## inkyvoyd Group Title 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. one year ago one year ago

1. inkyvoyd Group Title

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.

2. Algebraic! Group Title

what did you get for x?

3. inkyvoyd Group Title

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

4. inkyvoyd Group Title

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.

5. KingGeorge Group Title

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.

6. inkyvoyd Group Title

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

7. KingGeorge Group Title

Yes.

8. KingGeorge Group Title

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.

9. inkyvoyd Group Title

Uh, what about trig equations?

10. inkyvoyd Group Title

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

11. inkyvoyd Group Title

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

12. KingGeorge Group Title

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.

13. inkyvoyd Group Title

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

14. KingGeorge Group Title

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

15. inkyvoyd Group Title

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

16. KingGeorge Group Title

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