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 $\sin \frac{n\pi}{2}$.
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$\begingroup$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 :
$$ \cos\left( \frac{n\pi}{2}\right) = (-1)^{\lfloor \frac{n}{2} \rfloor} \left( \frac{1+(-1)^n}{2} \right) $$ $$ \sin\left( \frac{n\pi}{2}\right) = (-1)^{\lfloor \frac{n}{2} \rfloor} \left( \frac{1-(-1)^n}{2} \right) $$
$\endgroup$ 3 $\begingroup$There are two cases:
- $n$ is even, write it as $2k$ and then you have $\cos(k\pi)$ and $\sin(k\pi)$ which you already know.
- $n$ is odd, write it as $2k+1$ and then you have $\cos(k\pi+\frac\pi2)$ and $\sin(k\pi+\frac\pi2)$. Recall that $\sin(x)=\cos(x+\frac\pi2)$, and deduce from the previous case what the values are.
$\displaystyle \cos\frac{n\pi}{2}=\{1,0,-1,0\}=2\left\lfloor\frac{n}{4}\right\rfloor+2\left\lfloor\frac{n+1}{4}\right\rfloor+1-n$
$\displaystyle \sin\frac{n\pi}{2}=\{0,1,0,-1\}=n-2\left\lfloor\frac{n+1}{4}\right\rfloor-2\left\lfloor\frac{n+2}{4}\right\rfloor$
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