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Math Help - trigo>angle

  1. #1
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    trigo>angle

    sinA+3cosA=2

    Angle between 0* to 360*

    Can use this method Rsin(A+a) to find , or??


    THX FOR WHO HELPING
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  2. #2
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    Quote Originally Posted by sanikui View Post
    sinA+3cosA=2

    Angle between 0* to 360*

    Can use this method Rsin(A+a) to find , or??


    THX FOR WHO HELPING
      \sin A + 3\cos A = 2 \hfill \\

     \Rightarrow 3\cos A = 2 - \sin A \hfill \\

      \Rightarrow 3\sqrt {1 - \sin ^2 A}  = 2 - \sin A{\text{    (because }}\cos^2 A = 1 - \sin ^2 A) \hfill \\

    {\text{squaring both sides,}} \hfill \\

     \Rightarrow 9\left( {1 - \sin ^2 A} \right) = \left( {2 - \sin A} \right)^2  \hfill \\

      \Rightarrow 9 - 9\sin ^2 A = 4 - 4\sin A + \sin ^2 A \hfill \\

       \Rightarrow 10\sin ^2 A - 4\sin A - 5 = 0 \hfill \\

    {\text{This is a quadratic equation in sinA, solve by using quadratic formula, }}

    \sin A = \frac{{ - b \pm \sqrt {b^2  - 4ac} }}<br />
{{2a}} \hfill \\

     \sin A = \frac{{4 \pm \sqrt {\left( { - 4} \right)^2  - 4\left( {10} \right)\left( { - 5} \right)} }}<br />
{{2\left( {10} \right)}} \hfill \\

     \sin A = \frac{{4 \pm \sqrt {216} }}<br />
{{20}} = \frac{{4 \pm 6\sqrt 6 }}<br />
{{20}} \hfill \\

    \sin A = \frac{{2 \pm 3\sqrt 6 }}<br />
{{10}} = 0.934846922,{\text{  }} - 0.534846922 \hfill \\

     \Rightarrow {\text{For positive value,  }}A = 69^\circ ,{\text{  }}111^\circ  \hfill \\

     \Rightarrow {\text{For negative value, }}A = 212^\circ ,{\text{  }}328^\circ  \hfill \\
    Last edited by Shyam; September 11th 2008 at 09:37 PM.
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  3. #3
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    why cosA not same as cosA= 1-2sin^2 (A/2)??
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  4. #4
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    Hello, sanikui!

    \sin A+3\cos A\:=\:2,\quad 0^o \leq A < 360^o

    Divide by \sqrt{10}\!:\quad\frac{1}{\sqrt{10}}\sin A + \frac{3}{\sqrt{10}}\cos A \:=\:\frac{2}{\sqrt{10}} .[1]


    Let \theta be an angle in a right triangle with: . opp = 3,\:adj = 1,\;hyp = \sqrt{10}
    . . Then: . \sin\theta = \frac{3}{\sqrt{10}},\;\cos\theta = \frac{1}{\sqrt{10}}

    Then [1] becomes: . \cos\theta\sin A + \sin\theta\cos A \:=\:\frac{2}{\sqrt{10}}
    . . and we have: . \sin(A + \theta) \:=\:\frac{2}{\sqrt{10}}


    Then: . A + \theta \;=\;\sin^{-1}\left(\frac{2}{\sqrt{10}}\right) \quad\Rightarrow\quad A \;=\;\sin^{-1}\left(\frac{2}{\sqrt{10}}\right) - \theta

    Hence: . A \;=\;\sin^{-1}\left(\frac{2}{\sqrt{10}}\right) - \sin^{-1}\left(\frac{3}{\sqrt{10}}\right)


    . . \sin^{-1}\left(\frac{2}{\sqrt{10}}\right) \;=\;39.23^o \quad\Rightarrow\quad\begin{Bmatrix}140.77^o \\ 399.23^o \end{Bmatrix}

    . . \sin^{-1}\left(\frac{3}{\sqrt{10}}\right) \;=\;71.57^o


    Therefore: . A \;=\;\begin{Bmatrix}140.77^o - 71.57^o &=& \boxed{69.2^o} \\ \\[-3mm] 399.23^o - 71.57^o &=& \boxed{327.66^o} \end{Bmatrix}

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  5. #5
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    Quote Originally Posted by sanikui View Post
    why cosA not same as cosA= 1-2sin^2 (A/2)??
    If you use that formula, and split it in half-angle formula, then you have to split sinA also in half-angle formula. After doing this you still have both sin and cos in the equation and you cannot solve it.

    because you have to make cosA in terms of sinA, so that you can make an equation which has only one variable, so that you can solve it.

    so you have to use, \cos^2 A = 1 - \sin ^2 A
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  6. #6
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    Quote Originally Posted by Shyam View Post
    If you use that formula, and split it in half-angle formula, then you have to split sinA also in half-angle formula. After doing this you still have both sin and cos in the equation and you cannot solve it.

    because you have to make cosA in terms of sinA, so that you can make an equation which has only one variable, so that you can solve it.

    so you have to use, \cos^2 A = 1 - \sin ^2 A
    I understood. Thank you~!V^^
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