AP DSC School Assistant Math Model Test 2

Concept:When three expressions involving sine of different angles are equal, we set them equal to a common constant $K$, express $x$, $y$, $z$ in terms of $K$ and the sines, then compute $xy+yz+zx$ and simplify using sine addition formulas.Explanation:Let $x\sin\theta = y\sin\!\left(\theta+\frac{2\pi}{3}\right) = z\sin\!\left(\theta+\frac{4\pi}{3}\right) = K$.Rewrite the second and third arguments using $\frac{2\pi}{3}=\pi-\frac{\pi}{3}$ and $\frac{4\pi}{3}=\pi+\frac{\pi}{3}$:$y\sin\!\left(\theta+\pi-\frac{\pi}{3}\right)=y\bigl[-\sin(\theta-\frac{\pi}{3})\bigr] = -y\sin\!\left(\theta-\frac{\pi}{3}\right)$$z\sin\!\left(\theta+\pi+\frac{\pi}{3}\right)=z\bigl[-\sin(\theta+\frac{\pi}{3})\bigr] = -z\sin\!\left(\theta+\frac{\pi}{3}\right)$Thus we have:$x\sin\theta = -y\sin\!\left(\theta-\frac{\pi}{3}\right) = -z\sin\!\left(\theta+\frac{\pi}{3}\right) = K$From this, solve for $x$, $y$, $z$:$x = \frac{K}{\sin\theta}$, $\quad y = \frac{-K}{\sin(\theta-\frac{\pi}{3})}$, $\quad z = \frac{-K}{\sin(\theta+\frac{\pi}{3})}$Now compute $xy+yz+zx$:$$\begin{aligned} xy &= \frac{K}{\sin\theta}\cdot\frac{-K}{\sin(\theta-\frac{\pi}{3})} = \frac{-K^2}{\sin\theta\,\sin(\theta-\frac{\pi}{3})}\\ yz &= \frac{-K}{\sin(\theta-\frac{\pi}{3})}\cdot\frac{-K}{\sin(\theta+\frac{\pi}{3})} = \frac{K^2}{\sin(\theta-\frac{\pi}{3})\,\sin(\theta+\frac{\pi}{3})}\\ zx &= \frac{-K}{\sin(\theta+\frac{\pi}{3})}\cdot\frac{K}{\sin\theta} = \frac{-K^2}{\sin(\theta+\frac{\pi}{3})\,\sin\theta} \end{aligned}$$Add the three terms:$$xy+yz+zx = -K^2\left[\frac{1}{\sin\theta\sin(\theta-\frac{\pi}{3})} - \frac{1}{\sin(\theta-\frac{\pi}{3})\sin(\theta+\frac{\pi}{3})} + \frac{1}{\sin(\theta+\frac{\pi}{3})\sin\theta}\right]$$Combine the fractions over the common denominator $\sin\theta\sin(\theta-\frac{\pi}{3})\sin(\theta+\frac{\pi}{3})$:$$xy+yz+zx = \frac{-K^2\bigl[\sin(\theta+\frac{\pi}{3}) - \sin\theta + \sin(\theta-\frac{\pi}{3})\bigr]}{\sin\theta\sin(\theta-\frac{\pi}{3})\sin(\theta+\frac{\pi}{3})}$$Use the sine addition formulas:$\sin(\theta+\frac{\pi}{3}) = \sin\theta\cos\frac{\pi}{3}+\cos\theta\sin\frac{\pi}{3} = \frac{1}{2}\sin\theta + \frac{\sqrt{3}}{2}\cos\theta$$\sin(\theta-\frac{\pi}{3}) = \sin\theta\cos\frac{\pi}{3}-\cos\theta\sin\frac{\pi}{3} = \frac{1}{2}\sin\theta - \frac{\sqrt{3}}{2}\cos\theta$Substitute into the numerator:$\bigl[\frac{1}{2}\sin\theta + \frac{\sqrt{3}}{2}\cos\theta\bigr] - \sin\theta + \bigl[\frac{1}{2}\sin\theta - \frac{\sqrt{3}}{2}\cos\theta\bigr] = \left(\frac{1}{2}+\frac{1}{2}-1\right)\sin\theta + \left(\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}\right)\cos\theta = 0$Hence the numerator is zero, so $xy+yz+zx = 0$.Answer:0 (Option C)