Question

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\( \Large (1-e^{iθ} )^{(-1/2)}+(1-e^{(-iθ)} )^{(-1/2)}=(1+cosec(θ/2) )^{(1/2)} \)

Answer 1

\(\large (1-e^{iθ} )^{(-1/2)}+(1-e^{(-iθ)} )^{(-1/2)}=(1+cosec(θ/2) )^{(1/2)}\)

Answer 2

started with right side, couldn't go too far...now trying to start with left side

Answer 3

Hmm I can't seem to figure it out D:
I also tried starting from the right side.
Starting by using this:\[\Large\rm \sin \theta=\frac{e^{i \theta}-e^{-i \theta}}{2i}\]And then after the calculations, turning it into (1+csc(theta/2))^(1/2) gives us,\[\Large\rm \left(1+\csc\frac{\theta}{2}\right)^{1/2}=\left(1+\frac{2i}{(e^{i \theta})^{1/2}-(e^{-i \theta})^{1/2}}\right)^{1/2}\]But I got stuck there >.<
Mmmm

Answer 4

\( \Large \dfrac{(2i \sin θ)}{((\sqrt{(1-e^{(-iθ)} )}-\sqrt {(1-e^{iθ} ))} \sqrt {(2-e^{(-iθ)}-e^{iθ} )}) } \)

me here

Answer 5

Hartnn someday you need to teach me 1 + 1, you're so intelligent o-o

Answer 6

sorry, i can't teach 1+1, i am too dumb for it :P

Answer 7

i am more used to 1+1 =10, instead of 1+1 =2 :P

Answer 8

Are you sure the denominator is not -1?

Answer 9

denominator of what where which side ? :O

Answer 10

Left. You are multiplying by the conjugate right? I get

\[ 1-e^{i\theta} - (1 - e^{-i\theta}) \]

Answer 11

Oh never mind

Answer 12

yep, my left side work, till now is correct, but stuck!

Answer 13

\( (2−e^{(−iθ)}−e^{iθ}) = 2-2 cos θ\)

Answer 14

any other approach... ?

Answer 15

I haven't learn complex number but could you square both sides together?

Answer 16

i need to prove left side = right side, so i can't already assume it to be true...

Answer 17

anyways, what good will squaring both sides do ?

Answer 18

\(\text{LHS}^2=\text{RHS}^2\) is a necessary but not sufficient condition for \(\text{LHS}=\text{RHS}\).

Answer 19

the only application of complex numbers here is
\(e^{ix } = cos x +i \sin x~,~~ so~~e^{-ix } = cos x -i \sin x \)

Answer 20

So theta is real?

Answer 21

yep, its an angle...like 30 degree, 60, 90, ...

Answer 22

Am I allowed to do binomial expansion on the left hand side if I take the square of both sides?

Answer 23

yes, you can use any math formula available

Answer 24

Something is wrong in here.
\[
\begin{align*}
&\left(\left(1-e^{i\theta}\right)^{-1/2}+\left(1-e^{-i\theta}\right)^{-1/2}\right)^2\\
=&\left(1-e^{i\theta}\right)^{-1}+\left(1-e^{-i\theta}\right)^{-1}+2\left(1-e^{i\theta}\right)^{-1/2}\left(1-e^{-i\theta}\right)^{-1/2}\\
=&\frac{1}{1-e^{i\theta}}+\frac{1}{1-e^{-i\theta}}+2
\\&\text{I couldn't think of a situation that }2\neq2\left(1-e^{i\theta}\right)^{-1/2}\left(1-e^{-i\theta}\right)^{-1/2}\\
=&\frac{1-e^{-i\theta}}{\left(1-e^{i\theta}\right)\left(1-e^{-i\theta}\right)}+\frac{1-e^{i\theta}}{\left(1-e^{i\theta}\right)\left(1-e^{-i\theta}\right)}+2\\
=&\frac{2-e^{i\theta}-e^{-i\theta}}{1-e^{i\theta}-e^{-i\theta}+1}+2\\
=&1+2\qquad????
\end{align*}
\]

Answer 25

From @zepdrix's work... I'm wondering if this gets you somewhere.
\[\begin{align*}1+\csc\frac{\theta}{2}&=1+\frac{2i}{\exp\left(\dfrac{i\theta}{2}\right)-\exp\left(-\dfrac{i\theta}{2}\right)}\\\\&=\frac{\exp\left(\dfrac{i\theta}{2}\right)+2i-\exp\left(-\dfrac{i\theta}{2}\right)}{\exp\left(\dfrac{i\theta}{2}\right)-\exp\left(-\dfrac{i\theta}{2}\right)}\\\\
&=\frac{\left(\exp\left(\dfrac{i\theta}{4}\right)+i\exp\left(-\dfrac{i\theta}{4}\right)\right)^2}{\exp\left(\dfrac{i\theta}{2}\right)-\exp\left(-\dfrac{i\theta}{2}\right)}\\\\
\sqrt{1+\csc\frac{\theta}{2}}&=\frac{\exp\left(\dfrac{i\theta}{4}\right)+i\exp\left(-\dfrac{i\theta}{4}\right)}{\sqrt{\exp\left(\dfrac{i\theta}{2}\right)-\exp\left(-\dfrac{i\theta}{2}\right)}}\\\\


\end{align*}\]

Answer 26

i got there too, but stuck

Answer 27

thomas, 1 is correct, but
\(2\neq2\left(1-e^{i\theta}\right)^{-1/2}\left(1-e^{-i\theta}\right)^{-1/2}\\\)

Answer 28

ok,
\(\sqrt {1+2 \sqrt{2-2\cos \theta }}\)
this might lead to right side..

Answer 29

\(\sqrt {1+2/[\sqrt 2 \sqrt {(2\sin^2 (\theta/2 ))}]} = \sqrt {1+1/\sin (\theta /2)} =\sqrt {1+cosec \theta/2}\)
thanks thomas! :D

Answer 30

Where did this \(\sqrt {1+2 \sqrt{2-2\cos \theta }}\) come from?

Answer 31

i meant
\(\sqrt {1+2/ \sqrt{2-2\cos \theta }}\)

Answer 32

because
\((1-e^{i\theta })(1-e^{-i\theta } ) = (2-e^{i\theta }-e^{-i\theta })=2-2\cos \theta \)

Answer 33

\(\Large\rm \sin \theta=\frac{e^{i \theta}-e^{-i \theta}}{2i} \\ \Large\rm \cos \theta=\frac{e^{i \theta}+e^{-i \theta}}{2}\)

Answer 34

you get those ^^
by adding and subtracting these
\(e^{ix } = cos x +i \sin x~,~~ so~~e^{-ix } = cos x -i \sin x\)

Answer 35

Finally understand what is happening. Is \(\sqrt{a}\sqrt{b}=\sqrt{ab}\text{,}\:a,b\in\mathbb{C}\) generally true?

Answer 36

Nothing new here. Just thinking aloud.

\[\large \begin{align} (1-e^{i\theta} )^{-{1 \over 2}}+(1-e^{-i\theta} )^{-{1 \over 2}}&=(1+\csc(\theta/2) ^{1 \over 2} \cr {1 \over {\sqrt{1-e^{i\theta}}}}+ {1 \over {\sqrt{1-e^{-i\theta}}}} &= \cr { { \sqrt{1-e^{-i\theta}} +\sqrt{1-e^{i\theta}} } \over {\sqrt{1-e^{i\theta}}\sqrt{1-e^{-i\theta}}}} &= \cr { { \sqrt{1-e^{-i\theta}} +\sqrt{1-e^{i\theta}} } \over { \sqrt{ (1-e^{i\theta})(1-e^{-i\theta} ) } }} &= \cr { { \sqrt{1-e^{-i\theta}} +\sqrt{1-e^{i\theta}} } \over { \sqrt{ (1-\cos\theta-i\sin\theta)(1-\cos\theta + i\sin\theta ) } }} &= \cr { { \sqrt{1-e^{-i\theta}} +\sqrt{1-e^{i\theta}} } \over { \sqrt{ (2-2\cos\theta ) } }} &= \end{align}\]

Answer 37

\[ \large \begin{align} { { \sqrt{1-e^{-i\theta}} +\sqrt{1-e^{i\theta}} } \over { \sqrt{ (2-2\cos\theta ) } }} &= \cr \left[ { { \sqrt{1-e^{-i\theta}} +\sqrt{1-e^{i\theta}} } \over { \sqrt{ (2-2\cos\theta ) } }} \right] ^2 &= 1+\csc{\theta \over 2} \cr { { (\sqrt{1-e^{-i\theta}} +\sqrt{1-e^{i\theta}})^2 } \over { 2-2\cos\theta }} &= \cr { { 1-e^{-i\theta} +2\sqrt{1-e^{-i\theta}}\sqrt{1-e^{i\theta}} + 1-e^{i\theta} } \over { 2-2\cos\theta }} &= \end{align}\]

Answer 38

\[\large \begin{align} { { 2-2\cos\theta +2\sqrt{1-e^{-i\theta}}\sqrt{1-e^{i\theta}} } \over { 2-2\cos\theta }} &= \cr {{2-2\cos\theta } \over { 2-2\cos\theta }}+{ { 2\sqrt{1-e^{-i\theta}}\sqrt{1-e^{i\theta}} } \over { 2-2\cos\theta }} &= \cr 1 +{ { 2\sqrt{1-e^{-i\theta}}\sqrt{1-e^{i\theta}} } \over { 2-2\cos\theta }} &= 1+\csc{\theta \over 2}\cr { { 2\sqrt{1-e^{-i\theta}}\sqrt{1-e^{i\theta}} } \over { 2-2\cos\theta }} &= \csc{\theta \over 2}\end{align}\]

Answer 39

you still on it ?
want to see how i solved it ?

Answer 40

Sure

Answer 41

clearly the way to go.

Answer 42

write
\[ \exp(i \theta) = \cos(\theta) + i \sin(\theta) \\ \exp(-i \theta) = \cos(\theta) - i \sin(\theta)\]
and
\[ 1-\exp(i \theta) = 1 -\cos(\theta) - i \sin(\theta) \\
1-\exp(-i \theta) = 1 -\cos(\theta) + i \sin(\theta) \]

let
\[ (1- 1 -\cos(\theta)= a \\ \sin(\theta)= b \]

the left-hand side is
\[ \frac{1}{\sqrt{a- bi}} + \frac{1}{\sqrt{a+bi}} \\
= \frac{\sqrt{a+bi}+\sqrt{a-bi}}{\sqrt{a^2+b^2}}
\]
we use \[ a^2 + b^2 = 1+\cos^2(\theta)- 2 \cos(\theta)+\sin^2(\theta) = 2(1- \cos(\theta)= 2a\]
to write \[ \frac{\sqrt{a+bi}+\sqrt{a-bi}}{\sqrt{2a}} \]

square this quantity
\[ \frac{a+bi + a - bi + 2\sqrt{a^2+b^2}}{2a} \\ =\frac{ 2a+2 \sqrt{2a} }{2a} \\
= 1 + \frac{2}{\sqrt{2a}}
\]

meanwhile, the square of the right -hand side is
\[ 1+ \frac{1}{\sin(\frac{x}{2})} \\
= 1 + \frac{1}{\sqrt{\frac{1}{2} (1-\cos(x))}}\\
= 1 + \frac{1}{\sqrt{\frac{1}{2} a}}\\
= 1 + \frac{2}{\sqrt{2a}}
\]

which matches the left-hand side

Answer 43

yep, no other way
squaring sides is necessary to prove this!

Answer 44

**typo
let \[ 1- \cos(\theta)= a \\ \sin(\theta)= b \]

Answer 45

yes, it took some putzing around to find a way out of the weeds.