1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Very difficult trig identity

  1. Nov 20, 2008 #1
    1. The problem statement, all variables and given/known data
    I need help proving this identitiy.

    tan3A = -------------- or tan3A = cos2A-cos4A/sin4A-sin2A

    there both the same, just different way of writing it. please help! :)

    2. Relevant equations

    3. The attempt at a solution

    I honestly don't have any idea where to start :(
  2. jcsd
  3. Nov 20, 2008 #2
    Hello! Do you feel comfortable to start with sin3A/cos3A ? My point is to prove the right side of the equation, so tg3A=sin3A/cos3A. My next step will be sin(2A+A)/cos(2A+A).
    It will look like:
    Now use the trigonometric identities (product-to-sum formulas).
    Consider the fact that this forum uses LaTeX. You can find the math expressions by clicking on the Σ button (on the right side of the tools).
  4. Nov 20, 2008 #3
    I honestly, don't understand what your doing here?
  5. Nov 20, 2008 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    What specifically don't you understand about the previous post?
  6. Nov 20, 2008 #5


    User Avatar
    Science Advisor

    He's using the sum formulas for sine and cosine. Do you know them?
  7. Nov 20, 2008 #6
    ok so now i'm stuck on

    2sinAcosA+cos^2A-Sin^2ASinA / cos^2A-Sin^2AcosA+2sinAcosAsinA
  8. Nov 20, 2008 #7
    I'm not quite sure where you get that expression. If you don't manipulating the tan3A side to prove the right side, you can try using the sum to product identities on the right side.
    [tex]cos(a) - cos(b) = -2sin(\frac{a + b}{2})sin(\frac{a - b}{2})[/tex]
    [tex]sin(a) - sin(b) = 2cos(\frac{a + b}{2})sin(\frac{a - b}{2})[/tex]

    By the way, I like that Latex thing, now.
    Last edited: Nov 20, 2008
  9. Nov 20, 2008 #8
    Latex is very confusing for me, so sorry... ok well i'm at

    sin2AcosA + cos2AsinA / cos2AcosA - sin2AsinA

    i manipulated tan3A using the sum product identity
  10. Nov 20, 2008 #9
    actually i didn't use the product sum i used what i was told to use up a few posts
  11. Nov 20, 2008 #10
    chaos2009, how did they derive the product and sum formulas from the addition formulas?
  12. Nov 20, 2008 #11
    Hmm, well if your talking about the trigonometric identities, you just have to be really tricky about how you rewrite things.
  13. Nov 20, 2008 #12
    also, i think u made a mistake, in the cos- cos identity, isn't it = -2sin((a+b)/2)...
  14. Nov 20, 2008 #13
    Yes, yes it is.
  15. Nov 20, 2008 #14
    so how do i solve this then :S
  16. Nov 20, 2008 #15
    I'm not quite sure where the other people were trying to go with this problem. But, you can try simplifying the right side of your original equation using the sum to product identities. Your numerator is a cos x - cos y and the denominator is a sin x - sin y.
  17. Nov 20, 2008 #16
    i'm pretty sure this gives me -tan3A, does it not?
  18. Nov 20, 2008 #17
    After the sum to product identities, you should get something like:
    The sine function is an odd function so:
    [tex]sin(-x) = -sin(x)[/tex]
    Plug that back into what we got:
    I don't think there should be a negative.
  19. Nov 20, 2008 #18
    ok, one last question, so to bug the hell outta u, can you explain how the sin function akes it back to positive
  20. Nov 20, 2008 #19
    It is that middle equation I have in my previous post. The sine function is an odd function which means:
    [tex]sin(-x) = -sin(x)[/tex]
    Because we got a sin(-A) in the numerator, we replace it with -sin(A).
    [tex]-2 * sin(3A) * sin(-A) = -2 * sin(3A) * -sin(A)[/tex]
    The two negative sign will kind of cancel each other out because a negative times a negative is a positive. Therefore, our new numerator is positive.
  21. Nov 20, 2008 #20
    why is sinA on the denominator positive, from what i understand sin((2A-4A)/2)= sin-A
  22. Nov 20, 2008 #21
    nvm :$
  23. Nov 21, 2008 #22

    I wondering if we can use LHS to prove RHS..
    i cant do it, i tried.
    Last edited: Nov 21, 2008
  24. Nov 21, 2008 #23
    Look at my second post. Maybe, it is a little bit messy to go with, but it is ok. :smile:

    Starting with the right side it looks straight away.
  25. Nov 21, 2008 #24
    yes. of cos i know how to start wif the LHS.
    but i cant derive with RHS, why not you try it too. (:
    i also did tried with tan3A=[tex]\frac{tan2A+tanA}{1-(tan 2A)(tan A)}[/tex]
    Last edited: Nov 21, 2008
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook