Question

A sine wave is shown in the attachment. Approximate the point shown on the graph. The unlabelled x-axis is on the same units as the y.

Answer 1

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.

Answer 2

what did you get for x?

Answer 3

Some value that was correct for the first intersection of the function and the x axis. However, I"m looking for point p...

Answer 4

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.

Answer 5

Well, the first solution is basically what you did. To get the next one, you find the x-coordinate 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 x-coordinate of that first x-intercept with arcsin. Take that value, and call it r. Now, 4-r is the x-coordinate of your second point.

Answer 6

So I take advantage of the fact that the sine graph is symmetrical?

Answer 7

Yes.

Answer 8

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.

Answer 9

Uh, what about trig equations?

Answer 10

for 3c I have
\(\large \cos^{-1}(\cos(x))=\pi/3\)

Answer 11

I mean I'm tempted to just cancel out the functions, but I know that's wrong.

Answer 12

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.

Answer 13

Wait but since we take the arccos second, we can cancel them out right?

Answer 14

Even then, I don't think you can cancel them unless you're in a very restricted domain.

Answer 15

I mean, cos always has that range, so the arccos would always yield the defined number that came from cos?

Answer 16

According to wolfram, that only works for \(-1\le x\le1\)