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    Hey, I'm having trouble proving that
    2cos3x - sin2xsinx = 2cos2x cosx

    I can get the left hand side down to
    = 2cos3x - (2sinxcosx)sinx
    = 2cos3x - 2sin2xsinxcosx

    And then the right hand side down to
    = 2(cos2x-sin2x)cosx
    = (2cos2x-2sin2x)cosx
    = 2cos3x-2sin2xcosx

    From here I'm stumped as despite being identical bar the extra sinx I can't figure out if it's there by mistake or if I've made a mistake higher up in my working, any help?
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    One too many sinx's on your second line of working.


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    • Thread Starter
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    (Original post by Krollo)
    One too many sinx's on your second line of working.


    Posted from TSR Mobile
    Thanks!
 
 
 
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