# decreasing sequence for sin

I want to show that a sequence $\sin(n\pi x)$ where $x$ is in $[0,1]$ is decreasing? is there any test I can apply and if there is no test can somebody please tell me how to show it?

since I think that i am doing wrong. what I did:

$\sin((n+1)\pi x)=\sin(n\pi x+\pi x)=\sin(n\pi x)\cos(n\pi)+\sin(n\pi)\cos(n\pi x)<\sin(n\pi x)$ but why???

thanks.

It's not. Let $x = \frac{1}{2}$. Then your sequence is $0, 1, 0, -1, \dots$.
As wckronholm says, you can see that the statement is wrong because $x=\frac12$ produces an oscillating sequence. In fact, any nonzero rational number will produce an oscillating sequence. Irrational numbers will exhibit similar up-down-up behavior but they will not actually repeat numerically.
As for the argument, it has a typographical error which turns out to be fatal: it states that $\sin(n\pi x + \pi x) = \sin(n\pi x)\cos (n\pi) + \sin(n\pi)\cos(n\pi x)$ but what is actually true is that $$\sin(n\pi x + \pi x) = \sin(n\pi x)\cos (\pi x) + \sin(\pi x)\cos(n\pi x)$$ from which the final inequality no longer follows.