How to prove \frac{e^{jx}-je^{-jx}}{je^{jx}-e^{-jx}}=\fra

Alan Smith 2022-01-03 Answered
How to prove \(\displaystyle{\frac{{{e}^{{{j}{x}}}-{j}{e}^{{-{j}{x}}}}}{{{j}{e}^{{{j}{x}}}-{e}^{{-{j}{x}}}}}}={\frac{{{\tan{{x}}}-{1}}}{{{\tan{{x}}}+{1}}}}\)
where \(\displaystyle{j}=\sqrt{{-{1}}}\)
I've tried to prove it from the right hand
\(\displaystyle{\tan{{x}}}={\frac{{{e}^{{{j}{x}}}-{e}^{{-{j}{x}}}}}{{{j}{\left({e}^{{{j}{x}}}+{e}^{{-{j}{x}}}\right)}}}}\) (2)
and thus
\(\displaystyle{\tan{{x}}}-{1}={\frac{{{e}^{{{j}{x}}}-{e}^{{-{j}{x}}}-{j}{\left({e}^{{{j}{x}}}+{e}^{{-{j}{x}}}\right)}}}{{{j}{\left({e}^{{{j}{x}}}+{e}^{{-{j}{x}}}\right)}}}}\) (3)
and
\(\displaystyle{\tan{{x}}}+{1}={\frac{{{e}^{{{j}{x}}}-{e}^{{-{j}{x}}}+{j}{\left({e}^{{{j}{x}}}+{e}^{{-{j}{x}}}\right)}}}{{{j}{\left({e}^{{{j}{x}}}+{e}^{{-{j}{x}}}\right)}}}}\) (4)
The right side of Eq. (1) can be written as
\(\displaystyle{\frac{{{\tan{{x}}}-{1}}}{{{\tan{{x}}}+{1}}}}={\frac{{{e}^{{{j}{x}}}-{e}^{{-{j}{x}}}-{j}{e}^{{{j}{x}}}-{j}{e}^{{-{j}{x}}}}}{{{e}^{{{j}{x}}}-{e}^{{-{j}{x}}}+{j}{e}^{{{j}{x}}}+{j}{e}^{{-{j}{x}}}}}}\) (5)
Compared with the left side of Eq. (1), the numerator has the same part \(\displaystyle{e}^{{{j}{x}}}−{j}{e}^{{−{j}{x}}}\), and the denominator has the same part \(\displaystyle{j}{e}^{{{j}{x}}}−{e}^{{-{j}{x}}}\), but the other parts don't seem to be zero constantly.
I've checked Eqs. (1) and (5), they are all correct numerically. Can anyone help to find where I am wrong? Thanks in advance!

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

jgardner33v4
Answered 2022-01-04 Author has 4359 answers
It can be done easily from the left hand side itself, if you use Euler's formula:
\(\displaystyle{e}^{{{j}{x}}}={\cos{{x}}}+{j}{\sin{{x}}}\)
Just substitute this in LHS and simplify.
Not exactly what you’re looking for?
Ask My Question
0
 
nghodlokl
Answered 2022-01-05 Author has 2254 answers
Assuming \(\displaystyle{j}=\sqrt{{-{1}}}\)
First of all, observe that:
\(\displaystyle{\frac{{{\tan{{x}}}-{1}}}{{{\tan{{x}}}+{1}}}}={\frac{{{\frac{{{\sin{{x}}}-{\cos{{x}}}}}{{{\cos{{x}}}}}}}}{{{\frac{{{\sin{{x}}}+{\cos{{x}}}}}{{{\cos{{x}}}}}}}}}={\frac{{{\sin{{x}}}-{\cos{{x}}}}}{{{\sin{{x}}}+{\cos{{x}}}}}}\)
Moving to complex numbers:
\(\displaystyle{\frac{{{\sin{{x}}}-{\cos{{x}}}}}{{{\sin{{x}}}+{\cos{{x}}}}}}={\frac{{{e}^{{{j}{x}}}-{e}^{{-{j}{x}}}-{j}{\left({e}^{{{j}{x}}}+{e}^{{-{j}{x}}}\right)}}}{{{e}^{{{j}{x}}}-{e}^{{-{j}{x}}}+{j}{\left({e}^{{{j}{x}}}+{e}^{{-{j}{x}}}\right)}}}}={\frac{{{\left({1}-{j}\right)}{e}^{{{j}{x}}}-{\left({1}+{j}\right)}{e}^{{-{j}{x}}}}}{{{\left({1}+{j}\right)}{e}^{{{j}{x}}}-{\left({1}-{j}\right)}{e}^{{-{j}{x}}}}}}=\)
\(\displaystyle{\frac{{{\left({1}-{j}\right)}{\left({e}^{{{j}{x}}}-{\frac{{{1}+{j}}}{{{1}-{j}}}}{e}^{{-{j}{x}}}\right)}}}{{{\left({1}-{j}\right)}{\left({\frac{{{1}+{j}}}{{{1}-{j}}}}{e}^{{{j}{x}}}-{e}^{{-{j}{x}}}\right)}}}}={\frac{{{e}^{{{j}{x}}}-{\frac{{{1}+{j}}}{{{1}-{j}}}}{e}^{{-{j}{x}}}}}{{{\frac{{{1}+{j}}}{{{1}-{j}}}}{e}^{{{j}{x}}}-{e}^{{-{j}{x}}}}}}\)
Now, observe that:
\(\displaystyle{\frac{{{1}+{j}}}{{{1}-{j}}}}={\frac{{{\left({1}+{j}\right)}^{{2}}}}{{{\left({1}-{j}\right)}{\left({1}+{j}\right)}}}}={\frac{{{1}+{j}^{{2}}+{2}{j}}}{{{1}^{{2}}-{j}^{{2}}}}}={\frac{{{1}-{1}+{2}{j}}}{{{1}-{\left(-{1}\right)}}}}={j}\)
Hence:
\(\displaystyle{\frac{{{\tan{{x}}}-{1}}}{{{\tan{{x}}}+{1}}}}={\frac{{{e}^{{{j}{x}}}-{j}{e}^{{-{j}{x}}}}}{{{j}{e}^{{{j}{x}}}-{e}^{{-{j}{x}}}}}}\)
0
Vasquez
Answered 2022-01-08 Author has 9499 answers

Note that:
\(RHS=\frac{\tan x-\tan \frac{\pi}{4}}{1+\tan x \tan \frac{\pi}{4}}=\tan(x-\frac{\pi}{4})=\frac{\sin(x-\frac{\pi}{4})}{\cos(x-\frac{\pi}{4})}=\frac{\frac{1}{2j}}{\frac{1}{2}}\frac{e^{j(x-\frac{\pi}{4})}-e^{-j(x-\frac{\pi}{4})}}{e^{j(x-\frac{\pi}{4})}+e^{-j(x-\frac{\pi}{4})}}\)
and since \(e^{\frac{j\pi}{4}}=\cos(\frac{\pi}{4})+j\sin(\frac{\pi}{4}),e^{\frac{j\pi}{4}}=\cos(\frac{\pi}{4})-j\sin(\frac{\pi}{4})\) by Euler's formula:
\(\frac{\frac{1}{2j}}{\frac{1}{2}} \frac{(\cos \frac{\pi}{4}-j \sin \frac{\pi}{4})e^{jx}-(\cos \frac{\pi}{4}+j \sin \frac{\pi}{4})e^{-jx}}{(\cos \frac{\pi}{4}-j \sin \frac{\pi}{4})e^{jx}+(\cos \frac{\pi}{4}+j \sin \frac{\pi}{4})e^{-jx}}\)
\(=\frac{1}{j} \frac{(1-j)e^{jx}-(1+j)e^{-jx}}{(1-j)e^{jx}+(1+j)e^{-jx}} \ \ \ \ \ \ \ \ \ \ (\cos \frac{\pi}{4}=\sin \frac{\pi}{4})\)
\(=\frac{(1-j)e^{jx}-(1+j)e^{-jx}}{(j+1)e^{jx}+(j-1)e^{-jx}}\)
and dividng top and bottom by \(1-j, \frac{1+j}{1-j}\frac{1+j}{1+j}=\frac{2j}{j}=j\) and \(\frac{j-1}{1-j}=\frac{-(1-j)}{1-j}=-1\),so we obtain \(\frac{e^{jx}-je^{-jx}}{je^{jx}-e^{-jx}}\) which is the LHS.

0

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Relevant Questions

asked 2022-01-02
How to prove this inequality in interval: \(\displaystyle{\left({0},{\frac{{\pi}}{{{2}}}}\right)}\ \ {2}{\cos{{x}}}+{{\sec}^{{2}}{x}}-{3}\geq{0}\)
asked 2022-01-03
How can I prove the identity
\(\displaystyle{\frac{{{1}}}{{{1}+{\sin{{\left({x}\right)}}}}}}\equiv{\frac{{{{\sec}^{{2}}{\left({\frac{{{x}}}{{{2}}}}\right)}}}}{{{\left({\tan{{\left({\frac{{{x}}}{{{2}}}}\right)}}}+{1}\right)}^{{2}}}}}\)
asked 2022-01-03
How can I prove that:
\(\displaystyle{\left({1}+{\sin{{x}}}+{\cos{{x}}}\right)}^{{2}}={2}{\left({1}+{\cos{{x}}}\right)}{\left({1}+{\sin{{x}}}\right)}\)
I've started like this:
\(\displaystyle{\left({1}+{\sin{{x}}}+{\cos{{x}}}\right)}^{{2}}={\left({\left({1}+{\sin{{x}}}\right)}+{\cos{{x}}}\right)}^{{2}}={\left({1}+{\sin{{x}}}\right)}^{{2}}+{{\cos}^{{2}}{x}}+{2}{\left({1}+{\sin{{x}}}\right)}{\cos{{x}}}\)
asked 2021-12-31
Prove that \(\displaystyle{\frac{{{\sin{{\left({x}\right)}}}}}{{{2}}}}+{{\sin}^{{2}}{\left({\frac{{x}}{{2}}}\right)}}{\tan{{\left({\frac{{x}}{{2}}}\right)}}}\to{\tan{{\left({\frac{{x}}{{2}}}\right)}}}\)
asked 2021-12-30
Hint to prove \(\displaystyle{{\sin}^{{4}}{x}}+{{\cos}^{{4}}{x}}={\frac{{{3}+{\cos{{\left({4}{x}\right)}}}}}{{{4}}}}\)
asked 2021-12-30
I need to prove trigonometric equation
\(\displaystyle{\tan{{2}}}{x}={\frac{{{2}{\sin{{x}}}\cdot{\cos{{x}}}}}{{{{\cos}^{{2}}{x}}-{{\sin}^{{2}}{x}}}}}\)
asked 2022-01-17
Prove that the series \(\displaystyle{\sum_{{{i}={1}}}^{{\infty}}}{\frac{{{1}}}{{\sqrt{{n}}}}}{\sin{{\left({\frac{{{1}}}{{\sqrt{{n}}}}}\right)}}}\) is divergent
...