Showing Tan(2x) in exponential form

[MathsOfOld]

Hello fellow mathematicians, my question is this, using the identity: e^(ix)=cos(x)+jsin(x) show that tan(2x)=−i(e^(4ix)−1)/(e^(4ix)+1)

First, I showed that cos(x)=(e^(ix)+e(-ix))/2
and sin(x)=(e^(ix)+e(-ix))/2j

and then divided sin/cos to get tan

Then tried to use the identity tan(2x)=2tan(x)/1-tan^2(x)

However, this is where I must be going wrong, as my working does not come out with the required answer

Please Help!!!!

Thank you

 

[Deleted member 4993]

Hello fellow mathematicians, my question is this, using the identity: e^(ix)=cos(x)+jsin(x) show that tan(2x)=−i(e^(4ix)−1)/(e^(4ix)+1)

To match rest of the problem, that should be an 'i'.

First, I showed that cos(x)=(e^(ix)+e(-ix))/2
and sin(x)=(e^(ix)+e(-ix))/2j .........................Incorrect

sin(x) = [e^(ix) - e^(-ix)]/[2i]

and then divided sin/cos to get tan

Then tried to use the identity tan(2x)=2tan(x)/1-tan^2(x)

However, this is where I must be going wrong, as my working does not come out with the required answer

Please Help!!!!

Thank you

tan(2x)

= sin(2x)/cos(2x)

= (-i) * [e^(2ix) - e(-2ix)]/[e^(2ix) + e(-2ix)]

= (-i) * [e^(2ix) - 1/e^(2ix)]/[e^(2ix) + 1/e^(2ix)]

Now continue.....

 

[Ishuda]

Hello fellow mathematicians, my question is this, using the identity: e^(ix)=cos(x)+jsin(x) show that tan(2x)=−i(e^(4ix)−1)/(e^(4ix)+1)

First, I showed that cos(x)=(e^(ix)+e(-ix))/2
and sin(x)=(e^(ix)+e(-ix))/2j

and then divided sin/cos to get tan

Then tried to use the identity tan(2x)=2tan(x)/1-tan^2(x)

However, this is where I must be going wrong, as my working does not come out with the required answer

Please Help!!!!

Thank you

You would have to show your work in order for us to see where you possibly made a mistake or need to go further. However, I think I might start at
$\displaystyle sin(2x)\, =\, \frac{e^{2ix}\, -\, e^{-2ix}}{2i}$
and
$\displaystyle cos(2x)\, =\, \frac{e^{2ix}\, +\, e^{-2ix}}{2}$