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Math Help - trigonometric equality

  1. #1
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    trigonometric equality

     <br /> <br />
\begin{array}{l}<br />
 \frac{{\sin x + \sin y + \sin z}}{{\sin (x + y + z)}} = \frac{{\cos x + \cos y + \cos z}}{{\cos (x + y + z)}} = 2 \\ <br />
  \\ <br />
 find \\ <br />
 \sin x\sin y + \sin y\sin z + \sin x\sin z \\ <br />
 \end{array}<br /> <br />
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  2. #2
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    This is a very interesting problem. It has either too much symmetry, or not enough, depending on how you're looking at it.

    One thing I would highly recommend: abbreviate! I'm going to abbreviate as follows:

    \sin(x)=s(x)
    \cos(x)=c(x).

    If you square both equations and add together in the appropriate way, you can get it down to

    \displaystyle{s(x)s(y)+s(x)s(z)+s(y)s(z)+c(x)c(y)+  c(x)c(z)+c(y)c(z)=\frac{1}{2}}.

    Unfortunately, if you plug in a test case, you find out that the sin terms do not necessarily make up half of the sum. Here's one test case:

    x=0.1

    y=1.011189

    z=-1.07805,

    approximately. All of these are in radians. The sin terms add up to -0.75, and the cosine terms add up to 1.25.

    So there's a specific case, plus some general information about the problem. Nothing more is coming to mind at the moment. Your result should be -3/4. But now you have to prove it!
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  3. #3
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    Hello, fxs12!

    Askbeet has an excellent game plan,
    . . but I can't finish the problem.



    \dfrac{{\sin x + \sin y + \sin z}}{{\sin (x + y + z)}} \;=\; \dfrac{{\cos x + \cos y + \cos z}}{{\cos (x + y + z)}} \;=\; 2


    \text{Find: }\,\sin x\sin y + \sin y\sin z + \sin x\sin z

    We have: . \begin{array}{cccc}<br />
\sin x + \sin y + \sin z &=& 2\sin(x+y+z) & [1] \\<br />
\cos x + \cos y + \cos z &=& 2\cos(x+y+z) & [2] \end{array}

    \begin{array}{ccccc}\text{Square [1]:} & (\sin x + \sin y + \sin z)^2 &=& 4\sin^2(x+y+z) & [3] \\<br />
\text{Square [2]:} & (\cos x + \cos y + \cos z)^2 &=& 4\cos^2(x+y+z) & [4]\end{array}


    Add [3] and [4]:

    \underbrace{\begin{Bmatrix}\sin^2x \\ \cos^2x\end{Bmatrix}}_{\text{This is 1}} + \underbrace{\begin{Bmatrix}\sin^2y \\ \cos^2y\end{Bmatrix}}_{\text{This is 1}} + \underbrace{\begin{Bmatrix}\sin^2z \\ \cos^2z\end{Bmatrix}}_{\text{This is 1}} + 2\begin{Bmatrix}\sin x\sin y + \sin y\sin z + \sin x\sin z \\ \cos x\cos y + \cos y\cos z + \cos x\cos z \end{Bmatrix}

    . . . . . . . =\; 4\underbrace{\bigg[\sin^2(x+y+z) + \cos^2(x+y+z)\bigg]}_{\text{This is 1}}


    We have:

    . . 1 + 1 + 1 + 2\begin{Bmatrix}\sin x\sin y +\sin y\sin z + \sin x\sin z \\ \cos x\cos y + \cos y\cos z + \cos x\cos z\end{Bmatrix} \;=\;4


    . . \begin{Bmatrix}\sin x\sin y + \sin y\sin z + \sin x\sin z \\ \cos x\cos y + \cos y\cos z + \cos x\cos z\end{Bmatrix} \;=\;\frac{1}{2}


    And I can't go any further . . .

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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    ...

    cos(x-y)+cos(y-z)+cos(x-z)=1/2
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  5. #5
    MHF Contributor Also sprach Zarathustra's Avatar
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    I also have proved:

    sin(x-y)/sinx siny + sin(y-z)/siny sinz + sin(x-z)/sinx sinz = 0
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  6. #6
    MHF Contributor Also sprach Zarathustra's Avatar
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    Algebra instead trigo.

    sinx=a
    siny=b
    sinz=c

    cosx=\sqrt{1-a^2}
    cosy=\sqrt{1-b^2}
    cosz=\sqrt{1-c^2}

    tan(x+y+z)=\frac{tanx+tany+tanz-tanxtanytanz}{1-tanxtany-tanytanz-tanztanx}<br />

    tanx=\frac{a}{\sqrt{1-a^2}}

    tany=\frac{b}{\sqrt{1-b^2}}

    tanx=\frac{c}{\sqrt{1-c^2}}


    Hence:

    \dfrac{{\sin x + \sin y + \sin z}}{{\sin (x + y + z)}} \;=\; \dfrac{{\cos x + \cos y + \cos z}}{{\cos (x + y + z)}} \;=\; 2

    becomes:

    \frac{a+b+c}{\sqrt{1-a^2}+\sqrt{1-b^2}+\sqrt{1-c^2}}=tan(x+y+z)=2

    when we need to find: ab+bc+ac
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