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Maximum value of function

Let p, q \in \mathbb{N}. Find the maximum vale of the function

    \[f(x) = \sin^p x \cos^q x\]

Solution

Using the AM – GM inequality we have

    \begin{align*} \frac{1}{p+q} &= \frac{\overbrace{\frac{\sin^2 x}{p}+ \frac{\sin^2 p}{p} + \cdots + \frac{\sin^2 p}{p}}^{p} + \overbrace{\frac{\cos^2 x}{q} + \frac{\cos^2 x}{q} + \cdots + \frac{\cos^2 x}{q}}^{q}}{p+q} \\ &\geq \sqrt[p+q]{\left ( \frac{\sin^2 x}{p} \right )^p \left ( \frac{\cos^2 x}{q} \right )^q} \\ &=\sqrt[p+q]{\frac{f^2(x)}{p^p q^q}} \end{align*}

Hence,

    \[f(x) = \sqrt{\frac{p^p q^q}{\left ( p+q \right )^{p+q}}}\]

Equality holds when

    \[\frac{\sin^2 x}{p} = \frac{\cos^2 x}{q} \Leftrightarrow \cos^2 x = \frac{p}{p+q}\]

Since \frac{p}{p+q} \in (0, 1) there exists an x \in \mathbb{R} such that the equation has at least one root. The minimum follows.

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