Maximum value of $\sin{2x}\tan^2{x}$ between $0 \leq x \leq \frac{\pi}{4}$

Question

I need other approach to solve this question.

Problem The maximum value of $f(x) = \sin{2x}\tan^2{x}$ for $x \in [0,\frac{\pi}{4}]$

What I have done so far, I find the $f'(x) = 4\sin^2{x} + 2 \tan^2{x}$, and think that $f'(x) > 0$ for all $x$ in $[0,\frac{\pi}{4}]$. I guess $f(x)$ increases on the interval mentioned. So, it has maximum value when $x = \frac{\pi}{4}$.

My question is this: '' Is there any solution more simple than my approach?'' Is there a way to simplify the $f(x)$ to get easier step?

Answer 1

Here's another way. You have

$$\begin{equation}\begin{aligned} f(x) & = \sin 2x \tan^2 x \\ & = (2\sin x\cos x)\left(\frac{\sin x}{\cos x}\right)^2 \\ & = \frac{2\sin^3 x}{\cos x} \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

Since in $0 \lt x \le \frac{\pi}{4}$, you have that $\sin x$ is a strictly increasing function (and since it's $\gt 0$, you also have $\sin^3 x$ is an increasing function), plus $\cos x$ is a strictly decreasing function. This means the numerator is strictly increasing while the denominator is strictly decreasing, so the overall fraction in \eqref{eq1A} is strictly increasing. As such, this means the maximum of $f(x)$ would be reached at the maximum point in the range, i.e., $x = \frac{\pi}{4}$.

Answer 2

$$f(x)=\frac{2\sin^3 x}{\cos x},f'(x) \implies f(x)=2(2\sin^2 x+\tan^2 x)>0$$ So $f(x)$ is an increasing function and so in $x \in [0,\pi/4]$, we have $$f(x)\le f(\pi/4)=1.$$

Answer 3

Since $\sin$ and $\tan$ increase and non-negatives on $\left[0,\frac{\pi}{4}\right]$, we obtain: $$2\sin^2x\tan{x}\leq2\sin^2\frac{\pi}{4}\tan\frac{\pi}{4}=1.$$ The equality occurs for $x=\frac{\pi}{4},$ which says that we got a maximal value.