Now, since \(0 < x < \frac\pi2\), \(u = \sin x + \cos x = \sqrt2 \sin\left(x + \frac\pi4\right)\), so \(u \in (1, \sqrt2]\) (maximum \(\sqrt2\) at \(x = \frac\pi4\)). - AMAZONAWS 📅 March 6, 2026 👤 scraface Mar 06, 2026
![Now, since \(0 < x < \frac{\pi}{2}\), \(u = \sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right)\), so \(u \in (1, \sqrt{2}]\) (maximum \(\sqrt{2}\) at \(x = \frac{\pi}{4}\)).](https://soloferat.biz.id/images/now-since-0--x--fracpi2-u--sin-x--cos-x--sqrt2-sinleftx--fracpi4right-so-u-in-1-sqrt2-maximum-sqrt2-at-x--fracpi4.jpg)
