problem with sinus :)

• January 12th 2009, 02:03 PM
asi123
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.
• January 12th 2009, 02:18 PM
lukaszh
$x=\frac{\pi}{2}$
• January 12th 2009, 02:47 PM
NonCommAlg
Quote:

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.

there's no such $x$ because if $\sin x = -1,$ then $\cos x =0$ and hence $\sin(2x)=2\sin x \cos x =0 \neq 1 =(-1)^2.$
• January 12th 2009, 11:21 PM
asi123
well, lets say x=pi/2, is there a way to write the result in one line? like cos(n*pi) = (-1)^n.
• January 12th 2009, 11:27 PM
Mathstud28
Quote:

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.

\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 $\cos(n\pi)=\sin\left(\frac{\pi}{2}+n\pi\right)$ to arrive alternatively at $(-1)^n=\sin\left(\pi\cdot\frac{2n+1}{2}\right)$
• January 12th 2009, 11:44 PM
asi123
Quote:

Originally Posted by Mathstud28
\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 $\cos(n\pi)=\sin\left(\frac{\pi}{2}+n\pi\right)$ to arrive alternatively at $(-1)^n=\sin\left(\pi\cdot\frac{2n+1}{2}\right)$

Hey Mathstud28.
This is not what I meant (thanks a lot due (Happy)).
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.
• January 12th 2009, 11:51 PM
Moo
Hello,
Quote:

Originally Posted by asi123
Hey Mathstud28.
This is not what I meant (thanks a lot due (Happy)).
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 :
$\sin \left(\frac{n\pi}{2}\right)=\left\{\begin{array}{l l} 0 \text{ if n is even} \\
(-1)^{(n-1)/2} \text{ if n is odd} \end{array} \right.$
• January 12th 2009, 11:56 PM
Mathstud28
$\sin\left(\frac{n\pi}{2}\right)=\frac{1}{2}(-1)^{\frac{n(n-1)}{2}}\left[(-1)^{n-1}+1\right]$

Hows that?
• January 13th 2009, 12:02 AM
asi123
Quote:

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

Hows that?

Thanks a lot.
• January 13th 2009, 12:05 AM
Mathstud28
Quote:

Originally Posted by asi123
Thanks a lot.

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