The given function is f(x)=sinx.
∴f′(x)=cosx
(a) Since for each x∈(0,π2,),cosx>0⇒f′(x)>0.
Hence, f is strictly increasing in (0,π2).
(b) Since for each x∈(π2,π),cosx<0⇒f′(x)<0.
Hence, f is strictly decreasing in (π2,π).
(c) From the results obtained in (a) and (b), it is clear that f is neither increasing nor
decreasing in (0,π).