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I just got to college, and right now Calculus is looking like it's going to kill me, despite math being my strong subject. Got my first test back and I think it was roughly a 72%, mainly because I was missing a lot of Trig problems. I think a lot of that is due to things that we didn't have to memorize in high school that now need to be memorized, but even still I know I need work/help with it. So I'm just going to use this thread to run off random questions from my math.

Not sure why I couldn't have just simply asked my questions without all that, but whatever, here we go:

Here's the expression:

(x-2)(x+5)
----------- ≤ 0
(x+3)

My professor has been stressing that we have two major ways to approach these problems, algebraically and geometrically. So far I've been focusing in on algebraically, because I understood that method and was lost on the geometric, but for this problem the number of cases involved makes geometric the way to go, so she said. I copied down her work to go over (I was working through a previous problem as she worked this) but am still lost on part of it. Here's how she did it:

[original problem]
x-2 = 0
x = 2

OR

x + 5 = 0
x = -5

OR

x+3 = 0
x=-3

Up to this point, I follow her. She's defining what the zeroes are. I'm not entirely sure why though. Any explanation there would be nice.

Then, she does this (she also mentioned some chart-method, but she doesn't use that so idk what she means or if that would be easier for me):

There's a number line with -5, -3, and 2 marked. -5 and 2 are closed circles, -3 is open. (ignore the periods)

<---(-5)--(-3)--(2)--->
....N| N |... N |...P
....N| P |... P |...P
....N| N |... P |...P
------------------------
....N| P |... N |...P

that has something to do with what is positive and what is negative, and from there she got the solution:

sol: (-infinity, -5] union (-3, 2]
which I understand to be "x is less than or equal to -5 or x is greater than -3 but less than 2"

My main problem is understand what the little number line/letters thing is and how I determine which is negative and which is positive.


Ok, now the next problem, we started with:
x² > 4

now the work:
x² > 4
x² - 4 > 0 (why did we do that?)
(x-2)(x+2) > 0

case 1:(x-2)(x+2) > 0
x-2 > 0 AND x+2 > 0
x>2 AND x>-2
THEREFORE x>2

case 1 I understand just fine, it's case 2 where she lost me:
case 2:(x-2)(x+2) > 0
x - 2 < 0 AND x+2 < 0 (and here I'm lost, how did we get from the first line of this case to this?)
x<2 and x<-2
sol: x{x|x<-2 OR x>2}

For the first problem, you're trying to find an x such that (x-2)(x+5)/(x+3) is less than or equal to zero, so that it's negative. We define the zeroes, because that's where our solutions change from positive to negative. We can include x=2 and x=-5 in our solution because they're zero, however we exclude x=-3, since that would give us division by zero.

The real numbers can be thought of as a line, and we look at intervals of that line. The places where the sign of the solution changes are at the zeroes. So we consider the values of x within these intervals: -∞<x≤-5, -5≤x<-3, -3<x≤2, 2≤x<∞. Now, you've missed some labels in your chart underneath your number line, but each line corresponds to each of the above intervals too.

So, consider x≤-5. That means that x+5<0, x+3<0, and x-2<0. The result on the bottom line of the chart indicates that the final result will be negative, since you're multiplying three negative numbers together.

For -5≤x<-3, you'll have x+5>0, but x+3<0 and x-2<0, which gives you a positive result, since you're multiplying a positive value with two negative values.

You keep on doing that for each interval, and you'll end up with a bunch of intervals that result in negative results and positive results. The solution to the problem is you want the intervals with the negative results, so that's whenever x≤-5 or -3<x≤2, or in set notation (-∞,-5] ∪ (-3,2].

For the next problem, we have x² > 4. You manipulate inequalities just like you would equalities. For instance, if x² = 4, you'd have x²-4 = 4-4, which gives you x²-4=0. The same principle applies. x² > 4 means x²-4 > 4-4 and so x²-4 > 0. Then you factor to get (x-2)(x+2) > 0. Here we want x such that (x-2)(x+2) is positive. Remember that you can get a positive number either by multiplying two positive numbers or two negative numbers. The first case determines the value of x if x-2 and x+2 are both positive, while the second case determines x if x-2 and x+2 are both negative.

Ok, thanks. I think I've got both of those (and inequalities in general) a lot better now.

One thing I just came up against in studying trig online was these derivatives:
sin'(x) = cos(x)

cos'(x) = -sin(x)

tan'(x) = sec2(x)

sec'(x) = sec(x)tan(x)

cot'(x) = -csc2(x)

csc'(x) = -csc(x)cot(x)

Two issues here:
1: What's the difference between sin(x) and sin'(x)? I've never seen the apostrophe in there so I don't know what it means.

2: When plugged into my calculator, none of these seem to work. Nor do the other derivatives I saw either, for instance:

sin(a+b) = sin(a)cos(b)+cos(a)cos(b). When I plug numbers in (15 for a+b, 5 for a, 10 for b) but they don't give the same answer.

Since you haven't come across that notation, I'll assume you're not looking for derivatives, which is a concept in calculus that's not limited to trigonometric functions, but you actually want trigonometric identities. If you haven't seen derivatives before, then just wait.

As for identities, make sure you have the right one. For instance, sin(a+b) is sin(a)cos(b) + cos(a)sin(b), which is not what you have.

Is it possible to explain Calculus over the internet? xD

blkmage (post: 1419616) wrote:Since you haven't come across that notation, I'll assume you're not looking for derivatives, which is a concept in calculus that's not limited to trigonometric functions, but you actually want trigonometric identities. If you haven't seen derivatives before, then just wait.

As for identities, make sure you have the right one. For instance, sin(a+b) is sin(a)cos(b) + cos(a)sin(b), which is not what you have.

Well, for that identity, what you typed is actually what I was using, I just typed it wrong into the reply box. I was having problems with sin(a)cos(b) + cos(a)sin(b).

Well, I don't know what's wrong, because it does look like sin 15 = sin 10 cos 5 + cos 10 sin 5.