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Math Help - Rewrite Acos(wt)+Bsin(wt) as Ccos(wt+D)

  1. #1
    Junior Member teuthid's Avatar
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    Rewrite Acos(wt)+Bsin(wt) as Ccos(wt+D)

    Ok, so I keep seeing textbooks handwave over this:

    Constants A, B, C, and D exist such that
    A \cos{(\omega t)}+B \sin{(\omega t)}
    can be rewritten as
    C \cos{(\omega t -D)}

    But I haven't bee able to find/figure out the derivation of this. Can someone help me fill in the details of this?

    P.S. I found one book that added C=\sqrt{A^2+b^2} and D=\tan^{-1}{(B/A)}.
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  2. #2
    Super Member
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    Quote Originally Posted by teuthid View Post
    Ok, so I keep seeing textbooks handwave over this:

    Constants A, B, C, and D exist such that
    A \cos{(\omega t)}+B \sin{(\omega t)}
    can be rewritten as
    C \cos{(\omega t -D)}

    But I haven't bee able to find/figure out the derivation of this. Can someone help me fill in the details of this?

    P.S. I found one book that added C=\sqrt{A^2+b^2} and D=\tan^{-1}{(B/A)}.
    Let A = C*cos(D) and B = C*sin(D)

    Then A^2 + B^2 = C^2

    C = \sqrt{A^2 + B^2}

    Acos(ωt) + Bsin(ωt) = (\sqrt{A^2 + B^2})(\frac{A}{\sqrt{A^2 + B^2}}\cos(\omega{t}) + \frac{B}{\sqrt{A^2 + B^2}}\sin(\omega{t}))

    Now proceed.
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