Proving Euler Equations and Solving Complex Numbers in Signal and Systems Course

[dervast]

Hi i am reading about signal and systems course . What i want to prove is not a problem that i have to solve is something that the books take for granted and i want to prove it so i ll be able at exams to reprove so i won't have to remember it, (if u don't believe me i can give u the course's pdf that states what i say. Btw i posted here cause i don't know where else i should post problems like that

Homework Statement

I want to prove why
e^(-j*(theta))=cos(theta)-jsin(theta) (1)
i know that e^j(theta)=cos(theta)+jsin(theta) Why the minus sign affects the sin and not the cos
also i want to find out why cos(theta)=1/2[e^(j*theta)+e^(-j*theta)] (2)
My teachers book say that these things exist. It proves number (2) using Eulers equation.
I know that the angle theta can be found from theta=1/sin(y/(sqrt(x^2+y^2))
theta:is the angle of the complex number

Homework Equations

z=x+jy
z=e*e^j(theta)

The Attempt at a Solution

I have tried to solve the euler equation like that
e^j*theta=cos(theta)+j(sin(theta))
cos(theta)=e^j*theta-j(sin(theta))
I have tried to convert the j(sin(theta)) to something that has inside e^j*theta or something like that but i have failed :(
I am really very week converting exponential numbers to equations that include cosins and sins. If u have a good book for that that i can read it online please suggest it to me

 

[Edgardo]

Euler equation:
e^j(theta)=cos(theta)+jsin(theta)

I want to prove why
e^(-j*(theta))=cos(theta)-jsin(theta) (1)
i know that e^j(theta)=cos(theta)+jsin(theta) Why the minus sign affects the sin and not the cos.

First of all:
e^(-j*(theta)) = e^(j*(-theta)) =... <--- what is the result here using the Euler equation?

Secondly, you will need to show
sin(-x) = -sin(x) and
cos(-x) = cos(x)
In order to show that, use the definition of sin(x) and cos(x), see here

also i want to find out why cos(theta)=1/2[e^(j*theta)+e^(-j*theta)] (2)
My teachers book say that these things exist. It proves number (2) using Eulers equation.

For getting equation (2):
Calculate e^(j*theta)+e^(-j*theta) using
equation (1) e^(-j*(theta))=cos(theta)-jsin(theta) and
the Euler equation e^j(theta)=cos(theta)+jsin(theta)

 

[HallsofIvy]

You can, by the way, derive the Euler equation
e^{i\theta}= cos(\theta)+ i sin(\theta)
By looking at the Taylor's series expansions for the functions

e^x= 1+ x+ \frac{x^2}{2}+ \frac{x^3}{3!}+ \cdot\cdot\cdot +\frac{x^n}{n!}+\cdot\cdot\cdot
Replace x by i\theta and remember that i²= -1, i³= -i, i⁴= 1, i⁵= -i, etc. You get
e^{i\theta}= 1+ i\theta- \frac{\theta^2}{2!}- i\frac{\theta^3}{3!}+ \frac{\theta^4}{4!}+ \cdot\cdot\cdot
Separate that into "real" and "imaginary" parts:
e^{i\theta}= (1-\frac{\theta^2}{2!}+ \frac{\theta^4}{4!}- \cdot\cdot\cdot+ (-1)^n\frac{\theta^{2n}}{(2n)!}+\cdot\cdot\cdot)+ i(\theta- \frac{\theta^3}{3!}+ \frac{\theta^5}{5!}+ \cdot\cdot\cdot+ (-1)^n\frac{\theta^{2n+1}}{(2n+1)!)}+\cdot\cdot\cdot)
and compare those to the Taylor's series for sine and cosine of \theta:
sin(\theta)= \theta- \frac{\theta}{3!}+ \frac{\theta^5}{5!}+\cdot\cdot\cdot+ (-1)^{2n}\frac{\theta^{2n+1}}{(2n+1)!}+ \cdot\cdot\cdot
and
cos(\theta)= 1- \frac{\theta^2}{2!}+ \frac{\theta^4}{4!}+\cdot\cdot\cdot+ (-1)^n \frac{\theta^{2n}}{(2n)!}+ \cdot\cdot\cdot

Sorry, but I just cannot force myself to write "j" instead of "i"!

 

[Hootenanny]

Sorry, but I just cannot force myself to write "j" instead of "i"!

It seems unnatural in some way doesn't it?

 

[dervast]

Euler equation:
e^j(theta)=cos(theta)+jsin(theta)



First of all:
e^(-j*(theta)) = e^(j*(-theta)) =... <--- what is the result here using the Euler equation?

e^(j*(-theta))=cos(-theta)+jsin(-theta) and now i have to prove the next step u mention

Secondly, you will need to show
sin(-x) = -sin(x) and
cos(-x) = cos(x)
In order to show that, use the definition of sin(x) and cos(x), see here

I think that this image found in the site u mentioned

show that sin(-x) =(...lot of stuff i don't type)(-x)^(2*n+1) (2*n+1 is odd which means that the minus sign gets out of the parenthesis = -...(x)^(2*n+1)
for the cosin now
x^2n which means that (-x)^2*n=(x)^2n
this is the way i have thought to prove that sin(-x)=-sinx and cos(-x)=cos(x)
Am i correct?

For getting equation (2):
Calculate e^(j*theta)+e^(-j*theta) using
equation (1) e^(-j*(theta))=cos(theta)-jsin(theta) and
the Euler equation e^j(theta)=cos(theta)+jsin(theta)

Ok thanks a lot let me try it
e^(j*theta)+e^(-j*theta)=cos(theta)+jsin(theta)+cos(theta)-jsin(theta) =2*cos(theta)=>
cos(theta)=[e^(j*theta)+e^(-j*theta)]/2 so i think i have done it with your help guys..please fix everything that i have done wrongly

 

[dervast]

You can, by the way, derive the Euler equation
e^{i\theta}= cos(\theta)+ i sin(\theta)
By looking at the Taylor's series expansions for the functions

e^x= 1+ x+ \frac{x^2}{2}+ \frac{x^3}{3!}+ \cdot\cdot\cdot +\frac{x^n}{n!}+\cdot\cdot\cdot
Replace x by i\theta and remember that i²= -1, i³= -i, i⁴= 1, i⁵= -i, etc. You get
e^{i\theta}= 1+ i\theta- \frac{\theta^2}{2!}- i\frac{\theta^3}{3!}+ \frac{\theta^4}{4!}+ \cdot\cdot\cdot
Separate that into "real" and "imaginary" parts:
e^{i\theta}= (1-\frac{\theta^2}{2!}+ \frac{\theta^4}{4!}- \cdot\cdot\cdot+ (-1)^n\frac{\theta^{2n}}{(2n)!}+\cdot\cdot\cdot)+ i(\theta- \frac{\theta^3}{3!}+ \frac{\theta^5}{5!}+ \cdot\cdot\cdot+ (-1)^n\frac{\theta^{2n+1}}{(2n+1)!)}+\cdot\cdot\cdot)
and compare those to the Taylor's series for sine and cosine of \theta:
sin(\theta)= \theta- \frac{\theta}{3!}+ \frac{\theta^5}{5!}+\cdot\cdot\cdot+ (-1)^{2n}\frac{\theta^{2n+1}}{(2n+1)!}+ \cdot\cdot\cdot
and
cos(\theta)= 1- \frac{\theta^2}{2!}+ \frac{\theta^4}{4!}+\cdot\cdot\cdot+ (-1)^n \frac{\theta^{2n}}{(2n)!}+ \cdot\cdot\cdot

Sorry, but I just cannot force myself to write "j" instead of "i"!

The way u have just mentioned is for remembering to prove why
e^j*theta=cos(theta)+jsin(theta)

 

[Edgardo]

dervast, your calculations are correct.
As an exercise, you could try expressing sin(theta) in terms of the e-functions.

 

[dervast]

dervast, your calculations are correct.
As an exercise, you could try expressing sin(theta) in terms of the e-functions.

Ok let me try it
e^(j*theta)-e^(-j*theta)=cos(theta)+jsin(theta)-cos(theta)+jsin(theta)=>
2jsin(theta)=e^(j*theta)-e^(-j*theta)
sin(theta)=[e^(j*theta)-e^(-j*theta)]/2j

 

[Edgardo]

That's correct.

 

[Gib Z]

It seems unnatural in some way doesn't it?

Yes it does! I thought I would be stupid in thinking that as well, and perhaps I only thought so because I was accustomed to that notation, but now that I know people agree with me I feel much better :)

Which physicist introduced j and disrespected Euler!