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Math Help - equation

  1. #1
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    equation

    We've got following equation:
    sin(x+y)/sin(x) = sin(2x+y+z)/sin(z)

    Find z.
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  2. #2
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    Let M=\sin(x+y)/\sin(x), and let N=2x+y. Then the equation becomes

    M=\sin(N+z)/\sin(z), or

    M\sin(z)=\sin(N+z), assuming \sin(z)\not=0.

    Do you have any ideas about the RHS?
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  3. #3
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    Hello, olaolaola!

    \text{Solve for }z\!:\;\;\dfrac{\sin(x+y)}{\sin x} \;=\; \dfrac{\sin(2x+y+z)}{\sin z}

    We have: . \sin(x+y)\cdot\sin z \;=\;\sin x\cdot\sin([2x+y]+z)

    . . \sin(x+y)\cdot\sin z \;=\;\sin x\bigg[\sin(2x+y)\cdot\cos z + \cos(2x+y)\cdot\sin z\bigg]

    . . \sin(x+y)\cdot\sin z \;=\;\sin x\cdot\sin(2x+y)\cdot\cos z + \sin x \cdot\cos(2x+y)\cdot \sin z

    . . \sin(x+y)\cdot\sin z - \sin x\cdot\cos(2x+y)\cdot\sin z \;=\;\sin x\cdot\sin(2x+y)\cdot\cos z

    . . \bigg[\sin(x+y)-\sin x\cdot\cos(2x+y)\bigg]\sin z \;=\;\sin x\cdot\sin(2x+y)\cdot\cos z

    . . \displaystyle \frac{\sin z}{\cos z} \;=\;\frac{\sin x\cdot\sin(2x+y)}{\sin(x+y) - \sin x\cdot\cos(2x+y)}

    . . \displaystyle \tan z \;=\;\frac{\sin x\cdot\sin(2x+y)}{\sin(x+y) - \sin x\cdot\cos(2x+y)}

    . . . . z \;=\;\tan^{-1}\left[\dfrac{\sin x\cdot\sin(2x+y)}{\sin(x+y) - \sin x\cdot\cos(2x+y)}\right]

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