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Math Help - proof with complex numbers

  1. #1
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    proof with complex numbers

    a, x, y, z are real numbers such that:

     \frac{cosx + cosy + cosz}{cos(x+y+z)}= \frac{sinx + siny + sinz}{sin (x+y+z)} = a

    Using the identity, a =  \frac{cosx + i sinx + cosy + i siny + cosz + i sinz}{cos(x+y+z) + isin(x+y+z)}

    show that:

     a = cos(y+z) + cos(x+z) + cos(x+y)

    thanks
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  2. #2
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    Quote Originally Posted by differentiate View Post
    a, x, y, z are real numbers such that:

     \frac{cosx + cosy + cosz}{cos(x+y+z)}= \frac{sinx + siny + sinz}{sin (x+y+z)} = a

    Using the identity, a =  \frac{cosx + i sinx + cosy + i siny + cosz + i sinz}{cos(x+y+z) + isin(x+y+z)}

    show that:

     a = cos(y+z) + cos(x+z) + cos(x+y)
    Let u = e^{ix} = \cos x + i\sin x, and similarly v = e^{iy} and w = e^{iz}. Then u, v, w all have absolute value 1, so their inverses are equal to their complex conjugates: u^{-1} = \overline{u} (and similarly for v and w).

    Now show that  a = \frac{u+v+w}{uvw} = v^{-1}w^{-1} + w^{-1}u^{-1} + u^{-1}v^{-1} = \overline{vw+wu+uv}. But vw = \cos(y+z) + i\sin(y+z), with similar expressions for wu and uv. Deduce that a = \cos(y+z) + \cos(z+x) + \cos(x+y) - i\bigl(\sin(y+z) + \sin(z+x) + \sin(x)y)\bigr). But a, on the left side of that equation, is real, so the imaginary part of the right side of the equation must be zero.
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