Verifying Trig Identities help!

  • Thread starter InFiNiTeX
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  • #1
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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!
 

Answers and Replies

  • #2
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 ?
 
  • #3
robphy
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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].
 
  • #4
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How about just combining the fractions?
 
  • #5
robphy said:
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].
There's actually a simpler way to solve this without resorting to complex identities that the original poster may not know. :smile:
 
  • #6
robphy
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hypermorphism said:
There's actually a simpler way to solve this without resorting to complex identities that the original poster may not know. :smile:
While this may be true, this exponential method never fails, of course. :smile:
 
  • #7
Integral
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In fact it is nearly trivial, obvious to casual inspection.

OP , What have you tried?
 
  • #8
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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..

i hope that is the right answer, if not could you please help me go in the right direction? thanks for your guys much appreciated help!
 
Last edited:
  • #9
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:
[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 ?
 
  • #10
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Yes, look at hypermorphism's last post for ideas. You seem to have made several errors in simplification.
 
  • #11
Integral
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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
 
  • #12
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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....
 
Last edited:
  • #13
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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?
 
Last edited:
  • #14
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OHH!!!!! see i told you it would probably be very easy.... thanks for your help!
 

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