Just-for-fun-Prove-that-there-are-no-real-numbers-A-and-B-that-satisfy-sinA-2-sinB-

Question Number 18015 by alex041103 last updated on 13/Jul/17

J u s t f o r f u n

P r o v e t h a t t h e r e a r e n o r e a l n u m b e r s A a n d B t h a t

s a t i s f y

s i n A = \frac{2}{s i n B}

Commented by alex041103 last updated on 13/Jul/17

i f y o u w a n t y o u c a n g e n e r a l i z e i t

f o r c o m p l e x n u m b e r s a n d s h o w t h a t

i f A, B \in C t h e r e a r e s u c h n u m b e r s

w h i c h s i t i s f y t h i s

Answered by Tinkutara last updated on 13/Jul/17

LHS \in [- 1, 1] but

RHS \in (- \infty, - 2] \cup [2, \infty) hence there

is no solution in real numbers .

Commented by alex041103 last updated on 14/Jul/17

W h a t a b o u t t h e c o m p l e x n u m b e r s ?

Commented by alex041103 last updated on 14/Jul/17

y e s s

Commented by alex041103 last updated on 14/Jul/17

f o r c o m p l e x n u m b e r s i^{'} l l p o s t a s o l u t i o n

a f t e r 1.5 h o u r s

Answered by alex041103 last updated on 14/Jul/17

T i n k u t a r a d i d a n i c e p r o o f f o r t h e

c a s e w h e r e A, B \in R

B u t w h a t a b o u t t h e c a s e w h e r e

A, B \in C (c o m p l e x n u m b e r s)

W e k n o w t h a t f o r z \in C, s i n z \in (- \infty, \infty)

S o c l e a r l y t h e r e a r e s o l u t i o n s .

I n f a c t t h e r e a r e i n f i n a t e n u m b e r o f s o l u t i o n s .

W e k n o w t h a t f o r z \in C

(1) e^{i z} = c o s z + i s i n z ({E u l e r}^{'} s f o r m u l a)

\Rightarrow e^{i (- z)} = c o s z - i s i n z

o r

(2) - e^{- i z} = - c o s z + i s i n z

T h e n w e a d d (1) a n d (2)

2 i s i n z = e^{i z} - e^{- i z}

\Rightarrow s i n z = \frac{e^{i z} - e^{- i z}}{2 i}

N o w {l e t}^{'} s d e f i n e {s i n}^{- 1} (z) u s i n g t h e

c o m p l e x d e f i n i t i o n f o r s i n z .

L e t s = s i n z

\Rightarrow 2 i s = e^{i z} - e^{- i z}

2 {i s e}^{i z} = {(e^{i z})}^{2} - 1

l e t t = e^{i z}

t^{2} - 2 i s t - 1 = 0

\Rightarrow t_{1; 2} = \frac{2 i s \pm \sqrt{- 4 s^{2} + 4}}{2} = i s \pm \sqrt{1 - s^{2}}

\Rightarrow l n (e^{i z}) = l n (i s \pm \sqrt{1 - s^{2}}) = i z

\Rightarrow z = \sin^{- 1} (s) = - i l n (i s \pm \sqrt{1 - s^{2}})

A n d y o u c a n s e e i f y o u s u b s t i t u t e

b a c k s = s i n z, a f t e r y o u s i m p l i f y

{y o u}^{'} l l g e t s {in}^{- 1} (\sin z) = z .

S o n o w

s i n A = \frac{2}{s i n B} \Rightarrow A = {s i n}^{- 1} (\frac{2}{s i n B})

w e i n s e r t t h a t i n t o t h e f o r m u l a a n d w e g e t

A = - i l n (\frac{2 i}{s i n B} \pm \sqrt{1 - \frac{4}{{s i n}^{2} B}})

A n d w e a d d 2 k π (k \in Z) b e c a u s e

s i n (A) = s i n (A + 2 π)

A = - i l n (\frac{2 i}{s i n B} \pm \sqrt{1 - \frac{4}{{s i n}^{2} B}}) + 2 k π

E x a m p l e :

s i n A = \frac{2}{s i n 45 °} = 2 \sqrt{2}

A = - i l n (\frac{2 i}{(\frac{\sqrt{2}}{2})} \pm \sqrt{1 - \frac{4}{(\frac{2}{4})}}) + 2 k π =

= - i l n (2 \sqrt{2} i \pm \sqrt{- 7}) + 2 k π =

= - i l n (2 \sqrt{2} i \pm i \sqrt{7}) + 2 k π =

= - i l n [i (\sqrt{8} \pm \sqrt{7})] + 2 k π =

= - i [l n (i) + l n (\sqrt{8} \pm \sqrt{7})] + 2 k π

W e n o w t h a t

z = a + b i = r (c o s θ + i s i n θ) = {r e}^{i θ}

\Rightarrow z = e^{l n (r) + i θ}

\Rightarrow l n (z) = l n (r) + i θ

w h e r e r = \sqrt{a^{2} + b^{2}} a n d θ = \tan^{- 1} (\frac{b}{a}) .

I n o u r c a s e

l n (i) = l n (1) + i \frac{π}{2} = \frac{π i}{2}

\Rightarrow A = - i \times \frac{π i}{2} - i l n (\sqrt{8} \pm \sqrt{7}) + 2 k π

= \frac{π}{2} - i l n (\sqrt{8} \pm \sqrt{7}) + 2 k π

\Rightarrow A = \frac{π}{2} - i l n (\sqrt{8} \pm \sqrt{7}) + 2 k π, k \in Z