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I've attempted to solve this question by using $$f(x) = 2\sin\frac{x}{2}\cos^2\frac{x}{2} \leq \frac{(2\sin\frac{x}{2}+\cos^2\frac{x}{2})^2}{2}$$ but it results in the wrong answer every time. Is there another way to solve this question and is there anything I can do to change my method to make it work.

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Since $\cos x = 1-2\sin^2 \frac{x}{2}$, $$ f(x)=\left(2-2\sin^2\frac{x}{2}\right)\sin \frac{x}{2} = 2\sin \frac{x}{2} - 2\sin^3\frac{x}{2}. $$ Thus you can find the maximum of $f(x)$ by finding the maximum of $g(x)=2x-2x^3$ defined on $(0,1)$.

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By AM-GM $$\left(1+\cos{x}\right)\sin\frac{x}{2}=2\cos^2\frac{x}{2}\sin\frac{x}{2}=2\left(\sin\frac{x}{2}-\sin^3\frac{x}{2}\right)=$$ $$=-2\left(-\sin\frac{x}{2}+\sin^3\frac{x}{2}+\frac{1}{3\sqrt3}+\frac{1}{3\sqrt3}\right)+\frac{4}{3\sqrt3}\leq$$ $$\leq-2\left(-\sin\frac{x}{2}+3\sqrt[3]{\sin^3\frac{x}{2}\cdot\frac{1}{3\sqrt3}\cdot\frac{1}{3\sqrt3}}\right)+\frac{4}{3\sqrt3}=\frac{4}{3\sqrt3}.$$ THe equality occurs for $\sin\frac{x}{2}=\frac{1}{\sqrt3}$, which says that the answer is $\frac{4}{3\sqrt3}$.

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When you have $2 \sin \frac{x}{2} (1-\sin^2 \frac x2) = 2\sin \frac x2 - 2\sin^3 \frac x2$, where $x$ varies between $0$ and $\pi$, hence we can let $y= \sin \frac x2$, then $y$ varies between $0$ and $1$, and we are trying to find the maximum value of $2y-2y^3$. Now, we can just use ordinary differentiation, $2-6y^2=0\implies y^2 = \frac 13$, and in this case, the value of $2y-2y^3$ is $2y(1-y^2)=\frac{2}{3} \times \frac{2}{\sqrt 3}$.

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  • $\begingroup$ The maximum of $fog$ is the maximum of $f$? $\endgroup$ – Nosrati Jan 10 '17 at 5:32
  • $\begingroup$ Yes, this is the case here. In general, I do not know whether that is true. I used a one-one map between $[0,\pi]$ and $[0,1]$ in the process, so I am not sure if it would work in general. $\endgroup$ – астон вілла олоф мэллбэрг Jan 10 '17 at 5:33

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