a-gt-b-gt-0-a-2-cos-b-2-sin-a-2-b-2-sin-cos-Find-in-terms-of-a-b-0-pi-2-

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 θcos θ Find θ in terms of a, b. θ ∈ (0,(π/2)) .

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: 2a^2 m(m^2 −1)=0 ∧ 2b^2 m(m^2 −1)=(b^2 −a^2 )(m^4 −1) you solved the second term but the first term is ≠0, so the whole equation is not solved

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

dear MJS! the 1#part of eq. is: 2a^2 .m(m^2 +1)=0 ,and it has solved.

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 (3^2 /( (√(5^2 −2×3^2 ))))=arctan ((9(√7))/7) 25cos arctan ((9(√7))/7) −9sin arctan ((9(√7))/7)= =((25(√(154)))/(44))−((81(√(22)))/(44))≈−1.58368 16sin arctan ((9(√7))/7) × cos arctan ((9(√7))/7)=((18(√7))/(11))≈4.32941 so something went wrong 25cos θ −9sin θ=16sin θ cos θ has the approximate solution 0.93305, while arctan ((9(√7))/7)≈1.2849

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≈2.8557 while the solution is ≈.93305

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

dear MJS! please correct my typo or post correct answer. i know your answer will the best,finally.

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.

I {didn}^{'} t mean to offend you, I just wanted to

say {there}^{'} s some mistake . the solution must

solve the equation \dots 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=tan(θ/2) a^2 (((1−t^2 )/(1+t^2 )))−b^2 (((2t)/(1+t^2 )))=(a^2 −b^2 )((2t)/(1+t^2 ))×((1−t^2 )/(1+t^2 )) ((a^2 −a^2 t^2 −2b^2 t)/(1+t^2 ))=(((a^2 −b^2 )(2t−2t^3 ))/((1+t^2 )(1+t^2 ))) (1+t^2 )(a^2 −a^2 t^2 −2b^2 t)=(2a^2 t−2a^2 t^3 −2b^2 t+ 2b^2 t^3 ) a^2 −a^2 t^2 −2b^2 t+a^2 t^2 −a^2 t^4 −2b^2 t^3 = 2a^2 t−2a^2 t^3 −2b^2 t+2b^2 t^3 a^2 −a^2 t^4 −2b^2 t^3 −2a^2 t+2a^2 t^3 +2b^2 t−2b^2 t^3 =0 t^4 (−a^2 )+t^3 (2a^2 −4b^2 )+t(−2a^2 +2b^2 )+a^2 =0 t_1 +t_2 +t_3 +t_4 =((−(2a^2 −4b^2 ))/(−a^2 )) Σt_1 t_2 =0 Σt_1 t_2 t_3 =−(((−2a^2 +2b^2 ))/(−a^2 )) t_1 t_2 t_3 t_4 =(a^2 /(−a^2 )) wait...

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 \dots

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 \dots .

Commented by rahul 19 last updated on 22/Jun/18

��

Answered by ajfour last updated on 23/Jun/18

let for the ellipse of Q.38092 (x^2 /a^2 )+(y^2 /b^2 )=1 ⇒ (dy/dx)= −(x/y)((b^2 /a^2 )) slope of normal to circle at point of tangency (((r−y)/(r−x)))=(y/x)((a^2 /b^2 )) let r−x=rcos θ r−y=rsin θ ⇒ ((sin θ)/(cos θ))=(((1−sin θ)/(1−cos θ)))((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 θcos θ ...

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

\dots

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=(b^2 /a^2 )) cos θ −qsin θ −(1−q)sin θ cos θ =0 θ=2arctan t −((t^4 +2(2q−1)t^3 +2t−1)/(t^4 +2t^2 +1))=0 t^4 +2t+1=0 ⇒ t=±i ∉R ⇒ we can multiplicate without further trouble t^4 +2(2q−1)t^3 +2t−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+βi and α−βi and I hope the real ones are γ+δ and γ−δ. (t−α−βi)(t−α+βi)(t−γ−δ)(t−γ+δ)=0 t^4 − −2(α+γ)t^3 + +(α^2 +4ϑγ+β^2 +γ^2 −δ^2 )t^2 − −2(α^2 γ+αγ^2 −αδ^2 +β^2 γ)t+ +(α^2 +β^2 )(γ^2 −δ^2 )=0 ⇒ [1] 2(2q−1)=−2(α+γ) [2] 0=α^2 +4ϑγ+β^2 +γ^2 −δ^2 [3] 2=−2(α^2 γ+αγ^2 −αδ^2 +β^2 γ) [4] −1=(α^2 +β^2 )(γ^2 −δ^2 ) [1] α=−γ−2q+1 [2] β=(√(−α^2 −4αγ−γ^2 +δ^2 ))= =(√(2γ^2 +(4q−2)γ+δ^2 −(2q−1)^2 )) [3] 2=−4γ^3 +(6−12q)γ^2 −2(2γ+2q−1)δ^2 ⇒ ⇒ δ=(√(−((2γ^3 +3(2q−1)γ^2 +1)/(2γ+2q−1)))) ⇒ β=(√((2γ^3 +3(2q−1)γ^2 −2q(4q^2 −6q+3))/(2γ+2q−1))) [4] −1=((16γ^6 +48(2q−1)γ^5 +48(2q−1)^2 γ^4 +16(2q−1)^3 γ^3 +4(2q−1)γ^2 +4(2q−1)^2 γ−1)/((2γ+2q−1)^2 )) [γ≠(1/2)−q; this will be fulfilled with q=((1−(2)^(1/3) )/2)<0 so we don′t have to care] γ^6 +3(2q−1)γ^5 +3(2q−1)^2 γ^4 +(2q−1)^3 γ^3 +(q/2)γ^2 +((q(2q−1))/2)γ+((q(q−1))/4)=0 at this point we can only try to omit the γ^5 by setting γ=u−((3(2q−1))/6)=u−q+(1/2) which leads to u^6 −((3(2q−1)^2 )/4)u^4 +((48q^4 −96q^3 +72q^2 −16q+3)/(16))u^2 −(((8q^3 −12q^2 +6q+1)^2 )/(64))=0 we seem to have good luck set u=(√v) v^3 −((3(2q−1)^2 )/4)v^2 +((48q^4 −96q^3 +72q^2 −16q+3)/(16))v−(((8q^3 −12q^2 +6q+1)^2 )/(64))=0 omit v^2 by setting v=w−(((2q−1)^2 )/4) leads to w^3 +(q/2)w+((q(q−1))/4)=0 [x^3 +Px+Q=0 ⇒ ⇒ x_1 =((−(Q/2)+(√((P^3 /(27))+(Q^2 /4)))))^(1/3) +((−(Q/2)−(√((P^3 /(27))+(Q^2 /4)))))^(1/3) ] w_1 =((q)^(1/3) /2)(((1−q+(√((q−1)^2 +((8q)/(27))))))^(1/3) +((1−q−(√((q−1)^2 +((8q)/(27))))))^(1/3) ) (q−1)^2 +((8q)/(27))>0 ∀q>0 ⇒ w_1 ∈R ∀ 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...

{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

\dots now we have to set q and go all the way

back \dots solutions of polynomes of 4^{th} degree

usually are even worse \dots

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±βi will be real and γ±δ will be complex...

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 \dots

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 \dots

Commented by MJS last updated on 25/Jun/18

...I didn′t try w_2 and w_3 , if you like to: w_1 =(A)^(1/3) +(B)^(1/3) ⇒ w_2 =(−(1/2)+((√3)/2)i)(A)^(1/3) +(−(1/2)−((√3)/2)i)(B)^(1/3) w_3 =(−(1/2)−((√3)/2)i)(A)^(1/3) +(−(1/2)+((√3)/2)i)(B)^(1/3)

\dots 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 got it reduced to a cubic eq. quite easily, i shall share that soon..

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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