I was watching this video and at 5:30 $\cos \frac{\pi}{2}$ is substituted with $0$. Using my calculator the result I got was $-0.5$. Which one is correct?
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$\begingroup$$\cos\left(\frac{\pi}{2}\right)=0$. The reason you are getting $-0.5$ is because you are not putting brackets around $\pi/2$. Thus, you are obtaining the value of $(\cos\pi)/2=(-1)/2=-0.5$. You need to use the brackets as follows: $\cos\left(\frac{\pi}{2}\right)$.
$\endgroup$ $\begingroup$There is a difference between $(\cos \pi)/2 = \frac{-1}{2}$ and $\cos(\pi/2) = 0$
$\endgroup$ $\begingroup$There are a couple ways to show that $\cos (\pi/2)=0$
1) The value $\tan(\pi/2)$ is undefined. Since $\tan(x)=\frac{\sin(x)}{\cos(x)}$, and both $\sin(x)$ and $\cos(x)$ are continuous throughout the reals, this implies that it can only occur if $\cos (\pi/2)=0$.
2) The co-function identity $\cos(\pi/2-x)=\sin(x)$ implies that $\cos (\pi/2)=\cos(\pi/2-0)=\sin(0)=0$
3) Looking at the unit circle where $y=\sin(\theta)$ and $x=\cos(\theta)$. When $\theta= 90^0=\pi/2$, $\sin (\theta)=1$ and $\cos(\theta)=0$.
4) the derivative of $\sin(x)$ is $0$ only when $\sin(x)=1$, which only occurs when $x= \frac{n\pi}{2}$ for any $n\in \mathbb N$. Since the derivative of $\sin(x)$ is $\cos(x)$ this implies that $\cos(\pi/2)=\cos((1)\pi/2)=0$.
If you have any qustions let me know below.
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