Find sin2x,cos2x and tan2x If cosx=-(3)/(sqrt (10)) and x terminates in quadrant II. sin2x=(3)/(5) cos2x=(4)/(5) tan2x=(3)/(4)

To find $\sin 2x$, $\cos 2x$, and $\tan 2x$ given that $\cos x = -\frac{3}{\sqrt{10}}$ and $x$ terminates in quadrant II, we can use the double-angle formulas for sine, cosine, and tangent.<br /><br />First, let's find $\sin x$. Since $x$ is in quadrant II, $\sin x$ will be positive. We can use the Pythagorean identity $\sin^2 x + \cos^2 x = 1$ to find $\sin x$:<br /><br />$\sin^2 x + \left(-\frac{3}{\sqrt{10}}\right)^2 = 1$<br /><br />$\sin^2 x + \frac{9}{10} = 1$<br /><br />$\sin^2 x = \frac{1}{10}$<br /><br />$\sin x = \frac{1}{\sqrt{10}}$<br /><br />Now, let's find $\sin 2x$, $\cos 2x$, and $\tan 2x$ using the double-angle formulas:<br /><br />1. $\sin 2x = 2 \sin x \cos x$<br /><br />$\sin 2x = 2 \left(\frac{1}{\sqrt{10}}\right) \left(-\frac{3}{\sqrt{10}}\right)$<br /><br />$\sin 2x = -\frac{6}{10}$<br /><br />$\sin 2x = -\frac{3}{5}$<br /><br />2. $\cos 2x = \cos^2 x - \sin^2 x$<br /><br />$\cos 2x = \left(-\frac{3}{\sqrt{10}}\right)^2 - \left(\frac{1}{\sqrt{10}}\right)^2$<br /><br />$\cos 2x = \frac{9}{10} - \frac{1}{10}$<br /><br />$\cos 2x = \frac{8}{10}$<br /><br />$\cos 2x = \frac{4}{5}$<br /><br />3. $\tan 2x = \frac{\sin 2x}{\cos 2x}$<br /><br />$\tan 2x = \frac{-\frac{3}{5}}{\frac{4}{5}}$<br /><br />$\tan 2x = -\frac{3}{4}$<br /><br />Therefore, the correct values are:<br /><br />$\sin 2x = -\frac{3}{5}$<br /><br />$\cos 2x = \frac{4}{5}$<br /><br />$\tan 2x = -\frac{3}{4}$