## Precalculus (6th Edition)

x is in quadrant II, where its sine is positive, and cosine is negative. Furthermore, $90^{o} < \theta < 180^{o}\qquad/\div 2$ $45^{o} < \displaystyle \frac{\theta}{2} < 90^{o}$, so $\displaystyle \frac{\theta}{2}$ is in quadrant I, where $\displaystyle \tan\frac{\theta}{2}$ is positive. The Half-Angle Identity: $\displaystyle \tan\frac{\theta}{2}=\pm\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$ .. has the plus sign, we need $\cos\theta$ ... Pythagorean identity, ($\cos x$ is negative): $\cos\theta=-\sqrt{1-\sin^{2}\theta}=-\sqrt{1-\frac{9}{25}}$ $=-\displaystyle \sqrt{\frac{16}{25}}=-\frac{4}{5}=-0.8$ So, $\displaystyle \tan\frac{x}{2}=+\sqrt{\frac{1-(-0.8)}{1+(-0.8)}}=\sqrt{\frac{1.8}{0.2}}$ $=\sqrt{9}=3$