What is $\cos(k \pi)$?
I want to ask question for which I have been finding answer for.
Please could anyone explain me why $\cos(k \pi) = (-1)^k$ and also explain me same for $\sin(k \pi)$?
$\endgroup$ 74 Answers
$\begingroup$The cosine of a integer times $\pi$ is always $\pm 1$. It happens that when k is even, $\cos{k\pi}$ = 1, when k is odd, $\cos{k\pi}=-1$ and therefor $(-1)^k$.
The sine is always zero for every $k$. So $\sin{k\pi}=0$
edit: k is an integer. That is convention when using sine or consine
$\endgroup$ 2 $\begingroup$$\cos 0 = 1$ and $\cos \pi = -1$. Also, $\cos$ is a $2\pi$-periodic function, so the remaining cases follow immediately by the periodicity.
Same for $\sin$.
$\endgroup$ $\begingroup$Let $k\in\mathbb Z$. Then
$$\cos(0)=1,~~\text{for}~~k=0,$$ $$\cos(\pi)=\cos(-\pi)=-1,~~\text{for}~~k=\pm 1,$$ $$\cos(2\pi)=\cos(-2\pi)=1,~~\text{for}~~k=\pm 2,$$
and so on, where the first equalities hold as $\cos(\cdot)$ is an even function. Every time $k$ is even then we get $\cos(k\pi)=1$. When $k$ is odd, then $\cos(k\pi)=-1$. You can summarize these considerations into $$\cos(k\pi)=(-1)^k. $$
Can you apply the same lines to $\sin(k\pi)$ now?
$\endgroup$ $\begingroup$I will do it the right way :
This is a circle of radius $1$ unit. We complete the rectangle as shown and define $\sin$ and $\cos$ to be algebraic length on the respective axes as shown. Note that algebraic distance means that it can be negative also.
Now put that $\theta=\pi$
I hope you know the definition of angle. Clearly, $x-\text{coordinate}=-1$ Hence $cos\pi=-1$. What if it is $2\pi$? This time it is on positive side and it is $-1$. You can generalize it by using odd $k$ for former and even $k$ for latter.
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