If \(\frac{\pi}{2}\) \(\leq\) \(\theta\) \(\leq\) 2\(\pi\), find the maximum value of f(\(\theta\)) = \(\frac{4}{6 + 2cos \theta}\)
To find the maximum of \(f(\theta) = \frac{4}{6 + 2\cos \theta}\) for \(\frac{\pi}{2} \leq \theta \leq 2\pi\), simplify to \(f(\theta) = \frac{2}{3 + \cos \theta}\).
The maximum occurs when \(3 + \cos \theta\) is minimized. In the interval, \(\cos \theta\) ranges from \(-1\) to \(1\), with minimum \(-1\) at \(\theta = \pi\).
Thus, minimum denominator is \(3 - 1 = 2\), so maximum \(f(\theta) = \frac{2}{2} = 1\).
There is an explanation video available below.
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