错题详情

例1. 已知节日的顶点与原点重合，始边与轴的非负半轴重合，且终边过点 $(-3m, 4m)$。（$m \neq 0$）。 \begin{align*} &\therefore \quad \frac{2\sin\theta - \cos\theta}{\sin\theta + 2\cos\theta} = \\ &解：$R = \sqrt{(-3m)^2 + (4m)^2} = 5|m|$ \quad \frac{4}{5} \\ &\therefore \quad \sin\theta = \frac{y}{r} = \frac{4m}{5|m|} \quad \tan\theta = \frac{y}{x} = \frac{4}{3} \\ &\therefore \quad \cos\theta = \frac{x}{r} = \frac{-3m}{5|m|} \quad \therefore \quad \sin\theta = \frac{4}{5} \\ &\because \quad m > 0 \therefore \quad \sin\theta = \frac{4}{5} \quad \therefore \quad \tan\theta = \frac{4}{3} \\ &\therefore \quad 2 \times \frac{4}{5} + \frac{3}{5} = \frac{11}{5} \quad \therefore \quad \tan\theta + \frac{4}{5} = \frac{11}{5} \\ &\because \quad m < 0 \therefore \quad \sin\theta = -\frac{4}{5} \quad \therefore \quad \tan\theta = -\frac{4}{3} \\ &\therefore \quad \sin\theta = -\frac{4}{5} \quad \therefore \quad \cos\theta = -\frac{3}{5} \\ &\therefore \quad \frac{2 \times (-\frac{4}{5}) + (-\frac{3}{5})}{\frac{4}{5} - \frac{6}{5}} = \frac{-\frac{11}{5}}{-\frac{2}{5}} = -\frac{11}{2} \\ &\therefore \quad 2 \times (-\frac{4}{5}) - (-\frac{3}{5}) = -\frac{11}{5} \quad \therefore \quad \tan\theta = -\frac{11}{5} \\ \end{align*}