$cosleft(fracpi2 ight)=0$. The reason you are getting $-0.5$ is since you space not placing brackets roughly $pi/2$. Thus, you room obtaining the value of $(cospi)/2=(-1)/2=-0.5$. You have to use the brackets together follows: $cosleft(fracpi2 ight)$.
There room a couple ways to display that $cos (pi/2)=0$
1) The worth $ an(pi/2)$ is undefined. Due to the fact that $ an(x)=fracsin(x)cos(x)$, and also both $sin(x)$ and also $cos(x)$ are continuous throughout the reals, this implies that it have the right to only occur if $cos (pi/2)=0$.
You are watching: Cos(-pi/2)
2) The co-function identification $cos(pi/2-x)=sin(x)$ means that $cos (pi/2)=cos(pi/2-0)=sin(0)=0$
3) Looking in ~ the unit circle wherein $y=sin( heta)$ and $x=cos( heta)$. Once $ heta= 90^0=pi/2$, $sin ( heta)=1$ and $cos( heta)=0$.
4) the derivative the $sin(x)$ is $0$ only when $sin(x)=1$, which just occurs as soon as $x= fracnpi2$ for any type of $nin 4476mountvernon.combb N$. Because the derivative that $sin(x)$ is $cos(x)$ this indicates that $cos(pi/2)=cos((1)pi/2)=0$.
If you have any qustions permit me recognize below.
answer Mar 3 "17 in ~ 23:21
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