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Math Help - Trigonometry double angle question

  1. #1
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    Trigonometry double angle question

    Hey Im totally stuck on this question if somebody could show me a few pointers it would be well appreciated, thank you very much!


    Find expressions in sin(nx) and cos(mx) for:

    Sin^3(x)
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  2. #2
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    Re: Trigonometry double angle question

    Hi Khalif27,

    you have to apply (a+b)^3=a^3+3*a^2*b+3a*b^2+b^3
    and i^2=-1, i^3=-i,
    from the formula Moivre's you have:

    (cosx+isinx)^3=(cos3x+isin3x)
    (cosx+isinx)^3= cos^3(x)+3icos^2(x)*sin(x)-3cos(x)*sin^2(x) -icox^3(x)
    = cos^3(x)-3cos(x)*sin^2(x) + i*[3cos^2(x)*sin(x)-sin^3(x)]

    ==> cos(3x)= cos^3(x)-3cos(x)*sin^2(x) = cos^3(x)-3cos(x)*(1-cos^2(x))=4cos^3(x)-3cos(x)
    and sin(3x)= 3cos^2(x)*sin(x)-sin^3(x) =3(1-sin^2)*sin(x)-sin^3(x)=3sin(x)-4sin^3(x)
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