Question
Is the function $cos2x$ decreasing on ($0, \frac{\pi}{2}$)?
✓
Answer
Let f(x) = cos 2x
$\therefore$ $\mathrm{f}^{\prime}(\mathrm{x})$ = -2 sin 2x
Now, $0<x<\frac{\pi}{2}$
$\Rightarrow$ $0<2 x<\pi$
$\Rightarrow$ sin 2x > 0
$\Rightarrow$ -2 sin 2x < 0
$\therefore$ $\mathrm{f}^{\prime}(\mathrm{x})$ = -2 sin 2x < 0 on $\left(0, \frac{\pi}{2}\right)$
Therefore, f(x) = cos 2x is strictly decreasing in interval $\left(0, \frac{\pi}{2}\right)$.
$\therefore$ $\mathrm{f}^{\prime}(\mathrm{x})$ = -2 sin 2x
Now, $0<x<\frac{\pi}{2}$
$\Rightarrow$ $0<2 x<\pi$
$\Rightarrow$ sin 2x > 0
$\Rightarrow$ -2 sin 2x < 0
$\therefore$ $\mathrm{f}^{\prime}(\mathrm{x})$ = -2 sin 2x < 0 on $\left(0, \frac{\pi}{2}\right)$
Therefore, f(x) = cos 2x is strictly decreasing in interval $\left(0, \frac{\pi}{2}\right)$.
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