Question

maximum value of 4sinxcosx

Answer 1

$4\sin x\cos x=2\sin2x$

$-1\le\sin ax\le1$, which means $-2\le2\sin ax\le2$, which means the maximum value is 2.

Answer 2

The maximum value of $f(x) = 4\cdot \sin x\cdot \cos x$ is 2 for $x = \frac{\pi}{4}+\pi\cdot i$, $\forall \,i\in \mathbb{Z}$.

In this question we should use the following trigonometric identity to simplify the expression described on statement:

$\sin 2x = 2\cdot \sin x \cdot \cos x$ (1)

Then, the expression can be rewritten as following:

$f(x) = 2\cdot \sin 2x$ (2)

Sine is a periodic function bounded between 0 and 1. Hence, we must find the set of solutions such that sine reaches its maximum. That is to say:

$\sin 2x = 1$

$2\cdot x = \sin^{-1} 1$

$2\cdot x = \frac{\pi}{2}+2\pi\cdot i$, $\forall \,i\in \mathbb{Z}$

$x = \frac{\pi}{4}+\pi\cdot i$, $\forall \,i\in \mathbb{Z}$

Hence, the maximum value of $f(x) = 4\cdot \sin x\cdot \cos x$ is 2 for $x = \frac{\pi}{4}+\pi\cdot i$, $\forall \,i\in \mathbb{Z}$.

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