In the function f(t) = sin(t), what is the effect of multiplying t by a coefficient of 2?
The period is reduced by 2.
a. changed amp to 2 b. changed amp to 1/2 c. changed period to pi d. changed period to 4pi
The period of f(t)=sin(t) is 2*Pi, now the period of f(2t)=sin(2t) is reduced by 2, which is ....
drawar is right and if you know hat the one period of \( \sin(x)\) is \(2\pi\) you can do the magic.
I'm not so bright with this. I've got this answer wrong serveral times, hahah.
So do I need to plug it into the calculator to figure out where it changes???
If you're not good with visualizing functions, i can recommend geogebra. It's a rather easy to use function plotter which is free. You can get it at http://www.geogebra.org/cms/ Now back to your problem. The normal period is \(2\pi\) so if it is halved by \(2\) you can simply calculate it by \(\frac{2\pi}{2} = \pi\) as the two in the denominator and enumerator cancels.
So basically it's asking what will happen when you add pi to it?
So if i get your problem right you got \(f(x) = \sin(x)\) as your function you have to explain what changes if you have a look \(g(x) =\sin(2x)\), is that right?
O: I think so
Hang on i'll do a quick plot to make things clearer.
NO WAIT IT'S D
Right
... Right?
No, actually D is wrong
What...
No because it goes from 9pi/4 to 13pi/4
So the blue curve is \(\sin(x)\) and the black one \(\sin(2x)\)
You can see that both amplitudes are equally high so there's no change in that.
If we have a look at the period we can see that it takes \(2\pi\) for \(\sin(x)\) to start over. Whereas \(\sin(2x)\) starts repeating after it reaches \(\pi\)
I saw...
Is it clear so far or do you still have trouble understanding it?
i dont understandhow im wrong.
Tell me how you got answer D
I did.
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