proof required...

**nahduma** · February 8th 2010, 02:25 AM

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.

**Opalg** · February 8th 2010, 06:17 AM

Originally Posted by nahduma

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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