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Homework Help: Trig identities: does this equal zero?

  1. Mar 7, 2010 #1
    1. The problem statement, all variables and given/known data

    This is actually many steps through a calculus problem involving trig functions. I have not included the problem because I'm trying very hard to figure it out on my own (at least as far as it's possible). I've found the answer I'm looking for, but it's attached to a bunch of random junk. I'll be fine as long as the following identity is true:

    2(tanx(sinx)^2-cotx(cosx)^2) + 2(tanx+cotx) + (tanx)^2 + (cotx)^2 + 2 = 0

    Problem is, I just can't seem to make it work. I've triple-checked all the steps up to this point, so I'm pretty confident that the numbers here are correct. Is there any way that the above identity is true?

    2. Relevant equations

    (cosx)^2+(sinx)^2 = 1
    (tanx)^2 + 1 = (secx)^2
    (cotx)^2 + 1 = (cscx)^2

    3. The attempt at a solution

    Working piecemeal, trying to find hidden identities within the above equation. For instance:
    dividing 2sinxcosx from (2cotx-cotx(cosx)^2) produces 2sinxcosx((cscx)^2-(cotx)^2=2sinxcosx(1)=sin2x
    I can also find both (1+(tanx)^2) and (1+cotx)^2) in the equation, producing (secx)^2 and (cscx)^2. However, none of these seem to be steps in the right direction.

    It seems like what I need is to find some way to pull -2 out of an identity that equals 1, which would allow me to get rid of that 2 on the end. Then I need to arrange the rest of the identities so that they equal zero. I'm beginning to suspect this isn't possible, but my brain just can't let go.
  2. jcsd
  3. Mar 7, 2010 #2
    Nevermind. I started the problem over, taking the derivative using quotient rule rather than product rule and discovered that I had two signs flipped around. This is exactly the problem I kept slamming up against with the above, so I'm hoping this will solve everything. If'n it does, let me just say I really appreciate the service you guys have here, keep up the great work!
  4. Mar 7, 2010 #3
    So I've redone the entire equation, and have now become stuck at a different point. Here's what I have left:

    [ ((cotx)^2 - (tanx)^2) + 8(cosxsinx) +8(sinx)^2]/(1+tanx)^2(1+cotx)^2= -cos2x

    I'm pretty sure that denominator just equals 4, so what I'm looking for is the numerator to equal -4((cosx)^2-(sinx)^2) and then I've got my answer. But I can't find that in here anywhere. Am I missing something?
  5. Mar 7, 2010 #4
    Well, I'm really losing hope now. Of course, I found the denominator does not equal zero, but is in fact just another giant mess. I simplified the numerator by dividing (sinx)^2 and (cotx)^2 by (sinx)^2 to get (sinx)^2*(1+(secx)^2)=(sinx)^2(tanx)^2, then divided that and my original (tanx)^2 by -(tanx)^2 to get -tanx(1-(sinx)^2) = -(tanx)^2(cosx)^2 = -(sinx)^2. So after that, the numerator becomes:

    8(cosxsinx) + 6(sinx)^2

    And looks even more impenetrable than before. Ugh.
  6. Mar 7, 2010 #5
    If I substitute x = pi/4, I get 1/2 = 0, so I think you made an earlier mistake.
  7. Mar 7, 2010 #6

    Thank you. For some reason, such an obvious test never occurred to me :)
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