Show that f(x)= - is an increasing function on (- 4 , 4 ).

Question

Show that $\text{f}(\text{x})=\sin\text{x}-\cos\text{x}$ is an increasing function on $\Big(-\frac{\pi}{4},\frac{\pi}{4}\Big).$

✓

Answer

$\text{f}(\text{x})=\sin\text{x}-\cos\text{x}$
$\therefore\ \text{f}'(\text{x})=\cos\text{x}+\sin\text{x}$
$=\sqrt2\Big(\frac{1}{\sqrt2}\cos\text{x}+\frac{1}{\sqrt2}\sin\text{x}\Big)$
$=\sqrt2\Big(\frac{\sin\pi}{4}\cos\text{x}+\frac{\cos\pi}{4}\sin\text{x}\Big)$
$=\sqrt2\sin\Big(\frac{\pi}{4}+\text{x}\Big)$
Now,
$\text{x}\in\Big(-\frac{\pi}{4},\frac{\pi}{4}\Big)$
$\Rightarrow-\frac{\pi}{4}<\text{x}<\frac{\pi}{4}$
$\Rightarrow0<\frac{\pi}{4}<\text{x}<\frac{\pi}{2}$
$\Rightarrow\sin0^{\circ}<\sin\Big(\frac{\pi}{4}+\text{x}\Big)<\sin\frac{\pi}{4}$
$\Rightarrow0<\sin\Big(\frac{\pi}{4}+\text{x}\Big)<1$
$\Rightarrow\sqrt2\sin\Big(\frac{\pi}{4}+\text{x}\Big)>0$
$\Rightarrow\text{f}'(\text{x})>0$
Hence, f(x) is increasing function on $\Big(-\frac{\pi}{4},\frac{\pi}{4}\Big).$

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