How to find the absolute maximum of $f(x) = (\sin 2\theta)^2 (1+\cos 2\theta)$ for $0 \le \theta \le \frac{\pi}2$?

Question

How to find the absolute maximum of $f(x) = (\sin 2\theta)^2 (1+\cos 2\theta)$ for $0 \le \theta \le \frac{\pi}2$?

I found the derivative to be $f'(x)= (2\sin 4\theta)(1+\cos 2\theta) + (\sin 2\theta)^2 (-2\sin 2\theta)$.

Now I know that I have to set this equal to zero and solve for $\theta$ but that is where I get stuck, can I solve for $\theta$ or is there a different way that is easier?

Answer 1

I’d start by rewriting $f(x)$, using the identities $\sin2\theta=2\sin\theta\cos\theta$ and $\cos^2\theta=\frac12(1+\cos2\theta)$:

$$f(x)=\sin^22\theta(1+\cos2\theta)=8\sin^2\theta\cos^4\theta\;.$$

In fact, let’s go a step further and get rid of the sine:

$$f(x)=8\sin^2\theta\cos^4\theta=8(1-\cos^2\theta)\cos^4\theta=8\cos^4\theta-8\cos^6\theta\;.$$

Differentiating and factoring yields

$$f\,'(x)=16\sin\theta\cos^3\theta(3\cos^2\theta-2)\;.$$

Now set this to $0$ and solve. If you want, you can further simplify a little by noting that

$$3\cos^2\theta-2=3(1-\sin^2\theta)-2=1-3\sin^2\theta\;.$$

Answer 2

Here $$\displaystyle f(\theta) = \sin^2 2\theta\cdot (1+\cos 2\theta)\;\;,$$ Where $\displaystyle 0\leq \theta \leq \pi$

So $$f(\theta) = (1-\cos^2 2\theta)\cdot (1+\cos 2\theta)\;\;,$$ Now Put $\cos 2\theta = t\;\;,$ Then $-1\leq \cos 2\theta \leq 1$

So we get $$f(t) = (1-t^2)\cdot (1+t) = (1+t)^2\cdot (1-t)\;\;,$$ Where $-1\leq t\leq 1$

Answer 3

Let $\cos(2\theta)=:u$. We then have to find the maximum of the function $$g(u):=(1-u^2)(1+u)$$ in the interval $-1\leq u\leq 1$. One immediately checks that $g(-1)=g(1)=0$ and that $g(u)>0$ for $-1< u< 1$. The maximum is therefore taken at a zero of the derivative $$g'(u)=1-2u-3u^2=(1-3u)(1+u)$$ in $\ ]-1,1[\ $. There is one such zero, namely $u_*={1\over3}$, which leads to $g(u_*)={32\over27}$. This is also the global maximum of $\theta\mapsto f(\theta)$ in the interval $0\leq\theta\leq{\pi\over2}$, the corresponding $\theta$-value being $\theta_*={1\over2}\arccos{1\over3}$.

Answer 4

In order to find the stationary points of $g^2$ it is enough to find the roots of $g$ and the stationary points of $g$. Since:

$$ \sin^2(2\theta)(1+\cos(2\theta)) = 2\sin^2(2\theta)\cos^2(\theta) $$ we just need to find the stationary points of: $$ 2\sin(2\theta)\cos(\theta) = \sin(3\theta)+\sin(\theta) $$ that occur for $\cos\theta = 0$ or $3\cos^2\theta-2=0$. If $\cos^2\theta=\frac{2}{3}$, $\sin^2\theta=\frac{1}{3}$. That gives:

$$ \sin^2(2\theta)(1+\cos(2\theta))\leq 8\cdot\frac{1}{3}\cdot\frac{4}{9} = \frac{32}{27}.$$ That also follows from the AM-GM inequality.