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    Hi can someone please illustrate to me how to dervive the half angle formulae from the compound angle formulae (or double angle formulae)? thanks
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    sin(a+b) = sin(a)cos(b) + cos(a)sin(b)
    let b=a
    sin(2a) = sin(a)cos(a) + cos(a)sin(a)
    so
    sin(2a) = 2sin(a)cos(a)
    For the half angle formula. Let 2a = x
    sin(x) = 2sin(0.5x)cos(0.5x)

    cos(a+b) = cos(a)cos(b) - sin(a)sin(b)
    let b=a
    cos(2a) = cos²(a) - sin²(a)
    = 2cos²(a) - 1
    = 1 - 2sin²(a)
    For the half angle formula. Let 2a = x
    cos(x) = cos²(0.5x) - sin²(0.5x)
    = 2cos²(0.5x) - 1
    = 1 - 2sin²(0.5x)
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    (Original post by SsEe)
    sin(a+b) = sin(a)cos(b) + cos(a)sin(b)
    let b=a
    sin(2a) = sin(a)cos(a) + cos(a)sin(a)
    so
    sin(2a) = 2sin(a)cos(a)
    For the half angle formula. Let 2a = x
    sin(x) = 2sin(0.5x)cos(0.5x)

    cos(a+b) = cos(a)cos(b) - sin(a)sin(b)
    let b=a
    cos(2a) = cos²(a) - sin²(a)
    = 2cos²(a) - 1
    = 1 - 2sin²(a)
    For the half angle formula. Let 2a = x
    cos(x) = cos²(0.5x) - sin²(0.5x)
    = 2cos²(0.5x) - 1
    = 1 - 2sin²(0.5x)
    so which bit do you actually use?

    sin(x) = 2sin(0.5x)cos(0.5x)
    1 - 2sin²(0.5x)

    i understand how you got these, but how do you then use them?
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    Can you just use the double angle formula and divide the answer by 4?
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    What unit is this?
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    (Original post by melbourne)
    so which bit do you actually use?

    sin(x) = 2sin(0.5x)cos(0.5x)
    1 - 2sin²(0.5x)

    i understand how you got these, but how do you then use them?
    We had a thread about the half-angle formulae a while ago, and decided, I think, by some kind of majority, that you don't really need them.

    I can't remember having to use them.

    Aitch
 
 
 
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