query 9

query 9

As I think I mentioned before, your precalculus course should have fully prepared you for the review chapter you're working on, but due to the passage of time and other factors that is apparently not the case. So you're getting hung up on this chapter.

Don't let that continue. Keep working at this chapter, but move with the qa's and more on into Chapter 2.

I've got an entire series of qa's on the trig functions for my Precalculus II class, including lots of pictures, if you have the time and inclination to go through it.

As far as your work on this section, I do recommend that you follow my instructions to make those tables. Once you understand the tables it's much easier to understand the trig functions and the solutions to trigonometric equations.

You might also, or as an alternative, check out http://www.vhcc.edu/pc2fall9/frames%20pages/class_notes.htm . The links #01, #02, ..., #11 lead you to explanations of a lot of these ideas, including pictures. The CDs or DVDs for my second-semester precalculus course, which include video clips explaining the pictures, should also be available in the bookstore.

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RESPONSE --> 1 = 8 cos(2x+1) - 3 4=8 cos(2x+1) 32=cos(2x+1) ??

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20:38:48 ** 2x + 1 = `pi/3, or `pi / 3 + 2 `pi, or `pi / 3 + 4 `pi, ... or `pi / 3 + 2n `pi, ... with n = an integer. So x = (`pi / 3 - 1) / 2. Or x = (`pi / 3 + 2 `pi - 1) / 2 = (`pi / 3 - 1) / 2 + `pi. Or x = (`pi / 3 + 4 `pi - 1) / 2 = (`pi / 3 - 1) / 2 + 2 `pi. … Or x = (`pi / 3 + 2 n `pi - 1) / 2 = (`pi / 3 - 1) / 2 + n `pi ... x = (`pi / 3 - 1) / 2 + n `pi, n an integer, is the general solution. Since 0 (`pi / 3 - 1) / 2 < `pi he first two solutions are between 0 and 2 `pi, and are the only solutions between 0 and 2 `pi. We generally want at least the solutions between 0 and 2 `pi. ** RESPONSE i still don't understand how you get 2x + 1 = `pi/3 out of 1 = 8 cos(2x+1) - 3 does'pi mean the dirivitive of pi? please explain in simple terms. i hate this computer learning bullshit, i need some damn pictures, and need to get this torture over with so i can take calculus 2 in a classroom next spring. i have put this off as long as i could, but i am ready to finish this.

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RESPONSE --> ok. arccos? fndescribe? i had trig about 7 years ago, so i need help

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 = .5.

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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20:41:48 ** To get the graph of arccos(x) we take the inverse of the graph of cos(x), restricting ourselves to 0 < = x < = `pi, where y = cos(x) takes each possible y value exactly once. The graph of this portion of y = cos(x) starts at (0, 1), decreases at an increasing rate (i.e., concave down) to (`pi/2, 0) the decreases at a decreasing rate (i.e., concave up) to (`pi, -1). The inverse function reverses the x and y values. Since y values of the cosine function range from -1 to 1, the x values of the inverse function will run from -1 to 1. The graph of the inverse function therefore runs from x = -1 to x = 1. The graph of the inverse function starts at (-1, `pi) and runs to (0, `pi/2), reversing the values of y = cos(x) between (`pi/2, 0) and (`pi, -1). Since the graph of y = cos(x) is decreasing at a decreasing rate (i.e., concave up) between these points, the graph of the inverse function will be decreasing at a decreasing rate (i.e., concave up) over the corresponding region. The graph of the inverse function continues from (0, `pi/2) to (1, 0), where it decreases at an increasing rate. The values of the inverse cosine function range from 0 to `pi; the domain of this function is [-1, 1].**

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RESPONSE --> help? To start with think of a table for the cosine function from x = 0 to x = +pi. We use these values because the value of the cosine goes from 1 to -1, i.e., through its full range, between these x values.

Somehow the lines got run together.

arccos fndescribe the graph of the arccos or cos^-1 function, including a statement of its domain and range, increasing/decreasing behavior, asymptotes if any, and concavityexplain why the domain and range are as you describe should include 3 line breaks:

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

query 9

query 9

As I think I mentioned before, your precalculus course should have fully prepared you for the review chapter you're working on, but due to the passage of time and other factors that is apparently not the case. So you're getting hung up on this chapter.

Don't let that continue. Keep working at this chapter, but move with the qa's and more on into Chapter 2.

I've got an entire series of qa's on the trig functions for my Precalculus II class, including lots of pictures, if you have the time and inclination to go through it.

As far as your work on this section, I do recommend that you follow my instructions to make those tables. Once you understand the tables it's much easier to understand the trig functions and the solutions to trigonometric equations.

You might also, or as an alternative, check out http://www.vhcc.edu/pc2fall9/frames%20pages/class_notes.htm . The links #01, #02, ..., #11 lead you to explanations of a lot of these ideas, including pictures. The CDs or DVDs for my second-semester precalculus course, which include video clips explaining the pictures, should also be available in the bookstore.

......!!!!!!!!...................................

RESPONSE --> 1 = 8 cos(2x+1) - 3 4=8 cos(2x+1) 32=cos(2x+1) ??

.................................................

......!!!!!!!!...................................

20:38:48 ** 2x + 1 = `pi/3, or `pi / 3 + 2 `pi, or `pi / 3 + 4 `pi, ... or `pi / 3 + 2n `pi, ... with n = an integer. So x = (`pi / 3 - 1) / 2. Or x = (`pi / 3 + 2 `pi - 1) / 2 = (`pi / 3 - 1) / 2 + `pi. Or x = (`pi / 3 + 4 `pi - 1) / 2 = (`pi / 3 - 1) / 2 + 2 `pi. … Or x = (`pi / 3 + 2 n `pi - 1) / 2 = (`pi / 3 - 1) / 2 + n `pi ... x = (`pi / 3 - 1) / 2 + n `pi, n an integer, is the general solution. Since 0 (`pi / 3 - 1) / 2 < `pi he first two solutions are between 0 and 2 `pi, and are the only solutions between 0 and 2 `pi. We generally want at least the solutions between 0 and 2 `pi. ** RESPONSE i still don't understand how you get 2x + 1 = `pi/3 out of 1 = 8 cos(2x+1) - 3 does'pi mean the dirivitive of pi? please explain in simple terms. i hate this computer learning bullshit, i need some damn pictures, and need to get this torture over with so i can take calculus 2 in a classroom next spring. i have put this off as long as i could, but i am ready to finish this.

......!!!!!!!!...................................

RESPONSE --> ok. arccos? fndescribe? i had trig about 7 years ago, so i need help

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 = .5.

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.

.................................................

......!!!!!!!!...................................

20:41:48 ** To get the graph of arccos(x) we take the inverse of the graph of cos(x), restricting ourselves to 0 < = x < = `pi, where y = cos(x) takes each possible y value exactly once. The graph of this portion of y = cos(x) starts at (0, 1), decreases at an increasing rate (i.e., concave down) to (`pi/2, 0) the decreases at a decreasing rate (i.e., concave up) to (`pi, -1). The inverse function reverses the x and y values. Since y values of the cosine function range from -1 to 1, the x values of the inverse function will run from -1 to 1. The graph of the inverse function therefore runs from x = -1 to x = 1. The graph of the inverse function starts at (-1, `pi) and runs to (0, `pi/2), reversing the values of y = cos(x) between (`pi/2, 0) and (`pi, -1). Since the graph of y = cos(x) is decreasing at a decreasing rate (i.e., concave up) between these points, the graph of the inverse function will be decreasing at a decreasing rate (i.e., concave up) over the corresponding region. The graph of the inverse function continues from (0, `pi/2) to (1, 0), where it decreases at an increasing rate. The values of the inverse cosine function range from 0 to `pi; the domain of this function is [-1, 1].**

......!!!!!!!!...................................

RESPONSE --> help? To start with think of a table for the cosine function from x = 0 to x = +pi. We use these values because the value of the cosine goes from 1 to -1, i.e., through its full range, between these x values.

Somehow the lines got run together.

arccos fndescribe the graph of the arccos or cos^-1 function, including a statement of its domain and range, increasing/decreasing behavior, asymptotes if any, and concavityexplain why the domain and range are as you describe should include 3 line breaks:

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