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Question
Show that f(x) = sin x − cos x is an increasing function on (−π/4, π/4)?
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Solution 1
\[f\left( x \right) = \sin x - \cos x\]
\[f'\left( x \right) = \cos x + \sin x\]
\[ = \cos x\left( 1 + \frac{\sin x}{\cos x} \right)\]
\[ = \cos x\left( 1 + \cot x \right)\]
\[\text { Here, } \]
\[\frac{- \pi}{4} < x < \frac{\pi}{4}\]
\[ \Rightarrow \cos x > 0 . . . \left( 1 \right)\]
\[\text { Also, } \]
\[\frac{- \pi}{4} < x < \frac{\pi}{4} \Rightarrow - 1 < \cot x < 1\]
\[ \Rightarrow 0 < 1 + \cot x < 2\]
\[ \Rightarrow 1 + \cot x > 0 . . . \left( 2 \right)\]
\[\cos x\left( 1 + \cot x \right) > 0, \forall x \in \left( \frac{- \pi}{4}, \frac{\pi}{4} \right) \left[ \text { From eqs }. (1) \text { and }(2) \right]\]
\[ \Rightarrow f'\left( x \right) > 0, \forall x \in \left( \frac{- \pi}{4}, \frac{\pi}{4} \right)\]
\[\text { So,}f\left( x \right) \text { is increasing on }\left( \frac{- \pi}{4}, \frac{\pi}{4} \right).\]
Solution 2
f(x) = sinx − cosx
We differentiate f(x) with respect to x:
`f'(x) = d/dx (sinx-cosx) = cosx + sin x`
f′(x) = cosx + sinx
I `(-pi/4, pi/4)`, both sinx and cosx are positive.
Therefore, f′(x) = cosx + sinx > 0 throughout that interval.
This implies that f(x) is strictly increasing on `(-pi/4, pi/4)`
f(x) = sinx − cosx is an increasing function on `(-pi/4, pi/4)`
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