Question
Prove that the function $\text{f}(\text{x})=\cos\text{x}$ is:
Strictly decreasing in $(0,\pi).$

Answer

$\text{f}(\text{x})=\cos\text{x}$
$\text{f}'(\text{x})=-\sin\text{x}$
Here,
$0<\text{x}<\pi$
$\Rightarrow\sin\text{x}>0$ $[\because$ sine function is positive in first and second quadrent$]$
$\Rightarrow-\sin\text{x}<0$
$\Rightarrow\text{f}'(\text{x})<0,\forall\ \text{x}\in(0,\pi)$
So, f(x) is strictly decreasing on $(0,\pi).$

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