Question

is there an answer for (0/1)times(1/0)?

Answer 1

what is 1/0 ?

Answer 2

undefined

Answer 3

so i think the whole thing is undefined

Answer 4

so a number times something undefined is...?

Answer 5

undefined?

Answer 6

correct

Answer 7

ok here is why I asked that question

Answer 8

I solved the equation cos(x) tan(x) -2 cos ^2 (x) = -1
and I am positive my answers are correct, and the answers I got are 30 150 and 270 degrees

Answer 9

But when I check my answers, i get that tan is undefined at 270 but I solved the equation by
using the identity tan (x) = (sin x) / (cos x)

Answer 10

so do I exclude 270 from my solution?

Answer 11

yep :)

Answer 12

one more question

Answer 13

I solved the equation 2 cos (2x) +3 = 2 three different ways and they are:
1) Using the inverse method. From this method I got 60 and 120
2) I used the double angle formula to solve the same equation and got 60 120 240 300
3) I used the TI 83 and the same answers that I got from method 2

I am inclined to believe that method 2 works because the calculator confirmed my answers
however I dont understand why. Please explain

Answer 14

looks like I lost you for a moment
still there?

Answer 15

yes

Answer 16

ok, so when you solve with the inverse method you get... (I'm going to do this in radians because degrees are ugly)

Answer 17

ok radians is fine

Answer 18

btw the question wanted answers between 0 and 2pi

Answer 19

No No No wait wait wait

Answer 20

1/0 is complex infinity

Answer 21

what? hba

Answer 22

\[2\cos(2x)+3=2\]\[\cos(2x)=-\frac12\]\[2x=\cos^{-1}(-\frac12)=\{\frac{2\pi}3+2\pi n,\frac{7\pi}6+2\pi n\}\]\[x=\{\frac{\pi}3+\pi n,\frac{7\pi}{12}+\pi n\},~~~~n\in\mathbb{Z}\]now normally you are accustomed to ignoring the solutions that repeat because the \(2\pi\) part takes you out of the range of 0 to 2pi but in this case, notice that after dividing by 2 and solving for x, both \(\frac\pi3\) and \(\frac\pi3+\pi\) are both in the range 0 to 2pi, same goes for \(\frac{\pi7}{12}\), hence you have 4 answers within the range 0 to 2pi

Answer 23

Yes you know the concept of limit?

Answer 24

and yeah, what @hba ?
I don't think we need to play with limits here, this is just trig

Answer 25

But still if you see the concept of limit 1/0 is actually complex infinity.

Answer 26

what the is the difference between regular and complex infinity?

Answer 27

so what should I say to my trig students when I present them with this concept you just explained to me TuringTest?

Answer 28

what happens when you solve an equation that is like 3sin (2x) - 1 = 1
does that explanation you gave me works all the time?

Answer 29

yes, that explanation should work all the time

Answer 30

I guess I made a mistake when I showed my students the inverse method without giving them this explanation

Answer 31

damn it

Answer 32

you should make a point to your students that when you do the inverse method you will get something like \(ax=\theta+2\pi n\) where n=0,1,2... and 'a' is some constant, and that even though normally if you are working in the range 0 to 2pi you can ignore the extra solutions that come from the +2pi part, but since we have to divide by a to get the actual answer you wind up with is\[x=\frac\theta a+\frac{2\pi n}a\] adding \(\frac{2\pi}a\) may not take you out of the range 0 to 2pi (depending on what theta and a are), so you have to check if there are some n's that you can insert that will still give you an answer in the range 0 to 2pi. In the question I explained n=0 and n=1 give valid solutions

Answer 33

reading your explanation

Answer 34

hmm

Answer 35

I get what you are saying, so what happens when its sin (3x) and so on, works there too I guess

Answer 36

I am still confuse to when should I use this idea you just presented to me...its bc its a little difficult to understand.

Answer 37

i get that the answers will repeat because sin cos functions are continuous and thus their waves repeat.

Answer 38

yep, exactly the same
the solution will be\[3x=\sin^{-1}(y)=\theta\pm2\pi n,~~~~n\in\mathbb{Z}\]\[x=\frac\theta3\pm\frac{2\pi n}3,~~~~n\in\mathbb{Z}\]so you have to check to see if n other than n=0 give solutions that are still within [0,2pi], so try n=1, n=2, etc. until you are beyond 2pi

Answer 39

oh i get it

Answer 40

bc the sin (2x) means that the complete curve is made within pi and therefore by 2pi the answer is repeated again.

Answer 41

exactly

Answer 42

great. thank you so much, you gave me the algebraic understanding

Answer 43

btw can i ask you what you do?

Answer 44

any time you take an inverse trig function you get infinite answers, because\[\sin(\theta)=\sin(\theta\pm 2\pi n)\] so that is the origin of the idea...
I am actually a self-taught math/science enthusiast. I am in first semester physics in uni, but I have studied independently far beyond my schooling :)

Answer 45

I am just a high school math teacher who also loves math but not an enthusiast because I still have a lot more to learn in math.

Answer 46

Here is another question if you are interested....
where did the formula for volume of the sphere came from? I know its calculus but not sure where. LOL

Answer 47

Do you know about triple integrals and polar coordinates?

Answer 48

Just a warning, in the future you might be getting some more questions from me.

Answer 49

yes the concept of polar coordinates is to be able to convert rectangular coordinate graphs to polar.

Answer 50

triple integral my understanding is none. lol

Answer 51

are you from the USA?

Answer 52

yep, but I currently live in Mexico
without the triple integral, I can only say that the volume of a sphere is the integral of the surface area with respect to the radius\[V=\int_0^rAdr=\int_0^r4\pi r^2dr=\frac43\pi r^3\]if I think of a way to show that the surface area of a sphere is 4pir^2 without using sherical coordinates I will let you know

Answer 53

another way to show it would be to do a volume of revolution for the equation of a circle.... I don't know if you are good with volumes of revolution in calculus, and this one would be a bit pain

Answer 54

painful*

Answer 55

I'm happy to help, and you're welcome
if you want my attention you can tag me @TuringTest
in future please only ask one question per post please though :P

Answer 56

i can understand volume using revolution

Answer 57

i am will definitely ask you future questions and will limit it to 1

Answer 58

ok, so get a circlerevolve that sucker around x=0 and you should get your answer

Answer 59

yes

Answer 60

revolving it gives you the sphere

Answer 61

I meant we only need the upper half of the circle...

Answer 62

right i understand so far

Answer 63

so the area of each disk is\[\pi r^2=\pi(r^2-x^2)\]integrate that from x=-r to x=r with respect to x and out comes the volume of a sphere