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Question
Show that the function f given by f(x) = tan–1 (sin x + cos x) is decreasing for all \[x \in \left( \frac{\pi}{4}, \frac{\pi}{2} \right) .\]
Solution
\[f\left( x \right) = \tan^{- 1} \left( \sin x + \cos x \right)\]
\[f'\left( x \right) = \frac{1}{1 + \left( \sin x + \cos x \right)^2}\left( \cos x - \sin x \right)\]
\[ = \frac{1}{1 + 1 + 2 \sin x \cos x}\left( \cos x - \sin x \right)\]
\[ = \frac{\left( \cos x - \sin x \right)}{2 + \sin 2x}\]
Here,
\[\frac{\pi}{4} < x < \frac{\pi}{2}\]
\[ \Rightarrow \frac{\pi}{2} < 2x < \pi\]
\[ \Rightarrow \sin 2x > 0\]
\[ \Rightarrow 2 + \sin 2x > 0 . . . \left( 1 \right)\]
Also,
\[\frac{\pi}{4} < x < \frac{\pi}{2}\]
\[\cos x < \sin x\]
\[ \Rightarrow \cos x - \sin x < 0 . . . \left( 2 \right)\]
\[f'\left( x \right) = \frac{\left( \cos x - \sin x \right)}{2 + \sin 2x} < 0, \forall x \in \left( \frac{\pi}{4}, \frac{\pi}{2} \right) \left[ \text { From eqs . (1) and (2) }\right]\]
Therefore, f(x) is decreasing for all
\[x \in \left( \frac{\pi}{4}, \frac{\pi}{2} \right) .\]
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