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    I feel embarrassed even asking this but I can't find it on the net... I know it'll be there somewhere but whatever I Google I'm either getting really basic stuff, or really advanced stuff...

    What do I get if I multiply, say, cos2x by cos3x? Does it just stay as cos2xcos3x or can it be simplified further?

    Thanks
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    I'm not sure what you mean by simplify here. You can't straight away say that e.g. cos2xcos3x=cos6x but you can reduce it into powers of cos (do you want this) using the angle sum formula

    \cos (a+b) = \cos a \cos b - \sin a \sin b

    Note that cos(2x)=cos(x+x) and cos(3x)=cos(2x+x).

    Have a go at reducing and post what you get.
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    cosA*cosB = 2 * cos((A+B)/2) * cos((A-B)/2)
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    (Original post by FieldLeftBlank)
    cosA*cosB = 2 * cos((A+B)/2) * cos((A-B)/2)
    Yes that's another way to simplify it if you know the formula. I'm not sure if it's in the formula booklet currently.
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    (Original post by notnek)
    I'm not sure what you mean by simplify here. You can't straight away say that e.g. cos2xcos3x=cos6x but you can reduce it into powers of cos (do you want this) using the angle sum formula

    \cos (a+b) = \cos a \cos b - \sin a \sin b

    Note that cos(2x)=cos(x+x) and cos(3x)=cos(2x+x).

    Have a go at reducing and post what you get.
    Ah right, that makes sense, and rings a bell, not sure if this will give me what I want

    cos(2x) = cos(x)^2 - sin(x)^2
    cos(3x) = cos(x)cos(2x) - sin(x)(sin2x)
    cos(3x) = cos(x)(cos(x)^2 - sin(x)^2)...

    Okay I can just see this going in circles Don't think this is the direction the question expects me to go in

    I'll give this another shot, thanks for the help

    EDIT: got it, I was missing something else that was really obvious... whoops :o:
 
 
 
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