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Math Help - Trig solution

  1. #1
    Newbie
    Joined
    Mar 2009
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    1

    Red face Trig solution

    1. The problem statement, all variables and given/known data
    Given: sec \theta= \sqrt{10} where 0< \theta <90

    and \sqrt{10}sin(A- \theta)=sinA-3cosA



    Determine the solution of
    6cosA +3 = 2sinA


    for A \in [-180; 180], rounded off to one decimal digit.

    2. Relevant equations



    3. The attempt at a solution

    3=2sinA - 6cosA

    3=2(sinA-3cosA)

    \frac{3}{2}=sinA-3cosA

    \frac{3}{2}= \sqrt{10}sin(A- \theta)

    Now I can't get rid of the theta
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  2. #2
    Senior Member I-Think's Avatar
    Joined
    Apr 2009
    Posts
    288

    [MATH]\theta[/MATH]

    Good evening

    If sec\theta=\sqrt{10}, then

    \sqrt{10}sin(A-\theta)=\frac{sin(A-\theta)}{cos\theta}
    \frac{sinAcos \theta-sin\theta cosA}{cos\theta}
    sinA-tan\theta cosA=sinA-3cosA

    Hence tan\theta=3
    \theta=tan^{-1}3

    From your equation: =sin(A-)

    Convert to: \frac{3}{2\sqrt{10}}=sin(A-(tan^{-1}3))

    It should be possible to finish from here.
    Hope this helps

    Ciao
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