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Show that $f(x)=\sin x$ is an increasing function on $(-\pi / 2, \pi / 2)$.


Given:- Function $f(x)=\sin x$

Theorem:- Let $f$ be a differentiable real function defined on an open interval $(a, b)$.

(i) If $f^{\prime}(x)>0$ for $a l l_{x} \in(a, b)$, then $f(x)$ is increasing on $(a, b)$

(ii) If $f^{\prime}(x)<0$ for all $x \in(a, b)$, then $f(x)$ is decreasing on $(a, b)$


(i) Obtain the function and put it equal to $f(x)$

(ii) Find $f^{\prime}(x)$

(iii) Put $f^{\prime}(x)>0$ and solve this inequation.

For the value of $x$ obtained in (ii) $f(x)$ is increasing and for remaining points in its domain it is decreasing.

Here we have,

$f(x)=\sin x$

$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}(\sin \mathrm{x})$

$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\cos \mathrm{x}$

Now, as given

$x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

That is $4^{\text {th }}$ quadrant, where

$\Rightarrow \cos x>0$

$\Rightarrow f^{\prime}(x)>0$

hence, Condition for $f(x)$ to be increasing

Thus $f(x)$ is increasing on interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

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