Increasing and decreasing function doubt This is a question of increasing and decreasing functions.
$f(x)=\sin x + \cos x$, $x$ belongs to $[0, 2\pi]$
Derivative of this function $f'(x) = \cos x - \sin x$
For increasing function we put $f'(x) > 0$.
I tried to solve it this way:
$\cos x - \sin x > 0$
$\cos x > \sin x$
$\tan x < 1$
$x$ belongs to $\left(0, \dfrac{\pi}{4}\right)\cup \left(\dfrac{\pi}{2}, \dfrac{5\pi}{4}\right) \cup \left(\dfrac{3\pi}{2}, 2\pi\right)$
But the answer given in the book is $x$ belongs to $\left(0, \dfrac{\pi}{4}\right)\cup \left(\dfrac{5\pi}{4}, 2\pi\right)$.
Any help. 

What you did up to$$\cos x>\sin x\tag1$$is fine. After this point, you need to consider two possibilities:


*

*$x\in\left(0,\frac\pi2\right)\cup\left(\frac{3\pi}2,2\pi\right)$: then $\cos x>0$ and what you did is fine;

*$x\in\left(\frac\pi2,\frac{3\pi}2\right)$: then $\cos x<0$ and what you deduce from $(1)$ is that $\tan x>1$.Another possibility is to use the fact that\begin{align}\cos(x)+\sin(x)&=\sqrt2\left(\cos\left(\frac\pi4\right)\cos(x)+\sin\left(\frac\pi4\right)\sin(x)\right)\\&=\sqrt2\cos\left(x-\frac\pi 4\right).\end{align}Since you know where $\cos$ is increasing and decreasing…

