# Thread: problem with sinus :)

1. ## problem with sinus :)

Hey guys.
cos(n*x) when n is natural and x=pi equals to (-1)^n as I wrote in the pic.
What I'm looking for is the same thing in sinus, lets say again that n is natural, is there a x that can make sinus to behave something like cos? and by that I mean an expression that can be written in on line like (-1)^n (x can be anything).
I hope that was understood.
Thanks in advance.

2. $\displaystyle x=\frac{\pi}{2}$

3. Originally Posted by asi123
Hey guys.
cos(n*x) when n is natural and x=pi equals to (-1)^n as I wrote in the pic.
What I'm looking for is the same thing in sinus, lets say again that n is natural, is there a x that can make sinus to behave something like cos? and by that I mean an expression that can be written in on line like (-1)^n (x can be anything).
I hope that was understood.
Thanks in advance.
there's no such $\displaystyle x$ because if $\displaystyle \sin x = -1,$ then $\displaystyle \cos x =0$ and hence $\displaystyle \sin(2x)=2\sin x \cos x =0 \neq 1 =(-1)^2.$

4. well, lets say x=pi/2, is there a way to write the result in one line? like cos(n*pi) = (-1)^n.

5. Originally Posted by asi123
well, lets say x=pi/2, is there a way to write the result in one line? like cos(n*pi) = (-1)^n.
\displaystyle \begin{aligned}\cos(n\pi)&=\sin\left(\frac{\pi}{2}-n\pi\right)\\ &=\sin\left(\pi\left(\frac{1}{2}-n\right)\right)\\ &=\sin\left(\pi\cdot\frac{1-2n}{2}\right)\\ &=-\sin\left(\pi\cdot\frac{2n-1}{2}\right)\\ &=(-1)^n\end{aligned}

Actually more aestheically pleasing we can use $\displaystyle \cos(n\pi)=\sin\left(\frac{\pi}{2}+n\pi\right)$ to arrive alternatively at $\displaystyle (-1)^n=\sin\left(\pi\cdot\frac{2n+1}{2}\right)$

6. Originally Posted by Mathstud28
\displaystyle \begin{aligned}\cos(n\pi)&=\sin\left(\frac{\pi}{2}-n\pi\right)\\ &=\sin\left(\pi\left(\frac{1}{2}-n\right)\right)\\ &=\sin\left(\pi\cdot\frac{1-2n}{2}\right)\\ &=-\sin\left(\pi\cdot\frac{2n-1}{2}\right)\\ &=(-1)^n\end{aligned}

Actually more aestheically pleasing we can use $\displaystyle \cos(n\pi)=\sin\left(\frac{\pi}{2}+n\pi\right)$ to arrive alternatively at $\displaystyle (-1)^n=\sin\left(\pi\cdot\frac{2n+1}{2}\right)$
Hey Mathstud28.
This is not what I meant (thanks a lot due ).
I gave the cos just as an example.
I have the function f(x) = sin(nx), and I need to find any x that will give me the ability to write the result in one line.
Lets take again x = pi/2 f(x) = sin(n*p/2) = {0 when n=0, 1 when n=1, 0 when n=2, -1 when n=3 and so on...} is there a way to write this expression in one line that will be depend in n? (just like cos(nx) = (-1)^n)

Again, thanks a lot.

7. Hello,
Originally Posted by asi123
Hey Mathstud28.
This is not what I meant (thanks a lot due ).
I gave the cos just as an example.
I have the function f(x) = sin(nx), and I need to find any x that will give me the ability to write the result in one line.
Lets take again x = pi/2 f(x) = sin(n*p/2) = {0 when n=0, 1 when n=1, 0 when n=2, -1 when n=3 and so on...} is there a way to write this expression in one line that will be depend in n? (just like cos(nx) = (-1)^n)

Again, thanks a lot.
No, but in two lines it's possible :
$\displaystyle \sin \left(\frac{n\pi}{2}\right)=\left\{\begin{array}{l l} 0 \text{ if n is even} \\ (-1)^{(n-1)/2} \text{ if n is odd} \end{array} \right.$

8. $\displaystyle \sin\left(\frac{n\pi}{2}\right)=\frac{1}{2}(-1)^{\frac{n(n-1)}{2}}\left[(-1)^{n-1}+1\right]$

Hows that?

9. Originally Posted by Mathstud28
$\displaystyle \sin\left(\frac{n\pi}{2}\right)=\frac{1}{2}(-1)^{\frac{n(n+1)}{2}}\left[(-1)^n+1\right]$

Hows that?
Thanks a lot.

10. Originally Posted by asi123
Thanks a lot.
Note! I made a typo...the thing you quoted gives the values of $\displaystyle \sin\left(\frac{\pi(n+1)}{2}\right)$...to get the values of $\displaystyle \sin\left(\frac{n\pi}{2}\right)$ see my revised post.