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Math Help - New trigo-system

  1. #1
    Super Member dhiab's Avatar
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    New trigo-system

    Solve :
     <br />
sin(x)cos(y)=\frac{1}{4}<br />
     <br />
3Tan(x)=Tan(y)<br />
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  2. #2
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    Quote Originally Posted by dhiab View Post
    Solve :
     <br />
sin(x)cos(y)=\frac{1}{4}<br />
     <br />
3Tan(x)=Tan(y)<br />
    Infinitely many solutions.
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  3. #3
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     \sin(x) \cos(y) = \frac{1}{4}

     3\tan(x) = \tan(y) or

     3\sin(x)\cos(y) = \sin(y)\cos(x) = \frac{3}{4}

    we have

    \sin(y)\cos(x) = \frac{3}{4} and

     \sin(x) \cos(y) = \frac{1}{4}

     \sin(x+y) = \sin(x) \cos(y) + sin(y)\cos(x) = 1

     \sin(y-x) = sin(y)\cos(x) - \sin(x) \cos(y) = \frac{1}{2}

     x+y = \frac{4n+1}{2} \pi

     y-x = m\pi + (-1)^m \frac{\pi}{6}

    Therefore

    x = \frac{1}{2}(\frac{4n+1}{2} \pi - m\pi - (-1)^m \frac{\pi}{6} )

     y = \frac{1}{2}(\frac{4n+1}{2} \pi + m\pi + (-1)^m \frac{\pi}{6} )
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  4. #4
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    Hello, dhiab!

    I will assume that x and y are acute angles.
    You can generalize them later.


    Solve: .  \begin{array}{cccc}<br />
\sin x\cos y &=& \frac{1}{4} & [1] \\ 3\tan x&=& \tan y & [2] \end{array}

    From [2], we have: . 3\frac{\sin x}{\cos x} \:=\:\frac{\sin y}{\cos y} \quad\Rightarrow\quad \cos x\sin y \:=\:3\sin x\cos x

    Substitute [1]: . \cos x\sin y \:=\:3\left(\tfrac{1}{4}\right)


    . . \begin{array}{cccccc}\text{We have:} & \cos x\sin y &=& \frac{3}{4} \\ \\[-3mm]<br />
\text{Add [1]:} & \sin x\cos y &=& \frac{1}{4} \end{array}


    And we have: . \sin x\cos y + \cos x \sin y \:=\:\frac{3}{4} + \frac{1}{4} \quad\Rightarrow\quad \sin(x+y) \:=\:1 \quad\Rightarrow\quad x + y \:=\:\frac{\pi}{2}

    . . x and y are complementary . . \Rightarrow\quad \cos y \:=\:\sin x


    Substitute into [1]: . \sin x\cdot\sin x \:=\:\frac{1}{4} \quad\Rightarrow\quad \sin x \:=\:\frac{1}{2}


    Therefore: . x \:=\:\frac{\pi}{6},\;\;y \:=\:\frac{\pi}{3}

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