If sin A + sin B + sin $C = cos A + cos B + cos C = 0$ prove that sin²A + sin²B + sin² $C = cos²A + cos²B + cos²C = 3/2$.

Step 1: Square the given equations

We are given two equations: $\sin A + \sin B + \sin C = 0$ and $\cos A + \cos B + \cos C = 0$. To introduce squared terms, we square both of these equations.

Step 2: Expand the squared equations

Expanding the squared sums $(x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+zx)$, we get two new equations. Let's call them equation (1) and equation (2).

Step 3: Add the expanded equations

Now, we add equation (1) and equation (2). We group the terms $\sin^2X + \cos^2X$ and the product terms.

Step 4: Apply trigonometric identities

Using the identity $\sin^2X + \cos^2X = 1$ and the cosine addition formula $\cos(X-Y) = \cos X \cos Y + \sin X \sin Y$, we simplify the sum. This gives us equation (3).

Step 5: Isolate the sum of cosines

From equation (3), we can isolate the sum of the cosine difference terms.

Step 6: Substitute back into expanded equations

From equations (1) and (2), we can express $\sin^2A + \sin^2B + \sin^2C$ and $\cos^2A + \cos^2B + \cos^2C$ in terms of the product sums.

Step 7: Final Proof

We know that $2(\sin A \sin B + \sin B \sin C + \sin C \sin A) = (\sin A + \sin B + \sin C)^2 - (\sin^2A + \sin^2B + \sin^2C)$. Since $(\sin A + \sin B + \sin C)^2 = 0$, we have $2(\sin A \sin B + \sin B \sin C + \sin C \sin A) = -(\sin^2A + \sin^2B + \sin^2C)$. Substituting this into equation (1) gives $2(\sin^2A + \sin^2B + \sin^2C) = 3$. Thus, $\sin^2A + \sin^2B + \sin^2C = \frac{3}{2}$. Similarly, for the cosine terms, we can use the identity $2(\cos A \cos B + \cos B \cos C + \cos C \cos A) = (\cos A + \cos B + \cos C)^2 - (\cos^2A + \cos^2B + \cos^2C)$. Since $(\cos A + \cos B + \cos C)^2 = 0$, we have $2(\cos A \cos B + \cos B \cos C + \cos C \cos A) = -(\cos^2A + \cos^2B + \cos^2C)$. Substituting this into equation (2) gives $2(\cos^2A + \cos^2B + \cos^2C) = 3$. Thus, $\cos^2A + \cos^2B + \cos^2C = \frac{3}{2}$.

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