How Do You Prove Trigonometric Identities in Complex Number Equations?

[chaoseverlasting]

Homework Statement

I have an exam coming up on Monday, and I can't seem to solve this question. Please point me in the right direction.

$$x=e^{i\alpha}$$, $$y=e^{i\beta}$$ $$z=e^{i\gamma}$$.

If $$x+y+z=xyz$$, prove that,

$$cos(\beta -\gamma) + cos(\gamma -\alpha) + cos(\alpha -\beta) + 1=0$$

Homework Equations

$$e^{ix}=cosx +i sinx$$
$$2cos\theta=e^{i\theta}+e^{-i\theta}$$

The Attempt at a Solution

I tried to go from x+y+z=xyz by using the Euler form but couldn't get anywhere.

Then I multiplied the result by two to get $$2cos(\beta -\gamma) + 2cos(\gamma -\alpha) + 2cos(\alpha -\beta) + 2=0$$.

Splitting the cosine terms into the Euler form and then rearranging the equation, I got:

$$\frac{e^{i\alpha}+e^{i\beta}+e^{i\gamma}}{e^{i(\alpha+\beta+\gamma)}}(e^{2i\alpha}+e^{2i\beta}+e^{2i\gamma}-(e^{3i\alpha}+e^{3i\beta}+e^{3i\gamma}))+2=0$$

Now, if I can prove
$$e^{2i\alpha}+e^{2i\beta}+e^{2i\gamma}-(e^{3i\alpha}+e^{3i\beta}+e^{3i\gamma}=-2$$, then by implication, x+y+z=xyz as

$$e^{i\alpha}+e^{i\beta}+e^{i\gamma}=x+y+z$$ and $$e^{i(\alpha+\beta+\gamma)}=xyz$$.

How do I go about this?

 

[learningphysics]

The way I'd approach it... I'd try to simplify (y/z) + (z/x) + (x/y)... and see what comes out... the real part of this sum is:

$$cos(\beta -\gamma) + cos(\gamma -\alpha) + cos(\alpha -\beta)$$...

 

[chaoseverlasting]

I just tried that, here's where I am getting stuck with it:

$$\frac{xy^2+yz^2+zx^2+xyz}{xyz}$$. Where do I go from here?

 

[learningphysics]

Yeah, I wasn't able to get anywhere with that either...

using x + y + z = xyz

you know that... cos(alpha) + cos(beta) + cos(gamma) = cos(alpha + beta + gamma)

you also know that sin(alpha) + sin(beta) + sin(gamma) = sin(alpha + beta + gamma)

square both equations and add... with the cos(A-B) identity... this will give you the result.

 

[chaoseverlasting]

Thank you. Thats excellent! You're a lifesaver!