Question

Show that f(x) = sin x − cos x is an increasing function on (−π/4, π/4)?

Answer 1

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

\[f'\left( x \right) = \cos x + \sin x\]

\[ = \cos x\left( 1 + \frac{\sin x}{\cos x} \right)\]

\[ = \cos x\left( 1 + \cot x \right)\]

\[\text { Here, } \]

\[\frac{- \pi}{4} < x < \frac{\pi}{4}\]

\[ \Rightarrow \cos x > 0 . . . \left( 1 \right)\]

\[\text { Also, } \]

\[\frac{- \pi}{4} < x < \frac{\pi}{4} \Rightarrow - 1 < \cot x < 1\]

\[ \Rightarrow 0 < 1 + \cot x < 2\]

\[ \Rightarrow 1 + \cot x > 0 . . . \left( 2 \right)\]

\[\cos x\left( 1 + \cot x \right) > 0, \forall x \in \left( \frac{- \pi}{4}, \frac{\pi}{4} \right) \left[ \text { From eqs }. (1) \text { and }(2) \right]\]

\[ \Rightarrow f'\left( x \right) > 0, \forall x \in \left( \frac{- \pi}{4}, \frac{\pi}{4} \right)\]

\[\text { So,}f\left( x \right) \text { is increasing on }\left( \frac{- \pi}{4}, \frac{\pi}{4} \right).\]

Answer 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 sin⁡x and cos⁡x are positive.

Therefore, f′(x) = cos⁡x + sin⁡x > 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)`