# proof required...

• Feb 8th 2010, 02:25 AM
nahduma
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.
• Feb 8th 2010, 06:17 AM
Opalg
Quote:

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.

Let $\displaystyle u = e^{iA}\ (=\cos A + i\sin A)$, and similarly $\displaystyle v = e^{iB}$ and $\displaystyle w = e^{iC}$. Then $\displaystyle u+v+w = 0$. Also, u, v and w all have absolute value 1, so their inverses are equal to their complex conjugates: $\displaystyle u^{-1} = \overline{u}$, and similarly for v and w.
Then $\displaystyle uv+vw+wu = uvw\bigl(u^{-1} + v^{-1} + w ^{-1}\bigr) = uvw\bigl(\overline{u+v+w}\bigr) = 0$. Thus $\displaystyle 0 = (u+v+w)^2 = u^2+v^2+w^2 + 2(uv+vw+wu)$, and it follows that $\displaystyle u^2+v^2+w^2 = 0$. Take the real part of that to see that $\displaystyle \cos(2A) + \cos(2B) + \cos(2C) = 0$.
The results then follow from the facts that $\displaystyle \cos^2A = \tfrac12\bigl(1+\cos(2A)\bigr)$ and $\displaystyle \sin^2A = \tfrac12\bigl(1-\cos(2A)\bigr)$.