#### Question

Show that the function f(x) = sin (2x + π/4) is decreasing on (3π/8, 5π/8) ?

#### Solution

\[f\left( x \right) = \sin \left( 2x + \frac{\pi}{4} \right)\]

\[f'\left( x \right) = 2 \cos \left( 2x + \frac{\pi}{4} \right)\]

\[\text { Here, } \]

\[\frac{3\pi}{8} < x < \frac{5\pi}{8}\]

\[ \Rightarrow \frac{3\pi}{4} < 2x < \frac{5\pi}{4}\]

\[ \Rightarrow \pi < 2x + \frac{\pi}{4} < \frac{3\pi}{2}\]

\[ \Rightarrow \ cos \left( 2x + \frac{\pi}{4} \right) < 0 \left[ \because \text { Cos function is negative inthird quadrant } \right]\]

\[ \Rightarrow 2 \cos \left( 2x + \frac{\pi}{4} \right) < 0\]

\[ \Rightarrow f'\left( x \right) < 0, \forall x \in \left( \frac{3\pi}{8}, \frac{5\pi}{8} \right)\]

\[\text { So },f\left( x \right) \text { is decreasing on }\left( \frac{3\pi}{8}, \frac{5\pi}{8} \right).\]