ns was watching this video clip and at 5:30 $\cos \frac\pi2$ is substituted with $0$. Utilizing my calculator the result I acquired was $-0.5$. I m sorry one is correct?  $\cos\left(\frac\pi2\right)=0$. The factor you are gaining $-0.5$ is since you are not putting brackets approximately $\pi/2$. Thus, you room obtaining the value of $(\cos\pi)/2=(-1)/2=-0.5$. You need to use the brackets together follows: $\cos\left(\frac\pi2\right)$.  There room a pair ways to present that $\cos (\pi/2)=0$

1) The value $\tan(\pi/2)$ is undefined. Because $\tan(x)=\frac\sin(x)\cos(x)$, and both $\sin(x)$ and also $\cos(x)$ are continuous throughout the reals, this suggests 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 at the unit circle where $y=\sin(\theta)$ and also $x=\cos(\theta)$. When $\theta= 90^0=\pi/2$, $\sin (\theta)=1$ and also $\cos(\theta)=0$.

4) the derivative of $\sin(x)$ is $0$ only as soon as $\sin(x)=1$, which only occurs once $x= \fracn\pi2$ for any type of $n\in \sirhenryjones-museums.orgbb N$. Because the derivative the $\sin(x)$ is $\cos(x)$ this suggests that $\cos(\pi/2)=\cos((1)\pi/2)=0$.

If girlfriend have any qustions let me recognize below.

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reply Mar 3 "17 at 23:21 Sentinel135Sentinel135
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