Prove that the function given by `f(x) = cos x` is
(a) strictly decreasing in `(0,pi)`
(b) strictly increasing in `(pi,2pi)`, and
(c) neither increasing nor decreasing in `(0,2pi)`

Explanation

`f(x)=cosx`
`f ′(x)=−sinx`
`(i) x in(0,pi)
⇒sinx gt 0⇒−sinx lt 0`
=`f ′(x) lt 0`so f(x) is strictly decreasing on `(0,pi)`

(ii) `x in(pi,2pi)⇒sinx<0
⇒−sinx gt 0`
⇒`f ′(x) gt 0`
so f(x) is strictly increasing on `(pi,2pi)`
(iii) As `f ′(x) lt 0 for x∈(0,pi)`
and `f ′(x) gt 0 for x∈(0,2pi)`
Hence f(x) is neither increasing nor decreasing on `(0,2pi)`.