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A sine wave is shown in the attachment. Approximate the point shown on the graph. The unlabelled xaxis is on the same units as the y.
 one year ago
 one year ago
A sine wave is shown in the attachment. Approximate the point shown on the graph. The unlabelled xaxis is on the same units as the y.
 one year ago
 one year ago

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inkyvoydBest ResponseYou've already chosen the best response.0
This is a sine wave shifted down 3 units, with an amplitude five times that of the parent function, and a period of 4. Thus the equation of the function is f(x)=5 sin( (pi*x)/4)3 0=5 sin( (pi*x)/4)3 Then, the x coordinate is 3/5=sin((pi/4)x) (pi*x)/4=arcsin(3/5) X=(4arcsin(3/5))/pi My work. The last two lines are wrong because of the arcsin. Please explain to me how to determine the second intersection p and not the first intersection.
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.0
what did you get for x?
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.0
Some value that was correct for the first intersection of the function and the x axis. However, I"m looking for point p...
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.0
I'm really just having trouble understanding how to find each solution in this problem, and not just the 0<x<pi/2 solution that the inverse trig functions yield.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
Well, the first solution is basically what you did. To get the next one, you find the xcoordinate for when your function is 3 for the second time. So first, at x=0, f(x)=3. Second, at x=4, f(x)=3. Now, you can find the xcoordinate of that first xintercept with arcsin. Take that value, and call it r. Now, 4r is the xcoordinate of your second point.
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.0
So I take advantage of the fact that the sine graph is symmetrical?
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
That's what I would do in this case anyways. Once you get a more complex graph, I'd have to think about it some more.
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.0
Uh, what about trig equations?
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.0
for 3c I have \(\large \cos^{1}(\cos(x))=\pi/3\)
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.0
I mean I'm tempted to just cancel out the functions, but I know that's wrong.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
It's wrong because the functions are periodic. So \[\cos^{1}(\cos(x))=\cos^{1}(\cos(x+2\pi))\]So strictly speaking, it's not really an actual inverse.
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.0
Wait but since we take the arccos second, we can cancel them out right?
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
Even then, I don't think you can cancel them unless you're in a very restricted domain.
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.0
I mean, cos always has that range, so the arccos would always yield the defined number that came from cos?
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
According to wolfram, that only works for \(1\le x\le1\)
 one year ago
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