precalculus

[harsh cipher]

I believe it is but confirmation would be helpful.

second question is where did pi/2 come from?

It's because p = 2pi/4 which is equal to pi/2

and I divide pi into 4 equal parts to get pi/8

On the first grid, the graph starts at pi, then add pi/8 to pi is 9pi/8 and so on..

I answered my own question...

[rigid dune]

I don’t know how to move on with this

I need to find width and length

[harsh cipher]

divide 2 first then solve

oh no it's a quadratic equation

[rigid dune]

wouldn't that give me 2 widths?

[harsh cipher]

honestly i don't know. I'm learning as well

[rigid dune]

I feel like this is supposed to be simple but I'm just missing something

[harsh cipher]

but from what I see is if you use quad formula you have +,- root

and width cannot be neg

[rigid dune]

exactly

Okay so I asked another friend, apparently yes you do factor it and just use the positive answer not the negative one

[harsh cipher]

Nice you got it then

[viscid thistle]

71 help plz

[clever inlet]

What have you tried?

[viscid thistle]

Making a triangle

And trying it

[fleet yew]

,rotate

[clever inlet]

Do you know your double angle identities?

[harsh cipher]

wow

AMD how did you do that?

[viscid thistle]

Yea but that’s not a double angle

[harsh cipher]

,rotate

[harsh cipher]

oh wow

[clever inlet]

🤔

[harsh cipher]

,rotate

[harsh cipher]

sorry my bad

,rotate x 2

[clever inlet]

In what way is it not?

[harsh cipher]

,rotate x 3

[harsh cipher]

i just guessed the command ha!

[viscid thistle]

So

Anyone know

[clever inlet]

Answer my question

[viscid thistle]

What

[harsh cipher]

do your your double angle identities

[viscid thistle]

Double angle is sin2x

[clever inlet]

?????

[viscid thistle]

2sinx is something else

[clever inlet]

Do you know your double angle identity for cos(2x)?

[viscid thistle]

Yea

[clever inlet]

Then

Apply that?

[viscid thistle]

Bruh

U supposed to use a triangle

[clever inlet]

?????

Well

Good luck with that

[viscid thistle]

Plz:,(

[harsh cipher]

www.google.com

[viscid thistle]

I did

No bueno

[clever inlet]

I mean

I keep trying to help you

And you reject help

So idk

Work it out

[harsh cipher]

well try symbolab or purplemath or anything similar

[viscid thistle]

Ok

[harsh cipher]

or listen to Kangaroux

[viscid thistle]

Ok

[harsh cipher]

is anyone awake to help me

I have a question I'm having trouble with

Why do I have to use pi/2 to find the shift between pi/2 and 9pi/8?

where did pi/2 come from?

same in the 2nd question too

maybe my question is too stupid....

i'm so frustrated I'm not going to bed till i figure this out

[fading token]

pi/2 is a quarter period of the regular sine function

[harsh cipher]

@fading token thanks that's what I was thinking too at first

need to try it with different question

[shut sequoia]

I just started pre cal and I'm already confused... were doin arithmetic sequencing and a question i got is "The numbers represented by x, y, and z are the first three terms of an arithmetic sequence. Express Z in terms of x and y" could someone explain this?

[serene heath]

the difference between any 2 consecutive terms is constant

[mossy kestrel]

how do i find a number d such that the line containing the points (d, 4) and (-2,9) has slope -3

[rigid beacon]

@mossy kestrel do you know how to calculate slope given two points?

[mossy kestrel]

i think i got 11/3

-11/3

yes

[rigid beacon]

So you use that formula

And plug in what you know

And solve for the unknown

[neon garden]

Im confused on 33, can anyone help me?

[fluid shore]

Write out f(x) explicitly in both cases, where deg(f(x)) is even and odd

What do you notice about it?

@neon garden

[neon garden]

Uh

I’m not sure what you mean

Is this what you mean?

[fluid shore]

Nope. A general nth degree polynomial is written as follows:

$f(x) = a_0 + a_1x + a_2x^2 +... + a_nx^n$, where $n \in \bN$

[obsidian monolith]

Abhijeet Vats:

[fluid shore]

So consider the cases when n is even and when it is odd

Then prove the given assertion

[neon garden]

Do you mean the formula for even and old polynomials? f(-x) = f(x) and f(-x) = -f(x)

Well when n is even the graph resembles a parabola

and when it is odd it resembles a S kinda

[fluid shore]

No. I mean when n is even and when it is odd.

So suppose that n is even. Then, we can see that $a_1 = a_3 = a_5 =...= a_{n-1} = 0$.

[obsidian monolith]

Abhijeet Vats:

[fluid shore]

So your polynomial reduces to:

$f(x) = a_0 + a_2x^2 + a_4x^4 +...+a_nx^n$

[obsidian monolith]

Abhijeet Vats:

[fluid shore]

And clearly, f(-x) = f(x)

So it’s even

Now, do the same for it when n is odd

@neon garden

[neon garden]

Ok so

I’m not sure how to use the bot so I’ll just send a picture xD

Like that?

@fluid shore

[fluid shore]

f(x) does not have infinitely many terms

No okay, did you understand what I wrote above?

[neon garden]

I think I get the basics

[fluid shore]

No

Did you understand what I wrote above?

[neon garden]

you write a general odd degree polynomial

[fluid shore]

Okay, so where’s your general odd degree polynomial

Write it out explicitly and state that n is odd

You gotta be clear when writing up the proof of anything

[neon garden]

Do you include a subscript 0 for odd?

[fluid shore]

Wot

[neon garden]

$f(x) = a_0

huh

[fluid shore]

That’s not a general odd degree polynomial

Nope. A general nth degree polynomial is written as follows:

$f(x) = a_0 + a_1x + a_2x^2 +... + a_nx^n$, where $n \in \bN$

[obsidian monolith]

Abhijeet Vats:

[fluid shore]

Now let n be odd

[neon garden]

Ik I’m saying do you include that since 0 isn’t even or odd?

[fluid shore]

That’s literally all you need

0 is even and a_0 = a_0*x^0

[neon garden]

Oh I’ve always been told 0 isn’t even or odd

[hard hornet]

33, every coefficient of ODD power of x is 0

$f(x) = a_0 + a_1 x^1 + a_2 x^2 + a_3 x^3 + ....$

[obsidian monolith]

Cytis:

[hard hornet]

that means if every coefficient of an odd power of x is 0

$a_1 = a_3 = a_5 = ... = 0$

[obsidian monolith]

Cytis:

[hard hornet]

then do math

that's it lol

[neon garden]

Isn’t that what I had?

[fluid shore]

I don’t like being interrupted @hard hornet

Okay anyways

Okay anyways

[hard hornet]

You went the right direction, but you have to prove it for any polynomial of any power

[neon garden]

ignore the bottom part

[fluid shore]

No you see

That’s wrong. You need to write out f(x) explicitly

Cos what you’ve written up there doesn’t even resemble a proof

[hard hornet]

it can be an extremely informal ass proof

i dont think he's understanding the question

[neon garden]

Yeah sorry

[fluid shore]

At best, it’s an informal proof

[hard hornet]

his f(x) is wrong to begin with

[fluid shore]

Yes, that’s literally what I’ve said. @neon garden what’s the issue with writing out the general nth degree polynomial as I did?

[neon garden]

I thought you said do it for odd

[hard hornet]

read the question

[neon garden]

I can make a general one if you want

[hard hornet]

READ THE QUESTION PLEASE

[fluid shore]

No. I told you to do it for two cases

When n is even and when n is odd

[hard hornet]

abhijeet you're overcomplicating it

the question only asked for even powers, ONLY EVEN POWERS

m8 no wonder he's confused as hell

legit just read the question, and write out f(x)

and then we can go from there

[fluid shore]

No. The question is saying that the coefficients of the odd terms are 0. It says nothing about f(x) being an even degree polynomial.

[neon garden]

Ok like that?

[hard hornet]

you're giving him too much to do, just help him with his question, that's it

ok yes that's an example of a polynomial

[fluid shore]

@neon garden that’s a specific polynomial

[hard hornet]

now make all the odd power coefficients equal to 0

[fluid shore]

It has to be more general

[hard hornet]

what do you end up getting

abhijeet you're legit making this way harder than it needs to be

[hard hornet]

good

now, try write another different polynomial, and do the same thing

after you've done that, begin generalizing what f(x) really is

if you need help generalizing it, hit me up

[blissful forum]

yawn

[fluid shore]

I mean, it’s not really correct to write up a rough proof like that

But sure

Whatever you say

[hard hornet]

THIS IS PRECALCULUS, NOT UNIVERSITY BRO

[blissful forum]

yawn

[fluid shore]

Fine fine

[neon garden]

When you do the same thing with a different polynomial do you mean make the polynomial less general?

Like a specific example?

[hard hornet]

do a couple more examples

[neon garden]

Like that?

[hard hornet]

and then see if you can write f(x) in a more general form

odd powered coefficients = 0

what you did before this was fine

[neon garden]

Oh ok

[hard hornet]

i know this might be a bit hard, but when i mean generalize, I mean f(x) has some form where it represents all possible polynomials in your question

$f(x) = a_0 + a_2 x^2 + a_4 x^4 + .... $

[obsidian monolith]

Cytis:

[neon garden]

Yeah this is a bit confusing cause I’m not even taking pre-calc this year, just trying to prepare for next year

[hard hornet]

all good all good man

what I typed above is a polynomial with even numbered coefficients

right?

[neon garden]

Yeah

[hard hornet]

notice the .... near the end

it means it can end on x^6, or it can end on x^100

the only thing you need to know is that f(x) is some polynomial

it's just some polynomial, we don't know exactly what it is, but it's just a polynomial

[neon garden]

Ok

[hard hornet]

now, a neat trick we can do to help us generalize this is use some sort of substitution

this might be new to you, so i'm going to help you a bunch

we can transform what we had above to this

$f(x) = a_0 + a_2 x^2 + a_4 x^4 + .... + a_{2n} x^{2n}$

[obsidian monolith]

Cytis:

[hard hornet]

where n is some number

we don't care what number it is

we just know it's an integer

it could be 1, it could be 4, it could be w/e

now notice that this function, generalizes all polynomials with even power

because any even polynomial can be represented by this function, by plugging in a specific value of n

[neon garden]

Oh ok

[hard hornet]

if we wanted a polynomial to the 10th power, we just choose n = 5

does the generalization make sense?

[neon garden]

Kinda

so how would you do odd then?

[hard hornet]

an even number is always in the form of 2n

we can say that any even number is equivalent to 2n

i.e plug in any value of n, and you get an even number right?

what would an odd number be?

give it some thought

come up with an equation that always spits out an odd number

my equation y = 2n, always spits out an even number for ANY n

n = 3, y = 6, 6 is even
n = 10, y = 20, 20 is even

[neon garden]

2n + 1?

[hard hornet]

well done

how'd you come up with that?

[neon garden]

oh wow xD

it was kinda a guess

[hard hornet]

put some faith in urself hahaha

educated guess?

[neon garden]

yeah you could say that

[hard hornet]

there had to be some intuition behind that guess

tell me, how'd you get that

[neon garden]

I choose 6 for an example, and thought of ways to make it odd. If I doubled it and added one it worked

then tried it with 8

and it worked

So I just went with it

[hard hornet]

good job

that's one way to do it

now, knowing that any odd number is in the form of 2n+1, can you give me a polynomial with odd numbered power only?

generalized

take your time, i won't give u the soln this time

ping me back when u got something

[neon garden]

@hard hornet

[hard hornet]

good fucking job

[neon garden]

Thanks

[hard hornet]

now, back to your question

[neon garden]

Alright

[hard hornet]

the even powered polynomial

$f(x) = a_0 + a_2 x^2 + a_4 x^4 + ... + a_{2n} x^{2n}$

how do you know if a polynomial is even or odd function?

[obsidian monolith]

Cytis:

[neon garden]

The equations f(-x) = f(x) and f(-x) = -f(x)

[hard hornet]

one of them shows even function, the other shows odd function, which is which?

[neon garden]

the first one is even

[hard hornet]

good

now, in the general even number powered function, plug in f(-x)

and see what you can do from there

write out the eqn

and show me your work

[hard hornet]

Ill brb in like 5 mins

Battery dyin

[neon garden]

Alright

I’ll be right back too, gotta eat quick just ping me if that’s right

@hard hornet

[hard hornet]

for the last line, explicitly plug in the -x, i.e

$f(-x) = a_0 + a_2 (-x)^2 + a_4 (-x)^4 + ...$

[obsidian monolith]

Cytis:

[hard hornet]

@neon garden

and then from there, use logic and math to evaluate all the individual x's, and prove that f(x) = f(-x)

[neon garden]

Alright sounds good, thanks

[viscid thistle]

hey i got this sat question and it asks

P=S/(S+F)
A tennis league player uses the formula above to determine a player's first serve practice score, P, based on the number of successful first serves, S, and the faults on first serves, F. Which of the following expresses the number of successful serves, in terms of F and P?
A. S=FP/(1-P)
B. S=FP/(P-1)
C. S=F/(P-1)
D. S=F/(1-P)

if anyone could help me

and maybe explain id

be grateful

[clever inlet]

🤔

[kindred valve]

i'm really confused with the whole idea of doing synthetic division with complex numbers

[full garden]

hey guys I am doing this half life equation and I am required to divide 0.6931/the half time

my teacher said 0.6931 is the natural log

but what is this number the natural log of

because I was trying to get that number 0.6931 from my calculator and I couldn't

can someone please tell me how to get it from the calculator

[pale kettle]

This is the natural log of 2

[vagrant stirrup]

how do i maximize voulme while maintaing surface area

for optimization

[vernal moon]

Maintain surface area?

Sphere oft

Ify

[full garden]

yeah bro I am so stupid idk how to do natural log on my calclator

[vagrant stirrup]

clyinder

@vernal moon

[vernal moon]

Ok

[vagrant stirrup]

i found a video

i wil lwatch it now

ty tho

i have a group project im going to have to carry :((

[full garden]

guys how do I do this on my Casio calculator

[vernal moon]

Do

Ln(2)

Or log base e of 2

[full garden]

thank you bro

[vagrant stirrup]

how do i maintain the volume and minimize cost of a package

my volume has a radius of 2 and height of 4 of a clyinder

[fleet yew]

what volume

[vagrant stirrup]

volume of my clyinder

[fleet yew]

ik but what volume

[vagrant stirrup]

= 75.398223686155 centimeters3

r is 2 and height is 6 not 4

so u would get about 75

[fleet yew]

so we're dealing only with cylinders

[vagrant stirrup]

yeah

Top/bottom of cylinder is .05 cents
Sides of cylinder is .03 cents

[fleet yew]

and you are trying to maximize volume relative to ?

[vagrant stirrup]

no maintain voulme while minizming cost

[fleet yew]

what does minimizing cost mean in this scenario

[vagrant stirrup]

oh ok

let me give some context

so basically i have a math project and i need to do 3 re designs and see which is the best one for my project

in this case i am using can soup

[fleet yew]

uh that's not enough info to help you

what exactly are you trying to accomplish

[vagrant stirrup]

can i just send u the pic, im not sure wht u mean

[fleet yew]

sure yeah do that

[vagrant stirrup]

ok i will dm it to u

[kindred valve]

my math teacher every once in a while talks about the dreaded chapter 7, and i always got confused on what it was. I found out its trigonometry but I havent done that yet so i dont even know how hard it is

like is it harder sohcahtoa

[rigid sun]

its soh cah toa except easier

because you can use a calculator

and literally program every single function/ operation you need on there

sss triangle? done

sas triangle? done

equivalent sin angle? done

the only hard part is correctly coding your stuff

but once that is done you're set

literally press 7 buttons and get the answer instantly

the nice thing about all of it is that you have to know how to do the problem in order to program the computer to make solutions

so its good practice too, you're not missing out on the content

[harsh cipher]

Hi

Question

similar to yesterdays question....

why cant I write the equation y= -15cos 2pi?

instead we have to find "b" value

oh...if I want to write the answer in minutes i can leave it as 2pi and if I wanted to write it in seconds I have to do pi/30.

because period = 60 seconds in this question.

ha!

[elfin cradle]

hi is anyone here to help me really quickly?

i'm studying for a test and i'm stuck on one problem

"find all the solutions to the following equations."

x^4-i=0

i believe i'm supposed to change it to trig form and use demoivre's theorem?

[valid violet]

Do you know what roots of unity are

[elfin cradle]

nope, didn't learn that

[valid violet]

Oh

Do you know how to define sqrt(i)

[elfin cradle]

yeah

i think so

[valid violet]

How would you?

[elfin cradle]

i'm not exactly sure what define means

sorry

[valid violet]

$\sqrt{i}$

[obsidian monolith]

gfauxpas:

[valid violet]

Do you know what that means?

[elfin cradle]

i'm just guessing it is sqrt of -1 inside a sqrt

unless i am completely wrong

[valid violet]

Could you tell me its value without a calculator?

I'm assuming you're not allowed a calculator

[elfin cradle]

none of our problems in this chapter use imaginary numbers

other than this trick problem

so we get to use a calculator

[valid violet]

I see

[elfin cradle]

this unit is on vectors which i have down, trig form, demoivres

oh nevermind, scratch that, we have to ochange complex numbers into trig form as well

[valid violet]

I'm not sure how to answer it if you dont know roots of unity and you don't know sqrt(i)

Im thinking

[vernal moon]

Can I ask

[elfin cradle]

okay so the teacher gave us the answers

[vernal moon]

Why the fuck would you know how to work out the sqrt if i in precalc

[elfin cradle]

but none of my class or i know how to get it

[vernal moon]

Like I’m serious

[elfin cradle]

cos(pi/8)+isin(pi/8)

is the first one

we're supposed to throw that in a calculator for decimal answer

[valid violet]

I know the answer, I'm trying to figure out how you can know the answer

[vernal moon]

Reasonably speaking, the reasons for the trigonometric representations of imaginary numbers and shit

That is only probable

Provable

With calculus

[valid violet]

Because do you agree that

[elfin cradle]

my classmate is joining here now

[valid violet]

$x^4-i=(x^2+\sqrt{i})(x^2-\sqrt{i})$

Is this true?

[vernal moon]

Look

Look

I agree

But

[elfin cradle]

yes

i believe so

[valid violet]

Eer no

Mistake

[elfin cradle]

oh

yeah

[vernal moon]

I just feel imaginary numbers don’t belong in precalc

[obsidian monolith]

gfauxpas:

[vernal moon]

I’m not a teacher

[elfin cradle]

yeah thats true

my bad

[vernal moon]

I mean

You can

Work out the sqrt of i

[valid violet]

Okay so this is easier to solve now that it's factored ... but you don't know how to find sqrt(i)

[vernal moon]

hmm

[elfin cradle]

yeah we never learned that

[vernal moon]

Ya

[valid violet]

So idk how to help you

[elfin cradle]

like double sqrt

[vernal moon]

You gonna use algebra

Ik

What u can do

The sqrt of i will be a complex number with form a+bi

(a+bi)(a+bi)=i

[elfin cradle]

okay

[vernal moon]

Therefore

[valid violet]

Actually hold up

[vernal moon]

a^2-b^2+2abi=i

[valid violet]

Aleph stop

[vernal moon]

Ok

[valid violet]

You know de moivre?

[elfin cradle]

yeah

[valid violet]

So

You can find out

[vernal moon]

Uhh

I may know it under a different name

[valid violet]

This is a bit non rigorous

[vernal moon]

Yeah ik it

[valid violet]

But you don't have enough knowledge for full rigor

[vernal moon]

Just didn’t know it’s name

[elfin cradle]

i'm confused

[valid violet]

Sqrt(i)=i^(1/2)

So find the polar form of i

[elfin cradle]

i don't know what that is

the polar form

[valid violet]

Trig form

[elfin cradle]

ah

would that be 1

[vernal moon]

Uhh

[valid violet]

Is 1²=i?

[elfin cradle]

my Friend made me say that

-1

[vernal moon]

Lmao

[elfin cradle]

hes on vc with me

blocked on this server

[valid violet]

Is (-1)²=i?

[elfin cradle]

yes

no

no

[valid violet]

Lmao

[elfin cradle]

-i

[harsh cipher]

lol

[vernal moon]

Why is your friend banned

[elfin cradle]

he just joined less than 10 min ago

okay but back to math

[valid violet]

When you suggest-i, what am i going to ask you

Based on your knowledge of what i have asked you in the past 2 minutes

[vernal moon]

(-i)^2 does equal 1 tho

[elfin cradle]

fax

[valid violet]

We want sqrt(i)

[elfin cradle]

okay

[vernal moon]

Yes

I still feel like

My method would work

[valid violet]

It might

You can try it

[vernal moon]

Nah I’m lazy

And on the bus

[elfin cradle]

could we try the next problem

it's similar

x^5+243=0

[vernal moon]

Ok

You’ll have 5 roots

Each 2/5pi radians apart

[elfin cradle]

would i use demoivres for this one?

or the complex root formula

[valid violet]

If you find the trig form of i

[kindred valve]

really quick question guys

if (i) is a zero

[valid violet]

Then use de moivre to find i^(1/2)

[kindred valve]

what is the root for that

[valid violet]

What does that question mean

[kindred valve]

for example

it wants me to make an equation that is degree 4 with the zerios being i and 1 + 2i

[valid violet]

If a is a root, (z-a) is a factor

For polynomials

[kindred valve]

it doesnt want complex numbers in the factor

so like x^2+4x+4 is the same as 2+i

[elfin cradle]

|z|(cos(θ)+isin(θ))

[valid violet]

So you want a polynomial with real coefficients

?

[kindred valve]

yeah

[elfin cradle]

oh i think i got it

[kindred valve]

sorry i dont really know how to explain it

[valid violet]

Recall those have the property that

If a is a root

[elfin cradle]

cos(pi/2)+isin(pi/2)

[valid violet]

a* is also

Conjugate

[kindred valve]

oh yeah

[valid violet]

What happens if you square that cat

[kindred valve]

cat?

[valid violet]

Catify

[elfin cradle]

i^2

[kindred valve]

oh

[elfin cradle]

so 1

[kindred valve]

-1

[elfin cradle]

-1

[valid violet]

Not what we want then

I'll help you out

[elfin cradle]

what if we took the square root of the whole equation

[valid violet]

cos(pi/2)+i sin(pi/2)=i

[elfin cradle]

yeah

[valid violet]

1/2 power?

[elfin cradle]

yeah

[valid violet]

Is

?

[elfin cradle]

yeah

oh

hold on

pi/4

[steel wigeon]

yeah so it would be like 1(cos (pi/4 + sin(pi/4)i)

yuh

[valid violet]

Is this your friend

[elfin cradle]

no

[steel wigeon]

hello

[valid violet]

Okay

Back to this then

$x^4-i=(x^2+\sqrt{i})(x^2-\sqrt{i})$

[obsidian monolith]

gfauxpas:

[valid violet]

Now

This has 4 solutions because

[steel wigeon]

would you plug it in

[vernal moon]

Plug what in?

[valid violet]

These are the difference and sum of squares

[steel wigeon]

sqrti for what we just did

or something

[vernal moon]

Uhh

[valid violet]

You will need to figure out the 4th roots of i

[elfin cradle]

no

[valid violet]

To figure out sqrt(sqrt(i))

Think about it

[elfin cradle]

pi/8?

oh nvm

[vernal moon]

Well

[valid violet]

That's theta for i^(1/4) sure

[vernal moon]

Close

[valid violet]

How do you factor (x²-a²)?

[elfin cradle]

)x+a()x-a(

[valid violet]

Close enough

So that's how you will factor (x²-sqrt(i))

How do you factor

(x²+a²)?

[elfin cradle]

(x+isin(pi/4))(x-isin(pi/4)

[vernal moon]

Uhh

[valid violet]

I like your spirit

But it's incorrect

[elfin cradle]

am i hot or cold

[valid violet]

Warmish

(x²+a²)=(x+ai)(x-ai)

Here still a²=sqrt(i)

This method will get you the 4 roots without using roots of unity

It's kind of unfun

[elfin cradle]

would x^5+243=0 be done in a similar way

[valid violet]

You know how to find the real root

[elfin cradle]

possibly

[valid violet]

So you divide that by (x-a)

To get a degree 4 polynomial

[elfin cradle]

what does a mean

[valid violet]

Then it becomes the same

a is the name i gave to the real root, sorry

[elfin cradle]

oh okay

[valid violet]

The real root is -(243)^(1/5),

Cant do it without a calculator. I'm calling it a

Then (x^5+243)/(x-a)= a degree 4

[vernal moon]

Man this is why doing it after precalc makes sense tbh

You can use all the tools calc gives u

[valid violet]

Hopefully the next time you need to do these will be after you learn roots of unity bc

[elfin cradle]

i'm too bad at math to be in precalc

[valid violet]

^

[vernal moon]

Uhh

[valid violet]

This isn't even really precalc

[elfin cradle]

thank you for helping us out tonight

[vernal moon]

Yeah

[valid violet]

Yw

[steel wigeon]

ok

so

x=(cos5pi(k)/8 + isin5pi(k)/8)

is what i got

for x^4 - i = 0

[elfin cradle]

that's the second root

right

[steel wigeon]

k is what root u wanna find so if u wanna find 2 you plug in k for 2

[elfin cradle]

but how do you get cospi/8+isinpi/8

that should be what u get when plugging in 1

[steel wigeon]

from the complex root formula

x^4 = the trig form of i

then you solve for x using the complex root formula

[elfin cradle]

yeah no i can't do this

[valid violet]

It's unfair to ask you to do this without roots of unity, it's so much easier

I did my best without it but its not the correct way to do these problems

The correct way is to use roots of unity

I tried

It is indeed much harder this way

[steel wigeon]

yeah well

this is all we got

trig form demoivres and the complex root equation

[valid violet]

the answers are going to be in terms of sqrt(i), there's no way around it

and you'll have to find sqrt(sqrt(i)) with my method

which is i^(1/4)

you can try quadratic formula twice but it will be the same

quadratic in x^2 and then again quadratic in x

but the answers will be the same

in terms of sqrt(sqrt(i))

what's the complex root equation

[steel wigeon]

i dont know how to write that here

its like

r to the nth root(cos((theta + 2pi(k)/n)+isin((theta + 2pi(k)/n))

[valid violet]

that's kind of the roots of unity method except harder

[steel wigeon]

well

[valid violet]

try it

[steel wigeon]

yeah i did thats the answer i got above

like 5pi(k)/8

apparently its right

[valid violet]

actually is the roots of unity method

just a little roundabout

[steel wigeon]

huh

interesting

[valid violet]

roots of unity method is

any number^n = number^n 1^n

so the nth root will be a constant times the solutions to z^n = 1

these soluitions are called the nth roots of unity

[steel wigeon]

yeah i have no idea what that is

i guess i dont really need to for the time being anyway

[valid violet]

:<

[steel wigeon]

[maiden furnace]

Hello I need some help the problem asks to factor complety give zeros and the multiplexing how would i go around solving this

https://gyazo.com/fa99c65e135fb1c21ae5476d1eb44f99

[rigid sun]

tf

multiplexing?

as in like

telecomm?

we don't even use analog anymore

[coarse storm]

@maiden furnace , I always recommend performing a quick substitution of x=1 and -1

[rigid sun]

@maiden furnace

if even coefficeints add up to opposite values

[maiden furnace]

multiplicity my bad

[rigid sun]

then +1 is a solution

you can approach by u sub

or factoring

or

quartic formula

jk

[maiden furnace]

can we go by the factoring way

[rigid sun]

yes

[coarse storm]

You can do synthetic division

It is essentially factorising, however it is easier

[rigid sun]

@maiden furnace this trick will help u in the long run

if the odd and even power coefficients add up to opposite values 1 is a zero

if the odd and even power coefficients add up to the same value, -1 is a zero

so lets look at this problem

x^4, x^2, and -4 are all even powers

x^4 has a 1 coefficient

x^2 has a 3

-4 has a -4 obivously

1+3-4=0

[viscid thistle]

how do i solve cos6x=cos4x?

[rigid sun]

bruh

ok so the even coefficients add up to 0

but there are no odd coefficients

[viscid thistle]

wait wot

[rigid sun]

in other words, the coefficient on all the odd terms are 0

[viscid thistle]

there are no coefficients?

[rigid sun]

no u

na ge

bi zue jager

[viscid thistle]

tf

[rigid sun]

anyway, since both the even term sums and odd term sums are equal

[viscid thistle]

plz help me too mathman?

[rigid sun]

you can be sure that 1 is a solution to the polynomial

[coarse storm]

@viscid thistle

[viscid thistle]

ye

[maiden furnace]

so -1 is a zero thefore x+1 is a factor?

[coarse storm]

Please wait until PhD is done

[rigid sun]

no

[viscid thistle]

yes sir

[rigid sun]

the factor is (x-1)

and the solution is +1

[maiden furnace]

but it asks for all zeros

[rigid sun]

yes

so divide that polynomial by (x-1)

and then make sure you write down +1 as a zero and save it for you final answer

[viscid thistle]

u doone phd?

done*

so cos6x=cos4x idk hwo to approach this

[rigid sun]

reeeeeeeeeeeeee

[viscid thistle]

maybe a double angle formula

or sum and difference

[rigid sun]

move to the questions channel

[viscid thistle]

phd

can I fucking call u

instead of hopping around a fuck ton of channels and typing a lot

@willow bear plz help ur a math queen

[willow bear]

hhh

what do you need help with

[harsh cipher]

🤣

[willow bear]

@viscid thistle

[viscid thistle]

plz

i nee to solve it

cos6x=cos4x @willow bear

idk what tto do

i used rocket science

and I used missle codes

nothing worrks my guy

[vernal moon]

@harsh cipher why do u got an aleph in ur name

[willow bear]

@viscid thistle sorry for the sudden 10 minute disappearance, but i'm not a guy and would prefer not being called one

[viscid thistle]

my ma'am*

I actually have like 4 of these problems and I need help on all of them

can I dm u? or u want it here

[willow bear]

let's keep it here

so are you not able to use the formula $$\cos(u) - \cos(v) = -2\sin\paren{\frac{u+v}{2}}\sin\paren{\frac{u-v}{2}}?$$

[obsidian monolith]

Ann:

[viscid thistle]

oh fuck those

omg those suck to remember

lemme try

I get -2sin5xsinx

what I do now?

@willow bear

[willow bear]

no, you get -2sin(5x)sin(x) = 0.

[viscid thistle]

Yea that’s what I meant

[willow bear]

after this, all that remains is to solve two smaller equations: sin(5x) = 0 and sin(x) = 0, and take the union of their solution sets.

which i hope is obvious.

[viscid thistle]

Is C just 0?

X*

Cause any radian answer would fuck up sin(5x)

@willow bear

[willow bear]

no, 0 is by far not the only solution

[viscid thistle]

Uhh

I’m brain dead rn

[willow bear]

and 0 isn't even the only value of x which makes sin(x) = 0

[viscid thistle]

Well I know pi/2 and 3pi/2

But like sin(5x)

[willow bear]

no

neither sin(pi/2) nor sin(3pi/2) is zero

tell me, are you sleep-deprived right now

[viscid thistle]

Uhh

Kinda

But not really

[willow bear]

...

this is going to be difficult

but okay, i guess

let's focus JUST on sin(x) = 0 for now

can you solve this equation

on the entire real number line obviously

[viscid thistle]

Yes

Real

So pi/2 and 3pi/2

Make it 0

Right

[willow bear]

no they do not

[viscid thistle]

Oh FUCk that’s cos

[willow bear]

since when is 1 = 0

or -1 = 0

[viscid thistle]

I mean 2pi and 0

Shit

[willow bear]

i want you to solve the equation ON THE WHOLE NUMBER LINE. there are INFINITELY MANY SOLUTIONS.

[viscid thistle]

2pi(k) k=integer

[willow bear]

no

you're still missing out

[viscid thistle]

Pi(k)

[willow bear]

in fact you're missing out on infinitely many solutions

[viscid thistle]

K=integer

Right

[willow bear]

yes. $\sin(x) = 0$ has solutions $x = \pi k$, where $k$ runs over the integers

[obsidian monolith]

Ann:

[willow bear]

great

now

waht about sin(5x) = 0

you've already almost solved this equation too

[viscid thistle]

Uhhh 0

[willow bear]

no!

look.

this is

sin(something) = 0

[viscid thistle]

SHIT!

[willow bear]

you already know what the input of sin must be for the output to be 0

it must be pi times an integer

[viscid thistle]

Yea

But there’s a 5

[willow bear]

$\sin(5x) = 0 \ 5x = \pi k$

[obsidian monolith]

Ann:

[willow bear]

so what if there's a 5?!

[viscid thistle]

Pik/5

[willow bear]

there we go

[viscid thistle]

Wait is that the answer

Wooo

[rigid sun]

k

[willow bear]

yes, because anything of the form πk can also be expressed in the form πk/5

[viscid thistle]

Is that so🤔

[willow bear]

πk = π(5k)/5

[viscid thistle]

So sin(x)=0 has the solutions pik/5 technically

[willow bear]

no

it doesn't

[viscid thistle]

Ya

So like

[willow bear]

sin(1x) = 0 only has solutions πk

sin(5x) = 0 has solutions πk/5

[viscid thistle]

What’s the solution to the whole thing then

[willow bear]

it's just that any solution of the former is also a solution of the latter

which in general need not happen

but it does happen here

[viscid thistle]

So

@willow bear

I got three more 😅

[willow bear]

the solution set of $\sin(x) = 0$ is ${ \pi k \mid k \in \bZ }$

the solution set of $\sin(5x) = 0$ is ${ \tfrac15\pi k \mid k \ in \bZ }$

the solution set of $\sin(x)\sin(5x) = 0$ is the union of the two sets above, except that the former is a subset of the latter, so the union is equal to the latter, and so the answer is $x = \frac15\pi k, k \in \bZ$.

[obsidian monolith]

Ann:

[viscid thistle]

I see

I rmeber something like this in class

[willow bear]

ok so what other problems do you have

[viscid thistle]

The math ones?

[willow bear]

while i'm still here and available to help you

[viscid thistle]

Ok lemme get picture

We already did the first one

[stuck lark]

,rotate 180

[willow bear]

,rotate 180

[willow bear]

HSGUISDJKFGHKJLSDFJHKLSDJFGL

[stuck lark]

[viscid thistle]

Can we do 34?

[willow bear]

k

do you mind if i replace theta with t for ease of typing

[viscid thistle]

Ok

[willow bear]

so you have

-2 sin(2t) = sqrt(3) sin(t) - 3 sin(2t)

and there's a simple algebraic move that should jump out at you first thing

[viscid thistle]

Factoring

After u bring it one side

[willow bear]

no

well

ok

"bring it to one side"

ugh

but fine

[viscid thistle]

😮

[willow bear]

i don't like the wording "bring <blah> to <side>"

like

no, you're adding shit to both sides, you're just having things cancel out on one of them afterwards

anyway whatever

[viscid thistle]

Lol

[willow bear]

this will yield sin(2t) - sqrt(3)sin(t) = 0

and now the presence of sin(t) along with that sin(2t) should be a BIG hint as to what one can do with sin(2t) to make it more amenable to manipulation

[viscid thistle]

Wait what

[willow bear]

what

[viscid thistle]

I don’t think those are multiplied

[willow bear]

you don't think what is multiplied

[viscid thistle]

That’s what I got

[willow bear]

you fucked up the subtraction and haven't answered my question

-(a-b) is not the same as -a-b

[viscid thistle]

Oh wait I forgot a plus sign

[willow bear]

-2sin(2t) + 3sin(2t) = ?

[viscid thistle]

Is that it then

[willow bear]

-2sin(2t) + 3sin(2t) = ?

[viscid thistle]

sint

Sin2t*

[willow bear]

yes so

this will yield sin(2t) - sqrt(3)sin(t) = 0

[viscid thistle]

Ye

[willow bear]

and now the presence of sin(t) along with that sin(2t) should be a BIG hint as to what one can do with sin(2t) to make it more amenable to manipulation

[viscid thistle]

Double angle formula

[willow bear]

well then

why don't you tell me what applying that will result in

[viscid thistle]

Uhh 1 sec

That

[willow bear]

why do you insist on running ahead

[viscid thistle]

Fun?

I also ran around 7km today

[willow bear]

no, not that kind of running

i'm trying to take you through this problem step by step but you insist on going way ahead and inevitably fucking something up

at least let me know you're planning on doing that

[viscid thistle]

Ok

😢

[willow bear]

you've fucked up again in two ways

first off

[viscid thistle]

I meant one of them to be 11pi/6

[willow bear]

sin(t) = 0 does not have only 0 and 2pi as solutions

and cos(t) = sqrt(3)/2 does not have only pi/6 and 11pi/6 as solutions

[viscid thistle]

The instructions ask for only answers restricted to (0,2pi]

[willow bear]

(0, 2pi]? not [0, 2pi)?

[viscid thistle]

I mean the second one

[willow bear]

why do you write (0,2pi] when you mean [0,2pi)

anyway

there is STILL a solution you missed

[viscid thistle]

:0

[willow bear]

AND a solution you included that is outside the range you just specified

sin(pi) = ??????????

[viscid thistle]

Wot?

[willow bear]

can you tell me what sin(pi) is

[viscid thistle]

Shit I meant that

Not 2pi ://

My trig test is tomorrow 😰

[willow bear]

well gee oh my

yknow what your biggest problem is

it's rushing ahead

[viscid thistle]

No focus

[willow bear]

your INSISTENCE on rushing ahead

take this shit slowly

one step at a time

[viscid thistle]

Ok

[willow bear]

you keep making careless mistakes

you're perfectly capable of doing all these problems, i know that for a fact

you're just going too fast and that's what is fucking you up

[viscid thistle]

U want me to try 35?

[willow bear]

sure go right ahead

[viscid thistle]

B

@willow bear

[willow bear]

bad

right from the get go you have fucked yourself up with potential extraneous solutions

[viscid thistle]

😅

I mean I square it so it goes away

[willow bear]

no

squaring both sides of an equation almost always produces extraneous sols

[viscid thistle]

Damn

I should check for those

[willow bear]

no

in all honesty that'll just be a giant waste of time

why not write sin(t) as 2sin(t/2)cos(t/2)

[viscid thistle]

Wtf

What formula is that again

[willow bear]

that's

the double angle formula

the very same that you mentioned earlier

t = 2(t/2) after all

[viscid thistle]

It’s weird to read it through text

[willow bear]

...

how else would you possibly read it

[viscid thistle]

Handwritten

Wait but my solutions were right

How come I didn’t get any extraneous solutions?

[willow bear]

maybe you just lucked out

but you shouldn't risk it

[viscid thistle]

What should I do instead

[willow bear]

why not write sin(t) as 2sin(t/2)cos(t/2)

[viscid thistle]

I mean it’s not one of the formulas she gave us

It look like a fraction double angle formula for sin2t

[willow bear]

t = 2(t/2) after all

take the formula sin(2x) = 2sin(x)cos(x) and replace x with t/2

[viscid thistle]

Ok

Like that?

[willow bear]

no

first off

you can't divide by cos(t/2)

you'll lose solutions

GUARANTEED

and second, even if you DID divide by cos(t/2)... what would remain on the other side is not 0

[viscid thistle]

Oh shit

I forgot

Ignore that

That then?

[willow bear]

there we go

[viscid thistle]

But like

I have half angles now

I need to sleep soon can we do 36 real quick too

Plz 😢

[willow bear]

ugh