# Question #a8660

There are two maximum points
$\left(\frac{\pi}{6} , \frac{5}{4}\right) = \left(0.523599 , 1.25\right) \text{ " }$ and $\left(\frac{5 \pi}{6} , \frac{5}{4}\right) = \left(2.61799 , 1.25\right)$
There is one minimum point $\left(\frac{\pi}{2} , 1\right) = \left(1.57 , 1\right) \text{ }$

#### Explanation:

Let the given by $y = \sin x + {\cos}^{2} x$

Determine the first derivative $\frac{\mathrm{dy}}{\mathrm{dx}}$ then equate to zero, that is $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$

Let us begin

from the given

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

$\frac{d}{\mathrm{dx}} \left(y\right) = \frac{d}{\mathrm{dx}} \left(\sin x\right) + \frac{d}{\mathrm{dx}} {\left(\cos x\right)}^{2}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \cos x \cdot \frac{\mathrm{dx}}{\mathrm{dx}} + 2 \cdot {\left(\cos x\right)}^{\left(2 - 1\right)} \cdot \frac{d}{\mathrm{dx}} \left(\cos x\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \cos x \cdot 1 + 2 \cdot {\left(\cos x\right)}^{1} \cdot \left(- \sin x\right) \cdot \frac{\mathrm{dx}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \cos x - 2 \cdot \sin x \cdot \cos x \cdot 1$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \cos x - 2 \cdot \sin x \cdot \cos x$

Equate $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$\cos x - 2 \cdot \sin x \cdot \cos x = 0$

solve by factoring

$\cos x \left(1 - 2 \sin x\right) = 0$

Equate each factor to zero

$\cos x = 0 \text{ " }$ the first factor
$\arccos \left(\cos x\right) = \arccos 0$

$x = \frac{\pi}{2}$

find $y$, using the original equation
$y = \sin x + {\cos}^{2} x$
$y = \sin \left(\frac{\pi}{2}\right) + {\cos}^{2} \left(\frac{\pi}{2}\right)$
$y = 1 + {\left(0\right)}^{2}$
$y = 1$

solution $\left(\frac{\pi}{2} , 1\right) = \left(1.57 , 1\right) \text{ }$the Minimum Point
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$1 - 2 \sin x = 0 \text{ " " }$ the second factor

$2 \cdot \sin x = 1$

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

$\arcsin \left(\sin x\right) = \arcsin \left(\frac{1}{2}\right)$

$x = \frac{\pi}{6}$ also $x = \frac{5 \pi}{6}$

find $y$, using $x = \frac{\pi}{6}$ in the original equation

$y = \sin x + {\cos}^{2} x$
$y = \sin \left(\frac{\pi}{6}\right) + {\cos}^{2} \left(\frac{\pi}{6}\right)$

$y = \frac{1}{2} + {\left(\frac{\sqrt{3}}{2}\right)}^{2}$

$y = \frac{1}{2} + \frac{3}{4}$
$y = \frac{5}{4}$

solution $\left(\frac{\pi}{6} , \frac{5}{4}\right) = \left(0.523599 , 1.25\right) \text{ " }$the Maximum Point

the other Maximum Point is at $\left(\frac{5 \pi}{6} , \frac{5}{4}\right) = \left(2.61799 , 1.25\right)$

because $\sin \left(\frac{\pi}{6}\right) = \sin \left(\frac{5 \pi}{6}\right)$. That is why there are two maximum points.
Kindly see the graph and locate the critical points
graph{y=sin x+(cos x)^2 [-1, 5, -1, 1.5]}

God bless....I hope the explanation is useful.