题目详情 - Q20260425174340275

【分析】利用三角函数的基本关系式结合$\alpha\in\left(0,\frac{\pi}{2}\right)$可求得$\sin\alpha$和$\cos\alpha$的具体值，则$\sin\alpha-\cos\alpha$可求. 【详解】因为$\tan\alpha=\frac{1}{2}=\frac{\sin\alpha}{\cos\alpha}$， 由$\begin{cases}\frac{\sin\alpha}{\cos\alpha}=\frac{1}{2},\\\sin^2\alpha+\cos^2\alpha=1,\end{cases}$ 解得$\begin{cases}\sin\alpha=\frac{\sqrt{5}}{5},\\\cos\alpha=\frac{2\sqrt{5}}{5},\end{cases}$或$\begin{cases}\sin\alpha=-\frac{\sqrt{5}}{5},\\\cos\alpha=-\frac{2\sqrt{5}}{5},\end{cases}$, 又$\alpha\in\left(0,\frac{\pi}{2}\right)$， 所以$\sin\alpha=\frac{\sqrt{5}}{5},\ \cos\alpha=\frac{2\sqrt{5}}{5}$， 所以$\sin\alpha-\cos\alpha=\frac{\sqrt{5}}{5}-\frac{2\sqrt{5}}{5}=-\frac{\sqrt{5}}{5}$. 故答案为：$-\frac{\sqrt{5}}{5}$.