Prove that \(f(\theta)\) = \({4sin \theta\over 2 + cos\theta} – \theta\) is an increasing function of \(\theta\) in \([0, {\pi\over 2}]\).

Solution :

We have, \(f(\theta)\) = \({4sin \theta\over 2 + cos\theta} – \theta\)

\(\implies\) \(f'(\theta)\) = \((2 + cos\theta)(4 cos\theta) + 4 sin^2\theta\over (2 + cos\theta)^2\) – 1

\(\implies\) \(f'(\theta)\)  = \(8 cos\theta + 4\over (2 + cos\theta)^2\) – 1

\(\implies\) \(f'(\theta)\) = \(4\cos\theta – cos^2\theta\over (2 + cos\theta)^2\)

\(\implies\) \(f'(\theta)\) = \(cos\theta(4 – cos\theta)\over (2 + cos\theta)^2\) > 0  for all \(\theta\) \(\in\) \((0, {\pi\over 2})\).

[ \(\because\)   \(cos\theta\) > 0 , 4 – \(cos\theta\) > 0 and 2 + \(cos\theta\) > 0 ]

Hence, \(f(\theta)\) is increasing on \([0, {\pi\over 2}]\).


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