#### Question

Show that f(x) = cos x is a decreasing function on (0, π), increasing in (−π, 0) and neither increasing nor decreasing in (−π, π) ?

#### Solution

\[\text { Here }, \]

\[f\left( x \right) = \cos x\]

\[\text { Domain of cos x is }\left( - \pi, \pi \right).\]

\[ \Rightarrow f'\left( x \right) = - \sin x\]

\[\text { For } x \in \left( - \pi, 0 \right), \sin x < 0 \left[ \because \text { sine function is negative in third and fourth quadrant } \right]\]

\[ \Rightarrow - \sin x > 0\]

\[ \Rightarrow f'\left( x \right) > 0\]

\[\text { So, cos x is increasing in} \left( - \pi, 0 \right) . \]

\[\text { For x } \in \left( 0, \pi \right)),\sin x > 0 \left[ \because sine \text { function is positive in first and second quadrant } \right]\]

\[ \Rightarrow - \sin x < 0\]

\[ \Rightarrow f'\left( x \right) < 0\]

\[\text { So,f(x) is decreasing on }\left( 0, \pi \right).\]

\[\text { Thus,f(x) is neither increasing nor decreasing in }\left( - \pi, \pi \right).\]