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Math Help - A Trigonometry III Math Question

  1. #1
    leo
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    Arrow A Trigonometry III Math Question

    Given that sin A = 4/5 and cos B = -1/3, and that A and B are in the same quadrant, without finding the angles A and B, find the value of sin (A+B). Thanks a lot
    Last edited by leo; June 9th 2009 at 10:30 PM.
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  2. #2
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    Hello, Leo!

    Given that: . \sin A = \tfrac{4}{5},\;\cos B = \text{-}\tfrac{1}{3} and that A and B are in the same quadrant,
    without finding the angles A and B, find the value of \sin(A+B)
    Sine is positive in Quadrants 1 and 2.
    Cosine is negative in Quadrants 2 and 3.
    . . Hence, angles A and B are in Quadrant 2.


    We have: . {\color{blue}\sin A = \frac{4}{5}} in Quadrant 2.

    Since \sin^2\!A + \cos^2\!A \:=\:1, we have: . \left(\tfrac{4}{5}\right)^2 + \cos^2\!A \:=\:1 \quad\Rightarrow\quad \cos^2\!A \:=\:1-\tfrac{16}{25}
    . . \cos^2\!A \:=\:\tfrac{9}{25} \quad\Rightarrow\quad \cos A \:=\:\pm\tfrac{3}{5}
    Since A is in Quadrant 2: . {\color{blue}\cos A \:=\:-\frac{3}{5}}


    We have: . {\color{blue}\cos B = -\frac{1}{3}} in Quadrant 2.

    Since \sin^2\!B + \cos^2\!B \:=\:1, we have: . \sin^2\!B + \left(\text{-}\tfrac{1}{3}\right)^2 \:=\:1 \quad\Rightarrow\quad \sin^2\!B \:=\:1 - \tfrac{1}{9}
    . . sin^2\!B \:=\:\tfrac{8}{9}\quad\Rightarrow\quad \sin B \:=\:\pm\tfrac{2\sqrt{2}}{3}
    Since B is in Quadrant 2: . {\color{blue}\sin B \:=\:\frac{2\sqrt{2}}{3}}


    We know that: . \sin(A + B) \;=\;\sin A\cos B + \sin B\cos A

    Substitute: . \sin(A + B) \;=\;\left(\frac{4}{5}\right)\left(-\frac{1}{3}\right) + \left(\frac{2\sqrt{2}}{3}\right)\left(-\frac{3}{5}\right) \;=\;\frac{-4 - 6\sqrt{2}}{15}

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