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Question Number 174271 by mnjuly1970 last updated on 28/Jul/22

$$$$$$max(sin\left(x\right).{cos}^3\left(x\right))=?$$$$x\in [0,\frac{\pi }{2}]$$

Commented by blackmamba last updated on 28/Jul/22

$$f\left(x\right)=\sin x(1-\sin {}^{2}x)\cos x$$$$f\left(x\right)=\frac{1}{2}\sin 2x(1-\left(\frac{1-\cos 2x}{2}\right))$$$$f\left(x\right)=\frac{1}{4}\sin 2x(1+\cos 2x)$$$$f\left(x\right)=\frac{1}{4}\sin 2x+\frac{1}{8}\sin 4x$$$$f{}^{\prime }\left(x\right)=\frac{1}{2}\cos 2x+\frac{1}{2}\cos 4x=0$$$$\Rightarrow 2\cos 3x\cos x=0$$$$\{\begin{array}{l}\cos 3x=0\Rightarrow 3x=\frac{\pi }{2};x=\frac{\pi }{6}\\ \cos x=0\Rightarrow x=\frac{\pi }{2}\end{array}$$$$\{\begin{array}{l}forx=\frac{\pi }{6}\Rightarrow f\left(\frac{\pi }{6}\right)=\frac{1}{4}\sin \left(\frac{\pi }{3}\right)[1+\cos \left(\frac{\pi }{3}\right)]\\ forx=\frac{\pi }{2}\Rightarrow f\left(\frac{\pi }{2}\right)=\frac{1}{4}\sin \pi [1+\cos \pi ]=0\end{array}$$$$max=f\left(\frac{\pi }{6}\right)=\frac{1}{4}\times \frac{1}{2}\sqrt{3}\times \frac{3}{2}=\frac{3\sqrt{3}}{16}$$

Commented by mnjuly1970 last updated on 28/Jul/22

$$thxalotSir$$

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