Question

Interval of strictly increasing and decresing of function sinx + cosx

Answer 1

${\text{Here, we have }}f\left( x \right) = \sin x + \cos x \hfill \\ \Rightarrow f'\left( x \right) = \cos x - \sin x \hfill \\ {\text{Function }}f\left( x \right){\text{should be strictly increasing}} \hfill \\ \Leftrightarrow f'\left( x \right) > 0 \hfill \\ \Leftrightarrow \left( {\cos x - \sin x} \right) > 0 \hfill \\ \Leftrightarrow \sin x - \cos x < 0 \hfill \\ \Leftrightarrow \sin x < \cos x \hfill \\ \Leftrightarrow \tan x < 1 \hfill</p><p> \Leftrightarrow x \in \left( { - \frac{{3\pi }}{2}, - \frac{{3\pi }}{4}} \right) \cup \left( { - \frac{\pi }{2},\frac{\pi }{4}} \right) \cup \left( {\frac{\pi }{2},\frac{{5\pi }}{4}} \right) \cup \left( {\frac{{3\pi }}{2},2\pi } \right) \hfill \\</p><p>{\text{Function }}f\left( x \right){\text{should be strictly decreasing}} \hfill \\</p><p>\Leftrightarrow f'\left( x \right) < 0 \hfill \\</p><p>\Leftrightarrow \left( {\cos x - \sin x} \right) < 0 \hfill \\</p><p>\Leftrightarrow \sin x - \cos x > 0 \hfill \\</p><p> \Leftrightarrow \sin x > \cos x \hfill \\</p><p> \Leftrightarrow \tan x > 1 \hfill \\</p><p> \Leftrightarrow x \in \left( { - \frac{{7\pi }}{4}, - \frac{{3\pi }}{2}} \right) \cup \left( { - \frac{{3\pi }}{4}, - \frac{\pi }{2}} \right) \cup \left( {\frac{\pi }{4},\frac{\pi }{2}} \right) \cup \left( {\frac{{5\pi }}{4},\frac{{3\pi }}{2}} \right) \hfill$