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Question Number 48396 by behi83417@gmail.com last updated on 23/Nov/18

Commented by MJS last updated on 23/Nov/18

$\frac{2 a \pm \sqrt{4 (a^{2} + b^{2}) - {(a^{2} + b^{2})}^{2}}}{a^{2} + b^{2} + 2 b}$ $I put x = 2 \arctan u \land y = 2 \arctan v . {it}^{'} s possible$ $to solve the equations, but I have got no time$ $to post now$
Commented by behi83417@gmail.com last updated on 23/Nov/18

$t h a n k s s i r . w a i t i n g f o r . . . . . . .$
	Answered by ajfour last updated on 23/Nov/18

	$s_{1}^{2} + s_{2}^{2} + 2 s_{1} s_{2} = a^{2}$ $c_{1}^{2} + c_{2}^{2} + 2 c_{1} c_{2} = b^{2}$ $\Rightarrow 2 + 2 \cos (x - y) = a^{2} + b^{2}$ $\Rightarrow 4 \cos^{2} (\frac{x - y}{2}) = a^{2} + b^{2}$ $\Rightarrow 1 + \tan^{2} (\frac{x - y}{2}) = \frac{4}{a^{2} + b^{2}}$ $\Rightarrow x - y = {2 \tan}^{- 1} \sqrt{\frac{4}{a^{2} + b^{2}} - 1}$ $. . . . . (i)$ $A l s o$ $2 \sin (\frac{x + y}{2}) \cos (\frac{x - y}{2}) = a$ $2 \cos (\frac{x + y}{2}) \cos (\frac{x - y}{2}) = b$ $D i v i d i n g$ $\tan \frac{x + y}{2} = \frac{a}{b}$ $\Rightarrow x + y = {2 \tan}^{- 1} \frac{a}{b} . . . . (i i)$ $F r o m (i) & (i i)$ $\tan \frac{x}{2} = \tan [\frac{1}{2} (\tan^{- 1} \frac{a}{b} + \tan^{- 1} \sqrt{\frac{4}{a^{2} + b^{2}} - 1})]$ $\tan \frac{y}{2} = \tan [\frac{1}{2} (\tan^{- 1} \frac{a}{b} - \tan^{- 1} \sqrt{\frac{4}{a^{2} + b^{2}} - 1})]$ $_________________________.$
	Commented by behi83417@gmail.com last updated on 23/Nov/18
	$[tg((x+y)/2)=(a/b),4cos^2 ((x−y)/2)=a^2 +b^2 ]⇒ { ((cos(( x−y)/2)=((√(a^2 +b^2 ))/2))),((cos(( x+y)/2)=(b/(√(a^2 +b^2 ))))) :} ⇒ { ((2cos(x/2)cos(y/2)=((√(a^2 +b^2 ))/2)+(b/(√(a^2 +b^2 )))=((a^2 +b^2 +2b)/(2(√(a^2 +b^2 )))))),((2sin(x/2)sin(y/2)=((√(a^2 +b^2 ))/b)−(b/(√(a^2 +b^2 )))=((a^2 +b^2 −2b)/(2(√(a^2 +b^2 )))))) :} so: tg(( x)/2).tg(( y)/2)=((a^2 +b^2 −2b)/(a^2 +b^2 +2b)). and:tg(x/2)+tg(y/2)=tg((x+y)/2)(1−tg(x/2)tg(y/2))= =(a/b)(1−((a^2 +b^2 −2b)/(a^2 +b^2 +2b)))=((4a)/(a^2 +b^2 +2b)) i.e:tg(x/2) and tg(y/2),are the roots of: t^2 −((4a)/(a^2 +b^2 +2b))t+((a^2 +b^2 −2b)/(a^2 +b^2 +2b))=0 ⇒(a^2 +b^2 +2b)t−4at+(a^2 +b^2 −2b)=0 t=((2a±(√(4a^2 −[(a^2 +b^2 +2b)(a^2 +b^2 −2b))))/(a^2 +b^2 +2b))= =((2a±(√(4a^2 −[(a^2 +b^2 )^2 −4b^2 ])))/(a^2 +b^2 +2b))= =((2a±(√(4(a^2 +b^2 )−(a^2 +b^2 )^2 )))/(a^2 +b^2 +2b)) .$
	$[t g \frac{x + y}{2} = \frac{a}{b}, 4 {c o s}^{2} \frac{x - y}{2} = a^{2} + b^{2}] \Rightarrow$ ${\begin{cases} c o s \frac{x - y}{2} = \frac{\sqrt{a^{2} + b^{2}}}{2} \\ c o s \frac{x + y}{2} = \frac{b}{\sqrt{a^{2} + b^{2}}} \end{cases}$ $\Rightarrow {\begin{cases} 2 c o s \frac{x}{2} c o s \frac{y}{2} = \frac{\sqrt{a^{2} + b^{2}}}{2} + \frac{b}{\sqrt{a^{2} + b^{2}}} = \frac{a^{2} + b^{2} + 2 b}{2 \sqrt{a^{2} + b^{2}}} \\ 2 s i n \frac{x}{2} s i n \frac{y}{2} = \frac{\sqrt{a^{2} + b^{2}}}{b} - \frac{b}{\sqrt{a^{2} + b^{2}}} = \frac{a^{2} + b^{2} - 2 b}{2 \sqrt{a^{2} + b^{2}}} \end{cases}$ $s o : tg \frac{x}{2} . tg \frac{y}{2} = \frac{a^{2} + b^{2} - 2 b}{a^{2} + b^{2} + 2 b} .$ $a n d : tg \frac{x}{2} + tg \frac{y}{2} = tg \frac{x + y}{2} (1 - tg \frac{x}{2} tg \frac{y}{2}) =$ $= \frac{a}{b} (1 - \frac{a^{2} + b^{2} - 2 b}{a^{2} + b^{2} + 2 b}) = \frac{4 a}{a^{2} + b^{2} + 2 b}$ $i . e : t g \frac{x}{2} a n d t g \frac{y}{2}, a r e t h e r o o t s o f :$ $t^{2} - \frac{4 a}{a^{2} + b^{2} + 2 b} t + \frac{a^{2} + b^{2} - 2 b}{a^{2} + b^{2} + 2 b} = 0$ $\Rightarrow (a^{2} + b^{2} + 2 b) t - 4 at + (a^{2} + b^{2} - 2 b) = 0$ $t = \frac{2 a \pm \sqrt{4 a^{2} - [(a^{2} + b^{2} + 2 b) (a^{2} + b^{2} - 2 b)}}{a^{2} + b^{2} + 2 b} =$ $= \frac{2 a \pm \sqrt{4 a^{2} - [{(a^{2} + b^{2})}^{2} - 4 b^{2}]}}{a^{2} + b^{2} + 2 b} =$ $= \frac{2 a \pm \sqrt{4 (a^{2} + b^{2}) - {(a^{2} + b^{2})}^{2}}}{a^{2} + b^{2} + 2 b} .$
	Answered by MJS last updated on 23/Nov/18

	$x = 2 \arctan u$ $y = 2 \arctan v$ $(1) \frac{2 (u + v) (u v + 1)}{(u^{2} + 1) (v^{2} + 1)} = a$ $(2) \frac{2 (u v - 1) (u v + 1)}{(u^{2} + 1) (v^{2} + 1)} = b$ $(1) u^{2} = \frac{2 {u v}^{2} + 2 u - {a v}^{2} + 2 v - a}{{a v}^{2} - 2 v + a}$ $(2) u^{2} = \frac{2 - b - {b v}^{2}}{(b + 2) v^{2} + b}$ $(1) - (2) \Rightarrow u = \frac{{a v}^{2} - 2 v + a}{(b + 2) v^{2} + b}$ $using this in (2) we get$ $v^{4} - \frac{4 a}{a^{2} + b (b + 2)} v^{3} + \frac{2 (a^{2} + b^{2})}{a^{2} + b (b + 2)} v^{2} - \frac{4 a}{a^{2} + b (b + 2)} v + \frac{a^{2} + b (b - 2)}{a^{2} + b (b + 2)} = 0$ $now this is what I found :$ $1 - \frac{2 (a^{2} + b^{2})}{a^{2} + b (b + 2)} = - \frac{a^{2} + b (b - 2)}{a^{2} + b (b + 2)}$ $\Rightarrow \pm i solves above equation \Rightarrow$ $(v^{2} + 1) (v^{2} - \frac{4 a}{a^{2} + b (b + 2)} v + \frac{a^{2} + b (b - 2)}{a^{2} + b (b + 2)}) = 0$ $which leads to the posted solution$
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