Question

@callisto

Answer 1

I have to review the content first for like 10 minutes, but I will show you the assignment in advance.

Answer 2

@callisto, tangents

Answer 3

Okay!~

Answer 4

When I graph tan of (x), do I input sin(x) / cos(x)?

Answer 5

you can... but can't you just put y=tan(x)?!

Answer 6

I get error, Im using my T1 83

Answer 7

Even when i input sinx/cosx, I get error:(

Answer 8

Wait. Let me scan you something

Answer 9

This may help you a bit

Answer 10

Wow. That is really useful:) Thank you

Answer 11

That's a page of my book. Hope it helps

Answer 12

But only serve for revision. Not exactly the same for your question.

Answer 13

i know:) Thank you

Answer 14

So this one is basically the same as the previous assignment, but it is tan instead of sin and cos

Answer 15

I bet you should input y=tan(x)

Answer 16

Its strange in my computer I can, but in my caluclator, it wont work.

Answer 17

I tried mode radian, mode degrees, nothing works. It says invalid dimension

Answer 18

In calculator?

Answer 19

Yeah, my t1-83+

Answer 20

Sorry, don't know about that :(

Answer 21

you need to know what for 90, 270 ... tanx = undefined

Answer 22

For tangent, its period is 180
So tan(x+180) = tanx understand this?

Answer 23

yes

Answer 24

I just read it right now:)

Answer 25

Similarly, tan(3x+180) = tan (3x) , understand?

Answer 26

yes

Answer 27

Just now, I've mentioned that tan90 = undefined
so solve tan90 = tan3x
the angle at x would be the one with give undefined value for tan ratio

Answer 28

Oh ok

Answer 29

Do you really understand this point?

Answer 30

yes, Im also reading about it now and completing a self-exercise chart

Answer 31

Oh.. are you doing 2 things at the same time?

Answer 32

Basically, In order to complete the chart, I have to read:)

Answer 33

Technically, im doing 3 things simultaneously, Im also typin in OS:)

Answer 34

Perhaps you can do the reading first :)

Answer 35

The design of my course is a bit odd. They place explanations along exercises, not before them

Answer 36

Not in the entire course, but in some sections

Answer 37

Idk why they designed the course in such manner

Answer 38

Ok, Im just going to write a little quiz for 5 minutes and then we can start with this:)
Btw, i want to know something. Is this interrupting the preparation for your Math exam tomorrow?

Answer 39

o.o ...... -no comment- :P Go!!! take the test and get full marks!!!

Answer 40

If this is in any way interrupting your preparations, I do not want to place a burden on you

Answer 41

lol. They dont mark the test (Otherwise Id take my time and not finish it in 5 minutes:)), Its just a review quiz

Answer 42

I'm revising while teaching, so don't worry. If I don't want to teach, then ... I'll tell you :)

Answer 43

ok. Thank you:) I really appreciate you doing this:). If during any time, you need to do your own stuff, just let me know so that I can leave you alone:)

Answer 44

Tell me if you've finished the quiz

Answer 45

Ok. Im almost donw

Answer 46

:)

Answer 47

Ok. Im finished:)

Answer 48

Got 78% :(

Answer 49

o.o Get a better result next time :) You can do it!

Answer 50

I know,Im trying:)

Answer 51

Ok. Il try to solve Q1

Answer 52

Vertical asymptotes are x= when (tanx) is undefined

Answer 53

nvm. Lol i got confused

Answer 54

To solve for VA, I need to consider tanx = sinx/cosx and then set cosx = 0 and solve for x no?

Answer 55

I don't suggest you to do so. That would be complicated

Answer 56

you know tan90 = undefined. so in normal case, x=90 is the VA

Answer 57

so is the VA always teh same as the angle?

Answer 58

Because the VA of -270 is x = -270 right?

Answer 59

Not really. In your case, the equation is y=-2tan(3x+180)+3
That would be the same as y=-2tan(3x)+3 as the period of tan is 180degrees
Consider tan(3x), when tan3x=tan90 , there would be an VA
So, solve it and you'll get when x=30, there is an VA
Since the period of tan is 180, when x=180 +30 , there would be another VA. and so on.

Answer 60

Ohhh. Okok

Answer 61

Sorry, I have to go now. I'll come back and check as soon as possible!!

Answer 62

Oh sure. Thank you

Answer 63

You're welcome :)