help-10

[round nova]

NP GLAD I COULD HELP

[dusty relic]

.close

[obtuse pebble]

Channel closed

Closed by @dusty relic

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[finite stirrup]

I'm currently reading a section of a book on graded modules, and I'm a bit stuck. I have a ring R and two R-modules M and G. I'm told that I have a structure of G-module on Hom(M,G), but I don't see it.

For Hom(M,G) to have a structure of G-module, I need to define a group morphism from G to Aut(Hom(M,G)). At first, I thought "well, that's easy":

1. Let us denote the operations on M and G as +.
2. For two R-module morphisms f,g: M -> G, let f+g be the morphism such that (f+g)(m) = f(m) + g(m). Then, because G is abelian, f+g is still a R-module morphism, so we now have a structure of group on Hom(M,G).
3. For an element h of G and a R-module morphism f: M -> G, let g·f be the morphism such that g·f(m) = g + f(m). Then, 0·f = f and (g+h)·f(m) = g+h+f(m) = g + (h·f)(m) = (g·(h·f))(m), so G acts on Hom(M,G).
4. There's where problems start… For two R-module morphisms f,g: M -> G, h·(f+g)(m) = h + (f+g)(m) = h+f(m)+g(m), which is not equal to (h·f)(m) + (h·g)(m).

I thought of using the action by conjugaison (g·f)(m) = g + f(m) - g, but as a module is an abelian group, that's just the identity… I feel like I'm missing something obvious here

[vernal wigeon]

can u send that part of the book?

[vernal wigeon]

i think the confusion comes from « from \Lambda modules … we can form the G-module … », its a typo it should be a \Lambda-module and not a G-module

or maybe i’m missing smth

[finite stirrup]

Yeah, this is obviously a \Lambda-module, and then I learned that you can make a module with a group instead of a ring, and now I'm confused…

[vernal wigeon]

in your book G isn’t a group acting on anything, it’s just a Λ-module and your step 4 already shows the additive structure of G doesn’t give an action so i think yeah it’s a typo

[finite stirrup]

okay, thanks ^^

.close

[obtuse pebble]

Channel closed

Closed by @finite stirrup

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[zenith crest]

${A \subseteq \mathbb{R} : A$ is countable}

[warm shale]

toast
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[zenith crest]

Trying to find the cardinality of this set

[zenith crest]

is the idea to maybe make injections?

again lower bound is kind of easy to solve

m

i think this help channel broke

[obtuse pebble]

@zenith crest Has your question been resolved?

Channel closed

Closed by @zenith crest

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[celest aspen]

help,is there anyone who have practise about sin cos tan?

[frail vigil]

Do you have a specific question?

[grave quail]

do you meanyou want practice questions?

[celest aspen]

y

[dense yoke]

you should probably do a quick Google search for basic trigonometry practice questions in that case.

[celest aspen]

basic i can do but i can't find tricky one

[dense yoke]

what kind of tricky one

[thick oracle]

you dont need tricky ones bro you barely know what sin is

[celest aspen]

i don't know if there is a good site or place to find

[merry plume]

like idk the level you are at, because some trig questions need extra formulas to solve

[dense yoke]

just search. there's bound to be plenty

[regal hazel]

yea whts up

oh so u need practice material?

[grave quail]

how much trig have you done sean? do you know anything else other than sin, cos, tan?

[celest aspen]

i am at middle school grade 2

[frail vigil]

Try school textbook questions

[regal hazel]

yea do you need inverse trig?

or basic trig

[regal hazel]

or ask ur teacher for extra material

local books nd publishers

[celest aspen]

i done all the question on my text book

[regal hazel]

depends on the region

[regal hazel]

Try RD Sharma, heard its a good book

[celest aspen]

my teacher told me to do all of them today at school

[mystic folio]

There're many videos online about it

[merry plume]

do u mind sending the tricky question from ur book so i can see what level it is

[frail vigil]

If a textbook does not have questions, id be concerned

[merry plume]

u would be surprised to see some textbooks

[celest aspen]

ok

[regal hazel]

I think he needs mediocre practice with trignometry, so he gets fluent with it, am I right @celest aspen ?

[celest aspen]

also

[regal hazel]

Hmm

[grave quail]

do you know the course and chapter name?

[regal hazel]

Basic trignometry i think

[grave quail]

what does the front cover look like?

[vital herald]

hmm

[regal hazel]

do you know any formulas? @celest aspen

what formulas you know about trignometry

[grave quail]

they learned sin was a function half an hour ago

[celest aspen]

i know what is sin cos and tan

[regal hazel]

yk these ones?

[merry plume]

i believe this is too far

[celest aspen]

the second day of this chapter at school

[grave quail]

sean what are you studying for? what is the final exam called?

[regal hazel]

Yea i think so asw, it seems like the basic introduction

[merry plume]

u learnt that?

[regal hazel]

I think he has jst learnt abt trig angles

[grave quail]

he's just learned how to find sin cos tan

[merry plume]

ye i dont think it even involved pythagoras yet

maybe it does

im not sure but i dont see the questions needing that

[grave quail]

@celest aspen do you know pythagoras theorem?

[celest aspen]

yes

[celest aspen]

learned a month ago

[merry plume]

have u ever used pythagoras theorem in one of those questions about trig

[celest aspen]

yes

[regal hazel]

yk sin²x + cos²x = 1?

[celest aspen]

i an in dse course of hk

[merry plume]

i was also

[regal hazel]

alr whts tht

[merry plume]

i graduated 6 years ago

its a secondary school like board exam, called hkdse

[regal hazel]

ah

wait yk boards?

[merry plume]

basically same as a levels in other countries or ib

[merry plume]

wdym

[regal hazel]

Ur the first person im seeing who knows 'board' exams

most are clueless

[merry plume]

its because i have friends who took this

[regal hazel]

I see

anyways @celest aspen I can give u basic sums ifydm

[merry plume]

so basically you are form 2 or form 4? if you let me know i can suggest you a book

[celest aspen]

form 2

[grave quail]

dont confuse them with other trig formulas they barely know the basics

[regal hazel]

yea yea ill keep it to triangles nd sides only

[celest aspen]

do you have tricks on using calculator?

[regal hazel]

theta nd stuff

[regal hazel]

Scientefic calculator

but you dont need it

rn

the trig you're studying rn, is rlly simple

yk domains of the functions you're studying?

[merry plume]

if im not wrong they can use calculator

[regal hazel]

trig graphs?

[celest aspen]

i use CASIO fx-3650p super-fx

[regal hazel]

idk man in my country calculators are banned

[merry plume]

but the calculator is gd enough for basic

[regal hazel]

I have not seen the face of a calculator since years

[merry plume]

i just know cause i did hkdse same as what he will do

[regal hazel]

i see

[merry plume]

bro that seems hard

[regal hazel]

cuz it is 😭

you get to know abt nuclear phy nd atoms nd shi

when ur 17

or 16

you get to know abt generators, transformers, Permu nd combi, binomial theo, calculus, semi conductors

when ur 15-17

idk if thts tuff but yea thts most of the syllabus

nd thts only phy nd math we study

chem is pretty simple

chemical kinetics, biomolecules, electrochemistry

and organic

with tonnes of reactions

[merry plume]

i suggest since you are form 2, just go get those exercise books, because i also couldnt find online materials in that form, basically like those summer exercise books from like jp books

[merry plume]

is that math or like phy and chem

[regal hazel]

thts math nd phy

generators transformers semiconductors, nuclear phy, atoms are phy

and there's lot more

faraday's laws

maxwell nd ampere

[merry plume]

i remember if you go to the counter u can ask the person working there for like summer exercise books

[regal hazel]

induction of current nd stuff

[grave quail]

you should probably take this convo somewhere else, its not really contributing to sean's request

[regal hazel]

yea sry mb

[celest aspen]

i leave first

[merry plume]

yea i just recommend like jp books or like any like stores for textbooks, they have like specific materials u cant find online

[celest aspen]

i try them before, but they are too expensive

it is like 150~200 a book

$

[merry plume]

yea bro i went through that, i suggest carousell

its mostly half the price

[celest aspen]

that sounds great, i try it later

i have to do my eng hw now , let add a fri

then i closed now

.solved

[obtuse pebble]

Channel closed

Closed by @celest aspen

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[high grotto]

3rd ques

My progress

[ionic oar]

you could probably get some parallel sides

[high grotto]

Oh

[obtuse pebble]

@high grotto Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

[thick oracle]

[surreal plover]

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[onyx current]

[onyx current]

huh

[round nova]

??

[onyx current]

oh

[round nova]

u opened the channell..

oop-

[onyx current]

ohhhh

[round nova]

-# .close should close it :3

[onyx current]

wait I do have doubts tho

[round nova]

oh ask ahead!!

[onyx current]

class 11 trigonometry tripping me out

[round nova]

PERFECT

i love trig!!!

what do u no understand about it??

so far atleast-

[onyx current]

wait a second

i need to ss the question

[round nova]

tyt!!

[inner sierra]

they getting excited lol

[round nova]

YESSSSS- (alsong as its geometry trig stuff and not just a huge line of trig identties then i am happy :3)

[onyx current]

how does this work

if tan 225 isnt even in the question

how is it being used

[round nova]

okay let me explain

[onyx current]

i swear they pullin shi out they ahh

🙏

[onyx current]

ok

https://tenor.com/view/sleep-sleeping-sleepy-cat-cat-cute-gif-5647373625706459185

[round nova]

so so questions if i told you that i need to have 225 apples

and you only know the price of 100 apples and 125 apples

[onyx current]

Son 🙏

[round nova]

can u use that fact to tell me how much 225 apples would cost?

[onyx current]

yes

[round nova]

makes sense SO here is the cool thing

WE know WHAT tan (100) is

and we know what tan (125) is

so we can say that tan (225)

[onyx current]

i swear all these identities is tripping me out

[round nova]

is bassicayl tan(100 + 125)

[onyx current]

pause

[round nova]

does that make sense?

[onyx current]

pause

no wait

[round nova]

tyt tyt

[onyx current]

we DONT know tan100 and tan125

we KNOW tan225

becuz it is

180+45

so basically tan45

which is 1

[short crescent]

what? no

[onyx current]

from the freaky ahh cartesian plane

[round nova]

well yes u could do it that way^

[onyx current]

woahhhhh

twin

same name

[round nova]

wai wdym-

[onyx current]

thats how it works

my teacher is so cheeks

[short crescent]

$\tan (A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$

[warm shale]

Krish

[onyx current]

10th grade teacher had a passion, this teacher comes for the pay 🙏

[round nova]

yeah! we would then ofc need to do the identity

[onyx current]

i have this memorized

[round nova]

i am just epxlaing that any angle an be broken down into two angles

[onyx current]

but THERE ISNT A WAY TO APPLY son

[round nova]

and then we apply the identtity

[onyx current]

my friend said

that

only in tan cases

[short crescent]

oh oops i read this as "its equal to tan100 + tan125"

[onyx current]

u can add 2 tans

[short crescent]

my bad

[onyx current]

to make a tan

[round nova]

ah yeah oops my mistake i should have clarfied

[round nova]

well not really- you need to use the identity...

[short crescent]

no you did im just blind

[onyx current]

but there isnt a tan(a+b) there 😭

am i too dumb for this

[short crescent]

you can set A = 100 and B = 125

[round nova]

oaky oaky let me clarfiy some stuff

A and B are ANGLES

so go on ur calculator and write

tan(225)

and then go and write

[onyx current]

i know but we can only do that when its given tan 225 then u can write it as tan 100 +125 in brack

[round nova]

tan (100+125)

u will notice it gives the same answer

because this is true because they are both in the same tan

BUTTT- be warned

if u try to do tan(100) + tan(125) its wrong

[onyx current]

same answer

ye

i get that much

[willow flax]

tanX = something something
break X into A and B
so tan(A+B) = same something
now apply the identity

[short crescent]

you are given 225, no?

[round nova]

YES THIS ^

[onyx current]

in the question

no

[obtuse pebble]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[round nova]

^ @short crescent

okay so @onyx current here is the important thing to understand

in trig any angle inside any trig function can be split into as many parts as u want at any numbers u want AS LONG as it stays inside

[onyx current]

OOOO

WAIT

[round nova]

so sin(120) can become sin(100+20) or sin (90+30) or anything

[onyx current]

I THINK IM GETTING IT

PAUSE

[round nova]

YES YES CMON

[onyx current]

so

[hardy widget]

.pin

[onyx current]

no pause

[onyx current]

i know

that much

but then it isnt given like that

its given separately

thats what is stopping my brain

im smart enough to know tan125 + tan100 = tan225

but then how did he group it into tan(125+100)

[round nova]

tan(125) + tan(100) is not tan(225)

[onyx current]

is there a basic concept im missing

[short crescent]

this is not true

[onyx current]

WAIT

WIAT

I MEAN

I MEANT

NO

[round nova]

tyt tyt its okay..

[onyx current]

I MEANT

i meant to put the

not equal to sign

[short crescent]

so you know this identity, right? we are basically rearranging it by solving for tanA + tanB

[onyx current]

not the equal to sign

[round nova]

YES

that is correct!!

[willow flax]

X = A+B
then trigfunction(X) = trigfunction(A+B)

do you get this?

[onyx current]

yes

obv

i know that much

[round nova]

so you meant to say tan125 + tan100 =/= tan225

[willow flax]

so tan225 = tan(100+125)

are we on the same page?

[onyx current]

I loved maths till 10th but 11th 🙃

[onyx current]

hyes

yes

[onyx current]

yes

yes

[round nova]

okay okay perfect perfect!!

so lets look at the orginal question!!

[onyx current]

tan125 + tan100 + tan125.tan100

[willow flax]

now you need figure tanA+tanB + tanAB (as in the question)
you remember that
tan(A+B) = tanA + tanB/ 1 - tanAtanB

[round nova]

then where did he get the 225 from?

[onyx current]

yes

[onyx current]

THATS WHAT IM SAYING

HE PULLED IT OUT HIS AHH

🙏

[short crescent]

hold on

[willow flax]

so now calculate tan225. what is that?

[short crescent]

read this

[onyx current]

how tho

we can do

tanA+ tanB =

WAIT

WAIT

PAUSE

[short crescent]

do you understand that $\tan A + \tan B = \tan {(A+B)} (1- \tan A \tan B)$

[onyx current]

ohhhhhhhhhhhhhhhhh

[willow flax]

tan225 = 1
tan225 = tan(100+125)
so tan(100+25) = 1
are we on the same page?
@onyx current

[onyx current]

hmmm

i get it

i get it

[warm shale]

Krish

[short crescent]

we have this now

our question is in the form tanA + tanB + tanAB

yes?

[onyx current]

Maybe the real krish is you

not me

[short crescent]

sorry

tanAtanB at the end

[onyx current]

i get the answer twin

I get it

[willow flax]

!done

[onyx current]

Basicall

[obtuse pebble]

If you are done with this channel, please mark your problem as solved by typing .close

[onyx current]

.close

[obtuse pebble]

Channel closed

Closed by @onyx current

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[wise silo]

need help on my sparx

[obtuse pebble]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[dire moss]

First of all, do you even know what length you're trying to solve for?

[wise silo]

this might sound crazy but i just finished the question…

need help here though if that’s okay

[dire moss]

Ok, well to start, do you know whether you want to use the Law of Sine, or the Law of Cosine to solve this

[wise silo]

i’m honestly not sure, which one suits best?

[dire moss]

Before I answer, do you know what the sine law, and cosine law both say?

[wise silo]

not really unfortunately

[dire moss]

And before I give you the formulas, do you know how to label a triangle, (sides and angles)

[wise silo]

yes

Hypotenuse, opposite, adjacent

[dire moss]

I mean, in terms of "A, B, C" and "a, b, c"

[wise silo]

oh, i don’t

[dire moss]

First off, Capital letters represent angles, lowercase represent sides

The side opposite to each angle, get's the lowercase of that letter

Does that kind of make sense?

[wise silo]

yeah i get that

[dire moss]

If so, draw a random triangle, label it, and send a pic here

Just to see if you get it

[wise silo]

i need to get an app to draw one real quick if that’s okay, sorry for the wait

[dire moss]

Take you're time buddy

[wise silo]

is this correct?

[dire moss]

Technically, yes

But it's convention to write the angle letters on the verticies (corners)

[wise silo]

ahh okay

[dire moss]

And it doesn't HAVE to be a right angle triangle

[wise silo]

ohhhh alright

[dire moss]

Because if it was a right triangle, you would just use your regular trig ratios...

SOHCAHTOA

[wise silo]

yeah

[dire moss]

Ok, so....

Now, to what each law says

Sine Law: a/sin(A) = b/sin(B) = c/sin(C) OR sin(A)/a = sin(B)/b = sin(C)/c

Cosine Law: a² = b² + c² - 2bc*cos(A)

[wise silo]

right

[dire moss]

So, first thing I usually look for

Do I know an angle and it's opposite side length? If so, I'll use the law of sines.

[wise silo]

i can’t see any opposite side lengths in the question so would i use the law of cosines?

[dire moss]

Are you sure?

[wise silo]

i’m pretty sure

[dire moss]

Well, look at the 25 degree angle

Do you know the side length opposite it?

[wise silo]

40 cm?

[dire moss]

Bingo

So, do you know an angle and its opposite side?

[wise silo]

angle = A, side = a

[dire moss]

But I mean, do you have actual measurments for both?

[wise silo]

25° and 40 cm?

[dire moss]

Yessir

So, since you know an angle, and its oposing side, what law can we use?

[wise silo]

law of sines

[dire moss]

And here's a tip for the law of sines

Which ever type of measurement you're solving for, use the version that has that as the numerator

Are we solving for a side, or an angle?

[wise silo]

angle

[dire moss]

So, will angle go on top, or will a side length go on top?

[wise silo]

angle on top

[dire moss]

Bingo!!

So...

sin(A)/a = sin(B)/b

So, fill out some numbers

[wise silo]

ummm

so 25/40 for A/a

[dire moss]

but remember, it's sin(A)/a

[wise silo]

40/25 ?

[dire moss]

Let's take a step back

What is A?

[wise silo]

angle

[dire moss]

What measurement?

[wise silo]

°

[dire moss]

That's units

I want the value

The measurment

The number

[wise silo]

25

[dire moss]

Ok, what is a

[wise silo]

40

[dire moss]

what is sin(A)/a

[wise silo]

sin(25)/40

[dire moss]

Ok, amazing

Do the same thing for sin(B)/b

[wise silo]

i don’t have an angle B

[dire moss]

That's what we're trying to solve for

So, it's ok

sin(B)/73

Do you agree?

[wise silo]

yeah

[dire moss]

Ok, now, solve for B

[wise silo]

so i do sin(25)/40 to start?

[dire moss]

Well, let's come up with a single expression that you can enter in your calculator

B = .....

[wise silo]

i’m confused

[dire moss]

Just simple rearranging

if

sin(25)/40 = sin(B)/73

Would you agree then....

[wise silo]

is b 25?

[dire moss]

sin(B) = 73sin(25)/40

[dire moss]

No

[dire moss]

Do you see how I got to this point?

[dire moss]

b is the length. B is the angle

[wise silo]

i meant is B 25

[dire moss]

No. B is the angle we're solving for

Notice how it's trapped inside the sin?

[wise silo]

yeah

[dire moss]

That means it's an angle

[wise silo]

right

[dire moss]

Can you make it up to this point?

[wise silo]

no

[dire moss]

What about here?

[wise silo]

yeah

[dire moss]

Ok.

Do you believe me that you can do anything you want to an equation as long as I do it to both sides?

[wise silo]

yep

[dire moss]

What would I get if I multiplied both sides by 73?

[wise silo]

multiplied both numbers?

[dire moss]

Wdym by that?

[wise silo]

multiplying 25 and 40 by 73

[dire moss]

NO!!!

[wise silo]

oh okay

[dire moss]

So, let's go slowly here

You can't multiply the 25 by the 40 because it's trapped inside the sin. Yes?

[wise silo]

yes

[dire moss]

You cannot multiply two numbers if one of them is stuck in a function

[wise silo]

ohhhhhhh okay

[dire moss]

Would you agree that this is nonsense

4 * √3 = √12

[wise silo]

yes

[dire moss]

So, does

73* sin(25) = sin(73* 25)

[wise silo]

yeah

[dire moss]

It does not bro

[dire moss]

You just agreed with me that this was nonsense

[wise silo]

right

[dire moss]

You can't just multiply it in there

So, as ugly as it looks

[dire moss]

You just have to leave it as

73sin(25)

And then all that is divided by 40

so...

sin(B) = 73sin(25)/40

[wise silo]

so i multiply sin(25)/40 by 73

[dire moss]

Yessir

[willow flax]

you can figure out sin25 if you want to

sin(30-5)

and for 5 degrees, use small angle approximation

[wise silo]

i put 73 * sin(25)/40 on a calculator and got 0.77, is that correct?

[dire moss]

Bro, there is no need for that

[willow flax]

it asks for answer to 1 decimal point

[dire moss]

I'm aware

[dire moss]

It does. But don't round just yet

Let's keep the whole thing in that ugly form 'till the end. Ok?

[wise silo]

ok

[willow flax]

wasn't this the answer?

[dire moss]

so, do you agree that we have

sin(B) = 73sin(25)/40

[wise silo]

yes

[dire moss]

How do you undo sine?

[wise silo]

no clue to be honest with you

[dire moss]

That's ok

look for smth on your calculator that looks like

sin⁻¹

[wise silo]

ohhhhh i know that

[dire moss]

That stands for "inverse sine"

[wise silo]

inverse sine

yeah

[dire moss]

So, you're taking the inverse sine of all that nasty junk

sin⁻¹(73sin(25)/40)

[wise silo]

got a syntax error

[dire moss]

Send a pic on how you entered it

[dire moss]

You're missing a bracket for the sin(25)

You wrote "sin(25"

[wise silo]

shit i didn’t even notice

i got 50.4

[dire moss]

But it rounds up to....

[wise silo]

50.5

[dire moss]

Bingo

[wise silo]

tysm!

[dire moss]

No worries dude

[obtuse pebble]

@wise silo Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[sturdy bridge]

How can I rewrite the series with a to look like the series with b while changing the limit from 0 to 7
The series converges to 0 but is not absolutely convergent and should thus not have a set sum

[sturdy bridge]

I think I understand the general idea of how the order plays a role in the result / limit of the series but these are very generic series and I’m not sure how I could define them to give me 7 specifically

One method seems to be splitting a series into positive and negative elements (which is already a problem because I don’t know which elements would be positive/negative)
And then adding them up until you surpass the desired limit before you subtract stuff to fall under the limit and repeat the process

[obtuse pebble]

@sturdy bridge Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[scenic zenith]

is the best way to solve such type of questions is by counting degree of vertices?

[snow trout]

well for the negative eulerian path/circuit thing there is that theorem so that would be how you justify

[scenic zenith]

is there any such theorem for hamiltonian path and circuit as well?

[snow trout]

there are but there aren't any strong theorems that reduce the problem down to some simple numbers

hamiltonian paths are hard

these graphs are small though so try doodling

[scenic zenith]

aight

thanks

.close

[obtuse pebble]

Channel closed

Closed by @scenic zenith

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[swift depot]

help

[swift depot]

part a is 5pi/12 and pi/12

part b

im lost

i know to find the areas

with the formula

but like

theres no way

i can isolate r

[restive gorge]

what

[swift depot]

how

do u do b

idk how

ms is making ZERO sense

[restive gorge]

I would split it

[swift depot]

yh but hpow

[swift depot]

WOW

wait

omdz

OH NO

I WAS THINKING ABT IT BEING CURVED AS WELL

[restive gorge]

First integrate the the small part of C1, then the circle then C1 again basically

[swift depot]

basically

the triangle

and the 2 small little bits

man thats smart

broooooooooooooooooooooo thats acc sp smart

😭

bro

im too tired to acc work that out

its like

11:50

are u 100% thats right

bro tbh

it shld be

[restive gorge]

it's js common sense

integrating from 0 to pi/12 C1 and then continuing with C2 else you'd integrate passed the region R

[obtuse pebble]

@swift depot Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[dire moss]

<@&268886789983436800>

[obtuse pebble]

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[whole wind]

Hello, I have a maths Olympiad coming up soon, are there any tips to prepare for it? Or are there any range of questions that they usually ask?

[hollow ocean]

you might find better answers to this question in the #competition-math channel.

[inner sierra]

this is only for specific questions

[gloomy vector]

like they said, #competition-math is probably better. but theres contest collections in AoPS so you can get a feel of the oly problems

[whole wind]

Okay thank you guys

.close

[obtuse pebble]

Channel closed

Closed by @whole wind

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[steel night]

how does this represent a **plane **in R3, isnt it just a line?

[restive gorge]

why would it be a line

[steel night]

true, it doesnt follow the y=mx+b

[restive gorge]

For a line you need two independent equations of the form ax+by+cz=d

You can think of it like this, in IR³ we have points with 3 coordinates, now if you introduce a constraint like x+2y+6z=19, then that equation takes away one degree of freedom

so for example (x,y,z) becomes (19-2y-6z,y,z) since x=19-2y-6z, so you only have two free variables and a plane is 2-dimensional

[obtuse pebble]

@steel night Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[pine sequoia]

How would I start this problem? I am stuck. I need the left side to equal the right side. I need to use my reference sheet too. (The denominator on the right is sin^3x)

[alpine bloom]

Sorry but I don't know what the equation to be simplified is (i'm bad at reading other people's handwriting). Would you mind typing it?

[pine sequoia]

Nws one sec

cscx/1-cosx = 1+cosx/sin^3x

I’m trying to get the left side to equal the right

[naive atlas]

So what have you tried?

[pine sequoia]

I thought of multiplying the denominator on the left side on the top and bottom since I saw something similar done in class, but I didn’t think it was the right move

[naive atlas]

What does this mean

[pine sequoia]

Like

1-cos times the cscx

And 1-cos times 1-cosx

[naive atlas]

Oh so you multipled both the numerator and denominator by (1-cos(x))

[pine sequoia]

I tried that yea but like I wasn’t sure if that was right

[willow flax]

convert cosecx to 1/sinx

see if it leads you anywhere

coz it will

another hint : you will need to somehow use sin^2 + cos^2 = 1

[naive atlas]

You are supposed to multiply and divide with the conjugate of the denominator which would be 1+cosx and not 1-cosx , since multiplying with 1+cosx would lead to (a+b)(a-b) in the denominator

[pine sequoia]

wait which am i supposed to do

u both said two different things

😓

[willow flax]

no we did not

have you converted

cosecx into 1/sinx

?

[pine sequoia]

how would I covert cscx then?

split it?

[willow flax]

ok. idk what you did there. but scrap that. i am asking you to only convert cosecx into 1/sinx

nothing else

what do you end up with?

[pine sequoia]

sinx/1 times 1/1-cosx?

[willow flax]

$\frac{1}{\sin x(1-\cos x)}$

[warm shale]

TheAstorPastor

[willow flax]

this right?

[pine sequoia]

ohh I see

[willow flax]

can you multiply the denom with it's conjugate?

[pine sequoia]

Yeah

[willow flax]

what do you end up with?

[pine sequoia]

1/sin^3x

how do I get the top tho?

[willow flax]

why won't you multiply the same expression in the numerator as well
a/b = a times something / b times something

if you multiply something in the denom, you must multiply it in the numerator as well

[pine sequoia]

im confused

[willow flax]

so if you multiply 1+cosx in the denom. you will do it in the num as well

[pine sequoia]

but I multiplied 1-cosx?

[willow flax]

why will you multiply it 1-cosx?

you obtained sin^3x

right?

you already had a sinx

so you needed a sin2x

to get sin2x

you will do 1-cos2x

which is nothing but

(1-cosx)(1+cosx)

[pine sequoia]

I got it

[willow flax]

!done

[obtuse pebble]

If you are done with this channel, please mark your problem as solved by typing .close

[pine sequoia]

.close

[obtuse pebble]

Channel closed

Closed by @pine sequoia

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[sullen hemlock]

https://tenor.com/view/cat-this-is-a-cry-for-help-help-help-me-lord-help-me-gif-1631346409026690561

[drifting wraith]

,rccw

[alpine bloom]

3% of what?

[sullen hemlock]

Ok let me explain

[drifting wraith]

3% of 270

[sullen hemlock]

I need to calculate the lenght of a pipe that needs to go down 3%

And x needs to be 270 cm

[drifting wraith]

you can tell by the way th arrow almost points away from 270

[sullen hemlock]

Like it tilts downwords

[alpine bloom]

So 3% as a angle?

[sullen hemlock]

Yes

[round nova]

are u familar with sine?

[sullen hemlock]

Idk how to use it practiclly

[drifting wraith]

i can't solve it

[round nova]

okay so what is sine?

sine is opp/adj correct?

[drifting wraith]

but i assure you it's 1/10 easy

[sullen hemlock]

Like you put 90⁰ and ptrss sine and it says 1

In calculator

[round nova]

okay do u understand sine? i can explain it if uwant

[alpine bloom]

Law of sines works

[drifting wraith]

i don;t know law of sines

[sullen hemlock]

Sure

[round nova]

https://tenor.com/view/matematica-gif-7322246

okay so take a look at this gif here

ik it looks *veryyyy complicated but bear with me-

[sullen hemlock]

So im just looking for seno here

How do i find it or can i just put it in the calculator

[round nova]

so i want you to imagne holding ur fingers up like so ✌️

[drifting wraith]

both hands!

[round nova]

and notice how whenu open ur dingers more

the distance between ur fingwers tips

increases

[sullen hemlock]

Uh hu

[round nova]

and the less u open them

the smaller the distance

so Bassicly we know that distance probably means something in proprtion to the side length

and a bunch of smart ppl figured this out long ago

and because a 90 right angle only has 3 angles

and it adds up to 180

if we know one angle

we know all angles

and we we would be able to know how much each side is big in comparsion to the others

does that make sense?

[sullen hemlock]

The only angle we have is 0 from flat line with 3% downwards no?

[round nova]

well isnt there also the 90 degrees angles in the corner?

[sullen hemlock]

I mean yea

[round nova]

okay lets assume that angle on the left is 30 for example

we we would know that the angle on the top right

is 60

makes sense right?

[sullen hemlock]

Shouldnt it be 90⁰?

The one above X

[alpine bloom]

.... I might be wrong but the upper right should be the 90 one right?

[round nova]

yeah oops-

ur right both of you

i am sorry i mis drew it-

[sullen hemlock]

So the one most away is 30 so it adds up to 180

[round nova]

sorry i mean bottom left-

[sullen hemlock]

But what does that have to do with calculating the lenght of the bottom line

[round nova]

okay so here is the thing..

lets take an example

notice how we have an angle of 40 on the left

(this is just an example btw)

[sullen hemlock]

Ok

[round nova]

SO what sin does when u do sin(40)

is bassicly go ahead and asks if i have a hyptonuse of 1? in a traingle with angles 40 and 50 degrres-

[sullen hemlock]

It says 0.642 ...

[round nova]

how much would the opposite side be compared to the hyptonuse of 1?

SO that means that THE ORANGE side in the image above

is 0.642 For ever 1 length of the hyptonuse

IT GIVUES US the ratio

between the opposite side AND the hyptonuse

FUNFACT this is why sin(90) = 1

https://tenor.com/view/matematica-gif-7322246

look at the gif

notice how when the angle becomes 90

BOTH the hieght (the opp side to the angle) AND THE HYP (black line)

BECOME THE SAME LENGTH-

-# if it still doesnt make sense its okay u can tell me

[sullen hemlock]

A moment im trying to grasp this

I mean i get it

But whats the angle if its a flat 0 -3%

[round nova]

okay so we want a pipe that get smaller by 3%

so just to maeke sure i understand thie right-

u want the pipe to get smaller by 3% every 1 distance to the left

or just 3% slope?

[sullen hemlock]

No pipe is same size its just the Y axis that changes

A slope yea

[round nova]

okay so 3% as the angle

[sullen hemlock]

Yea

[round nova]

so if total angle is 360-

we would need to do 3/100 * 360 first then to convert to angles-

so at an angle of 10.8

[sullen hemlock]

So that means the angle is 10.8 right

[round nova]

well yes if what u mean is 3% of the total angle?

[sullen hemlock]

And we use sin on 10.8

[round nova]

i am still unsure what u mean by the 3% tbh-

[round nova]

yes and sin gives us the ratio of how big the side opp is compared to the hyptonuse-

[sullen hemlock]

Like the pipe is constantlly in the angle of 3% so it works

[round nova]

-# although in thie scenario we dont have hyptonuse so ithink we will prob use tan

tan = opp/adj for refrnece..

[sullen hemlock]

Ok so we just go 10.8+270 and that the length of the hypothonuse

[round nova]

wwwwhwwhaha-

wai? wdym-

10.8 is an angle not a length??

[sullen hemlock]

Wait

Yes the angle is 10.8

So the side is that and sin

And its 0.187

[round nova]

okay so we have the side adjacent to our angle of 10.8 right?

[round nova]

wait- wait-

[sullen hemlock]

Nah everything got messed up

Let me write it down

Ok thats it

[round nova]

question- which side is the hyponuse the one on top or bottom?

[sullen hemlock]

The longer one?

Bottom

[round nova]

okay the onee we dont know the side of correct?

[sullen hemlock]

👍

[round nova]

okay so since we dont know the side on the bottom it wouldnt really be usefull to use sin-

so here is the thing right

all of the function of trig

sin and cos and tan

[sullen hemlock]

Is that ctgx?

[round nova]

what they bassicly do is they tell you how to split your food for example on ur two siblings-

so bassily sin tells you how much food should u give to silbing 1 in comparsion to how much TOTAL food u have

and cos tells u how much u should give to sibling 2 in comparsion to how much TOTAL food u have

BUT tan tells you how much should sibling one get IN COMPARSION to sibling TWO

so they all are desrcibing ratios-

here of sibling one is the side opp to the angle-

and sibling two is the aide adjancet (next to) the angle

does that make sense so far?

[sullen hemlock]

Give me a second i need to take a look in the notebook

Ok so the formula is a/sin(10.8)

And then when you put in the calculator it says its long 1440 cm

Unbelivable

[round nova]

...

okay so that is using sines law and sines law bassicle says is-

side (a) / sin(A) = b / sin(B)

this is another way to solve it i was going to solve it using tan-

but if u would like we can also solve it using sines law-

[round nova]

-# also i think tis number might be a little off-

[sullen hemlock]

Ok so i used the tan and the result says 51.5

Thats the side ig

Or is that 270+51.5 is the entier hypotoneuse

[round nova]

OKAY so tan bassicly gives us the ratio of how much food silbing one got vs sibling 2

if we know silbing 2 got 270 food

how much will sibling 1 get?

[sullen hemlock]

So its 51.5 of other 2 sides

[round nova]

also what how-

its just tan(10.8)

,w tan(10.8)

[sullen hemlock]

0.1907

Ok

[round nova]

okay so that means for every 1 food sibling 2 gets
we get 0.1907

can u calculate x now?

[round nova]

OH WAIT-

did u mean final answer??

[sullen hemlock]

Yea

[round nova]

i am so sorry i didnt notice

yes that is correct-

[sullen hemlock]

But what did i even calculte then

The hypothenuse?

No

[round nova]

no no-

u calculated the OTHER side no the hyp-

[sullen hemlock]

The side?

[round nova]

YES

[sullen hemlock]

Thank god

[round nova]

if u want the hypt u can do cos(10.8)

[sullen hemlock]

Ok the rest is easy

[round nova]

and cos would give u how much sibling 2 is in comparsion to the total (hyp)

so we know that for every one length of hyp- the upper side gets cos(10.8) which is 0.98

so if the side on the top is 270

that means-

the hyp is around 274.8 i belive

anyways i hope that answere ur question!

[sullen hemlock]

So the result is 274.86

[round nova]

no no the Hypotnuse is 274-

X is 51.5

[sullen hemlock]

Yes that

Finally bro

[round nova]

funfact u can check with pythagroerm therom!!

anyways i hope this helpedd!!

[sullen hemlock]

Yes thats what i did

It did thanks

[round nova]

-# also dont forget to .close when ur done

[sullen hemlock]

.close

[obtuse pebble]

Channel closed

Closed by @sullen hemlock

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[carmine sluice]

Hi! I have a question that has been plagueing me recently when it comes to a personal cryptography project.
To set the scene for my question, what I'm working on solving is a type of cryptography problem that effectively serves as a more generalized form of the solitaire cipher.
When you build the ciphertext, you have a deck of cards labeled 0-82, and you have an instruction sheet. On the instruction sheet are "swaps" which each have an assigned letter, and a "base permutation". When you start writing your plaintext, for each letter, you first shuffle the deck based on the base permutation, and then you take and apply the letter specific swaps. There are a couple important rules to these swap constructions. They all have a unique effect on index 0 of the deck, they have equally as many swaps as eachother, and that number of swaps ("k") is no more than 5. After you apply both the base permutation, and the swaps correspondant to that specific letter, you now write down which exact card comes up at index 0, which will be your first ciphertext character. You keep the deck in this new reshuffled order and repeat from letter selection until your message is completely written.

Now, my question. This space at first glance appears rugged and completely resistant to analysis. I am very curious, however, about that top card uniqueness constraint. When starting from any position, applying the same plaintext letters will always create an identical repeat pattern no matter what the underlying state of the deck is, because that top card's pathway through the deck is fixed. Is there a way to reinterpret this problem such that you can generate a deck cipher targeting specific repeats? If you could, would the result be identical after relabeling the deck and applying substitution to the plaintext?
Apologies if this explanation is unclear! I have more details, further explanation, and an example, but I wanted to send this as a single message.

[obtuse pebble]

@carmine sluice Has your question been resolved?

[carmine sluice]

sorry, i believe this goes more in the "broad conceptual questions" category, and this isnt a good channel!

-close

.close

[obtuse pebble]

Channel closed

Closed by @carmine sluice

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[halcyon shuttle]

do i have to use beggar's method for this?

[halcyon shuttle]

cuz i did not understand how i can begin

.close

[obtuse pebble]

Channel closed

Closed by @halcyon shuttle

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[dusty relic]

how would I solve

x/2-1 = 3-x/5

it wouldn't let me send a picture sry

[dusty relic]

I don't rlly get it (if I was in an exam I'd skip it)

[vale ermine]

$\frac{x}{2}-1=3-\frac{x}{5}$?

[warm shale]

JohnR

[high lily]

multiply both sides by the lcd (lcm of the denominators)
to rid yourself of fractions

[obtuse pebble]

@dusty relic Has your question been resolved?

[dusty relic]

yes

[dusty relic]

huh

[high lily]

can you give a more detailed response

[dusty relic]

I don't understand what you mean

care to explain

or show what we multiply

[high lily]

can you identify the denominators of all fractions present

[dusty relic]

2
5

2 and 5

[high lily]

and can you determine the lcm (or any multiple of those)

[dusty relic]

idk

10?

[high lily]

yes

now that you have the lcm,
multiply both sides of the equation by 10

[dusty relic]

huh

wdym

what get * by 10

[high lily]

each side of the equation

Both the left side of the equal sign
And right side of the equal sign

[obtuse pebble]

@dusty relic Has your question been resolved?

Channel closed

Closed by @dusty relic

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[oak solar]

i cant seem to understand how to solve this

[short crescent]

have you learned u-substitution?

[oak solar]

sort of i still cant wrap my head around that

[short crescent]

so if we think about this without actually writing anything out, does the derivative of one of the functions cancel out another?

lets say we set u = cosx. what would du/dx be?

[oak solar]

wait just to be clear du is the derivative of u right

[short crescent]

yes

same as u'

[oak solar]

cause whats really bugging me is the du and dx and that stuff its really confusing me

still uhh du should be -sin i think

[short crescent]

du/dx is just another way to write out u'(x)

[short crescent]

good

so du/dx = -sinx. we want to "replace" dx with du. this means we should solve this equation for dx.

if we solve for dx and replace it, will anything cancel out?

[lone echo]

u should not be cosx

u should be sinx

[restive gorge]

to add on: you can think of du/dx as a useful notation, because it allows you to make sort of algebra

[short crescent]

i know this, i'm just solidifying the general idea of u-sub.

just as we need to know how to do it, we should also know why the other way doesn't work

[lone echo]

fair enough

[oak solar]

dx should be du/-sin

[short crescent]

good!

so lets rewrite the original integral

$\int u \cdot e^{\sin x} \frac{\dd u}{-\sin x}$

[warm shale]

Krish

[short crescent]

does it make sense how we got here so far?

[oak solar]

yes it makes sense for now

[short crescent]

okay great

now here, is there any simple way you can think of cancelling terms or integrating?

[oak solar]

wym

i know what u mean i just cant think of one

[short crescent]

like if you had 2x/2, you could cancel the 2's. could you cancel any terms out here?

[short crescent]

right, there isn't anything that could cancel out

so setting u = cosx does not work for this problem.

[oak solar]

oh right

[short crescent]

what about setting u = sinx?

can you solve for dx again but now with u = sinx?

[oak solar]

du/cosx = dx

that should be it

[short crescent]

yes!

lets rewrite the integral again with du/cosx = dx

$\int \cos x \cdot e^u \frac{\dd u}{\cos x}$

[warm shale]

Krish

[short crescent]

can anything cancel out here?

[oak solar]

the 2 cox

cosx

[short crescent]

yes exactly, and that is exactly what we need to cancel out since we are integrating with respect to u now

so let's rewrite this again after cancelling terms.

$\int e^u \dd u$

[warm shale]

Krish

[oak solar]

oh that makes so much more sense now

the integral of e is just e

[short crescent]

right!

[oak solar]

so then we switch back u which is

e^sinx

[short crescent]

yes exactly

so what does the integral evaluate to?

[oak solar]

e^sinx this

[short crescent]

very close! youre missing one thing

[oak solar]

oh right +c

[short crescent]

yep

so your answer is $e^{\sin x} + C$

[warm shale]

Krish

[oak solar]

thank you you really saved me i was genuinely struggling with it

[short crescent]

and do you understand how we can double check this?

[oak solar]

uhh lowkey no

[short crescent]

well, what is integration, really?

[oak solar]

apart from the derivative nvm

[short crescent]

it's the opposite of differentiating, right?

if we take the derivative of x^2 we get 2x, and if we integrate 2x we get x^2

[short crescent]

we are already given the derivative of this