Finding range of $\sin(x)\cos(2x)$
I have to find the range of $f(x)=\sin(x)\cos(2x)$. Here's what I have so far:
$f'(x)=-\cos(x)\cos(2x)-2\sin(2x)\sin(x)$. So $\cos(x)\cos(2x)+2\sin(2x)\sin(x)=\cos(x)(2\cos^2(x)-1)+4\sin^2x\cos x=0$. Hence $\cos(x)[2\cos^2(x)+4\sin^2(x)-1]=0$.
From $\cos(x)=0$, we get the minimum is $-1$, but I can't figure out how to solve the second equation.
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$\begingroup$As an alternative we can use the fact that $\cos2x=1-2\sin^2x$, which means that$$ f(x)=\sin x(1-2\sin^2x)=-2\sin^3x+\sin x. $$It remains to find the range of $-2u^3+u$ within $[-1,1]$ after making the substitution $u=\sin x$. This is quite easy to do; the range turns out to be $[-1,1]$. The minimum is achieved when $u=1$, and the maximum when $u=-1$.
$\endgroup$ $\begingroup$Alternative solution with no derivatives - is it a precalculus exercise?
Note that$$f(x)=\sin(x)\cos(2x)=\sin(x)(1-2\sin^2(x)).$$Therefore, since $\sin(\mathbb{R})=[-1,1]$, it follows that$$f(\mathbb{R})=g([-1,1])$$where $g(t)=t(1-2t^2)$.
Now for $t\in [-1,1]$, both factors $t$ and $(1-2t^2)$ are in $[-1,1]$, hence $g([-1,1])\subset [-1,1]$.
On the other hand $g(-1)=1$ and $g(1)=-1$.
So we may conclude that
$$f(\mathbb{R})=g([-1,1])=[-1,1].$$
There's a shortcut here. Since $\sin x,\,\cos 2x\in[-1,\,1]$, $\sin x \cos 2x\in[-1,\,1]$. The choice $x=\pm\frac{\pi}{2}$ obtains $\sin x \cos 2x=\mp 1$, so both extrema are obtainable. We want the range of $u(1-2u^2)$ for $u\in [-1,\,1]$, so the continuity of polynomials implies the range is the full interval $[-1,\,1]$.
$\endgroup$ $\begingroup$There are standard product-to-sum identities that give
$$\sin x\cos 2x = \frac{1}{2}(\sin 3x - \sin x).$$
(This is easy to derive by applying the sum and difference of angle formulae to $\sin(2x+x)$ and $\sin(2x-x).$)
Then it's pretty easy to find angles that make the last sum equal to $\frac{1}{2}(1-(-1))$ and $\frac{1}{2}(-1-1).$
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