Why $\sin(n\pi) = 0$ and $\cos(n\pi)=(-1)^n$?

Question

I am working out a Fourier Series problem and I saw that the suggested solution used
$\sin(n\pi) = 0$ and $\cos(n\pi)=(-1)^n$ to simply the expressions while finding the Fourier Coefficients $a_0$, $a_n$, $b_n$.

I am aware that the $\sin(x)$ has a period of $2\pi$. So I am thinking that every half of period, the graph of $\sin(x)$ has to cut through the $x$ axis thus giving us the value $0$. Am I right to think that way or is there some more important reason for that?

Also, how do they come up with $\cos(n\pi) = (-1)^n$?

Answer 1

As you can see from the plot you included $\sin(n\pi) = 0$ for any integer $n$.

Also, $\cos(0) = 1$, $\cos(\pi) = -1$, $\cos(2\pi) = 1$, etc. So $\cos(n\pi) = 1$ for $n$ even and $\cos(n\pi) = -1$ for $n$ odd, which is also true for $(-1)^n$.

Answer 2

For the sine case:

$n = \cdots -1, 0, 1, 2, 3, \cdots$

Gives us

$\cdots \sin(-\pi), \sin(0), \sin(\pi), \sin(2\pi), \sin(3\pi),\cdots$ Which is exactly where the sine function has its roots, so it is always equal to $0$.

For the cosine case, use the identity $\cos(x) = \cos(x + 2\pi) $ (period of the cosine function is $2\pi$) and plug $\cos(0)$ and $\cos(\pi)$ to verify this.

Answer 3

Since:

$sin(x) = \frac{exp(ix) - exp(-ix)}{2i} $

for x = n$\pi$:

$ => sin(n\pi) = \frac{exp(in\pi) - exp(-in\pi)}{2i}$

You can think of the positive exponential term as rotating counterclockwise, and the negative exponential term rotating clockwise. Since both exponential terms are of equal value (i.e. 1), but only moving in opposite directions, they cancel out at all points n

$cos(x) = \frac{exp(ix) + exp(-ix)}{2} $, the physical intuition follows from above

Answer 4

On a unit circle $x$ coordinate of any point on the circle is given by $\cos\theta$ and $y$ coordinate is given by $\sin\theta$

Now, $\sin(n\pi)$, where $n=0,1,2,3...$ is always the X-axis and on X-axis we have $y=0$

and $\cos(n\pi)$ assumes $x=1$ or $x=-1$ as

$\cos(0.\pi)=1=(-1)^0,\cos(1.\pi)=-1=(-1)^1,\cos(2.\pi)=1=(-1)^2...$

Answer 5

Recall that Euler's formula is $e^{ix}=\cos x+i\sin x$. When $x=\pi$, we have $e^{i\pi}=\cos\pi+i\sin\pi=-1\implies e^{i\pi}+1=0$.

From $e^{i\pi}+1=0\implies e^{i\pi}=-1\implies (e^{i\pi})^n=(-1)^n$. From $e^{ix}=\cos x+i\sin x$, when $x=n\pi$, $e^{in\pi}=\cos {n\pi}+i\sin {n\pi}=(-1)^n$.

This implies that $\cos {n\pi}=(-1)^n$ and $\sin {n\pi}=0$, for all $n\in \mathbb Z$.

Answer 6

The picture burned into my visual cortex is the unit circle, not the graphs of sine and cosine:

This together with $2\pi$-periodicity (which is geometrically obvious) immediately gives $$ \sin(n\pi) = 0,\quad \cos(n\pi) = (-1)^{n} $$ for all integers $n$.

Answer 7

If you go through the unit circle process, then you see how they are equivalent. But if you want a proof then:

You may observe the fact that $\sin(0)=0$. Then you may prove that $\sin(x+\pi)=\sin(x)$ and $\sin(x-\pi)=\sin(x)$ using angle addition formulas. Having done this you prove:

$$\sin(0 \pm \pi n)=0, n \in \mathbb{N}$$

Now for cosine you use similar logic. observe $\cos(0)=1$ and $\cos(x \pm \pi)=-\cos x$...