Proving an identity using de moivre's theorem

[Relz]

For homework I need to prove this statement using de Moivre's theorem. I'm not quite sure how to do that because there aren't any coefficients in the equation.

sin 2theta = 2 sin (theta)*cos (theta)

 

[soroban]

Hello, Relz!

$\displaystyle \text{Prove, using DeMoivre's theorem: }\:\sin2\theta \:=\:2\sin\theta\cos\theta$

From DeMoivre's theorem, we have: .$\displaystyle e^{i\theta} \:=\:\cos\theta + i\sin\theta\;\;[1]$

. . Hence:.$\displaystyle e^{i(2\theta)} \:=\:\cos2\theta + i\sin2\theta \:=\:e^{2i\theta}\;\;[2]$

Square [1]: .\(\displaystyle (e^{i\theta})^2 \:=\\cos\theta + i\sin\theta)^2 \:=\\cos^2\!\theta - \sin^2\!\theta) + (2\sin\theta\cos\theta)i \:=\:e^{2i\theta}\;\;[3]\)

Equate [2] and [3]: .\(\displaystyle \cos2\theta + i\sin2\theta \:=\\cos^2\!\theta - \sin^2\!\theta) + (2\sin\theta\cos\theta)i\)

Equate real and imaginary components: .$\displaystyle \begin{Bmatrix}\cos2\theta &=& \cos^2\!\theta - \sin^2\!\theta \\ \sin2\theta &=& 2\sin\theta\cos\theta \end{Bmatrix}$

 

[Relz]

Hello, Relz!



From DeMoivre's theorem, we have: .$\displaystyle e^{i\theta} \:=\:\cos\theta + i\sin\theta\;\;[1]$

. . Hence:.$\displaystyle e^{i(2\theta)} \:=\:\cos2\theta + i\sin2\theta \:=\:e^{2i\theta}\;\;[2]$

Square [1]: .\(\displaystyle (e^{i\theta})^2 \:=\\cos\theta + i\sin\theta)^2 \:=\\cos^2\!\theta - \sin^2\!\theta) + (2\sin\theta\cos\theta)i \:=\:e^{2i\theta}\;\;[3]\)

Equate [2] and [3]: .\(\displaystyle \cos2\theta + i\sin2\theta \:=\\cos^2\!\theta - \sin^2\!\theta) + (2\sin\theta\cos\theta)i\)

Equate real and imaginary components: .$\displaystyle \begin{Bmatrix}\cos2\theta &=& \cos^2\!\theta - \sin^2\!\theta \\ \sin2\theta &=& 2\sin\theta\cos\theta \end{Bmatrix}$

I'm sorry but I don't really understand what you're doing. Could you explain it?

 

[pka]

I'm sorry but I don't really understand what you're doing. Could you explain it?

Do you understand that if $\displaystyle a+bi=x+yi$ then $\displaystyle x=a~\&~y=b~?$

If you do then:
$\displaystyle (cos(\theta)+i\sin(\theta))^2=cos(2\theta)+i\sin(2\theta)$
OR
$\displaystyle cos^2(\theta)-\sin^2(\theta)+2cos(\theta)\sin(\theta)i=cos(2\theta)+i\sin(2\theta)$

So $\displaystyle \sin(2\theta)=2cos(\theta)\sin(\theta)$

 

[Relz]

Do you understand that if $\displaystyle a+bi=x+yi$ then $\displaystyle x=a~\&~y=b~?$

If you do then:
$\displaystyle (cos(\theta)+i\sin(\theta))^2=cos(2\theta)+i\sin(2\theta)$
OR
$\displaystyle cos^2(\theta)-\sin^2(\theta)+2cos(\theta)\sin(\theta)i=cos(2\theta)+i\sin(2\theta)$

So $\displaystyle \sin(2\theta)=2cos(\theta)\sin(\theta)$

I'm sorry but I just don't get how you go from the first equation to the next step. Also, it's 2sin (theta) cos (theta), not the other way around. Unless it doesn't matter..in which case, forget you read that last bit

 

[burakaltr]

I'm sorry but I just don't get how you go from the first equation to the next step. Also, it's 2sin (theta) cos (theta), not the other way around. Unless it doesn't matter..in which case, forget you read that last bit

FROM EULER'S THEOREM :


e to the power of (i times theta) = Cos(theta) + i Sin (theta)

Digest & Accept it as is please - It can be proven in Calculus by use of Taylor's Expansion

 

[Relz]

FROM EULER'S THEOREM :


e to the power of (i times theta) = Cos(theta) + i Sin (theta)

Digest & Accept it as is please - It can be proven in Calculus by use of Taylor's Expansion

Ohhhh okay! I've got it now, thanks. Sorry, I guess I just had a brain fart or something

 

[pka]

Also, it's 2sin (theta) cos (theta), not the other way around.

Are you really saying that you are trying to prove this, and you don't know that $\displaystyle \cos(x)\sin(x)=\sin(x)\cos(x)~?$.

Surely you are not questioning that multiplication is commutative?

 

[Relz]

Are you really saying that you are trying to prove this, and you don't know that $\displaystyle \cos(x)\sin(x)=\sin(x)\cos(x)~?$.

Surely you are not questioning that multiplication is commutative?

No, I'm not. I understand that it is commutative. I was just making sure that you had the same equation as I did