#2**0 **

This was answerd just a day or two ago:

remember cos(x+pi) = -cos(x)

d/dx (-cos(x)) = sinx and at pi sin(x) = **0**

ElectricPavlov Nov 27, 2018

#1**+1 **

The slope of the line tangent to a curve at a point is just the derivative of that curve evaluated at the point.

\(g(x) = \cos(x+\pi)\\ \dfrac{dg}{dx}(x) = -\sin(x+\pi)\\ \dfrac{dg}{dx}(\pi) = -\sin(\pi + \pi) = \\ \sin(2\pi) = 0\\ \text{hence the slope of the line tangent to }g(x) \text{ at }x=\pi \text{ equals }0.\)

.Rom Nov 27, 2018

#2**0 **

Best Answer

This was answerd just a day or two ago:

remember cos(x+pi) = -cos(x)

d/dx (-cos(x)) = sinx and at pi sin(x) = **0**

ElectricPavlov Nov 27, 2018