The maximum value of a trigonometric expression (without derivative)

Question

For how trivial it may appear, here I am with this question: find the value of $x$ for which this expression has its maximum.

$$f(x) = 1 + 4\sin^2(x) + 4\sin(x)\cos(x)$$

Without making use of derivatives, and only basing on trigonometric expressions / formulas.

What I have tried is to manipulate a bit the expression, but I couldn't be able to reach any interesting expression.

Answer 1

Basics: $$ \begin{align} % \sin^{2} x &= \frac{1}{2} \left( 1 - \cos 2 x \right) \\[3pt] % \cos x \sin x &= \frac{1}{2} \sin 2x % \end{align} $$

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The target function becomes $$ f(x) = 1 + 4 \sin^{2} x + 4 \sin x \cos x = 3 + 2 \left( \sin 2 x - \cos 2 x \right) \tag{1} $$

When does $\sin \theta - \cos \theta$ achieve maximal value? When $$ \theta = \frac{3}{4}\pi + 2\pi k, \quad k \in \mathbb{Z} $$

Therefore, the function (1) achieves maximum value at $$ \boxed{x = \frac{3}{8}\pi + \pi k, \quad k \in \mathbb{Z}} $$

Answer 2

Using

$2 \, \sin^2 x = 1-\cos(2x)$,

$\sin(2x)=2\,\sin x \, \cos x$,

and $\sin(\theta)-\cos(\theta) = \sqrt 2 \, \sin (\theta - \pi/4)$,

we get:

\begin{align} f(x) & = 1 + 4\sin^2(x) + 4\sin(x)\cos(x) \\ & = 1 + 2 - 2\cos(2x) + 2\sin(2x) \\ & = 3 + 2(\sin(2x)-\cos(2x)) \\ & = 3 + 2 \,\sqrt 2 \, \sin (2x-\pi/4) \end{align}

So the max of $f(x)$ should be $3+2\sqrt 2$.