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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 $\sin(k \pi)$?

Guy
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user3215228
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  • Someone voted to close saying question missing context? I disagree. This doesn't sound like homework either. – Guy Apr 07 '14 at 13:28
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    cos of k*pie... :P – Avitus Apr 07 '14 at 13:28
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    @Avitus I know. Editing. @ op , that should be pi not pie. :p – Guy Apr 07 '14 at 13:29
  • Okay, so as there are two votes now, would someone explain the reason(s) behind it? – Guy Apr 07 '14 at 13:30
  • @Sabyasachi You mean two votes to close? I think one could find this to easy or something that can be easily found on the internet (or any math book). I am not one of the voters anyway, everyone has to learn :) – Mathias711 Apr 07 '14 at 13:42
  • @Mathias711 Yes I meant two close votes. And yes, everyone has to learn. Saying something is too easy, is a bit elitist, imo. – Guy Apr 07 '14 at 13:48
  • Do you know the circle definition of $\sin$ and $\cos$? – AD - Stop Putin - Apr 07 '14 at 14:06

4 Answers4

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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

Mathias711
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  • what will be for cos k(2π)?? please can you explain me in general so that i can apply it in my problems – user3215228 Apr 07 '14 at 13:58
  • $\cos{2k\pi}$ will always be 1, for every integer k. Because $\cos{0}=1$, and the cosine repeats itself every $2\pi$. Therefor it will always be 1 – Mathias711 Apr 07 '14 at 15:59
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$\cos 0 = 1$ and $\cos \pi = -1$. Also, $\cos$ is a $2\pi$-periodic function, so the remaining cases follow immediately by the periodicity.

Same for $\sin$.

Yoni Rozenshein
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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?

Avitus
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I will do it the right way : enter image description here

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.

evil999man
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