Csdrhrt

I have to find the range of $f(x)=sin(x)cos(2x)$. Here's what I have so far:

$f'(x)=-cos(x)cos(2x)-2sin(2x)sin(x)$. So $cos(x)cos(2x)+2sin(2x)sin(x)=cos(x)(2cos^2(x)-1)+4sin^2xcos x=0$.
Hence $cos(x)[2cos^2(x)+4sin^2(x)-1]=0$.

From $cos(x)=0$, we get the minimum is $-1$, but I can't figure out how to solve the second equation.

Alternative solution with no derivatives - is it a precalculus exercise?

Note that
$$f(x)=sin(x)cos(2x)=sin(x)(1-2sin^2(x)).$$
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 $tin [-1,1]$, 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
$$f(mathbb{R})=g([-1,1])=[-1,1].$$

As an alternative we can use the fact that $cos2x=1-2sin^2x$, which means that
$$ f(x)=sin x(1-2sin^2x)=-2sin^3x+sin x. $$
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 $u=-1$.

There's a shortcut here. Since $sin x,,cos 2xin[-1,,1]$, $sin x cos 2xin[-1,,1]$. The choice $x=pmfrac{pi}{2}$ obtains $sin x cos 2x=mp 1$, so both extrema are obtainable. We want the range of $u(1-2u^2)$ for $uin [-1,,1]$, so the continuity of polynomials implies the range is the full interval $[-1,,1]$.

There are standard product-to-sum identities that give

$$sin xcos 2x = frac{1}{2}(sin 3x - sin x).$$

(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 $frac{1}{2}(-1-1).$