trigonometry
How to find the maximum value of $12\sin x -9\sin^2x$ ;
My approach :
This can be written as $-[(3\sin x -2)^2-4]$.
It means that the function will be maximum when $(3\sin x-2)^2 <4$ due to negative sign outside bracket.
But I am not getting how to proceed from here, please suggest. Thanks.
This is not an independent answer but a response to Lord Soth's speculation of a geometric proof.
Consider two right-handed triangles $ABC$ and $ABD$ with base $1$ and heights $\sqrt{n}$ and $\sqrt{n-1}$. Let $\theta$ be the angle $\measuredangle CBD$. The area of the triangle $BCD$ can be computed in two ways:
Equate them gives us:
$$\sin \theta = \frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n}\sqrt{n+1}}$$
On the other hand,
$$\begin{align} \theta &= \measuredangle CBD = \measuredangle CBA - \measuredangle DBA\\ &= \sin^{-1}\frac{|AC|}{|BC|} - \sin^{-1}\frac{|AD|}{|BD|} = \sin^{-1}\sqrt{\frac{n}{n+1}} - \sin^{-1}\sqrt{\frac{n-1}{n}} \end{align}$$
We obtain
$$\sin^{-1}\left( \frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n}\sqrt{n+1}}\right) = \sin^{-1}\sqrt{\frac{n}{n+1}} - \sin^{-1}\sqrt{\frac{n-1}{n}}$$
and we have turned the original series into a telescoping one. The partial sum of the first $n$ terms of original series becomes the angle $\measuredangle ABC$. When $n \to \infty$, the line $BC$ becomes vertical and this geometrically justify why the limit of the series is $\frac{\pi}{2}$.
You can evaluate this by using complex methods.
Let $\alpha=e^{2\pi i/7}$. Then $$\sin\frac{2\pi}{7}=\frac{\alpha-\alpha^{-1}}{2i}\ ,\quad \sin\frac{4\pi}{7}=\frac{\alpha^2-\alpha^{-2}}{2i}\ ,\quad \sin\frac{8\pi}{7}=\frac{\alpha^4-\alpha^{-4}}{2i}\ .$$ Now let $S$ be the sum of these three numbers. Write $S$ in terms of powers of $\alpha$ and calculate $S^2$. Initially it's a mess, but you can use the relations $$\alpha^7=1\quad\hbox{and}\quad \alpha^6+\alpha^5+\cdots+\alpha=-1$$ to show that it simplifies, amazingly, to $$S^2=\frac{7}{4}\ .$$ It's not hard to see that $S$ is positive, so $$S=\frac{\sqrt7}{2}\ .$$
Comment. You can also write the sum as $$S=\frac{1}{2i}\sum_{k=1}^6 \Bigl(\frac{k}{7}\Bigr)\alpha^k\ ,$$ where $(\frac{k}{7})$ is a Legendre symbol, and this connects the sum with some very interesting and often difficult mathematics.
Best Answer
Note $$ 12\sin x-9\sin^2x=3\sin x(4-3\sin x)\le \left(\frac{3\sin x+(4-3\sin x)}{2}\right)^2=4 $$ and the equal sign holds if and only if $3\sin x=4-3\sin x$ or $\sin x=\frac{2}{3}$. Thus the max is 4 when $\sin x=\frac{2}{3}$.