# How do you find points of inflection for y= sin x + cos x?

Jul 14, 2015

The point of inflexion are: $\left(\frac{3 \pi}{4} + 2 k \pi , 0\right) \text{ AND } \left(\left(- \frac{\pi}{2} + 2 k \pi , 0\right)\right)$

#### Explanation:

1 - First we have to find the second derivative of our function.

2 - Second, we equate that derivative$\left(\frac{{d}^{2} y}{{\mathrm{dx}}^{2}}\right)$ to zero

$y = \sin x + \cos x$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \cos x - \sin x$

$\implies \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - \sin x - \cos x$

Next, $- \sin x - \cos x = 0$

$\implies \sin x + \cos x = 0$

Now, we shall express that in the form $R \cos \left(x + l a m \mathrm{da}\right)$

Where $\lambda$ is just an acute angle and $R$ is a positive integer to be determined. Like this

$\sin x + \cos x = R \cos \left(x + \lambda\right)$

$\implies \sin x + \cos x = R \cos x \cos l a m \mathrm{da} - \sin x \sin l a m \mathrm{da}$

By equating the coefficients of $\sin x$ and $\cos x$ on either side of the equation,

$\implies R \cos l a m \mathrm{da} = 1$

and $R \sin \lambda = - 1$

$\frac{R \sin \lambda}{R \cos \lambda} = \frac{- 1}{1} \implies \tan \lambda = - 1 \implies \lambda = {\tan}^{-} 1 \left(- 1\right) = - \frac{\pi}{4}$

And ${\left(R \cos \lambda\right)}^{2} + {\left(R \sin \lambda\right)}^{2} = {\left(1\right)}^{2} + {\left(- 1\right)}^{2}$

$\implies {R}^{2} \left({\cos}^{2} x + {\sin}^{2} x\right) = 2$

But we know the identity , ${\cos}^{2} x + {\sin}^{2} = 1$

Hence, ${R}^{2} \left(1\right) = 2 \implies R = \sqrt{2}$

In a nut shell, $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - \sin x - \cos x = \sqrt{2} \cos \left(x - \frac{\pi}{4}\right) = 0$

$\implies \sqrt{2} \cos \left(x - \frac{\pi}{4}\right) = 0$

$\implies \cos \left(x - \frac{\pi}{4}\right) = 0 = \cos \left(\frac{\pi}{2}\right)$

So the general solution of $x$ is : $x - \frac{\pi}{4} = \pm \frac{\pi}{2} + 2 k \pi$ , $k \in \mathbb{Z}$

$\implies x = \frac{\pi}{4} \pm \frac{\pi}{2} + 2 k \pi$

So the points of inflexion will be any point that has coordinates :
$\left(\frac{\pi}{4} \pm \frac{\pi}{2} + 2 k \pi , \sqrt{2} \cos \left(\frac{\pi}{4} \pm \frac{\pi}{2} - \frac{\pi}{4}\right)\right)$

We have two cases to deall with,

Case 1

$\left(\frac{\pi}{4} + \frac{\pi}{2} + 2 k \pi , \sqrt{2} \cos \left(\frac{\pi}{4} + \frac{\pi}{2} - \frac{\pi}{4}\right)\right)$

$\implies \left(\frac{3 \pi}{4} + 2 k \pi , \sqrt{2} \cos \left(\frac{\pi}{2}\right)\right)$

$\implies \left(\frac{3 \pi}{4} + 2 k \pi , 0\right)$

Case 2

$\left(\frac{\pi}{4} - \frac{\pi}{2} + 2 k \pi , \sqrt{2} \cos \left(\frac{\pi}{4} - \frac{\pi}{2} - \frac{\pi}{4}\right)\right)$

$\implies \left(- \frac{\pi}{2} + 2 k \pi , \sqrt{2} \cos \left(- \frac{\pi}{2}\right)\right)$

$\implies \left(\left(- \frac{\pi}{2} + 2 k \pi , 0\right)\right)$