Question

Show that the function given by \(f(x) = sin x\) is (a) strictly increasing in \((0,\frac{π}{2})\) (b) strictly decreasing in \((\frac{π}{2},π)\) (c) neither increasing nor decreasing in \((0, π)\)

Answer 1

The given function is \(f(x) = sin x\).
\(∴ f'(x)=cos x\)
(a) Since for each \(x∈(0,\frac{π}{2}),cosx>0\), we have \(f'(x)>0\)
Hence, \(f\) is strictly increasing in \((0,\frac{π}{2})\)
(b) Since for each \(x∈(\frac{π}{2}),cos x<0\), we have \(f'(x)<0\)
Hence, \(f\) is strictly decreasing in \((\frac{π}{2π})\).
(c) From the results obtained in (a) and (b), it is clear that \(f\) is neither increasing nor decreasing in \((0, π)\).