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Math Help - ST and CT by vectors

  1. #1
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    ST and CT by vectors

    Using vectors prove that

    a) \sin(\alpha-\beta)= \sin \alpha\cos \beta-\cos \alpha\sin \beta
    b) \cos (\alpha-\beta)=\cos \alpha\cos \beta + \sin \alpha\sin \beta
    Last edited by ns1954; February 3rd 2010 at 05:56 AM.
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  2. #2
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    Hello ns1954
    Quote Originally Posted by ns1954 View Post
    Using vectors prove that

    a) \sin(\alpha-\beta)= \sin \alpha\cos \beta-\cos \alpha\sin \beta
    b) \cos (\alpha-\beta)=\cos \alpha\cos \beta + \sin \alpha\sin \beta
    First, note that the dot product of vectors x_1\,\textbf i+y_1\,\textbf j and x_2\,\textbf i+y_2\,\textbf j is:
    (x_1\,\textbf i+y_1\,\textbf j).(x_2\,\textbf i+y_2\,\textbf j)= x_1x_2 +y_1y_2
    =|x_1\,\textbf i+y_1\,\textbf j|\,|x_2\,\textbf i+y_2\,\textbf j|\cos \theta , where \theta is the angle between the vectors
    *****************

    Next, we'll prove the second result first, because it's more straightforward.

    Consider the points on the unit circle A (\cos \alpha, \sin\alpha), B (\cos\beta, \sin\beta). The radii OA, OB make angles \alpha, \beta with the positive x-axis, measured anticlockwise. So \angle BOA = \alpha - \beta, measured anticlockwise from OB to OA.

    Using vector notation, we have:
    \textbf{OA} = \cos\alpha\,\textbf i+\sin\alpha\,\textbf j

    \textbf{OB} = \cos\beta\,\textbf i+\sin\beta\,\textbf j
    Then, using the above result:
    \textbf{OA}.\textbf{OB} = \cos\alpha\cos\beta+\sin\alpha\sin\beta
    =|OA||OB|\cos(\alpha - \beta)

    =\cos(\alpha - \beta), since |OA| = |OB| = 1.
    \Rightarrow \cos(\alpha-\beta) = \cos\alpha\cos\beta+\sin\alpha\sin\beta
    ***************

    Then the proof of the first result. Note that:
    \sin(\alpha-\beta)=\cos\Big(\pi/2-(\alpha -\beta)\Big), using the result just proved
    =\cos(\pi/2+\beta - \alpha)
    Then, with the point A defined as above, consider the point B, (-\sin\beta, \cos\beta) - see the attached diagram. Then the radius OB makes an angle (\pi/2+\beta) with the x-axis. Therefore \angle AOB = \pi/2+\beta - \alpha.

    Then, as before:
    \textbf{OA}.\textbf{OB} = \cos\alpha(-\sin\beta)+\sin\alpha\cos\beta
    =|OA|.|OB|\cos(\pi/2+\beta - \alpha)

    =\sin(\alpha - \beta)
    \Rightarrow \sin(\alpha - \beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta
    Grandad
    Attached Thumbnails Attached Thumbnails ST and CT by vectors-untitled.jpg  
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  3. #3
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    Is it possible to prove this without using koordinate vectors? To say so, more evidence direct profe, like for cosine theorem.
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