1. **Problem statement:** Given $\tan x = -\frac{2}{3}$ with $-\frac{\pi}{2} < x < 0$, find the exact values of: 8. $\cos 2x$ 9. $\sin \frac{x}{2}$ 2. **Recall formulas and quadrant info:** - Since $\tan x = \frac{\sin x}{\cos x}$, and $\tan x$ is negative in the fourth quadrant ($-\frac{\pi}{2} < x < 0$), we know $\sin x < 0$ and $\cos x > 0$. - Double angle formula for cosine: $$\cos 2x = \cos^2 x - \sin^2 x = 1 - 2\sin^2 x = 2\cos^2 x - 1$$ - Half angle formula for sine: $$\sin \frac{x}{2} = \pm \sqrt{\frac{1 - \cos x}{2}}$$ The sign depends on the quadrant of $\frac{x}{2}$. 3. **Find $\sin x$ and $\cos x$ from $\tan x = -\frac{2}{3}$:** Let $\sin x = a$, $\cos x = b$. Since $\tan x = \frac{a}{b} = -\frac{2}{3}$ and $b > 0$, $a < 0$, set: $$a = -2k, \quad b = 3k$$ Using Pythagorean identity: $$a^2 + b^2 = 1 \Rightarrow (-2k)^2 + (3k)^2 = 1 \Rightarrow 4k^2 + 9k^2 = 1 \Rightarrow 13k^2 = 1 \Rightarrow k = \frac{1}{\sqrt{13}}$$ So: $$\sin x = -\frac{2}{\sqrt{13}}, \quad \cos x = \frac{3}{\sqrt{13}}$$ 4. **Calculate $\cos 2x$ using $\cos 2x = 2\cos^2 x - 1$:** $$\cos^2 x = \left(\frac{3}{\sqrt{13}}\right)^2 = \frac{9}{13}$$ $$\cos 2x = 2 \times \frac{9}{13} - 1 = \frac{18}{13} - 1 = \frac{18}{13} - \frac{13}{13} = \frac{5}{13}$$ 5. **Determine the sign of $\sin \frac{x}{2}$:** Since $x$ is in the fourth quadrant ($-\frac{\pi}{2} < x < 0$), $\frac{x}{2}$ is in the fourth quadrant divided by 2, so $\frac{x}{2}$ is in $(-\frac{\pi}{4}, 0)$, which is the fourth quadrant where sine is negative. Therefore, $\sin \frac{x}{2} < 0$. 6. **Calculate $\sin \frac{x}{2}$:** $$\sin \frac{x}{2} = - \sqrt{\frac{1 - \cos x}{2}} = - \sqrt{\frac{1 - \frac{3}{\sqrt{13}}}{2}}$$ Simplify inside the square root: $$1 - \frac{3}{\sqrt{13}} = \frac{\sqrt{13} - 3}{\sqrt{13}}$$ So: $$\sin \frac{x}{2} = - \sqrt{\frac{\frac{\sqrt{13} - 3}{\sqrt{13}}}{2}} = - \sqrt{\frac{\sqrt{13} - 3}{2\sqrt{13}}}$$ Rationalize denominator inside the root: $$\sin \frac{x}{2} = - \sqrt{\frac{(\sqrt{13} - 3) \sqrt{13}}{2 \times 13}} = - \sqrt{\frac{13 - 3\sqrt{13}}{26}}$$ **Final answers:** $$\boxed{\cos 2x = \frac{5}{13}}$$ $$\boxed{\sin \frac{x}{2} = - \sqrt{\frac{13 - 3\sqrt{13}}{26}}}$$