# What are the points of inflection of f(x)= sin^2x - cos^2x on x in [0, 2pi] ?

Sep 27, 2016

The points of inflection are :
$\left(\textcolor{b r o w n}{\frac{\pi}{4} , 0}\right)$Or$\left(\textcolor{b r o w n}{\frac{3 \pi}{4} , 0}\right)$

#### Explanation:

To find the inflection points we have to calculate the second derivative then set the equation to zero.

Knowing these properties:
color(blue)((sinu)'=u'cosu
color(red)((cosu)'=-u'sinu
$\textcolor{g r e e n}{{v}^{n} = n {v}^{n - 1}}$
color(orange)(sin2x=2sinxcosx

First let us find f''(x):

$f ' \left(x\right) = \textcolor{g r e e n}{2 \sin x \left(\sin x\right) ' - 2 \left(\cos x\right) \left(\cos x\right) '}$ using 3rd property
$f ' \left(x\right) = 2 \sin x \textcolor{b l u e}{\cos x} - 2 \cos x \left(\textcolor{red}{- \sin x}\right)$
$f ' \left(x\right) = 2 \sin x \cos x + 2 \cos x \sin x$
$f ' \left(x\right) = 4 \sin x \cos x$
$f ' \left(x\right) = 2 \textcolor{\mathmr{and} a n \ge}{\left(2 \sin x \cos x\right)}$
$f ' \left(x\right) = \textcolor{\mathmr{and} a n \ge}{2 \sin 2 x}$

Let us find $f ' ' \left(x\right)$:
$f ' ' \left(x\right) = \left(f ' \left(x\right)\right) '$
$f ' ' \left(x\right) = \left(2 \sin 2 x\right) '$
$f ' ' \left(x\right) = 2 \textcolor{b l u e}{\left(\sin 2 x\right)} '$
f"(x)=2*(color(blue)(2cos2x)) using first property above
f"(x)=4cos2x

Now set f"(x)=0
$4 \cos 2 x = 0$
$\cos 2 x = 0$
In the given interval $| 0 , 2 \pi |$ then :
$2 x = \frac{\pi}{2} \Rightarrow x = \frac{\pi}{4}$
$f \left(\frac{\pi}{4}\right) = {\sin}^{2} \left(\frac{\pi}{4}\right) - {\cos}^{2} \left(\frac{\pi}{4}\right) = {\left(\frac{\sqrt{2}}{2}\right)}^{2} - {\left(\frac{\sqrt{2}}{2}\right)}^{2} = 0$
The point is$\left(\textcolor{b r o w n}{\frac{\pi}{4} , 0}\right)$

Or
$2 x = \frac{3 \pi}{2} \Rightarrow x = \frac{3 \pi}{4}$
$f \left(\frac{3 \pi}{4}\right) = {\sin}^{2} \left(\frac{3 \pi}{4}\right) - {\cos}^{2} \left(\frac{3 \pi}{4}\right)$
$f \left(\frac{3 \pi}{4}\right) = {\left(\sin \left(\pi - \frac{\pi}{4}\right)\right)}^{2} - {\left(\cos \left(\pi - \frac{\pi}{4}\right)\right)}^{2}$
$f \left(\frac{3 \pi}{4}\right) = {\left(\sin \left(\frac{\pi}{4}\right)\right)}^{2} - {\left(- \cos \left(\frac{\pi}{4}\right)\right)}^{2}$
$f \left(\frac{3 \pi}{4}\right) = {\left(\frac{\sqrt{2}}{2}\right)}^{2} - {\left(\frac{\sqrt{2}}{2}\right)}^{2} = 0$
The second point is $\left(\textcolor{b r o w n}{\frac{3 \pi}{4} , 0}\right)$