precalculus

[timid jacinth]

and that'd give me cos(theta) = 3/5

[tired nest]

Yep, perfect.

[timid jacinth]

OH

MY

GOD.

that was like

a revelation

doesn't 1-(4/5)^2 become 1-(4/5) if I take the sqrt of the entire thing though?

[tired nest]

Nope.

Because the square root isn't distributive.

[timid jacinth]

hhmm

[tired nest]

It's like... if you square (a+b) you don't get a^2 + b^2

[timid jacinth]

oh yeah

[tired nest]

Because the square of a number is the number times itself, so (a+b)(a+b)

[timid jacinth]

you get 2a+2b

hang on

you get

a^2 +2ab +b^2

right?

[tired nest]

Yeah.

[timid jacinth]

okay

that makes sense now

so always evaluate to the simplest level before I take the sqrt of something?

[tired nest]

Yeah, if you can do it.

For this type of question though, yes.

[timid jacinth]

Alright

I have a similiar type of question; I'm going to try the other method here, can you guys interrupt me if I make a mistake?

Find exact values of remaining trig functions of theta.

Sin(theta)= -7/25, 180degrees < theta < 270degrees

From here, I can determine that theta must lie in quadrant 3, because sin = y, and sin is negative, therefore it must be in either quadrant 3 or 4, but I know theta is between 180 degrees and 270 degrees, therefore it must be in quadrant 3.

Now, by using the concept of a circle; I know through sin(theta) = opposite / hypotenuse, I can tell from sin(theta) = -7 / 25 that my radius of the circle will be 25.

From here I can apply pythagorean theorem to determine what cos(theta) is since cos(theta) = x on a circle.

sqrt(x^2+(-7)^2) = 25.

or x^2+(-7)^2 = 25^2.

x^2 +49 = 625.

x^2 = 576

sqrt(x^2) = sqrt(576)

x = + or - 24, since I know I'm in quadrant III, my cos(theta) must be negative, therefore x must be negative so I take x= -24, or cos(theta) = -24/25

@tired nest am I correct so far?

[tired nest]

Yeah, looks good so far.

[timid jacinth]

Awesome. So now, I just need to use my knowledge of the identities of the functions.

For example; tan(theta) = sin(theta)/cos(theta), so tan(theta) = (-7/25)/(-24/25), which is eqaul to -7/-24 = 7/24

[tired nest]

Yeah. 👌

[timid jacinth]

Can we go over even and odd trig functions? Why is it that only cos(theta) and sec(theta) are even functions and everything else is odd?

sorry

[tired nest]

They're not both even.

[clever inlet]

Only cos is even isn't it?

And the reciprocal of cos

[tired nest]

Yes.

[timid jacinth]

changed it

meant sec(theta) not sin(theta)

[clever inlet]

You saw the interactive unit circle right?

[tired nest]

Ohh.

[timid jacinth]

I've seen the unit circle

and I think I saw the interactive one

[clever inlet]

You can see how the graphs are made

As you go round the circle

[timid jacinth]

my problem is, I haven't seen the graphs of the functions

so I just need to know how the unit circle explains them to be even / negative functions

[clever inlet]

so back to the unit circle

notice

cos(0) = 1

if you go up to like say cos(45)

you'll get 1/sqrt2

and if you go down to cos(-45)

you'll get the same thing

basically imo

just play with the interactive unit circle

[tired nest]

Pick an angle and look on the unit circle at the positive and negative of that angle.

An even trig function will give you the same value for both.

As kangaroux demonstrated.

[timid jacinth]

Ah

gotcha

so if I take

sin(30) for example

I see sin(30) = 1/2

if I look for sin(-30), I'll see that sin(-30) = -1/2

[tired nest]

Yeah.

[timid jacinth]

and -sin(30) = sin(-30)

so the function is odd

[tired nest]

Exactly.

[timid jacinth]

if I take cos(30), cos(30) = sqrt(3)/2

cos(-30) = cos(330) = sqrt(3)/2, so since cos(x) = cos(-x), the function is even

[tired nest]

Yeah.

[timid jacinth]

okay

and that's going to be the case with all of the fuctions that aren't cos or the reciprocal of cos

It must be hilarious watching me on here, watching me learn step by step lmao

[tired nest]

And you can think of it a bit like this... the 0 point is on the x axis, so the x axis is like the line of symmetry for the unit circle. On one side is positive and one side is negative.

[willow bear]

wouldn't describe it as hilarious

[timid jacinth]

just a couple of weeks ago I was asking about basic functions and now I'm on to basic trig

infuriating?

[tired nest]

You're actually doing really well, chief.

[willow bear]

yeah

[clever inlet]

^

[willow bear]

honestly tho, sin is odd, cos is even, and the parity of everything else can be built up from there

[timid jacinth]

yeah, I just understand mathematical concepts better when they're broken down into words

[tired nest]

Yeah. So to finish what I was describing just now, if you look at that line of symmetry, you can see how sin is all positive on one side and all negative on the other. The cos is actually symmetric though.

So it just comes out even that way.

[timid jacinth]

so if i'm asked "Use the​ even-odd properties of the trigonometric functions to find the exact value of the given expression. Do not use a calculator." for cot(-60)

I just need to find cot(60) and see if it's equal to cot(-60) to determine if it's even, if it isn't I just need to take cot(-60) and verify it's = to -cot(60)

[willow bear]

cot = cos/sin = even/odd = odd

so cot(-60°) = -cot(60°)

also, "even-odd properties"? seriously? til "parity" is a term too complicated for college

[timid jacinth]

Yeah, never seen parity used in a mathematical context

hmm

[willow bear]

do you know how function parity interacts with operations on functions? outside of the context of trig

[timid jacinth]

Not a clue

cot(-60) = cot(300) right?

[willow bear]

degrees, yes

[timid jacinth]

sin(300) = -sqrt(3/2)?

[willow bear]

yes

okay so
for addition (and subtraction):
even + even = even, odd + odd = odd
for multiplication (and division):
even * even = even, odd * odd = even, even * odd = odd
for composition:
even(even) = even, odd(even) = even, even(odd) = even, odd(odd) = odd

(these are all easy to verify using the definitions of even and odd functions)

[timid jacinth]

Okay

I wonder when I'll be introduced to this

or if it was skipped over

sin(-270) = sin(90), so 1?

[willow bear]

yes

i'm a bit annoyed how you don't state your units there and like

== sin(90)

[granite stirrup]

0.89399666

[willow bear]

:p

[timid jacinth]

I would if I knew how to make the degree symbol

or else I'm typing sin(-270degrees)

[willow bear]

deg is okay

[timid jacinth]

ok fair enough

So, I'm getting a little confused here.

if sin(-270deg) = 1, and I'm asked to use even-odd properties to find the value of the expression; am I doing -sin(-270deg) = -(1) = -1, which would mean that it isn't odd right?

but I know sin is an odd function

[willow bear]

sin(-270°) = -sin(270°)

= -(-1) = 1

[timid jacinth]

oh shit

yeah

@willow bear how old are you if you don't mind me asking?

[willow bear]

18

[timid jacinth]

Wow

and you're already a math wizard

or

math witch

but that sounds

weird

and

an insult

[tired nest]

😄

[timid jacinth]

damn

cos(-pi) is throwing me off

[tired nest]

Why so?

[timid jacinth]

so cos(pi) = -1

[willow bear]

and so's cos(-pi)

cos is even

[timid jacinth]

cos(-pi) = cos(0) right?

[willow bear]

no

even functions let you ignore a -1 multiplier (aka negative sign) on their input

[timid jacinth]

okay

[willow bear]

pi and -pi are also the same point lel

[timid jacinth]

oh shit

yeah

let me draw that, and tell me if I'm correct in my understanding

https://gyazo.com/667dea6c6b22c6d5e7eede9124fb2482 based paint and gyazo

[willow bear]

yup

[timid jacinth]

Do you practice math as a hobby?

or were you just ulta receptive to your classes when you were younger?

because, aren't you advanced for your age?

[west furnace]

chiefqueef lmfao

i'm dying rn

[willow bear]

i do indeed do math as a hobby

[timid jacinth]

@west furnace why

[west furnace]

your name

[viscid thistle]

Im "advanced" for my age as well, I mean, advanced is more of a how much time you put into it atleast for me that is, I do math as a hobby, a job and what not @timid jacinth point being, to get to an advanced level you need to study!

[timid jacinth]

yeah

the career I want will involve a lot of math

my problem is I haven't done math for 7 years

[viscid thistle]

Take this to general?

[timid jacinth]

Is my reasoning correct?

(sin47deg)(csc47deg) = (sin47deg)(1/sin47deg) = 1

[willow bear]

yes

[timid jacinth]

Thanks

tan(theta) = tan(theta + k(pi)) right?

or am I expressing it incorrectly?

[willow bear]

yeah, tan(θ) = tan(θ+πk) for all integers k

[timid jacinth]

I'm onto range questions now, and range is just a concept that messes me up even moreso when it comes to trig functions

I understand the range of sin and cos

How does the range of tan work?

[crimson gate]

Have you looked at the graphs yet?

Or still just using unit circle

[timid jacinth]

Nope, trying to avoid them until I'm introduced to them in class so just the unit circle

[willow bear]

ok

[crimson gate]

So tan=sin/cos

when sin(theta)=1, cos(theta)=0

[timid jacinth]

which occurs at pi/2

[crimson gate]

Yep

[timid jacinth]

and every odd interger multiple of pi/2

[crimson gate]

So as you approach pi/2, sin(number that's almost pi/2) will be infinitely close to one and cos(that number) will be infinitely close to zero

When you divide it by that number that's infinitely close to 0 you get an infinitely large number

And when you take the cos(number that's right after pi/2 but infinitely close) you'll be dividing by that same thing as before, but since its after pi/2 it's negative, which means you're dividing a positive by a negative, so you get a negative infinitely large number

That's why tan has a range of (-inf,inf) but the domain doesn't include odd multiples of pi/2

Because if theta were an odd multiple of pi/2 you'd be dividing by 0

[timid jacinth]

right, because tan = sin/cos, cos is 0 at odd integer multiples of pi/2

[willow bear]

that is a rather convoluted description

[timid jacinth]

Here is where I got lost a little

And when you take the cos(number that's right after pi/2 but infinitely close) you'll be dividing by that same thing as before, but since its after pi/2 it's negative, which means you're dividing a positive by a negative, so you get a negative infinitely large number

I understood everything up to that point

[willow bear]

i can explain geometrically why the range of tan is R

[timid jacinth]

for example, .9999 / 0.000001 will give me a massive number, and as sin gets closer to one and cos gets closer to 0, but both remaining below 1 and above 0 respectively, that number will only get larger so +inf

[crimson gate]

Yep

[timid jacinth]

so, are we now talking about getting closer to -pi/2?

[crimson gate]

So after you cross pi/2 on the unit circle, the y coord stays positive but x becomes negative, since pi/2 is on the y-axis. You'll have .9999/-.0000001

[willow bear]

mh

[crimson gate]

Sorry, hard to spellcheck on mobile :x

[willow bear]

i feel like this is a bit too informal for my taste

[timid jacinth]

couldn't you just say the same for -pi/2 though?

like, y is now negative, and x is still positive

[willow bear]

steel, you're explaining the asymptotes of tan, which is not quite what chief was looking for

[timid jacinth]

as y approches -pi/2 it's something like -.99999 / 0.0000001

because in that quadrant, x is still positive right?

so that'd give me an infinitely large negative number

[crimson gate]

It approaches infinity and negative infinity, would that not show that the range would be (-inf,inf) in this case?

[timid jacinth]

I'm not getting the concept of crossing pi/2 into the second quadrant

[willow bear]

you don't need it, really

want a more geometric explanation for the range of tan?

[timid jacinth]

Sure, understanding both can't hurt though

[crimson gate]

Maybe a geometric explanation would help, verbality in math can only go so far

[timid jacinth]

I tend to be a bit more receptive to verbal descriptions 😛

[willow bear]

okay, so i'll want you to draw a unit circle on the xy plane, and the line x = 1

which touches your circle at (1,0)

aight?

[timid jacinth]

what do you mean the line x = 1 that touches the circle at 1,0?

[willow bear]

one moment

[timid jacinth]

if you could draw it on paint or something, which is what I was going to do and then send you a screen shot, that'd help immensely

Steel, can you explain that second part of your description a little differently?

[timid jacinth]

Okay

gotcha

[willow bear]

so

we're gonna make this line an axis of sorts

the tangent axis, as it were

[timid jacinth]

ok

[willow bear]

the point where it touches your circle is 0, and the units and direction are the same as on the y axis

[timid jacinth]

Ok

[willow bear]

now, this gives a nice visual representation of tan(θ)

to find the tan of an angle, draw a straight line through your angle's point and the center and look where that line crosses the tangent axis

that crossing point will be tan(θ)

[timid jacinth]

https://gyazo.com/ac346cc6284a2a1deea110c1fb1d6fc4

this?

[willow bear]

yeah

[timid jacinth]

Okay

[willow bear]

now, if cos(θ) = 0, then the line you draw will be vertical, and thus won't intersect the tan axis

make sense?

[timid jacinth]

you're talking about the terminal line right?

[willow bear]

the one i have in green

[timid jacinth]

right

yeah

[willow bear]

yeah ok so

[timid jacinth]

nvm

not necessary

[willow bear]

imagine θ changing gradually from 0 to π/2, and its terminal line with it

[timid jacinth]

ok

[willow bear]

the point on the tan axis will sweep out its entire upper half

[timid jacinth]

the half above 0 right?

[willow bear]

yeah

[timid jacinth]

Ok

[willow bear]

and a similar thing will happen if you let θ go from 0 to -π/2 instead

so basically, the entire tan axis gets covered

[timid jacinth]

right

[willow bear]

so the range of tan is R

[timid jacinth]

wouldn't it eventually hit the tan axis though?

like let's say, it's right by 1, where cos is almost 0

wouldn't it hit it eventually, just really really high up?

because it still has a slight gradient

[willow bear]

yeah, but almost 0 and exactly 0 are different things

[timid jacinth]

ah

gotcha

[tired nest]

Parallel lines never intersect.

[timid jacinth]

but it would only be parallel to the tan axis at pi/2 right? or where y = 1

[willow bear]

yep

[timid jacinth]

if it's just shy of y=1, wouldn't the line intersect eventually?

[willow bear]

yep

that corresponds to the fact that tan is only ever undefined at single points

[timid jacinth]

So why can't we assign a value other than infinity for when that terminal line intersects with the tan axis?

let's say it's as close to 1 as possible, wouldn't the point of intersection just be a ridiculously high number rather than infinity?

[tired nest]

You can. You only have infinity when it's parallel.

[timid jacinth]

so why do we write the range as (-inf, inf) instead of [ridiculously large negative number, ridiculously large positive number]

[willow bear]

there is no such thing as as close as 1 to possible

whatever number you claim to be closest to 1, i can find one that's closer by taking (your number + 1)/2

[timid jacinth]

but when it's parallel, it isn't defined I thought?

Okay

yeah

because

[willow bear]

precisely

[timid jacinth]

.999 is going to infinitely reoccur

or whatever it's called

.999 is larger than .9999

etc

[willow bear]

also, for every number x, there is an angle such that tan(θ) = x

[timid jacinth]

so it's just going to get infinitely closer to 1

but never actually touch

[tired nest]

That's why you have the curved brackets for the range. (-inf, inf) That means the actual values aren't included but everything possible in between is included.

[timid jacinth]

Ok

and the same applies for the opposite direction towards -1 or -pi/2

but the number is just going to be an infinitely negative number

which is expressed as -inf

[tired nest]

Yep.

[timid jacinth]

Thanks everyone

https://gyazo.com/d2146831b261bd6597f66cfb23b09879

wouldn't they eventually touch?

or will it keep getting closer and closer

but never actually touch?

[tired nest]

Yeah they touch, cause they're not parallel.

[timid jacinth]

so if they touch, let's say at -1000000

why isn't the point at which they touch considered the limit to the range?

so like [-10000000, 10000000]

[willow bear]

because there is an angle with a bigger tan

[tired nest]

Because it will also touch at -10000001 and -10000002 and so on. All the way until -inf

[willow bear]

tan is unbounded

[crimson gate]

Because you can always get slightly closer to pi/2 without getting to pi/2

[timid jacinth]

ohhhhh yeah

man

[crimson gate]

Trust me when you see the graphs it'll all click instantly

[timid jacinth]

both the verbal and the geometrical descriptions really helped me

yeah I'm actually excited to see them

I just don't want to look at them and base all of my knowledge off of them since I'm supposed to be understadning this stuff without them

[tired nest]

That's fine. Since you're going to be learning them soon anyway.

[timid jacinth]

How old are you when you typically learn this stuff? In England, where I was educated, we aren't introduced to this stuff at all in highschool

and we graduate at 16

you're only introduced to trig and calc etc if you elect to take math at the college and uni level

at least, that was the case when I was in school

[crimson gate]

Depends, usually senior year of high school in America I think, so about 17 or 18. Many people end up taking precalc before then so it could be from 15-18, depending on the math experience they went into high school with, if they take summer classes, etc

[timid jacinth]

ah ok

you go to college in the Uk between 16-18

and that's where you're introduced to this if you chose to continue with maths

obviously I didnt

[crimson gate]

I took precalc my sophmore year and then calc i and ii junior year, my high school didn't offer calc iii so I had to wait until now (first year of college) to take it

[clever inlet]

Oh

College starts at 16?

[timid jacinth]

Kind of Kanga

we call what you call college uni

[crimson gate]

You also cover some simple trig before precalc I think

[clever inlet]

I was wondering why they don't teach trig in high school

[timid jacinth]

for us it goes highschool < college < Uni

[clever inlet]

That makes more sense then

[timid jacinth]

college does not = uni

you go to uni after graduating college

My issue is, I'm doing uni in the U.S, and I left highschool in the U.K at 16, I'm 22 now and the extent of my knowledge was basic algebra

so I was never introduced to trig or anything of that sort

[crimson gate]

From what I've heard, education is a lot more focused in the uk, where Americans have to take tons of general stuff in hs whereas you get much more freedom in the uk, but idk if that's true

[timid jacinth]

absolutely true

for example, gen ed's in Uni don't exist in the UK

for example, if I choose to major in English in the UK at Uni

all I'd be doing is English classes

no need for general ed's in math / sci / humanities

so if you don't choose to take math's at a higher level past high school in the UK, you simply don't do mathematics past basic algebra

[clever inlet]

Over here, they have 3 math courses

With more content in each

And they pick stuff they feel like is useful for further education

Like how they do calculus before a lot of Alg 2 topics

Or they teach differentiation before quadratics

[timid jacinth]

Well, I'm in an Alg 2 class at the same time as a trig class

and I'm findining, what was review for the trig class is the material I'm learning at the same time in the alg 2 class

is the range of tan the same for cot? If so, why?

[crimson gate]

It is, and it's for the same reasons, except for even multiples of pi/2 instead of odd (since even mults if pi/2 will make sin=0 and the sin is on the bottom in cot)

[timid jacinth]

cot = cos/sin right?

so 2pi/2 = pi, = 1,0

y = 0 here

[crimson gate]

Mhm

[timid jacinth]

so it's the same logic right

just flipped?

[crimson gate]

Yeah

[timid jacinth]

as we approach 0 (y) but remain above 0, we'd still be dividing almost 1/0^+ or w/e the notation is

which would result in a massive positive number

alternatively, as we approach -1, but never actually hit -1, we'd be dividing almost -1 by a tiny number greater than 0

which would result in a massive negative number

[crimson gate]

Yes

[timid jacinth]

with the 1 and -1 being on the x axis this time right?

or no?

like at pi and 0

[crimson gate]

Yeah and when theta = pi or 0, it's at -1 or 1 o n the x axis

[timid jacinth]

Okay, I'm going to try and draw the same thing

Anne(?) drew earlier

but for cot

fuck

I don't think I can

[willow bear]

the cot axis is the horizontal line touching your circle at y = 1

[timid jacinth]

would cot's axis be vertical like tan's?

ah

thought so

had it right the first time then but second guessed myself

https://gyazo.com/fd57bdffba4531b39bc006a8992ef0b8 something like this?

[willow bear]

yup

[timid jacinth]

Cool

thanks!

can't believe microsoft are doing away with paint

the madmen

[willow bear]

what?

[crimson gate]

Nah they're not

[timid jacinth]

Yeah, paint is no longer a thing so I'm told

or they're not making new versions for future operating systems

[crimson gate]

It's like when they "got rid of internet explorer" and now they have "edge"

[timid jacinth]

Oh

they're just making a silkier program?

[crimson gate]

Pretty sure it's something along those lines

[timid jacinth]

I can live with that I guess

Alright, I have my first trig test here with me, I got a 72% on it

I'm going to go over a few questions I got wrong

For example;

One question was; given f(x) = sqrt(x-1)

(a) state the domain and range of f(x) (use interval notation)

I put (1, inf) and got it wrong, my answer should have been [1, inf) right?

for domain at least

[willow bear]

yes

f(1) exists

and is equal to 0

[timid jacinth]

fug

I have a problem with range

maybe I'm just foggy rn

how would I find the range of f(x)=sqrt(x-1)

[willow bear]

the range of sqrt is [0, inf)

[timid jacinth]

and that's because the outcome will always be greater than or equal to 0 since it's a square root?

[willow bear]

output, but yes

[timid jacinth]

that's what I meant, so the -1 doesn't effect anything?

[willow bear]

the -1 affects the domain

[timid jacinth]

Okay

so f(x)=sqrt(x-1), find f^-1(x)

[willow bear]

sqrt(y-1) = x, solve for y

[timid jacinth]

y = x+1?

[willow bear]

no

how did you get that?

[timid jacinth]

let me go back at it

Uh

so I took the original function f(x)=sqrt(x-1),

substituted y for f(x), so y=sqrt(x-1)

substituted x for y, so x=sqrt(y-1)

squared both sides, so x^2 = y-1

yeah I have no fucking clue tbh

[willow bear]

y = x**^2** + 1, surely?

[timid jacinth]

yeah

looking back at it

for some reason when I did sqrt(y-1)^2

I took y^2 + 1

or somehting

y^2 - 1

added the 1, so I had x^2 + 1 = y^2, took the sqrt of both sides

so I got x + 1 = y

that's where I went wrong

no idea why I did that

so the inverse of f(x)=sqrt(x-1) is f(x)=x^2+1?

so if f(x)=sqrt(x-1) then f^-1(x) = x^2 + 1, I can verify this by doing f^-1(f(x))?

[willow bear]

indeed

[timid jacinth]

So that evaluates to x

[willow bear]

however, you do have to explicitly restrict the domain of your f^-1

to [0, +inf)

since that was the range of the original function

[timid jacinth]

right

if I got x as my final output of f^-1(f(x))

does that prove it?

[willow bear]

yes

f compose f^-1 = id

[timid jacinth]

hmm

That's it for me tonight, thanks for all of your help

[viscid thistle]

I feel like one of the few people who uses set notation for domain and range,hahah

in Algebra that is

[gleaming schooner]

Good morning

[clever inlet]

morning

[gleaming schooner]

I have a quadratic equation that I need to find the roots of.

[clever inlet]

?

[gleaming schooner]

It's f(x) = -1/5(x-1)² + 1

So I set 0 = -1/5(x-1)² + 1

And I get -1 = -1/5(x-1)²

[clever inlet]

Yeah

Looks right so far

[gleaming schooner]

Divide both sides by -1/5

And I get 5 = (x+1)²

[tired nest]

x-1? Or x+1?

You changed it halfway.

[gleaming schooner]

x -1

Oops

Oh crap

I'm sorry

I messed up the equation entirely

I have to start over lol

I combined two different equations.

[tired nest]

No problem. 😄

[clever inlet]

If it was the right equation

[tired nest]

It happens.

[clever inlet]

Then you got really close

To the actual answers

Which is still good

Since you know what you're doing

[gleaming schooner]

Well I ran into an issue 😦

The actual equation is:

f(x) = -1/5(x + 4)² + 1

[clever inlet]

solve for roots?

[gleaming schooner]

0 = -1/5(x + 4)² + 1

-1 = -1/5(x + 4)²

Divide both sides by -1/5

5 = (x+4)²

[clever inlet]

yeah

you basically got it

[gleaming schooner]

5 = x² + 8x + 16

[clever inlet]

ehh

[gleaming schooner]

o,o

Did I mess up

[clever inlet]

if you already got it at 5 = (x+4)²

you can just root both sides

[gleaming schooner]

Ah..

[clever inlet]

so you get

x + 4 = sqrt5

and

x + 4 = -sqrt5

[gleaming schooner]

Hm

What's after that?

To get x

[clever inlet]

isolate x

subtract 4 from both sides

[gleaming schooner]

I never understood the whole +- thing with square roots

How does that work?

[clever inlet]

well

say you have

x^2 = 4

both (2)^2 and (-2)^2

= 4

you can also look at it in terms of the parabola graph

[timid jacinth]

If I'm told to find the exact value of all trig functions and I'm only given the fact that the point (3,-4) is on the terminal side of theta, do I assume cos(theta)=3 and sin(theta)=-4?

[tired nest]

No.

Do you know about sin = o/h and cos = a/h?

[timid jacinth]

Yeah

[gleaming schooner]

Ahh

I understand

@clever inlet How'd you graph that?

[clever inlet]

desmos

[gleaming schooner]

In Desmos

[clever inlet]

using your equation

[gleaming schooner]

Ohh

[clever inlet]

well

[gleaming schooner]

My equation

Alright

[clever inlet]

the simplified

[full elm]

Desmos

[clever inlet]

y = (x+4)^2

and

y = 5

[gleaming schooner]

@clever inlet So the final answer, for the roots would be x = ± sqrt(5) - 4?

[clever inlet]

yep

[gleaming schooner]

Alright, thanks! 😃

[tired nest]

Okay, @timid jacinth you basically want a triangle with the hypotenuse from (0,0) to (3,-4)

Right angle to the x axis, like in the unit circle.

That's what it's saying.

[gleaming schooner]

-1/5(0 + 4)^2 + 1 gives me -1/5(16) + 1

How do you get -11/5 from that ?

@clever inlet

[clever inlet]

?

=0 i assume?

actually nvm

not a quadratic

-1/5(16) + 1

-16/5 + 1

-16/5 + 5/5

-11/5

[gleaming schooner]

How do you get -16/5 + 5/5

I think I'm confused on that

[clever inlet]

-1/5(16) = -16/5 right?

[gleaming schooner]

YE

Yea*

I understand that part

[clever inlet]

1 is the same as 1/1 ?

[gleaming schooner]

Yea

[clever inlet]

so you need a common denominator

by multiplying top and bottom by 5

5/5

[gleaming schooner]

Ahh

Alright

Thanks

[timid jacinth]

I'm asked to find the exact value of all trig functions, where the point (3,-4) lies on the terminal side of theta,

Do I use x,y,r definitions here and the Pythagorean theorem to find the functions?

Like since so have the point (3,-4) I technically have x and y, so I can do x^2 + y^2 = r^2 to find cos and sin and from there the rest of the functions?

[willow bear]

i mean, you can draw a triangle with θ's terminal side as its hypotenuse

a "signed triangle" of sorts, in that its legs are allowed to be negative

the hypotenuse of that triangle will be 5, by the pythagorean theorem

and well yeah now that you have all three sides of it, all you need to do is take their ratios

[timid jacinth]

I can also consider it as (3)^2 + (-4)^2 = 25, so I have r=5, sin = -4/5, cos = 3/5 etc, or is this flawed reasoning?

[willow bear]

yeah, sin(θ) = -4/5 and cos(θ) = 3/5

and once you have those two, you have everything

[timid jacinth]

Right

I think that was the intended way I find it

These were questions on my test that I botched

I didn't even put an answer down for this one

But it's so simple looking back now

[signal mesa]

hey can anyone help me prove lim x->0 of f(x)=x^4 cos(2/x)

since the limit of a product is the same as the product of the limits can i just say since x^4 = 0 then the limit of the function must be 0

as x approaches 0

[mental maple]

that is only true if both limits separately exist

[signal mesa]

o

hmm

[mental maple]

https://i.imgur.com/e9nJJKN.png

this is the graph of cos(2/x)

that limit does not exist as x -> 0

but you can use for example -1 < cos(2/x) < 1

[signal mesa]

ohh squeeze theorem

[mental maple]

yep

you got it

[signal mesa]

christ how did someone come up with this lmao

do i have to put down any assumptions before the proof i dont think theyre expecting a rigorous proof or anything

[mental maple]

did you learn the formal definition of the limit?

then probably not

just state that -x^4 <= x^4 sin(2/x) <= x^4

and that both x^4 and -x^4 go to 0 as x -> 0

[signal mesa]

epsilon delta you mean

[mental maple]

ye

[gleaming schooner]

How can I graph the following quadratic equation without using a calculator: f(x) = -1/5(x + 4)² + 1 The roots are (-4 + sqrt(5), 0) and (4 + sqrt(5), 0), the y-intercept is (0, -11/5)

How can I graph those roots? o,o

[mental maple]

uhm, i would just draw a parabola with vertex (-4,1)

all parabola are the same anyway

[crimson gate]

Yeah so you know the vertex is (-4,1) because it's in vertex form, and you know the roots, so you can just plot the three points and graph from there

[signal mesa]

we went over the epsilon delta definition but i dont remember the professor using it in conjunction with squeeze theorem

but i guess a formal definition of a limit when youre trying to prove a limit exists in a proof would be necessary lol

[mental maple]

ye, if you take the squeezew theorem for granted it's not needed

you can prove usingthe definition that limits preserve non-strict inequalities

the squeeze theorem follows from that

[signal mesa]

thank-you

[signal mesa]

if im trying to take the limit at 2 and i factor the numerator of this equation into (x+3)(x-2) im not able to cancel the denominator am i

[gleaming schooner]

Thanks

I'm having trouble taking a quadratic equation written in standard form, and converting it to vertex form

I have f(x) = 3x² - 42x - 91

[mental maple]

@signal mesa consider what happens when x > 2 and when x < 2 separately

[gleaming schooner]

I have to factor out 3 and complete the square, rigth?

right*

[signal mesa]

it becomes negative when x is less than 2

[mental maple]

@gleaming schooner yes

[gleaming schooner]

So, I do that and I get:

y = 3(x² - 14x + 49) - 91

And I then have to take away 3 * 49 from the right side.

[mental maple]

let's use separate channels

@gleaming schooner go to algebra

[gleaming schooner]

Alright

[mental maple]

(x+3)(x-2)/|x-2| = x+3 if x > 2

(x+3)(x-2)/|x-2| = -x-3 if x < 2

do you see why?

[signal mesa]

are we assigning x = -x

[mental maple]

that's not really a good way to say it

|x| is a piecwise defined function

it's x when x > 0

and -x when x < 0

[signal mesa]

so that applies even to the x which isnt in the absolute value notation

[mental maple]

basically, the absolute value makes your whole function piecewise defined

one part when x-2 > 0, and one part when x-2 < 0

https://i.imgur.com/Tq2nqUf.png

it looks like this

one part is x+3

one part is -x-3

[signal mesa]

thank you thats a big help

[viscid thistle]

AAAAAAAAAAAAA

https://www.desmos.com/calculator/5lhdac966e

[timid jacinth]

https://gyazo.com/a3b6f49a8fc4b2dbe3af3202a54f91c6

for B I found 45.6deg, and I found sqrt(51) for b

is this correct?

[blazing raven]

yup

[timid jacinth]

how can b be greater than a though?

if a is a longer leg?

or is that just the result of poor drawing?

[blazing raven]

um?

oh I see. Yeah. If the problem didn't say b or a is longer then yeah. You are likely safe.

[timid jacinth]

Alright

[tired nest]

Always trust the maths over the drawing, unless explicitly told otherwise.

[timid jacinth]

gotcha

so if I'm told to find all functions of B

trig functions that is

[velvet sparrow]

U wot?

[timid jacinth]

my sin(45.6) = sqrt(51)/10 right?

[tired nest]

How did you get that, chief?

Oh, nevermind, I see.

Yep, looks good.

[timid jacinth]

Alright

so for tan I got 7/sqrt(51)

why can't I leave it like that again?

in one class I'm being told to rewrite 7/sqrt(51) to 7(sqrt(51))/51

[tired nest]

You want to avoid leaving roots in the denominator.

[timid jacinth]

and in the other I'm being told to leave it alone

[tired nest]

It's not incorrect if there's a root in the denominator, but usually it's preferred without it.

[timid jacinth]

Ok

so it's down to etiquette

[sonic heath]

It's usually better to put roots in the numerator for the sake of doing sums with fractions

You don't want to be finding 'root 51' as a common denominator.

😛

[timid jacinth]

ah

[sonic heath]

If forced to add.

[timid jacinth]

gotcha

[velvet sparrow]

What wouls be wrong wuth that?

[timid jacinth]

Hmm

this is a weird one

One sec, I'll put it up on paint and show where I'm at

So, the red is my working out so far

I'm confused how to advance beyond this

given the restriction

https://gyazo.com/2f1488494636f30281d2769f12e64603

[sonic heath]

Okay

[velvet sparrow]

Use the definitioms of trig functions

[sonic heath]

Do you know your trigonometric identities when a right angle is involved?

[timid jacinth]

Yep

[sonic heath]

Okay, what are they?

[timid jacinth]

SOHCAHTOA

[sonic heath]

Right

So, in this case - We'll start with finding c

You already have the length of b

And the angle B = 41 degrees

So, what identity involves c, b and an angle?

(Considering the location of the angle, of course)

[timid jacinth]

cos(A)=b/c, so cos(49deg)=4/c

[sonic heath]

Slight mixup there - cos is not (opposite/hypotenuse)

[timid jacinth]

I know, but for angle A

cos = 4/c

as b is adjacent to angle A

[sonic heath]

Oh I see what you're doing

My bad

And yes - You're correct.

And yeah you can go about it that way

Taking that cos as 180 - theta

So, yeah - your working there is correct

[timid jacinth]

so now I need to solve for c?

[sonic heath]

Okay, so now we need side lengths

[timid jacinth]

Okay

[sonic heath]

https://gyazo.com/2f1488494636f30281d2769f12e64603

So you mentioned trigonometric identities SOH CAH TOA

And you have internal angles

You gave me the relationship before cos(90-theta) = 4/c

Which is the same as

=tex \cos(49) = \dfrac{4}{c}

[sonic heath]

So, we need to rearrange for c

Can you make c the subject of that equation?

[timid jacinth]

c=4/cos(49)?

[willow bear]

=tex \cos(49^\circ)

[timid jacinth]

not sure

[sonic heath]

Ah, thank you

[willow bear]

yes, c = 4/cos(49°)

[sonic heath]

I didn't know how to do the degree sign 😛

And yes Chief, you're right.

[timid jacinth]

Okay

[sonic heath]

Now repeat for the missing side, using the trigonometric identities again.

[timid jacinth]

so I just leave c = 4/cos(49deg) for now

[willow bear]

what's the problem y'all are doing?

[timid jacinth]

https://gyazo.com/2f1488494636f30281d2769f12e64603

[velvet sparrow]

Trig

[willow bear]

c = 4/sin(41°), a = 4 cot(41°)

lel

[sonic heath]

"so I just leave c = 4/cos(49deg) for now"

You can do, for simplicity sake

If you want

[willow bear]

they're asked to leave answers as exact values

49° isn't one of those angles with an easy sin or cos

[sonic heath]

You could always leave it like that until you find 'a' as well, and then do the calculation

But yeah - She's right

Looks like you've got to give a precise answer here.

[willow bear]

seriously though this is the point where

i feel like sharing the way i tend to view trig functions geometrically

not as ratios, but as multipliers

[timid jacinth]

so how do I express c=4/cos(49deg) as an exact value?

[sonic heath]

Calculator 😛

[timid jacinth]

we're not allowed to use calculators

[willow bear]

no. it already is an exact value

leave it at that

[timid jacinth]

ok

[sonic heath]

Heh

[velvet sparrow]

Elaborate @willow bear

[sonic heath]

My bad Chiefqueef - apparently I'm still half asleep.

[timid jacinth]

so now I can move on to sin(49deg), which I think is sin(49deg)= a/(4/cos(49deg))

[sonic heath]

The reason why it is the exact value is because if you WERE to put it into a calculator

You would round off

And lose the 'exactness'

Hence why you must give that answer you just found.

[willow bear]

@timid jacinth you can figure out a from b directly

[velvet sparrow]

Not allowed

[willow bear]

what?

[timid jacinth]

Ok

that makes sense

[sonic heath]

"so now I can move on to sin(49deg), which I think is sin(49deg)= a/(4/cos(49deg))"

Yes.

And re-arrange for a

[timid jacinth]

okay

[willow bear]

you're overcomplicating things here!

b/a = tan(41°)!!!

[sonic heath]

Hahaha.

Yeah I'm stopping talking now

[willow bear]

anyway

this is what i meant by thinking of trig functions as multipliers

it's also a nice explanation for why sec and csc are named the way they are

[sonic heath]

That's really quite cool

[timid jacinth]

a=(sin49deg)/(4/cos(49deg))

[willow bear]

no

a = sin(49°) ** * ** 4/cos(49°)

if you insist on going that route

[timid jacinth]

I thought I was just rearranging for a?

[willow bear]

sin(49°) = a/(something)

a = (something) * sin(49°)

but once again holy shit chief you're making your own life harder here ;w;

[timid jacinth]

eh

[sonic heath]

=tex sin(49^{\circ})= \dfrac{a}{(\frac{4}{cos(49^{\circ})}}

[timid jacinth]

I'm a bit lost right now

I guess I'm supposed to figure this out with methods I'd have been taught up until this point

which, if I recall is complementary angles and x,y,r definitions

[willow bear]

hang on

let's start over

[willow bear]

you were given this, right?

[timid jacinth]

yea

[willow bear]

and you want to figure out side a

a.k.a. BC

so

4/a = tan(41°)

is that clear?

[timid jacinth]

a is the leg between B and C right?

[willow bear]

yes

[timid jacinth]

yeah that makes sense

[willow bear]

for clarity

[timid jacinth]

becuase tan = O/A, 4/a

[willow bear]

yeah so even though you've figured out c, you really do not need it to find a

[timid jacinth]

so right now I've got tan(41deg)=4/a

[willow bear]

yeah, and now rearrange for a

[timid jacinth]

a= 4/tan(41deg)

pls rspnd

[willow bear]

👍

[timid jacinth]

so I leave my a as that, now I need to solve for side c

[willow bear]

no you don't

you already found c earlier!

c = 4/cos(49°)

[timid jacinth]

so, I'm done?

[willow bear]

yes

[timid jacinth]

wow

seems a bit like a fast halt

[willow bear]

the problem isn't really hard or long tbh

at some point, trigonometric manipulations like these are going to become automatic

[timid jacinth]

A boy can dream

[sonic heath]

Just a case of practice

Honestly, the 'multiplier' thing just shown is news to me, because I was taught SOH CAH TOA via rote memorisation in school

Rather than shown the reason and actual relationships.

[willow bear]

eh

i mean it's okay to define sin, cos and tan as their respective ratios

[timid jacinth]

I have no fucking clue what the reason or relationships are

[sonic heath]

Of course sqrt{2}, but I'd prefer to have a more intuitive understanding of it

[willow bear]

i meant okay from a mathematical standpoint

[sonic heath]

Because I have no idea how secant etc are defined other than through calculating them each time or looking them up

[willow bear]

but yeah i feel like introducing them as multipliers, like in my picture, is better

[sonic heath]

Indeed.

It's better because you don't have to have the hypotenuse length to immediately get an answer

You can have any of the lengths

[willow bear]

oh yeah

[timid jacinth]

tbf

wtf is a multiplier

[willow bear]

a thing you multiply by

in my picture, in the first triangle you can see that the hypotenuse, which i marked with r, gets multiplied by sin(θ) and cos(θ) to produce the legs

[timid jacinth]

so

lets say

sin(theta) = 1/2

and

r = 1

to find the leg I'd just do 1/1 * 2/1 = 2?

[willow bear]

why'd you divide 1 by sin(θ)?

[timid jacinth]

oh

wtf

yeah

mb

1/1 * 1/2 = 1/2

so my leg is 1/2

[willow bear]

well

yes

(i mean, θ = π/6, so :P)

[timid jacinth]

o

;<

[willow bear]

?

[timid jacinth]

I didn't make that connection

for whatever reason

So back to that earlier problem, a=4/tan(41deg), c=4/sin(41deg)

just wanting to make sure I've got it correct for note taking purposes

I found c with theta = 41deg intead of 49 this time

and by using sin

[willow bear]

yup you're correct

i mean, sin(41°) and cos(49°) are the same thing anyway lol

[timid jacinth]

yeah

tru dat

[timid jacinth]

not very fond of biology

[viscid thistle]

Biology is good for one thing, application of exponents and exponent functions

[calm thicket]

^

[willow bear]

ayy

[viscid thistle]

lol

[smoky lance]

anyone about?

[clever inlet]

?

[late haven]

o/

[willow bear]

don't ask to ask
just post your damn problem

[smoky lance]

How would I find the complex zeros of

x^4 + 3x^3 - 19^x2 + 27x - 252

[willow bear]

rational root theorem

[viscid thistle]

^

[willow bear]

and hope that you can get it down to a quadratic

[smoky lance]

yeah I did that

and no dice

[willow bear]

then you're fucked

[smoky lance]

cool

thanks for the help

[willow bear]

all hail wolfram alpha, our robot overlord

[viscid thistle]

🙏

[subtle forum]

Anyone here that can help me with how to write 5i as exponential?

or, in exponential form

[desert thistle]

Try placing it in the complex plane

look for the angle, and distance to origin

[viscid thistle]

@GONNATAKEURBRAINS JOHNSON#9914 if you don't have access to wolfram or anything for degree 4 there is a formula, its disgusting though

[austere bramble]

$$for z\in\mathbb{C}$$ we can think of $$z=re^{i\theta}$$ where r becomes the norm and theta is the angle. 5i has distance 5 from 0, and i is on pi/2, so you get $$5i=5e^{i\pi/2}$$

[subtle forum]

Thanks @desert thistle

[viscid thistle]

could someone help me with transformations of sin and cos functions

[tired nest]

What's the question, @kurry ?

[tired nest]

Okay, so, try this first: what happens when you sub pi/6 into #1?

[agile tendon]

How do you prove Thurston's geometrization conjecture using Ricci flow with surgery to affirm as a corollary the Poincaré conjecture for 3-manifolds?

[austere bramble]

if you know how you'd be famous

[signal mesa]

what would be the domain in interval notation of a constant?

just [a] ?

[willow bear]

uh

what?

can you show what you're trying to do?

like, the context

[willow bear]

the domain of its derivative

i.e. the set of points where the derivative is defined

[signal mesa]

ohhh

so minus infinity to infinity

sheeit

[hollow cobalt]

how would i write (1 + i sqrt3)^3 in the form a + bi?

[willow bear]

you know how to expand (x+y)^3, right?

@hollow cobalt

[hollow cobalt]

oops, was doing other problems

i thought i did, but im failing right now

[willow bear]

(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3

(1 + sqrt(3)i)^3 = 1 + 3 * sqrt(3)i + 3 * (sqrt(3)i)^2 + (sqrt(3)i)^3

= 1 + 3 sqrt(3)i + 3 * (-3) + (sqrt(3))^3 * i^3

[hollow cobalt]

damn. . lost me on the last one

[willow bear]

what exactly are you confused about?

[hollow cobalt]

i understand the expansion and i see how the 2nd line is just the complex rendition of that

im not sure where its going from there

[willow bear]

i'm not done yet

[hollow cobalt]

ah ok

[willow bear]

do you understand how i got from the expansion to 1 + 3 sqrt(3)i + 3 * (-3) + (sqrt(3))^3 * i^3?

[dusk kettle]

does anyone actually teach the set of outputs of a function as its codomain, or does everyone say range?

[hollow cobalt]

uhh

hold on let me rewrite it on paper and look at it. . getting confused by the format i think

also havent slept

[dusk kettle]

=tex 1 + 3\sqrt{3}i + 3 (-3) + \sqrt{3}^3 i^3

[dusk kettle]

=tex (1 + \sqrt{3}i)^3 = 1 + 3 \sqrt{3}i + 3 (\sqrt{3}i)^2 + (\sqrt{3}i)^3\ =1 + 3\sqrt{3}i + 3 (-3) + \sqrt{3}^3 i^3

[hollow cobalt]

thanks

ok yeah duh

now i see it

though ,

why distribute at the end

?

with the sqrt3^3 i^3

[willow bear]

i was about to show why

[dusk kettle]

I think the plan is to factor out the i

[willow bear]

this whole thing equals $$1 + 3\sqrt{3}i - 9 + 3\sqrt{3} \cdot (-i)$$

[willow bear]

=tex = -8 + 3\sqrt{3}i - 3\sqrt{3}i

[willow bear]

which is just -8

[hollow cobalt]

aaaaah

ok

thanks a lot 😃

[hollow cobalt]

stuck on one last

how do i find f^-1 (inverse f) of f(x) = 1/1+e^-x

assuming that f is a one to one function

[willow bear]

=tex f(x) = \frac{1}{1 + e^{-x}}

[willow bear]

this?

[hollow cobalt]

yes

[naive hearth]

Hello all, I'm having trouble finding the equation of the line that is perpendicular to 3x + 4y = 7.

I start by converting this to slope intercept form to get the slope.

So then I have =tex y = -3/4x +7/4

[clever inlet]

Yeah

And for perpendicular

m1 * m2 = -1

And for point

I guess the y int

Actually nvm

I'm dumb

The y int probably isn't on the perpendicular line

[viscid thistle]

hopefully this helps (the ' thingy is the derivative but you can think of it as the slope of that line)

[naive hearth]

ok, so would m2 be just 3/4? or for a fraction do you invert it to be 4/3?

[clever inlet]

Wait

[viscid thistle]

that is the thought process u need to have, to see that for a parallel line the slope will be exactly the same

perpendicular line = slope is opposite and reciprocal

[clever inlet]

Is there even enough info for a single line equation?

[viscid thistle]

(in basic words, flip the sign and the denominator/numerator)

the answer to that therefore would be any line with a slope of -4/3

[clever inlet]

Ah yeah true

[viscid thistle]

don't try to just memorize that though, take a look at the drawing (sorry for it being so shitty) and try to realize how the slopes are affected

ms paint masterrace

[clever inlet]

👍

[naive hearth]

@tan thanks! so you flip the denominator/numerator and change the sign of both? I'm having trouble conceptualizing that but I think I can get there with some work.

m1 = -3/4 and m2 = -4/3

[clever inlet]

Positive

Cause if you Double negative

The product will be +1

[viscid thistle]

yes, my answer was incorrect on the sign

try to grasp the head around the concept though, let me see if i find some good explanations for it

http://www.algebralab.org/lessons/lesson.aspx?file=geometry_coordparallelperpendicular.xml this site is a bit ugly but it has nice examples displaying it

[naive hearth]

Thanks @tan and @clever inlet !

Now I need to find the the line with that slope(4/3) that passes through the points (-2/3, 7/8).

So I will use point-slope form, y- y1 = m(x - x1)

so now I have y - (7/8) = 4/3(x - (-2/3))

I need to get this in y = mx + b form, but I'm a bit lost...

[viscid thistle]

alright so you have this right now

=tex y - \frac{7}{8} = \frac{4}{3}(x-(\frac{-2}{3})

[viscid thistle]

right?

which is the same as

=tex y - \frac{7}{8} = \frac{4}{3}(x+(\frac{2}{3})

[naive hearth]

The book i'm using isn't very helpful, I don't understand why they multiply by 2.

[viscid thistle]

wait, isnt this technically calculus work?

this was on my precalculus book, on the review section @viscid thistle