Verifying Trig Identities help!

What have you tried and where are you stuck ?

 

Substitute [tex]\cos \theta = \displaystyle\frac{e^{i\theta}+e^{-i\theta}}{2}[/tex] and [tex]\sin \theta = \displaystyle\frac{e^{i\theta}-e^{-i\theta}}{2i}[/tex] with the understanding that [itex]i^2=-1[/itex].

 

How about just combining the fractions?

 

There's actually a simpler way to solve this without resorting to complex identities that the original poster may not know.

 

While this may be true, this exponential method never fails, of course.

 

In fact it is nearly trivial, obvious to casual inspection.

OP , What have you tried?

 

This isn't correct. The equation you may be attempting to reference is sin²(x) + cos²(x) = 1.
Regarding your original equation, try putting the fractions on opposite sides of the equation:
[tex]\frac{\cos x - \cos y}{\sin x + \sin y} + \frac{\sin x - \sin y}{\cos x + \cos y} = 0[/tex]
is the same as
[tex]\frac{\cos x - \cos y}{\sin x + \sin y} = -\frac{\sin x - \sin y}{\cos x + \cos y}[/tex]
or
[tex]\frac{\cos x - \cos y}{\sin x + \sin y} = \frac{\sin y - \sin x}{\cos x + \cos y}[/tex]
Do you see a way to get rid of the fractions from here ?

 

Yes, look at hypermorphism's last post for ideas. You seem to have made several errors in simplification.

 

I have bolded your error
(cos^2 x ) - (cos^2 y)/(sin x + sin y) (cos x + cos y) + (sin^2 x)( sin^2 y)/ (sin x + sin y) (cos x + cos y)

go back and double check where this came from, it is not correct. Everything else is ok

 

You have this equation, from hypermorphism's last post:

[tex]\frac{\cos x - \cos y}{\sin x + \sin y} = \frac{\sin y - \sin x}{\cos x + \cos y}[/tex]

Do you remember cross multiplication?