Eulers formula problem

1. Oct 5, 2011

metalscot

The question is: Using Eulers formula for e^±iθ , obtain the trigometric identities for cos(θ1, θ2) and sin(θ1, θ2)

I think I have completed the real and imaginary solutions for the base e^+iθ using the real part cos(θ1+θ2)and imaginary isin(θ1+θ2)

Gaining

cos(θ1 + θ2) = cosθ1 cosθ2 - sinθ1 sinθ2

sin(θ1 + θ2) = sinθ1 cosθ2 + cosθ1 sinθ2

My problem is how to proceed from here for the base e^-iθ

should my real and imaginary parts be in the form

cos(θ1-θ2) + isin(θ1-θ2)

or

cos(θ1+θ2) - isin(θ1+θ2)

Thanks in advance for any help

2. Oct 5, 2011

Mute

I'm not sure what it is you are trying to do. As I understand it, You want to derive the angle addition identities for sine and cosine using Euler's formula. So, the strategy is:

You know

$$e^{i(\theta_1 \pm \theta_2)} = \cos(\theta_1 \pm \theta_2) + i \sin(\theta_1 \pm \theta_2).$$

However, you also know $\exp(i(\theta_1 \pm \theta_2)) = \exp(i\theta_1)\exp(\pm i \theta_2)$. Do you think you can figure out the next step to get the identities?

Last edited by a moderator: Oct 5, 2011
3. Oct 5, 2011

metalscot

Thank you Mute

e^(θ1±θ2)=cos(θ1±θ2)+isin(θ1±θ2)

Gives the below for +

e^(θ1+θ2)=cos(θ1+θ2)+isin(θ1+θ2)

Using the rule cos(A+B) = cosAcosB - sinAsinB
sin(A+B) = sinAcosB + sinBcosA

I get real part cos(θ1 + θ2) = cosθ1 cosθ2 - sinθ1 sinθ2
and imaginary part sin(θ1 + θ2) = sinθ1 cosθ2 + cosθ1 sinθ2

For the negative however using the same method I get the same real part and I am unsure as to the imaginary part as follows

(e^(θ1±θ2)=cos(θ1±θ2)+isin(θ1±θ2)

for - gives

e^(θ1-θ2)=cos(θ1+θ2)+isin(θ1-θ2)

using the rule above for the real part I again get cos(θ1 + θ2)= cosθ1cosθ2 - sinθ1sinθ2

from here however i am stuck?

4. Oct 5, 2011

Mute

Hi Metalscot - could you clarify what the problem wants you to do? I get the impression that you are supposed to use Euler's formula to obtain the angle sum identities. Is this correct?

If so, you cannot use the identities during your work, as you are trying to prove them. If you take another look at my previous post, I said

$$e^{i(\theta_1\pm\theta_2)} = e^{i\theta_1}e^{\pm i\theta_2}$$

What happens if you use Euler's formula for both exponentials on the right hand side?

5. Oct 6, 2011

metalscot

Using Eulers formula for e^±iθ , obtain the trigometric identities for cos(θ1, θ2) and sin(θ1, θ2)

The above is the only information that I am given

Using eulers formula for both exponentials I get

eiθ1= cosθ1 + isinθ2

e-iθ2=cosθ2-isinθ2

eiθ2=cosθ2+isinθ2

combing i get cosθ1+isinθ2+2cosθ2

6. Oct 6, 2011

Mute

That's not what I was trying to get you do. I didn't want you to sum those forms together.
Say we do the plus case. What I wanted you to try was the following:

$$e^{i\theta_1}e^{\pm i\theta_2} = (cos\theta_1 + i\sin\theta_1)(\cos\theta_2 \pm i \sin\theta_2)$$

Expand the right hand side and collect the real parts together and the imaginary parts together so that after you expanded it it looks like (stuff) + i*(other stuff) (i.e., x + i*y form for complex numbers).

After you do that, you want to compare to the form you got when you just wrote $\exp(i(\theta_1 + \theta_2))$ out directly in terms of cosine and sine.

Now, if two complex numbers are equal, what does that tell you about their real and imaginary parts? This should get you the plus trig identities. If you redo the steps with the minus sign, you will get the minus trig identities.

So, try it again and let us know if you get stuck again.

Last edited: Oct 6, 2011
7. Oct 6, 2011

metalscot

Thanksfor your help Mute I think I understand now