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How are these two equations equal? trig identities possibly?

  1. Apr 28, 2013 #1
    1. The problem statement, all variables and given/known data
    Proof that (1/6)sin(3x)-(1/18)sin(9x) = (2/9)sin^3(3x)


    2. Relevant equations



    3. The attempt at a solution

    I am just curious exactly how the power on the sine function is cubic on one side. It obviously has to do with something that increases the power on the sin(9x) function after it becomes a sin(3x). In other words, higher multiple angles somehow increase the trig power and I am not exactly sure how or what formula I should use to prove their equality.
     
  2. jcsd
  3. Apr 28, 2013 #2

    SteamKing

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    I would use the multiple angle formulas for sin(nx) on the LHS of the equation to expand sin(9x).
     
  4. Apr 29, 2013 #3

    Mentallic

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    I'd expand sin(9x) by first expressing it as sin(6x+3x) and then expand both cos(6x) and sin(6x) terms with cos(2(3x)) and sin(2(3x)) - or equivalently, cos(3x+3x) and sin(3x+3x).

    However, if you know the expansion of cos(3u) and sin(3u) in terms of cos(u) and sin(u), then you can jump straight to expressing sin(9x) in terms of 3x.
     
  5. Apr 29, 2013 #4

    Mark44

    Staff: Mentor

    A couple of minor points. "How are these two equations equal?"

    You have only one equation, and the goal of the exercise is to prove that the equation is an identity.

    One equation can never be "equal" to another equation. An equation might be equivalent to another equation if the solutions sets for the two equations are the same.
     
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