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Question Number 38151 by ajfour last updated on 22/Jun/18

$a > b > 0$ $a^{2} \cos θ - b^{2} \sin θ = (a^{2} - b^{2}) \sin θ \cos θ$ $F i n d θ i n t e r m s o f a, b .$ $θ \in (0, \frac{π}{2}) .$
Commented by behi83417@gmail.com last updated on 23/Jun/18
$let: m=e^(iθ) ,i=(√(−1)) ⇒a^2 .((m^2 +1)/(2m))−b^2 .((m^2 −1)/(2im))=(a^2 −b^2 )((m^4 −1)/(4im^2 )) ⇒2a^2 .m(m^2 +1)+2mib^2 (m^2 −1)=i(b^2 −a^2 )(m^4 −1) ⇒ { ((2a^2 .m(m^2 +1)=0 (A))),((2b^2 .m(m^2 −1)=(b^2 −a^2 )(m^4 −1) (B))) :} ⇒^A (a=0)∨(m=0)∨(m^2 +1)=0 a=0⇒cosθ=1⇒θ=0 ( ×) m=0⇒e^(iθ) =0 (×) m^2 +1=0⇒m=±i⇒e^(iθ) =e^(±i(π/2)) ⇒θ=±(π/2) (×) ⇒^B 2b^2 .m(m^2 −1)=(b^2 −a^2 )(m^4 −1) ⇒ { ((m^2 −1=0⇒m=±1⇒e^(iθ) =±1=e^(±iπ) )),((⇒θ=±π (×))) :}(×) 2b^2 .m=(b^2 −a^2 )(m^2 +1)⇒((m^2 +1)/(2m))=(b^2 /(b^2 −a^2 )) ((m^2 +1+2m)/(m^2 +1−2m))=((b^2 +b^2 −a^2 )/(b^2 −b^2 +a^2 ))⇒(((m+1)/(m−1)))^2 =2(b^2 /a^2 )−1 ⇒((m+1)/(m−1))=((√(2b^2 −a^2 ))/a)⇒((m+1+m−1)/(m+1−m+1))=(((√(2b^2 −a^2 ))+a)/((√(2b^2 −a^2 ))−a)) ⇒m=((b^2 +a.(√(2b^2 −a^2 )))/(b^2 −a^2 )) ⇒e^(i𝛉) =((b^2 +a.(√(2b^2 −a^2 )))/(b^2 −a^2 )) ,(p=(b^2 /a^2 )) ⇒i𝛉=ln((b^2 +a(√(2b^2 −a^2 )))/(b^2 −a^2 )) ⇒i𝛉=ln((p+(√(2p−1)))/(p−1)) .■ =ln((p/(p−1))+((√(1−2p))/(p−1)).i)= =[(1/2)ln(((p^2 +(1−2p))/((p−1)^2 )))+i.arctg(((√(1−2p))/(p−1))/(p/(p−1)))]= =[(1/2)ln1+i.arctg((√(1−2p))/p)]=i.arctg((√(1−2p))/p) ⇒𝛉=arctg((√(1−2p))/p)=arctg((a.(√(a^2 −2b^2 )))/b^2 ).■ sir Ajfour! this is a real solution for your quistion.$
$l e t : m = e^{i θ}, i = \sqrt{- 1}$ $\Rightarrow a^{2} . \frac{m^{2} + 1}{2 m} - b^{2} . \frac{m^{2} - 1}{2 i m} = (a^{2} - b^{2}) \frac{m^{4} - 1}{4 {i m}^{2}}$ $\Rightarrow 2 a^{2} . m (m^{2} + 1) + 2 {m i b}^{2} (m^{2} - 1) = i (b^{2} - a^{2}) (m^{4} - 1)$ $\Rightarrow {\begin{cases} 2 a^{2} . m (m^{2} + 1) = 0 (A) \\ 2 b^{2} . m (m^{2} - 1) = (b^{2} - a^{2}) (m^{4} - 1) (B) \end{cases}$ $\overset{A}{\Rightarrow} (a = 0) \lor (m = 0) \lor (m^{2} + 1) = 0$ $a = 0 \Rightarrow c o s θ = 1 \Rightarrow θ = 0 (\times)$ $m = 0 \Rightarrow e^{i θ} = 0 (\times)$ $m^{2} + 1 = 0 \Rightarrow m = \pm i \Rightarrow e^{i θ} = e^{\pm i \frac{π}{2}} \Rightarrow θ = \pm \frac{π}{2} (\times)$ $\overset{B}{\Rightarrow} 2 b^{2} . m (m^{2} - 1) = (b^{2} - a^{2}) (m^{4} - 1)$ $\Rightarrow {\begin{cases} m^{2} - 1 = 0 \Rightarrow m = \pm 1 \Rightarrow e^{i θ} = \pm 1 = e^{\pm i π} \\ \Rightarrow θ = \pm π (\times) \end{cases} (\times)$ $2 b^{2} . m = (b^{2} - a^{2}) (m^{2} + 1) \Rightarrow \frac{m^{2} + 1}{2 m} = \frac{b^{2}}{b^{2} - a^{2}}$ $\frac{m^{2} + 1 + 2 m}{m^{2} + 1 - 2 m} = \frac{b^{2} + b^{2} - a^{2}}{b^{2} - b^{2} + a^{2}} \Rightarrow {(\frac{m + 1}{m - 1})}^{2} = 2 \frac{b^{2}}{a^{2}} - 1$ $\Rightarrow \frac{m + 1}{m - 1} = \frac{\sqrt{2 b^{2} - a^{2}}}{a} \Rightarrow \frac{m + 1 + m - 1}{m + 1 - m + 1} = \frac{\sqrt{2 b^{2} - a^{2}} + a}{\sqrt{2 b^{2} - a^{2}} - a}$ $\Rightarrow m = \frac{b^{2} + a . \sqrt{2 b^{2} - a^{2}}}{b^{2} - a^{2}}$ $\Rightarrow e^{i θ} = \frac{b^{2} + a . \sqrt{2 b^{2} - a^{2}}}{b^{2} - a^{2}}, (p = \frac{b^{2}}{a^{2}})$ $\Rightarrow i θ = l n \frac{b^{2} + a \sqrt{2 b^{2} - a^{2}}}{b^{2} - a^{2}} \Rightarrow i θ = l n \frac{p + \sqrt{2 p - 1}}{p - 1} . ◼$ $= l n (\frac{p}{p - 1} + \frac{\sqrt{1 - 2 p}}{p - 1} . i) =$ $= [\frac{1}{2} l n (\frac{p^{2} + (1 - 2 p)}{{(p - 1)}^{2}}) + i . a r c t g \frac{\frac{\sqrt{1 - 2 p}}{p - 1}}{\frac{p}{p - 1}}] =$ $= [\frac{1}{2} l n 1 + i . a r c t g \frac{\sqrt{1 - 2 p}}{p}] = i . a r c t g \frac{\sqrt{1 - 2 p}}{p}$ $\Rightarrow θ = a r c t g \frac{\sqrt{1 - 2 p}}{p} = a r c t g \frac{a . \sqrt{a^{2} - 2 b^{2}}}{b^{2}} . ◼$ $s i r A j f o u r! t h i s i s a r e a l s o l u t i o n$ $f o r y o u r q u i s t i o n .$
Commented by ajfour last updated on 23/Jun/18

$t h e r e h a s t o b e a r e a l s o l u t i o n . .$
Commented by MJS last updated on 23/Jun/18

$the mistake is simple but it took me quite$ $some time to see it :$ $2 a^{2} m (m^{2} - 1) = 0 \land 2 b^{2} m (m^{2} - 1) = (b^{2} - a^{2}) (m^{4} - 1)$ $you solved the second term but the first term$ $is \neq 0, so the whole equation is n o t solved$
Commented by behi83417@gmail.com last updated on 23/Jun/18

$You can't use 'macro parameter character #' in math mode$ $2 a^{2} . m (m^{2} + 1) = 0, a n d i t h a s s o l v e d .$
Commented by MJS last updated on 23/Jun/18

$sorry, typo .$ $but have you tried your solution with any$ $real numbers a and b ?$ $a = 5$ $b = 3$ $\arctan \frac{3^{2}}{\sqrt{5^{2} - 2 \times 3^{2}}} = \arctan \frac{9 \sqrt{7}}{7}$ $25 \cos \arctan \frac{9 \sqrt{7}}{7} - 9 \sin \arctan \frac{9 \sqrt{7}}{7} =$ $= \frac{25 \sqrt{154}}{44} - \frac{81 \sqrt{22}}{44} \approx - 1.58368$ $16 \sin \arctan \frac{9 \sqrt{7}}{7} \times \cos \arctan \frac{9 \sqrt{7}}{7} = \frac{18 \sqrt{7}}{11} \approx 4.32941$ $so something went wrong$ $25 \cos θ - 9 \sin θ = 16 \sin θ \cos θ has the$ $approximate solution 0.93305, while$ $\arctan \frac{9 \sqrt{7}}{7} \approx 1.2849$
Commented by MJS last updated on 23/Jun/18

$sorry I misread$ $still with this value the left side gives$ $- 26.523$ $but the righr side is$ $- 4.3294$ $and the new θ \approx 2.8557 while the solution$ $is \approx . 93305$
Commented by behi83417@gmail.com last updated on 23/Jun/18

$d e a r M J S! p l e a s e c o r r e c t m y t y p o o r$ $p o s t c o r r e c t a n s w e r .$ $i k n o w y o u r a n s w e r w i l l t h e b e s t, f i n a l l y .$
Commented by MJS last updated on 24/Jun/18

$I {didn}^{'} t mean to offend you, I just wanted to$ $say {there}^{'} s some mistake . the solution must$ $solve the equation . . . we all should test our$ $solutions with at least 2 or 3 sets of real$ $numbers before posting them .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 22/Jun/18

	$t = t a n \frac{θ}{2}$ $a^{2} (\frac{1 - t^{2}}{1 + t^{2}}) - b^{2} (\frac{2 t}{1 + t^{2}}) = (a^{2} - b^{2}) \frac{2 t}{1 + t^{2}} \times \frac{1 - t^{2}}{1 + t^{2}}$ $\frac{a^{2} - a^{2} t^{2} - 2 b^{2} t}{1 + t^{2}} = \frac{(a^{2} - b^{2}) (2 t - 2 t^{3})}{(1 + t^{2}) (1 + t^{2})}$ $(1 + t^{2}) (a^{2} - a^{2} t^{2} - 2 b^{2} t) = (2 a^{2} t - 2 a^{2} t^{3} - 2 b^{2} t +$ $2 b^{2} t^{3})$ $a^{2} - a^{2} t^{2} - 2 b^{2} t + a^{2} t^{2} - a^{2} t^{4} - 2 b^{2} t^{3} =$ $2 a^{2} t - 2 a^{2} t^{3} - 2 b^{2} t + 2 b^{2} t^{3}$ $a^{2} - a^{2} t^{4} - 2 b^{2} t^{3} - 2 a^{2} t + 2 a^{2} t^{3} + 2 b^{2} t - 2 b^{2} t^{3} = 0$ $t^{4} (- a^{2}) + t^{3} (2 a^{2} - 4 b^{2}) + t (- 2 a^{2} + 2 b^{2}) + a^{2} = 0$ $t_{1} + t_{2} + t_{3} + t_{4} = \frac{- (2 a^{2} - 4 b^{2})}{- a^{2}}$ $Σ t_{1} t_{2} = 0$ $Σ t_{1} t_{2} t_{3} = - \frac{(- 2 a^{2} + 2 b^{2})}{- a^{2}}$ $t_{1} t_{2} t_{3} t_{4} = \frac{a^{2}}{- a^{2}}$ $w a i t . . .$
	Commented by math khazana by abdo last updated on 22/Jun/18

	$b u t t h e t r a i n d o n t w a i t . . . .$
	Commented by rahul 19 last updated on 22/Jun/18
	��
	Answered by ajfour last updated on 23/Jun/18

	$l e t f o r t h e e l l i p s e o f Q . 38092$ $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ $\Rightarrow \frac{d y}{d x} = - \frac{x}{y} (\frac{b^{2}}{a^{2}})$ $s l o p e o f n o r m a l t o c i r c l e a t$ $p o i n t o f t a n g e n c y$ $(\frac{r - y}{r - x}) = \frac{y}{x} (\frac{a^{2}}{b^{2}})$ $l e t r - x = r \cos θ$ $r - y = r \sin θ$ $\Rightarrow \frac{\sin θ}{\cos θ} = (\frac{1 - \sin θ}{1 - \cos θ}) (\frac{a^{2}}{b^{2}})$ $a^{2} \cos θ (1 - \sin θ) = b^{2} \sin θ (1 - \cos θ)$ $a^{2} \sin θ - b^{2} \sin θ = (a^{2} - b^{2}) \sin θ \cos θ$ $. . .$
	Answered by MJS last updated on 25/Jun/18

	${it}^{'} s not nice and probably not useable but$ ${it}^{'} s possible$ $let a = 1 and b^{2} = q (q = \frac{b^{2}}{a^{2}})$ $\cos θ - q \sin θ - (1 - q) \sin θ \cos θ = 0$ $θ = 2 \arctan t$ $- \frac{t^{4} + 2 (2 q - 1) t^{3} + 2 t - 1}{t^{4} + 2 t^{2} + 1} = 0$ $t^{4} + 2 t + 1 = 0 \Rightarrow t = \pm i \notin R \Rightarrow we can multiplicate$ $without further trouble$ $t^{4} + 2 (2 q - 1) t^{3} + 2 t - 1 = 0$ $approximate solving with some values of q$ $leads to the conclusion that this has two$ $complex and two real solutions . the complex$ $ones must be α + β i and α - β i and I hope the$ $real ones are γ + δ and γ - δ .$ $(t - α - β i) (t - α + β i) (t - γ - δ) (t - γ + δ) = 0$ $t^{4} -$ $- 2 (α + γ) t^{3} +$ $+ (α^{2} + 4 ϑ γ + β^{2} + γ^{2} - δ^{2}) t^{2} -$ $- 2 (α^{2} γ + α γ^{2} - α δ^{2} + β^{2} γ) t +$ $+ (α^{2} + β^{2}) (γ^{2} - δ^{2}) = 0$ $\Rightarrow$ $[1] 2 (2 q - 1) = - 2 (α + γ)$ $[2] 0 = α^{2} + 4 ϑ γ + β^{2} + γ^{2} - δ^{2}$ $[3] 2 = - 2 (α^{2} γ + α γ^{2} - α δ^{2} + β^{2} γ)$ $[4] - 1 = (α^{2} + β^{2}) (γ^{2} - δ^{2})$ $[1] α = - γ - 2 q + 1$ $[2] β = \sqrt{- α^{2} - 4 α γ - γ^{2} + δ^{2}} =$ $= \sqrt{2 γ^{2} + (4 q - 2) γ + δ^{2} - {(2 q - 1)}^{2}}$ $[3] 2 = - 4 γ^{3} + (6 - 12 q) γ^{2} - 2 (2 γ + 2 q - 1) δ^{2} \Rightarrow$ $\Rightarrow δ = \sqrt{- \frac{2 γ^{3} + 3 (2 q - 1) γ^{2} + 1}{2 γ + 2 q - 1}}$ $\Rightarrow β = \sqrt{\frac{2 γ^{3} + 3 (2 q - 1) γ^{2} - 2 q (4 q^{2} - 6 q + 3)}{2 γ + 2 q - 1}}$ $[4] - 1 = \frac{16 γ^{6} + 48 (2 q - 1) γ^{5} + 48 {(2 q - 1)}^{2} γ^{4} + 16 {(2 q - 1)}^{3} γ^{3} + 4 (2 q - 1) γ^{2} + 4 {(2 q - 1)}^{2} γ - 1}{{(2 γ + 2 q - 1)}^{2}}$ $[γ \neq \frac{1}{2} - q; this will be fulfilled with$ $q = \frac{1 - \sqrt[3]{2}}{2} < 0 so we {don}^{'} t have to care]$ $γ^{6} + 3 (2 q - 1) γ^{5} + 3 {(2 q - 1)}^{2} γ^{4} + {(2 q - 1)}^{3} γ^{3} + \frac{q}{2} γ^{2} + \frac{q (2 q - 1)}{2} γ + \frac{q (q - 1)}{4} = 0$ $at this point we can only try to omit the γ^{5}$ $by setting γ = u - \frac{3 (2 q - 1)}{6} = u - q + \frac{1}{2}$ $which leads to$ $u^{6} - \frac{3 {(2 q - 1)}^{2}}{4} u^{4} + \frac{48 q^{4} - 96 q^{3} + 72 q^{2} - 16 q + 3}{16} u^{2} - \frac{{(8 q^{3} - 12 q^{2} + 6 q + 1)}^{2}}{64} = 0$ $we seem to have good luck$ $set u = \sqrt{v}$ $v^{3} - \frac{3 {(2 q - 1)}^{2}}{4} v^{2} + \frac{48 q^{4} - 96 q^{3} + 72 q^{2} - 16 q + 3}{16} v - \frac{{(8 q^{3} - 12 q^{2} + 6 q + 1)}^{2}}{64} = 0$ $omit v^{2} by setting v = w - \frac{{(2 q - 1)}^{2}}{4} leads to$ $w^{3} + \frac{q}{2} w + \frac{q (q - 1)}{4} = 0$ $[x^{3} + P x + Q = 0 \Rightarrow$ $\Rightarrow x_{1} = \sqrt[3]{- \frac{Q}{2} + \sqrt{\frac{P^{3}}{27} + \frac{Q^{2}}{4}}} + \sqrt[3]{- \frac{Q}{2} - \sqrt{\frac{P^{3}}{27} + \frac{Q^{2}}{4}}}]$ $w_{1} = \frac{\sqrt[3]{q}}{2} (\sqrt[3]{1 - q + \sqrt{{(q - 1)}^{2} + \frac{8 q}{27}}} + \sqrt[3]{1 - q - \sqrt{{(q - 1)}^{2} + \frac{8 q}{27}}})$ ${(q - 1)}^{2} + \frac{8 q}{27} > 0 \forall q > 0 \Rightarrow w_{1} \in R \forall q > 0$ $. . . now we have to set q and go all the way$ $back . . . solutions of polynomes of 4^{th} degree$ $usually are even worse . . .$
	Commented by ajfour last updated on 25/Jun/18

	$A w e s o m e! m a n y m a n y t h a n k s S i r .$
	Commented by MJS last updated on 25/Jun/18

	$you {won}^{'} t be able to use this for further$ $simplifications or transformations but at$ $least {it}^{'} s possible to calculate approximate$ $values using a calculator . . .$ $by the way, it seems β and δ are imaginary$ $numbers, so α \pm β i will be real and γ \pm δ will$ $be complex . . .$
	Commented by MJS last updated on 25/Jun/18

	$. . . I {didn}^{'} t try w_{2} and w_{3}, if you like to :$ $w_{1} = \sqrt[3]{A} + \sqrt[3]{B}$ $\Rightarrow w_{2} = (- \frac{1}{2} + \frac{\sqrt{3}}{2} i) \sqrt[3]{A} + (- \frac{1}{2} - \frac{\sqrt{3}}{2} i) \sqrt[3]{B}$ $w_{3} = (- \frac{1}{2} - \frac{\sqrt{3}}{2} i) \sqrt[3]{A} + (- \frac{1}{2} + \frac{\sqrt{3}}{2} i) \sqrt[3]{B}$
	Commented by ajfour last updated on 26/Jun/18

	$I g o t i t r e d u c e d t o a c u b i c e q .$ $q u i t e e a s i l y, i s h a l l s h a r e t h a t$ $s o o n . .$
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