course mth 172
ok. her'es more. i'll be back tommorrow. hopefully i'll be able to move on to the next lesson now.
Looks like you've just about got the basics here. See my notes.
If anything is not clear let me know.
You get 4 = 8 cos(2x+1), which when you divide both sides by 8 gives you cos(2x+1) = .5. the extended table I recommended below that you make for cos(x) using x = 0, pi/6, pi/3, pi/2, 2pi/3, 5pi/6 and pi will show you that cos(pi/3) = .5. From this you conclude that 2x + 1
The circular definition of the cosine function also shows you that cos(pi / 3 + 2 pi * n ) is also .5, for any value of n. This is because 2 pi corresponds to a complete trip around the circle, which puts you back where you started, and 2 pi * n corresponds to n trips around the circle.
If you extend that table to x = pi, 7 pi / 6, ..., 2 pi you will also see that cos(5 pi/3) = .5, and we can say that cos(5 pi / 3 + 2 pi * n) = .5.
From this we conclude that 2x + 1 can take the values `pi/3, or `pi / 3 + 2 `pi, or `pi / 3 + 4 `pi, ... or `pi / 3 + 2n `pi, ... with n = an integer.
We then solve 2x + 1 = pi / 3 + 2 pi n for x, subtracting 1 from both sides then dividing both sides by 2.
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ok. i remebered the radian format calc. mode the only thing i still don't understand is the + 2 pi n part of 2x + 1 = pi / 3 + 2 pi n
why is that there then, if it cancels itself out as 1 rev.?
Let's say you've got a cam driven by a circular wheel. If the motion continues the cam will pass the same points over and over.
If say the cam had radius 8 cm and the equation of its motion was 4 = 8 cos(2t + 1) then it would be at the 4 cm position and moving in the negative direction every time 2 t + 1 was equal to pi/3 + 2 pi n.
It gets even better; it would also be at that position moving in the positive direction every time 2 t + 1 was equal to 5 pi / 3 + 2 pi n.
To understand all this you've gotta really understand the circular model.
This also applies to harmonic waves and to lots and lots of other things.
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arccos fn describe the graph of the arccos or cos^-1 function, including a statement of its domain and range, increasing/decreasing behavior, asymptotes if any, and concavity explain why the domain and range are as you describe
ok. it is like a mirrored image of the non inverse with it's startpoint x laid on top of the first graph.
mirrored and rotated; right idea
it's range is 0-2 pi (radian) because it would start all over after 1 revolution. oh, does this work in the vegative, too, like you're going the opposite direction around the circle? how would you tell where the start/end are?
If it's a function it can't have two values, as it would if you let it go in the negative direction as well as the positive.
So we restrict the inverse cosine function to x values from -1 to 1, with y values going just from 0 to pi rad; to get all the way to 2 pi rad you would have to start repeating x values, which as I said wouldn't give you a function.