Question

Show that the function $f\left(x\right)={\cot }^{-1}(\sin x+\cos x)$ is decreasing on $(0,\frac{\pi }{4})$ and increasing on $(\frac{\pi }{4},\frac{\pi }{2})$.

Answer 1

$f\left(x\right)={\cot }^{-1}(\sin x+\cos x)$

$f^{\prime }\left(x\right)=\frac{-(\cos x-\sin x)}{1+(\sin x+\cos x{)}^2}=\frac{\sin x-\cos x}{2+\sin 2x}$

when

$0<x<\pi /4$ $0<x<\pi /4$

$sinx<cosx$

$\sin x-\cos x<0$

$0<2x<\pi /2$

$0<\sin 2x<1$

so, $2+\sin 2x>0$ so, $f^{\prime }\left(x\right)<0$ for $x\in (0,\pi /4)$

so $f\left(x\right)$ is a decreasing function on interval $(0,\pi /4)$

$\pi /4<x<\pi /2$ $\pi /4<x<\pi /2$

$\pi /2<2x<\pi$ $\sin x-\cos x>0$

$1<\sin 2x<-1$ so, $\frac{\sin x-\cos x}{2+\sin 2x}>0$

so, $2+\sin 2x>1$

so for $x\in (\pi /4,\pi /2)$ the $f\left(x\right)$ is increasing function.