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How do I draw this graph? Eg. $\cos\pi$ starts at y=1, flows down to x=$\pi$ as the mid point where y=-1 and goes back to y=1.

I'm not sure how to decipher the $2\pi10t$ here and put them into figures in a graph.

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Do you know how to graph $y=\cos x$? Do you know how to turn a graph of $y=f(x)$ into a graph of $y=f(ax)$ by stretching/shrinking? Can you tell what the period of $\cos(2\pi\cdot10 t)$ is? –  anon May 7 '12 at 11:21
    
Somebody should probably mention that $\cos2\pi10t$ is an extremely fishy way of denoting $\cos(20\pi t)$, hence this interpretation (which I share) must be checked by the OP. –  Did May 7 '12 at 12:03
    
@anon Not sure about y=$cosx$. I know $y=f(x)$ and $y=f(ax)$. Period of $cos 2\pi10t$ is 1/10 as given by Henry below? –  resting May 8 '12 at 11:10
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You probably know the basic shape of a cosine curve.

$\cos x$ is $1$ when $x=0$, is $0$ when $x=\pi/2$, is $-1$ when $x=\pi$, is $0$ when $x=3\pi/2$, and is $1$ when $x=2\pi$, and then the pattern repeats.

So $\cos (2\pi 10 t)$ is $1$ when $t=0$, is $0$ when $t=1/40$, is $-1$ when $t=1/20$, is $0$ when $t=3/40$, and is $1$ when $t=1/10$, and then the pattern repeats.

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How do you decipher the $2\pi10t$? $10t$ = 1/10 when x = 1? And $2\pi$ is disregarded? –  resting May 8 '12 at 11:05
    
@resting: No. When $t=\frac{1}{10}$ you get: $10t=1$; $2\pi 10 t=2\pi$; and $\cos (2\pi 10 t)=\cos(2\pi)=1$. –  Henry May 8 '12 at 20:04
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