Negative Angles

[destain]

I get a bit confused with these, can someone go through how you work negative angles out? something like -pi/4 or something simple,
because with normal ones you locate the quadrant etc, find the basic angle, and if its in quadrant two, you go pi - basic angle,
if it is quadrant 3, pi + basic angle......

With negatives, whats the rule etc?

2) Also why can't you differentiate at end points?  As in, the differentiated graph can't include end points

[Conic]

sin(-x) = -sin(x)
cos(-x) = cos(x)
tan(-x) = -tan(x)

A negative angle is pretty much 2π - the angle if that makes sense. So -π/4 = 2π-π/4 = 7π/4
Since the angle is in the 4th quadrant sine will be negative, cosine will be positive and tan will be negative.

[destain]

what if the negative angle is really big? like some of those that you get...156pi/5 or whatever, can't think of one atm :S

Also..-2sin(3(x-pi/4))
when you sub in 0, you get Root2
But..since -pi/4 is in brackets, shouldnt you do -pi/4 x 3 THEN x-2?

[abeybaby]

all you need to remember, is that positive means go anticlockwise, and negative means go clockwise. so -pi/4 means go pi/4 radians clockwise.

as for why you cant differentiate endpoints, do you have a good understanding of limits?

[destain]

what if the negative angle is really big? like some of those that you get...156pi/5 or whatever, can't think of one atm :S

Also..-2sin(3(x-pi/4))
when you sub in 0, you get Root2
But..since -pi/4 is in brackets, shouldnt you do -pi/4 x 3 THEN x-2?

I don't get how that doesn't work

and uhm i guess kinda? decent? LOL

[Conic]

what if the negative angle is really big? like some of those that you get...156pi/5 or whatever, can't think of one atm :S

It's best to simplify really large angles beforehand.

[Conic]

Also..-2sin(3(x-pi/4))
when you sub in 0, you get Root2
But..since -pi/4 is in brackets, shouldnt you do -pi/4 x 3 THEN x-2?

(at x=0)

[destain]

oh ok i understand that now!

and if anyone could help me...on this, bit difficult to ask but.. Insight 2007, techactive, multiple choice question  5
I don't quite get how they answer that? It's a cubic graph and it asks you to find area under the graph with three points or something, to me it should be anti diff with the terminals b on top and c on the bottom, since it's under the x-axes, they use a to c? Is it the same thing anyway or is there something im missing?

[barydos]

oh ok i understand that now!

and if anyone could help me...on this, bit difficult to ask but.. Insight 2007, techactive, multiple choice question  5
I don't quite get how they answer that? It's a cubic graph and it asks you to find area under the graph with three points or something, to me it should be anti diff with the terminals b on top and c on the bottom, since it's under the x-axes, they use a to c? Is it the same thing anyway or is there something im missing?

I think they're looking for the total area from x=a to x=c, not just the area enclosed between the curve and the x-axis. Just a bit of interpreting to do there. I may be wrong?

[destain]

but then, it doesn't say and the line x=-3 or whatever , it's not bounded so...I dunno!!!!!

ALSO
please someone help me with this, i got so frustrated before with this

How do you reexpress y= 2x+1/x-5 -------> 2+ 11/x-5
I can work it backwards, but with the first equation, I don't know how to get to the second, might sound weird but yeah.,.OMG

[barydos]

but then, it doesn't say and the line x=-3 or whatever , it's not bounded so...I dunno!!!!!

ALSO
please someone help me with this, i got so frustrated before with this

How do you reexpress y= 2x+1/x-5 -------> 2+ 11/x-5
I can work it backwards, but with the first equation, I don't know how to get to the second, might sound weird but yeah.,.OMG

if you actually divide the polynomial out you can get the answer that way


or you could do 2x+1 = 2x-10 + 11 and so  2x-10 = 2(x-5) you can cancel the x-5 out.. this is a little tricker so go with the division of polynomials.

edit: and i see what you mean now when it doesn't say that it's bounded by a line x=a.... :S

[Phy124]

How do you reexpress y= 2x+1/x-5 -------> 2+ 11/x-5

Easiest way is to just do "long" division (it really won't be all that long).

Alternatively for questions like these you can do it like this:



If you're wondering what happened in the second step, it is legitimate because holistically 0 is being added to the numerator, it is just expressed differently.

However this isn't all that time efficient for an example like this. On the other hand, this can be practical for other types of question involving simplifying expressions with powers e.g. Re: U-substitution of Indefinite Integrals

Yeah... beaten 

[Bhootnike]

How do you reexpress y= 2x+1/x-5 -------> 2+ 11/x-5

Easiest way is to just do "long" division (it really won't be all that long).

Alternatively for questions like these you can do it like this:



If you're wondering what happened in the second step, it is legitimate because holistically 0 is being added to the numerator, it is just expressed differently.

However this isn't all that time efficient for an example like this. On the other hand, this can be practical for other types of question involving simplifying expressions with powers e.g. Re: U-substitution of Indefinite Integrals

Yeah... beaten 

how do you know to choose 10 and 11 ?! any tricks or just logic?

[Conic]

how do you know to choose 10 and 11 ?! any tricks or just logic?

I seems to be due to the -5 in the denominator for the -10 term.

[barydos]

How do you reexpress y= 2x+1/x-5 -------> 2+ 11/x-5

Easiest way is to just do "long" division (it really won't be all that long).

Alternatively for questions like these you can do it like this:



If you're wondering what happened in the second step, it is legitimate because holistically 0 is being added to the numerator, it is just expressed differently.

However this isn't all that time efficient for an example like this. On the other hand, this can be practical for other types of question involving simplifying expressions with powers e.g. Re: U-substitution of Indefinite Integrals

Yeah... beaten 

how do you know to choose 10 and 11 ?! any tricks or just logic?

This is just how I see it.
For (2x+1)/(x-5), we try to convert the numerator into two parts where one part is a multiple of the denominator.
Looking at 2x+1, we'll try to convert it into a multiple of (x-5). 2x+a = 2(x+a/2). (x+a/2) = (x-5), so a = -10
So it's become 2x -10, GREAT! but we cannot change the value from 2x+1.
So we turn 2x+1 into 2x-10 +10 +1

or  (2x-10)/(x-5) + 11/(x-5)  =  2 + 11/(x-5)

Hope that helps. /bad at explaining