:=\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} };\:\mathrm{cos}\:\mathrm{2arctan}\:{t}\:=\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\right] \\ $$$$\frac{\mathrm{2}\left({u}+{v}\right)\left({uv}+\mathrm{1}\right)}{\left({u}^{\mathrm{2}} +\mathrm{1}\right)\left({v}^{\mathrm{2}} +\mathrm{1}\right)}={a} \\ $$$$−\frac{\left({uv}−\mathrm{1}\right)\left({uv}+\mathrm{1}\right)}{\left({u}^{\mathrm{2}} +\mathrm{1}\right)\left({v}^{\mathrm{2}} +\mathrm{1}\right)}={b} \\ $$$$\Rightarrow\:{u}=\frac{\mathrm{2}{a}\pm\sqrt{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left(\mathrm{4}−{a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{b}};\:{v}=\frac{\mathrm{2}{a}\mp\sqrt{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left(\mathrm{4}−{a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{b}} \\ $$$$\Rightarrow\:\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:+\mathrm{tan}\:\frac{{y}}{\mathrm{2}}\:={u}+{v}=\frac{\mathrm{4}{a}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{b}} \\ $$$$\mathrm{in}\:\mathrm{our}\:\mathrm{case} \\ $$$$\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:+\mathrm{tan}\:\frac{{y}}{\mathrm{2}}\:=\frac{\mathrm{8}\sqrt{\mathrm{3}}\left(\mathrm{7}\sqrt{\mathrm{5}}−\mathrm{10}\sqrt{\mathrm{2}}\right)}{\mathrm{9}}\approx\mathrm{2}.\mathrm{325321} \\ $$
Commented by aliesam last updated on 03/Jun/19
$${thank}\:{you}\:{sir} \\ $$