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Trigonometric Identity

  1. Dec 12, 2011 #1
    1. The problem statement, all variables and given/known data

    http://img829.imageshack.us/img829/3413/daumequation13237287425.png [Broken]

    Prove that LS=RS.

    2. Relevant equations

    There are no relevant equations.

    3. The attempt at a solution

    http://img829.imageshack.us/img829/3413/daumequation13237287425.png [Broken]
     

    Attached Files:

    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Dec 12, 2011 #2

    Simon Bridge

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    if LHS=RHS, then:

    [tex](\cos x-\sin y)(\cos x+\sin y)=(\cos y+\sin x)(\cos y-\sin x)[/tex]

    ... right?
    So why not expand it out and look for the usual identities?
     
    Last edited: Dec 12, 2011
  4. Dec 12, 2011 #3
    According to my teacher/school (I'm from Ontario), you're not allowed to work on both sides like that. You have to work on each side independent from the other. Why? I have no idea why... :(

    edit: My teacher says you can't do that, because we don't know if LS=RS. We're trying to see if LS=RS by proving it.
     
  5. Dec 12, 2011 #4
    http://img9.imageshack.us/img9/4701/daumequation13237312394.png [Broken]

    That's another identity I'm having trouble with. I have no idea how to change the exponents from 4 to 2. I'm totally lost with this one. I can't even think of a first step.
     
    Last edited by a moderator: May 5, 2017
  6. Dec 12, 2011 #5

    eumyang

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    Regarding your first post in this thread: do NOT multiply out the numerators. Leave them in factored form. So you have this:
    [tex]\frac{(\cos x - \sin y)(\cos y + \sin x)}{cos^2 y - sin^2 x}[/tex]
    Now use the Pythagorean identities to rewrite BOTH terms in the denominator.

    If you figure this out, you'll probably ask, how did I know NOT to multiply out the numerator? I noticed that when you multiplied top and bottom by (cos y + sin x), that is also the numerator of the RHS. You'll see what happens by leaving the numerator in factored form.
     
  7. Dec 12, 2011 #6

    eumyang

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    Note that on the LHS,
    sin2 x + cos4 x = sin2 x + (cos2 x)2
    Use the Pythagorean identity for what's inside the parentheses and see what happens.
     
    Last edited by a moderator: May 5, 2017
  8. Dec 12, 2011 #7
    Amazing! I feel so happy solving these. After that, I noticed that the denominator was a difference of square, and I cancelled out the factors.
     
  9. Dec 12, 2011 #8
    Haha. I got this identity too! Thanks!

    The interesting part about this one is that when I wrote it down on paper, I couldn't solve it. It wasn't until I was inputting into my equation editor to generate the images did I realize what I missed earlier.
     
  10. Dec 12, 2011 #9

    Simon Bridge

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    If LHS=RHS then I can propose (what I did) - and attempt to disprove it. If the proposition is false, then surely the equation will yield an inconsistency like 2=3? Of course, not all will reduce easily.

    You could then start by putting both sides over a common denominator ... then you only have to prove the numerators are the same. This will be mathematically equivalent to the above approach.

    On the other one - without looking: you can reduce the power by using the half-angles.

    [ah - beat me to it]
     
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