Let 
. Find the maximum vale of the function
![\[f(x) = \sin^p x \cos^q x\]](https://www.math.tolaso.com.gr/wp-content/ql-cache/quicklatex.com-ddc343daa0860d33c6bf5dea660a1569_l3.png "Rendered by QuickLaTeX.com")
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*}](https://www.math.tolaso.com.gr/wp-content/ql-cache/quicklatex.com-de72bfd158cb204eff23c34d71a034cc_l3.png "Rendered by QuickLaTeX.com")
Hence,
![\[f(x) = \sqrt{\frac{p^p q^q}{\left ( p+q \right )^{p+q}}}\]](https://www.math.tolaso.com.gr/wp-content/ql-cache/quicklatex.com-90d531065976427ce76522ebff31672c_l3.png "Rendered by QuickLaTeX.com")
Equality holds when
![\[\frac{\sin^2 x}{p} = \frac{\cos^2 x}{q} \Leftrightarrow \cos^2 x = \frac{p}{p+q}\]](https://www.math.tolaso.com.gr/wp-content/ql-cache/quicklatex.com-1b1f70122d8a170a8b5dd64048cf4ce5_l3.png "Rendered by QuickLaTeX.com")
Since 
there exists an 
such that the equation has at least one root. The minimum follows.
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Let . Find the maximum vale of the function
Solution
Using the AM – GM inequality we have
Hence,
Equality holds when
Since there exists an such that the equation has at least one root. The minimum follows.