Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Verifying Trig Identities help!

  1. Mar 24, 2005 #1
    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!
     
  2. jcsd
  3. Mar 24, 2005 #2
    What have you tried and where are you stuck ?
     
  4. Mar 24, 2005 #3

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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].
     
  5. Mar 24, 2005 #4
    How about just combining the fractions?
     
  6. Mar 24, 2005 #5
    There's actually a simpler way to solve this without resorting to complex identities that the original poster may not know. :smile:
     
  7. Mar 24, 2005 #6

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    While this may be true, this exponential method never fails, of course. :smile:
     
  8. Mar 24, 2005 #7

    Integral

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

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

    OP , What have you tried?
     
  9. Mar 24, 2005 #8
    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: Mar 24, 2005
  10. Mar 24, 2005 #9
    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 ?
     
  11. Mar 24, 2005 #10
    Yes, look at hypermorphism's last post for ideas. You seem to have made several errors in simplification.
     
  12. Mar 24, 2005 #11

    Integral

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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
     
  13. Mar 24, 2005 #12
    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: Mar 24, 2005
  14. Mar 24, 2005 #13
    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: Mar 24, 2005
  15. Mar 24, 2005 #14
    OHH!!!!! see i told you it would probably be very easy.... thanks for your help!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Verifying Trig Identities help!
  1. Trig Identities (Replies: 4)

  2. Trig Identities: (Replies: 2)

Loading...