tan-2-1-find-lets-solve-for-fun-

Question Number 19385 by NEC last updated on 10/Aug/17

{t a n}^{2} β = - 1

f i n d β \dots . l e t s s o l v e f o r f u n

Commented by NEC last updated on 10/Aug/17

p l e a s e h e l p

Commented by Tinkutara last updated on 10/Aug/17

There is no such β (real or complex) .

Commented by NEC last updated on 10/Aug/17

t h e β c o u l d a l s o b e r e p r e s e n t e d b y

x o r θ o r a n y v a l u e .

A f r i e n d a c t u a l l y g a v e m e t h e

q u e s t i o n i^{'} m s t o c k e d .

t h i s w a s w h a t i d i d

{t a n}^{2} x = - 1

{t a n}^{2} x + 1 = 0

\sec^{2} x = 0

\frac{1}{\cos^{2} x} = 0

from that point i became

confused \dots . .

i know it might lead to a complex

number but if its possible i^{'} l l l i k e

t o s e e t h e s o l u t i o n

Commented by Tinkutara last updated on 10/Aug/17

It is impossible . Let z = e^{i θ} = \cos θ +

i \sin θ and \bar{z} = e^{- i θ} = \cos θ - i \sin θ .

z + \bar{z} = 2 \cos θ = e^{i θ} + e^{- i θ}

\frac{1}{\cos θ} = \frac{2}{e^{i θ} + e^{- i θ}} = \frac{2 e^{i θ}}{e^{2 i θ} + 1} and this

equal to 0 implies that e^{i θ} = 0 which is

impossible .

Commented by NEC last updated on 10/Aug/17

t h a n k s

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