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Help with a Trigonometric identity

  1. Feb 2, 2008 #1
    Help with a Trigonometric identity....

    1. The problem statement, all variables and given/known data

    (sin x + sin 2x + sin 4x) / (cos x + cos 2x + cos 4x) = tan 2x

    2. Relevant equations

    sin 2x = 2sinxcosx; cos 2x = cos^2x - sin^2x

    3. The attempt at a solution

    solving left side,

    =[sin x + sin 2x + sin (2x + 2x)]/[cos x + cos 2x + cos (2x+2x)]
    =[sin x + sin 2x + 4 sin x*cos^3 x - 4 sin^3 x * cos x] / [cos x + cos 2x + 1 - 6 sin^2 x * cos^2 x]

    Cannot go beyond this!

    solving right side,
    =2 sin x cos x/(cos^2 x - sin^2 x)
    =sin 2x/cos 2x
     
  2. jcsd
  3. Feb 3, 2008 #2

    Defennder

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    I don't even think that identity holds. Try x=pi/4, pi/6. It doesn't add up.
     
  4. Feb 3, 2008 #3

    HallsofIvy

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    What exactly is the "problem statement"? You give an expression. What do want to do with it?
     
  5. Feb 3, 2008 #4

    Defennder

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    He wants to show how the left-hand side of the equation can be reduced to the right hand side. It's an O-level type of proving trigo identities question.
     
    Last edited: Feb 3, 2008
  6. Feb 3, 2008 #5
    This is one of those messed up identities. As Defennnder already pointed out, there are values that make the right side undefined, that clearly don't make the left side undefined. Its not a true identities. Many text books for some reason do this...
     
  7. Feb 4, 2008 #6

    HallsofIvy

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    And how do you know? Did questionn tell you personally or is that just your opinion. The fact that it is NOT an identity makes me wonder.
     
  8. Feb 4, 2008 #7

    HallsofIvy

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    And how do you know? Did questionn tell you personally or is that just your opinion. The fact that it is NOT an identity makes me wonder.
     
  9. Feb 4, 2008 #8

    Defennder

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    Well, that's the best intepretation I can give it. The equal sign is a big hint though. More importantly the "strategy" used to solve this type of problem is to either start with the LHS and reduce it to the RHS or start with RHS then reduce to LHS is evident in his post. The thread title also hints at this. But if he didn't make any typo error, then perhaps the question should be interpreted to mean "Prove the trigo identity, if possible"
     
  10. Feb 4, 2008 #9

    HallsofIvy

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    Why are you assuming it couldn't be "solve this equation"?
     
  11. Feb 4, 2008 #10

    Defennder

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    Yeah I guess you may be right. That's probably what it could mean, since I don't know what else it could be. Just have to wait for the OP to reply.
     
  12. Feb 4, 2008 #11
    sin4x = 2sin2xcos2x

    thats the way i was taught to do it
     
    Last edited: Feb 4, 2008
  13. Feb 4, 2008 #12

    rock.freak667

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    Usually the factor formulae are helpful when you have things like sinP+sinQ
     
  14. Feb 5, 2008 #13
    The topic of the title is Help with a Trigonometric identity.... Why would it become a solve the equation?
     
  15. Feb 6, 2008 #14
    I'm also going to assume it's a trig identity, since that's the title of this thread, not "help me solve this trig equation." If you inspect the graph of each function, they're sort of close together between about -pi/4 and pi/4. But, clearly, the graphs are not the same. (OP, that's one strategy to see if it is an identity.)

    I know I've seen a trig identity very similar to that problem. If I had more time, I'd play a little and see if I couldn't come up with one.
     
    Last edited: Feb 6, 2008
  16. Feb 6, 2008 #15
    Okay, I worked it backwards a bit
    (I know "you're not allowed to do this", but I was trying to see what the typo might have been.)

    multiply the right side by the denominator.
    tan2x (cosx + cos2x + cos4x)
    now, the tan2x times cos2x takes care of the sin2x term on the left hand side of the equation.

    Subtract the sin2x from both sides to see what we have left.
    Left: sinx + sin4x
    Right: tan2x(cosx+cos4x)
    Right: tan2x * cosx + tan2x*cos4x
    I'll look at the 2nd term first:
    tan2x is tan(4x/2) - use a half angle formula.
    [(1-cos4x)/sin4x]*cos4x

    oh crud, never mind; I wanted to badly to see 1 - cos^2 4x in that numerator, which would reduce that 2nd term to sin4x. Close, but no cigar. However, that hinted that there may be something relatively close to what the OP's original problem is... My kids are waiting in the car else I'd spend a couple more minutes on it.

    It might be worthwhile to attempt to start with tan2x written as tan(4x/2) and use the half angle identity = (1-cos4x)/sin4x then fiddle around with it to introduce some other terms.
     
    Last edited: Feb 6, 2008
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