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Math Help - proof required...

  1. #1
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    proof required...

    if cosA+cosB+cosC=0=sinA+sinB+sinC
    prove that
    cos^2 A + cos^2 B + cos^2 C = sin^2 A + sin^2 B + sin^2 C = 3/2

    I have no idea how to equate the two. This was part of the questions in the chapter on complex numbers, so perhaps I should have included it in number theory. But any proof is welcome.
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  2. #2
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    Quote Originally Posted by nahduma View Post
    if cosA+cosB+cosC=0=sinA+sinB+sinC
    prove that
    cos^2 A + cos^2 B + cos^2 C = sin^2 A + sin^2 B + sin^2 C = 3/2

    I have no idea how to equate the two. This was part of the questions in the chapter on complex numbers, so perhaps I should have included it in number theory. But any proof is welcome.
    The hint about using complex numbers is very helpful!

    Let u = e^{iA}\ (=\cos A + i\sin A), and similarly v = e^{iB} and w = e^{iC}. Then u+v+w = 0. Also, u, v and 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.

    Then uv+vw+wu = uvw\bigl(u^{-1} + v^{-1} + w ^{-1}\bigr) = uvw\bigl(\overline{u+v+w}\bigr) = 0. Thus 0 = (u+v+w)^2 = u^2+v^2+w^2 + 2(uv+vw+wu), and it follows that u^2+v^2+w^2 = 0. Take the real part of that to see that \cos(2A) + \cos(2B) + \cos(2C) = 0.

    The results then follow from the facts that \cos^2A = \tfrac12\bigl(1+\cos(2A)\bigr) and \sin^2A = \tfrac12\bigl(1-\cos(2A)\bigr).
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