It correct to say that \( \sin 2 x=2 \sin x \) ? otivate your answer by choosing any three values of \( x \) and coupleting the table below those values. Round your answers off to three decimal places where needed. \begin{tabular}{|c|c|c}\hline\( x \) & \( \sin 2 x \) & \( 2 \sin x \) \\ \hline\end{tabular}

To determine if it is correct to say that \( \sin 2x = 2 \sin x \), we'll evaluate both expressions for some chosen values of \( x \). However, we know from trigonometric identities that \( \sin 2x = 2 \sin x \cos x \), so it will likely show differing results. Let's choose \( x = 0 \), \( x = \frac{\pi}{6} \), and \( x = \frac{\pi}{4} \) for our calculations: 1. For \( x = 0 \): - \( \sin 2(0) = \sin 0 = 0 \) - \( 2 \sin(0) = 2 \cdot 0 = 0 \) 2. For \( x = \frac{\pi}{6} \) (approximately 0.524): - \( \sin 2\left(\frac{\pi}{6}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \approx 0.866 \) - \( 2 \sin\left(\frac{\pi}{6}\right) = 2 \cdot \frac{1}{2} = 1 \) 3. For \( x = \frac{\pi}{4} \) (approximately 0.785): - \( \sin 2\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{2}\right) = 1 \) - \( 2 \sin\left(\frac{\pi}{4}\right) = 2 \cdot \frac{\sqrt{2}}{2} = \sqrt{2} \approx 1.414 \) Now we can fill in our table: \begin{tabular}{|c|c|c}\hline \( x \) & \( \sin 2x \) & \( 2 \sin x \) \\ \hline 0 & 0 & 0 \\ \(\frac{\pi}{6}\) & 0.866 & 1 \\ \(\frac{\pi}{4}\) & 1 & 1.414 \\ \hline \end{tabular} From the calculations, we can see that for most values of \( x \) other than zero, \( \sin 2x \) does not equal \( 2 \sin x \). Therefore, it is incorrect to say that \( \sin 2x = 2 \sin x \) in general terms.