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Thread: 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; Jun 9th 2009 at 10:30 PM.
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  2. #2
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    Hello, Leo!

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


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

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


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

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


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

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

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