# Verifying Trig Identities help!

cos x - cos y sin x - sin y
sin x + sin y + cos x + cos y = 0

or to see better i guess....

(cos x - cos y)/(sin x + sin y) + (sin x - siny)/(cos x + cos y) = 0

can you guys help me? i'm really stuck on this!

InFiNiTeX said:
cos x - cos y sin x - sin y
sin x + sin y + cos x + cos y = 0

or to see better i guess....

(cos x - cos y)/(sin x + sin y) + (sin x - siny)/(cos x + cos y) = 0

can you guys help me? i'm really stuck on this!
What have you tried and where are you stuck ?

robphy
Homework Helper
Gold Member
Substitute $$\cos \theta = \displaystyle\frac{e^{i\theta}+e^{-i\theta}}{2}$$ and $$\sin \theta = \displaystyle\frac{e^{i\theta}-e^{-i\theta}}{2i}$$ with the understanding that $i^2=-1$.

How about just combining the fractions?

robphy said:
Substitute $$\cos \theta = \displaystyle\frac{e^{i\theta}+e^{-i\theta}}{2}$$ and $$\sin \theta = \displaystyle\frac{e^{i\theta}-e^{-i\theta}}{2i}$$ with the understanding that $i^2=-1$.
There's actually a simpler way to solve this without resorting to complex identities that the original poster may not know.

robphy
Homework Helper
Gold Member
hypermorphism said:
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.

Integral
Staff Emeritus
Gold Member
In fact it is nearly trivial, obvious to casual inspection.

OP , What have you tried?

well this is what i tried to do....

first i tried to combine them

(cos x - cos y) ( sin x - sin y)/ (sin x + sin y) (cos x + cos y)

then i got

(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)

and after that part i kinda just confused myself but i just thought of this....

that part changes to

(1 - sin^2 x) - (1 - sin^2 y) / (sin x + sin y) (cos x + cos y) + (1 - cos^2 x) (1 - cos y) / (sin x + sin y) (cos x + cos y)

which i can turn into

(1 - sin x) - (1 - sin y) / cos x + cos y + (1 - cos x) (1 - cos y) / (sin x + sin y)

now those fractions cancel out right? cause (1 - sin x) and (1 - sin y) is the same as cos x and cos y as (1 - cos x) and (1 - cos y) are the same as (sin x) and (sin y) right? so if all that cancels out i get 0!

while in the shower it struck me that 1 - sin x = cos x , so i went from there..

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InFiNiTeX said:
while in the shower it struck me that 1 - sin x = cos x , so i went from there..
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:
$$\frac{\cos x - \cos y}{\sin x + \sin y} + \frac{\sin x - \sin y}{\cos x + \cos y} = 0$$
is the same as
$$\frac{\cos x - \cos y}{\sin x + \sin y} = -\frac{\sin x - \sin y}{\cos x + \cos y}$$
or
$$\frac{\cos x - \cos y}{\sin x + \sin y} = \frac{\sin y - \sin x}{\cos x + \cos y}$$
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.

Integral
Staff Emeritus
Gold Member
(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

ok thanks for your help Integral! and hypermorphism, sorry i just can't see how to get rid of the fractions... i'm sure its very easy and i will feel really dumb though....

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

$$\frac{\cos x - \cos y}{\sin x + \sin y} = \frac{\sin y - \sin x}{\cos x + \cos y}$$

Do you remember cross multiplication?

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OHH!!!!! see i told you it would probably be very easy.... thanks for your help!