If $\sin A + \sin B + \sin C = \cos A + \cos B + \cos C = 0$, prove that $\sin^2 A + \sin^2 B + \sin^2 C = \cos^2 A + \cos^2 B + \cos^2 C = 3/2$.

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Step-by-Step Solution

Step 1: Set up the given equations

We are given two equations. Let's label them as equation (1) and equation (2) for easy reference. These equations state that the sum of the sines of angles $A$, $B$, and $C$ is 0, and similarly, the sum of their cosines is also 0.

Step 2: Square and add the equations

To relate the given sums to sums of squares, we square both equation (1) and equation (2). Then, we add the squared equations together. This step is crucial because it will allow us to use the identity $\sin^2 x + \cos^2 x = 1$.

Step 3: Expand and simplify using trigonometric identity

We expand the squared terms using the formula $(x+y+z)^2 = x^2+y^2+z^2+2xy+2yz+2zx$. Then, we group the $\sin^2 \theta + \cos^2 \theta$ terms, which simplify to $1$. We also use the cosine addition formula $\cos(X-Y) = \cos X \cos Y + \sin X \sin Y$ to simplify the cross-product terms.

Step 4: Simplify the equation

After simplifying the sum of the $\sin^2 \theta + \cos^2 \theta$ terms to $3$, we rearrange the equation to isolate the sum of the cosine difference terms. This gives us a key intermediate result.

Step 5: Express $\sin^2 A + \sin^2 B + \sin^2 C$ in terms of cosines

We use the identity $\sin^2 x = 1 - \cos^2 x$ for each term in the expression $\sin^2 A + \sin^2 B + \sin^2 C$. This allows us to express the sum of sine squares in terms of the sum of cosine squares.

Step 6: Calculate $\sin^2 A + \sin^2 B + \sin^2 C$

From equation (1), we have $\sin A + \sin B = -\sin C$. Squaring both sides gives $\sin^2 A + \sin^2 B + 2\sin A \sin B = \sin^2 C$. Rearranging, we get $\sin^2 A + \sin^2 B - \sin^2 C = -2\sin A \sin B$. Similarly, from equation (2), we get $\cos^2 A + \cos^2 B - \cos^2 C = -2\cos A \cos B$. Adding these two results and using $\sin^2 x + \cos^2 x = 1$ and $\cos(A-B) = \cos A \cos B + \sin A \sin B$, we find $\cos(A-B) = -1/2$. By symmetry, $\cos(B-C) = -1/2$ and $\cos(C-A) = -1/2.

Step 7: Substitute back and find the final result

We use the property that if $x+y+z=0$ and $|x|=|y|=|z|=1$, then $x^2+y^2+z^2=0$. By letting $x=e^{iA}$, $y=e^{iB}$, $z=e^{iC}$, we have $x+y+z = (\cos A + \cos B + \cos C) + i(\sin A + \sin B + \sin C) = 0 + i(0) = 0$. Thus, $e^{i2A} + e^{i2B} + e^{i2C} = 0$. Expanding this, we get $(\cos 2A + \cos 2B + \cos 2C) + i(\sin 2A + \sin 2B + \sin 2C) = 0$. This implies $\cos 2A + \cos 2B + \cos 2C = 0$. Using the double angle formula $\cos 2\theta = 2\cos^2 \theta - 1$, we substitute and solve for $\cos^2 A + \cos^2 B + \cos^2 C$. Similarly, using $\cos 2\theta = 1 - 2\sin^2 \theta$, we solve for $\sin^2 A + \sin^2 B + \sin^2 C.