Finding the values of $\cos \frac{n\pi}{2}$ and $\sin \frac{n\pi}{2}$. i know that the values of $\cos n\pi=(-1)^{n}$ and $\sin n\pi=0$. Now i want to know that what is the general expressions of $\cos \frac{n\pi}{2}$ and $\sin \frac{n\pi}{2}$.

Using identities $\sin(x)^2 = \frac{1-\cos(2x)}{2}$ and $\cos(x)^2 = \frac{1+\cos(2x)}{2}$ we get
$$ \left(\sin\left(\frac{n\pi}{2}\right)\right)^2 = \frac{1 - \cos(n\pi)}{2} = \frac{1 - (-1)^n}{2} $$
$$ \left(\cos\left(\frac{n\pi}{2}\right)\right)^2 = \frac{1 + \cos(n\pi)}{2} = \frac{1 + (-1)^n}{2}$$
Thus :


*

*$\sin\left(\frac{n\pi}{2}\right) = 0$ if $n$ is even and $\pm 1$ if odd

*$\cos\left(\frac{n\pi}{2}\right) = 0$ if $n$ is odd and $\pm 1$ if even


To conclude we have to solve $\pm 1$ cases. 


*

*$\cos$ case : $n$ is even, $n=2p$. Then 
$$\cos\left(\frac{n\pi}{2}\right) = \cos\left(p\pi\right) = (-1)^p$$
So the result is $1$ if $p$ is even, $-1$ if odd

*$\sin$ case : $n$ is odd, $n=2p+1$. Then 
$$\sin\left(\frac{n\pi}{2}\right) =\sin\left(p\pi+\frac{\pi}{2}\right) = \cos\left(p\pi\right) = (-1)^p$$
So $1$ if $p$ is even, $-1$ if odd


Finally we get the following general expressions :
$$ \cos\left( \frac{n\pi}{2}\right) = (-1)^{\lfloor \frac{n}{2} \rfloor} \left( \frac{1+(-1)^n}{2} \right) $$
$$ \sin\left( \frac{n\pi}{2}\right) = (-1)^{\lfloor \frac{n}{2} \rfloor} \left( \frac{1-(-1)^n}{2} \right) $$
