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4Sine(4X) = -8Sin(2x) Double angle identity

  1. Mar 4, 2015 #1
    I kind of understand what to do with this when there are no numbers in front of the expressions, I also kind of understand that you can rewrite 4Sine(4x) as 4Sine(2x+2x) hat do I do with the 4 and 8? In an algebra problem you could divide the 4 into the -8, then simplify that expression, am I correct in assuming that taking just sin(2x+2x) can be written as: 2Sin(x)Cos(x) + 2Sin(x)Cos(x)? Or am I way off base in all of my assumptions? The explanation in my math book doesn't explain the whole problem.
     
  2. jcsd
  3. Mar 4, 2015 #2

    haruspex

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    Seems reasonable. You could also consider 2x as y, simplifying the innards.
    No.
     
  4. Mar 4, 2015 #3
    What do I do with the 2?
     
  5. Mar 4, 2015 #4

    haruspex

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    You mean after dividing the 4 in to the 8? Leave it be for now. Concentrate on expanding the sin(4x).
     
  6. Mar 4, 2015 #5
    Expanding the (4x) this is the bit je ne comprends pas. (I don't understand)
     
  7. Mar 4, 2015 #6

    haruspex

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    How would you expand sin(2x)?
     
  8. Mar 4, 2015 #7
    I believe it is 2Sin(x)Cos(x) but you said that that was not how you did 4x, so now I'm thinking: Sin(2x)Cos(2x) Or it could be 2SinCos but that doesn't seem right. If the middle one is correct I think I may understand now.
     
  9. Mar 4, 2015 #8

    Mark44

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    No, and here's why. You cannot split up sin(2x + 2x) to sin(2x) + sin(2x) as if "sin" were multiplying (2x + 2x). You are confusing function notation with the distributive law, which says that a(b + c) = ab + ac.

    sin(2x + 2x) is no more equal to sin(2x) + sin(2x) than is ##\sqrt{a + b}## the same as ##\sqrt{a} + \sqrt{b}##.

    The gist of this problem, at least according to what you wrote in the thread title, is solve the equation 4sin(4x) = -8sin(2x).

    Can you solve the equation 4sin(2y) = -8sin(y)?
     
    Last edited: Mar 5, 2015
  10. Mar 5, 2015 #9
    Then I have absolutely no clue what is going on with that part of the problem, I know it relates somehow to the sum and difference identities, but the two with the x confuses me.
     
  11. Mar 5, 2015 #10
    I think Mr @Mark44 is true.
    Can you start solving this equation?
    The only identity you need is
    sin(2u)=2sin(u)cos(u)
    What is the first step now?
     
  12. Mar 5, 2015 #11

    SteamKing

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    The trigonometric identities for finding the sine and cosine of the sum of two angles are:

    sin (α + β) = sin (α) * cos (β) + cos (α) * sin (β)
    cos (α + β) = cos (α) * cos (β) - sin (α) * sin (β)

    If you want to expand sin (2x) = sin (x + x), which implies that α = β = x, then

    sin (2x) = sin (x) * cos (x) + cos (x) * sin (x)

    sin (4x) = sin (2x + 2x), which implies α = β = 2x

    I'll leave it to you to work out the rest of the expansion for sin (4x).
     
  13. Mar 5, 2015 #12

    haruspex

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    Yes, it's 2sin(x)cos(x), and you can do 4x 'that way', but the way you extended it to 4x was wrong. Think of it as sin(2(2x)).
     
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