# How do you find the maximum value of f(x) = sinx ( 1+ cosx) ?

Apr 28, 2018

#### Explanation:

The maximum value is calculated with the first and second derivatives.

The function is

$f \left(x\right) = \sin x \left(1 + \cos x\right) = \sin x + \sin x \cos x$

$= \sin x + \frac{1}{2} \sin 2 x$

The first derivative is

$f ' \left(x\right) = \cos x + 2 \times \frac{1}{2} \cos \left(2 x\right)$

$f ' \left(x\right) = 0$

When

$\cos x + \cos 2 x = 0$

$2 {\cos}^{2} x + \cos x - 1 = 0$

Solving this quadratic equation in $\cos x$

The solutions are

$\cos x = \frac{- 1 \pm \sqrt{{\left(- 1\right)}^{2} - 4 \left(2\right) \left(- 1\right)}}{4} = \frac{- 1 \pm 3}{4}$

$\left\{\begin{matrix}\cos x = - 1 \\ \cos x = \frac{1}{2}\end{matrix}\right.$

$\iff$, $\left\{\begin{matrix}x = \pi + 2 k \pi \\ x = \frac{\pi}{3} + 2 k \pi \\ x = \frac{5}{3} \pi + 2 k \pi\end{matrix}\right.$

The second derivative is

$f ' ' \left(x\right) = - \sin x - 2 \sin \left(2 x\right)$

Therefore,

$\left\{\begin{matrix}f ' ' \left(\pi\right) = - 0 - 0 = 0 \\ f ' ' \left(\frac{\pi}{3}\right) = - \frac{\sqrt{3}}{2} - \sqrt{3} < 0 \\ f ' ' \left(\frac{5}{3} \pi\right) = \frac{\sqrt{3}}{2} + \sqrt{3} > 0\end{matrix}\right.$

When $\left(x = \pi + 2 k \pi\right)$ corresponds to points of inflexions.

When $x = \frac{\pi}{3} + 2 k \pi$ corresponds to maximum

When $x = \frac{5}{3} \pi + 2 k \pi$ corresponds to minimum

graph{sinx+1/2sin(2x) [-2.08, 10.404, -2.845, 3.395]}