geometry-and-trigonometry

[fickle raptor]

well lets think

law of cosine state c^2 = a^2 -b^2 2abcosgamma

to isolate c we need the root

[spring lion]

Is this right so far?

No no wait.

Hmmm

Do I put the 1200²+1700² on the left?

Or?

[fickle raptor]

hit the little wrench and put on "degree" mode

[spring lion]

Dawg I straight up FORGOT ABOUT 52

Yeah this shit is what happens when you study trigonometry for 5 hours

The other day I was actually cooked

[spring lion]

Ahhh riiiight!

I know Sal puts the degree mode on too

Damnnn

I simply do an arccos and it just gives the degree anyway

Shouldn't it be - though?

Did I get it right?

[fickle raptor]

thats not what i got

[spring lion]

1348.369823

[spring lion]

Damn I made a mistake then.

[fickle raptor]

u got it

i messed up

[spring lion]

Oh

Btw I have a question

[fickle raptor]

i had a plus lol

[spring lion]

When we have 2 sides in cos

Law

Do we just subtract that from the unknown side?

Like

[fickle raptor]

well if you have c then its just square root but if its a or b than you minus the (a^2) and munis the c^2 and then multiply everyhing by -1

[spring lion]

c²=1200²+1700²-2*1200*1700 cos(52)

[spring lion]

Yeah I know about the multiplication by -1.

It always gives a negative for these anyway.

[spring lion]

Would c²-1200²-1700² work?

No?

Ahhh wait.

If we knew about c it would work.

But we're finding the c here.

Okay okay, soooo...

Is the answer 1348 then...?

[spring lion]

I don't wanna make a mistake here...this is the bonus question

[fickle raptor]

well your solving for c

[spring lion]

Yea

[orchid rampart]

,w sqrt(1200²+1700²-212001700 *cos(52degrees))

[fickle raptor]

got the same thing

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by sigma018989

[spring lion]

Yupppp

[fickle raptor]

thats alot of decimals

[spring lion]

You gotta teach me how to do these calculations with the bot man 😭

[spring lion]

😭😭😭😭😭😭😭😭

[spring lion]

Is there a guide?

[orchid rampart]

Its just ,w bla bla

[spring lion]

No I mean

[spring lion]

Seeing the answer visually like this?

Ohhhh wait

Right right I'm slow

,w

Alright.

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by twocenttalon

[fickle raptor]

what happened to my pie 😢

[orchid rampart]

lol

2*pi

[spring lion]

What in the world is that symbol 😭🙏

Integrals are soon for me btw

I'm doing precalculus

[orchid rampart]

,w int _0 ^R 2pirdr

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by sigma018989

[spring lion]

This is the third trigonometry section on Khan academy

[cunning lion]

what symbol

i count like 8

[spring lion]

The integral symbol, I'm really curious what integrals do. (Don't spoil, I want to be surprised)

I'm treating math as a show lol

Right now I'm barely in Season 1

The world of math is a vast ocean

And I know a raindrop rn

[orchid rampart]

Its a snake

[spring lion]

PYTHON?!

😱

[orchid rampart]

Probably not

[spring lion]

Dangit

[fickle raptor]

is a low hanging s

duh

[spring lion]

Alright gang I'm gonna stop studying now my head legit hurts

And I've been holding the bathroom for 30 minutes trying to solve these questions

(And succeeding!)

Although my worst nightmare is back

Sinusoidal equations...

I was awake until 3-4 am trying to solve that.

Let's see if this time things will be different with my formidable foe...

But everything aside WOW I am having FUN MAN!

5 videos back to back...am I cooked tomorrow?

[spring lion]

Lol

[civic forge]

yoooo
wait they spelled sign wrong

[spring lion]

Wdym

[civic forge]

sine

i used to spell it like that

but nah

[spring lion]

What's the correct one?

[celest tangle]

Soo can toa

[next mantle]

They didn't

Sign ≠ Sine

[dark sparrow]

nope

"sine" and "sign" are different words meaning different things

[lime crown]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question!

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.

[long geyser]

What is T_<5,2> read as?

[trail perch]

a Transformation that transforms an object by 5 pts of abscissa and 2pts of ordinate

[long geyser]

Thanks

[lime valley]

chat im dogwater at trigno metry

can any1 tell me some book or vids to absolutey master this

[spring lion]

Help

How do I even solve these??

I gotta revisit the interval videos ngl...

[spring lion]

Damn I couldn't find them.

Can someone remind me how to find the intervals?

Google gives nothing.

Ahhh it's -1 and 1.

Right right.

Huh? The video now says it's -pi/2, 0...bruh.

Which one is right?

God damn it

I chose B but it was false

I eliminated 2 other options

Oh WOW...I'm DUMB. I literally

Wow

Nevermind.

Ok I managed to finish it.

4/4

I got no clue how to solve these...

Help 🙏

I'm absolutely gonna need pen and paper for this.

[spring lion]

Ping me

[spring lion]

https://www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:trig/x9e81a4f98389efdf:sinus-eq/v/sine-solutions

[spring lion]

I really don't understand this^

[spring lion]

.

The video didn't have this range

I just don't get these ranges in general.

[spring lion]

Wait

I googled it wrong I think

Let me check again

Ohhhh.

Yeahhhh now I understand slightly more!!

[spring lion]

Okayyy I get this now.

[spring lion]

I just rewatched everything, let me think of a way to solve this...

cos(12x+2πn)=-1

cos(-12x+2πn)=-1

How do I continue from here?

(12x+2πn)=arccos(-1)?

Hmmmmm.

Arccos(-1)=180°...

Soooo...

180/12

15°-360°

Is one answer?

Damn it's not in the answers list

Rip

Oh wait.

360/12.

Ahhh.

God dang it it was false

For a second I had a lot of hope I got it right

Oh wow

I'm DUMB

I did get it right

But chose the wrong answer

😭😭😭

And arc cos range is 0, 180°

But wait.

Why was option D incorrect?

[spring lion]

I'm gonna stop here and wait for someone to reply to this I suppose.

[whole finch]

cos(12x) = -1 means
12x = 180 + 360n degrees.
Divide by 12 both sides to get your answer

[spring lion]

Why no cos(-12x) as well?

[spring lion]

I actually did get that answer yeah.

[whole finch]

You can do that, on which case you would get
x = -15 - 30n degrees, but shifting n to -(n+1) gives you the original answer anyway. They're the same thing

[spring lion]

Huh

[whole finch]

Sorry n to -(n+1)

[spring lion]

So

[whole finch]

Try substituting -(n+1) into n in the x = -15 -30n. Don't you get x = 15 + 30n? I suggest to do this with pen

[spring lion]

Right after I leave my room to take a break 😭

[spring lion]

Why +1?

[whole finch]

You'll see when you carry out the calculations. Another way to think about this is that cos(-12x) =cos(12x) anyway, so nothing changes

[spring lion]

OH

OHHHH

now I get it.

Even if I got -12x it'd be the same?

[whole finch]

Yeah

[spring lion]

Ok ok

My dum ahh 😭🙏

Thank you.

[whole finch]

Don't do this for sinx though.

np

[spring lion]

Yeah I know

Sinx is π-theta

[whole finch]

👍

[spring lion]

👍

[stone fractal]

what is value of Pie

π

[wary elm]

22/7 or 3.14
Or (2√2/9801) summation from n =0 to ∞ ( (4n)! (1103+26390n)/ (n!)⁴(369)^4n
Or its mainly ratio of circumference to the diameter

[upper echo]

It's 1/pi in Ramanujan's series

[shut thorn]

3.1415926535.......

[turbid monolith]

wrong.

[turbid monolith]

3.14159265358979323846264338327950288419716939937510...

[shut thorn]

OH CMON

YOU CANT SAY WRONG THEN JUST EXTEND FROM WHERE I LEFT OFF

😭

[amber bough]

engineering ending: 3

[languid cedar]

Both of you are wrong

Clearly π = π

[orchid rampart]

Wrong

🥧

[real mango]

guys can anyone

help me with this

[worthy eagle]

@real mango part a or b

[real mango]

b

[worthy eagle]

can't we just say that it's impossible because area can't change after reassembling

[real mango]

there is a difference of 1 sq unit in the 2 areas

[worthy eagle]

-# I did look up on google why it said to use part a and it was the same thing area difference of 1 unit but twisted as in above picture. It was to teach something but there could be better way to teach you this

[real mango]

is this correct

[exotic yarrow]

yeah that should be fine

[real mango]

and for a

[dark sparrow]

those first two are APPROXIMATIONS, it must be stressed

[worthy eagle]

open a help channel and ping me, I can help with that

[sly urchin]

First problem of Day 2 (Vietnam MO 2025-2026)

@hybrid peak

[edgy karma]

guys if i want to count the shape in a figure how to do it? and that too the figure is of rectangle which has 2 diagonal lines nd we have to count the traingles

is there any trick or formula?

[silent plank]

can you rephrase the question
and/or provide an image

its hard to understand what you mean

[worthy eagle]

yes there are formulas

let me send you link wait

https://youtu.be/ntMWGoYiO28?si=m4faSlIkkrQSRs87 @edgy karma

[edgy karma]

i can't thank you enough buddy thank you so much it has the foundation idea of how to solve it and i helped me a lot thanks i finally got the answer of the question i was trying from last 2 days

[civic forge]

dayum

thx dude

idk how i’m in geometry 🫤

[opal grove]

Guys I’m genuinely confused. Am I misunderstanding something here? (This is a question in the textbook I’m given)

[clever flint]

u do cosine rule right

you know how its c^2 = a^2 +b^2 -2abcos(C)

[opal grove]

Ja

[clever flint]

rearrange to solve for cos(C)

[opal grove]

Huh

Wdym rearrange

[clever flint]

ok so with this formula

your a, b, c are your sides

[opal grove]

Mhm

[clever flint]

and your A, B, C are your angles

in the question you are given the 3 sides a, b, c, which are 12, 6 and 8

and its asking you to find angle A (opposite to side a)

so when i say rearrange, you want to add/subtract and multiply/divide things by both sides to isolate A

so your formula is a^2 = b^2 + c^2 -2bcCos(A)

so subtract b^2 and c^2 from both sides

so now you have a^2 - b^2 -c^2 = -2bcCos(A)

now divide both sides by -2bc

so you have (b^2 + c^2 -a^2) / 2bc = Cos(A)

now take the inverse Cos of both sides to get
Cos^-1 ( (b^2 + c^2 -a^2) / (2bc) ) = A

and plug in all the values for a, b and c

put it in your calc and you will get your angle A

[opal grove]

My brain is exploding hold on

[clever flint]

haha all good take ur time

[opal grove]

Huuuu

This is what I’m getting at

How does ‘a’ turn from positive to negative in the third equation thingy

Aaaaaa

[clever flint]

ohhh ok

its because we divided by -2bc

so the negative flips the sign of all of them

notice how b^2 and c^2 become +ve and a^2 becomes -ve

and another thing

the cos^-1 surrounds the whole fraction

[opal grove]

My brain is forming wrinkles 😭

And why would the ‘a’ go on the other side

[clever flint]

wdym

your goal is to solve for A (the angle)

so you move all the sides (a b and c) to the other side of the equation

[opal grove]

The second equation thing, the a^2 is next to the -b, then as a fraction it’s next to the c^2

[clever flint]

well 5 + 3 is the same as 3 + 5

all that changed was that, since we divided by -2ab (notice the negative), the signs of b^2, c^2, and a^2 all flip

so instead of it being -b^2 -c^2 and +a^2

its now so instead of it being +b^2 +c^2 and -a^2

( A = cos^{-1}( \frac{b^2+c^2-a^2}{2bc}) )

[somber coyote]

WesStreet99

[clever flint]

this is your final formula

i can show you how i got it very slow and step by step if it still isnt clicking

[opal grove]

Hold on lemme put it in my calculator

[clever flint]

ok

[opal grove]

I wish I could email my math teacher but it’s summer holidays fahhh

[clever flint]

its ok let me type smth out

[opal grove]

This one gave me 90.5 degrees

But it’s 117.3

[clever flint]

( a^2 = b^2 + c^2 -2bccos(A) )

( a^2 - b^2 - c^2 = -2bccos(A) )

( \frac{a^2 - b^2 - c^2}{-2bc} = cos(A) )

( \frac{-(a^2 - b^2 - c^2)}{2bc} = cos(A) )

( \frac{-a^2 + b^2 + c^2}{2bc} = cos(A) )

( cos^{-1}( \frac{b^2 + c^2 -a^2}{2bc}) = A )

[somber coyote]

WesStreet99

[clever flint]

there

[opal grove]

HUH okay give me a sec

[clever flint]

read it line by line

the top is the original formula you have to youse

use

and you rearrange to solve for the angle A

[clever flint]

your problem here is that you dont square the b and c in the denominator

its just b and c

[opal grove]

Oh yeah I didn't see that thanks

OH OKAY YEAH I GOT THE ANSWER

But i still need this whole thing like verbally explained

I get it

But I don't get it

[clever flint]

ok im glad u got it

basically im sure you teacher gave you the cosine rule formula

which is typically expressed as ( c^2 = a^2 + b^2 -2abcos(C) )

[somber coyote]

WesStreet99

[opal grove]

Bc for the previous questions when solving for A, I just swapped the c parts for the a parts and it gave me the correct answer

Either that or sometimes the textbook is wrong

[clever flint]

yeah thats basically what ur doing

just call A your C

[opal grove]

But when I did
a²=b²+c²-2bcCos(A) it said "No solution found"

:]

[clever flint]

maybe u typed smth wrong

u gotta rearrange that formula to get A remember

[opal grove]

nSolve(12²=6²+8²-2×6×8×cos(a),a)

[clever flint]

i have no idea how that calculator works

is it because there is an n in front of the Solve?

[opal grove]

We use nSolve for cosine rule and sine rule

It stands for numerical solve

[clever flint]

wtffff ive never used a calculator like that before

so ur telling me u usually dont have to isolate the A?

[opal grove]

No

With b and d, I was able to just use nsolve

[clever flint]

what the fuck thats insane

this calculator is like a computer

[opal grove]

This is my older brother's old calculator he got in 2019. He put bloody pokemon emerald on it

[clever flint]

??????

HOW

[opal grove]

IDK

I TRIED ASKING HIM HOW I COULD PUT TETRIS ON IT AND HE SAID NO

[clever flint]

so r u telling me you never rearrange equations

because your calculator can solve stuff anyway

even if it isnt in the form x = blah blah blah

[opal grove]

Yes unless if the teacher says to do so before the calculation

[clever flint]

wowowow

thats a good shortcut

[opal grove]

But these calculators are very expensive

~$250-300 AUD

[clever flint]

damnnnn

ye well idk why the calc said no solution

but like at least it worked when u rearranged ig

[opal grove]

Lwk I need a verbal explanation as why everything is what it is

[clever flint]

like how the formula works

[opal grove]

Yes and the whole rearranging thing, like why is it the way it is

I get some of it but not all of it

[clever flint]

its just a regular algebraic manipulation

just like how if you have x + 5 = 14

you subtract 5 on both sides

to get x = 9

your trying to solve for x

in the case of the cosine rule

you trying to solve for the unknown angle 'A'

so put all the other values on the other side of the equation to isolate angle A

then u can put in the numbers and it comes out to A = 117.8

[opal grove]

So we're trying to remove the A from the cosine part

[clever flint]

well thats the last step yes

but the full thing is your trying to get A on the LHS and all the other stuff on the RHS

so you begin by subtracting b^2 and c^2 from both sides

then dividing by -2bc on both sides

then taking the inverse cosine of both sides

so A is left by itself

[opal grove]

I hate how I'm unable to understand this

Bc I learn best visually, step by step, and verbally 😭

[clever flint]

yeah i understand

but its no different to x +5 = 14

like if i had 4x -22 = 2x

how would u do this

[opal grove]

Fahh hold on

I haven't done algebra in two years uhhh hmm

Like this part of it

[clever flint]

( 4x - 22 = 2x ) solve for x

[somber coyote]

WesStreet99

[clever flint]

nice and big writing for u

[opal grove]

Hmm

Uhh idk 😭 can u pls explain

[clever flint]

yeah sure

so the question asks us to solve for x

the strategy to solve linear algebraic equations like this (linear meaning the unkown variable x has a degree of 1. not x^2 or x^5, only x) is to move the letters (variables) to the LHS and the numbers to the RHS

so we have 4x - 22 = 2x

lets move the letters to the left side

there is a 2x on the right so we need to move that

how can we do this?

we can subtract by 2x from both sides

what u do two one side YOU MUST DO TO THE OTHER

that keeps the LHS and RHS equal

so we subtract 2x from both sides so we have 4x - 22 - 2x = 2x -2x

now we can combine like terms

4x - 2x is 2x

and 2x - 2x is 0 of course

so now we have 2x - 22 = 0

now lets move the numbers to the right

add 22 on both sides

2x -22 + 22 = 0 +22

so 2x = 22

now we divide by 2 both sides to get x by itself

so x = 11

does that make sense?

we use inverse operations to manipulate the equation to get x on one side, and the number on the other side

[opal grove]

Girl hold on, I'll read and process this in like 30 min. I need to shower

[clever flint]

[opal grove]

My hair takes like 5-7 hours to dry and it’s almost 6pm my

Mb*

[clever flint]

5 - 7 hourse

hours?

[opal grove]

Yep

[opal grove]

I think I’m remembering

I think I get it

[clever flint]

okok good

its the same thing with cosine rule

or we're doing is manipulating the equation

to solve for A

[sly urchin]

Guys i need help, i know this term in my native language but not in English

[fervent stump]

Mind sharing a diagram? I did not understand reflection of BC or I don't know the term maybe then

[sly urchin]

What don't you understand about the reflection?

[fervent stump]

Like reflection of the side?

[sly urchin]

Of the line, yeah

[fervent stump]

Then I don't know it

[sly urchin]

Maybe i worded that badly

[gilded kayak]

what's your native lunguge?

[sly urchin]

Given ∆ABC. The points E on AC, F on AB. ∠AEF=∠ABC then EF is called the...

[sly urchin]

Vietnamese

[sly urchin]

Basically it's a line parallel to the tangent at A of (ABC)

[gilded kayak]

sry not mine , but you can translate the term by ggl translater

and am not that edvansed in math to understand whay your talking about

[sly urchin]

Well i tried, it sucked

[gilded kayak]

ask chat GPT

well somtimes he get it wors then google so ...

[orchid rampart]

that's Thale's

nah wait

[sly urchin]

That's the first thing i did

And it said that's not a term

Like wtf

[orchid rampart]

you mean this?

[sly urchin]

I found the term

[orchid rampart]

send it

[sly urchin]

Antiparallel

[orchid rampart]

oh

[sly urchin]

By looking at wiki pages

[sly urchin]

You know it?

[gilded kayak]

i don't now how to help sry bud

[orchid rampart]

no

[sly urchin]

Nah i got it

Well that's fine, anyways

[gilded kayak]

good for you

[sly urchin]

The term i was looking for is antiparallel

[gilded kayak]

alr have a good day mate

[nocturne pumice]

Hi

im new here

please can someone solve the first part

[tranquil iron]

Rookie/Dumb question: Can you do trigonometry without a calculator?

[shrewd peak]

ofc

[orchid rampart]

depends

[dark sparrow]

you can but calculations will be limited

[sly urchin]

#geometry-and-trigonometry message
Can anyone guide me on how to solve the first part using Dumpty point? I saw that T is the A-Dumpty of ∆AEF but couldn't prove it

[night sorrel]

Man, where did you get those questions?

[long geyser]

Is this good?

[cunning lion]

you cite pythagorean theorem, but that only applies to right triangles. how do you know they're right?

[long geyser]

Omg

I thought that applied to every triangle, I forgot

Is it clear from the image that it's 90 degrees? Or do I not have enough information?

[cunning lion]

no, you should only assume an angle is right if it is marked as such

[long geyser]

Ty!

I'm gonna play the video and see what he did, I don't know how to move forward.

Tysm for checking it! 🫶

[strange pond]

You may find it a good exercise to try to generalise the pythagorean theorem to non-right triangles (as in the cosine rule)

[tidal niche]

if you cannot do basic trigonometry values without a calculator, you cant do trigonometry

like

but trig in general

calculator is recommended

memorise some values tho

[devout gorge]

if someone is determined enough i believe they should be able to get approx values through sine series and coscie series iirc

[ebon heart]

How does calculator solve trigonometry? Like what if I have to solve for tan(0.028rad), how'd I do it?

[exotic yarrow]

https://math.stackexchange.com/questions/395600/how-does-a-calculator-calculate-the-sine-cosine-tangent-using-just-a-number

[lime valley]

gng i was having a lot of issues with when to add / subtract pi in inverse trigno metric functions

can any1 recommend somethin

[silent plank]

understand the unit circle
reference angles, supplementary identities
periodic properties

should be plenty of resources online

you should claim a help channel if you get stuck on a specific question

[atomic flame]

Does anyone know the geogebra app

If so, chat why isn't my circle thing closing

There's sum wrong with my sequences

[dark sparrow]

if only your screenshot showed what your sequences even are lmfao

cause rn it doesnt

do you have a laptop you could do this on, instead of a phone screen?

[atomic flame]

I was planning on showing them if someone that knew geogebra appeared

[atomic flame]

Unfortunately no but I'll try to screenshot it

Or just send it here

Can you open it?

[dark sparrow]

apparently i can't bc my version of geogebra is too old lmfao

oops

[atomic flame]

😭😭

[atomic flame]

https://www.geogebra.org/m/mq6cdmsj

How about now

[dark sparrow]

hm ok now i can open it but can't seem to figure out the issue

[atomic flame]

I would expect it to be closed but there's that gap for some reason. It's okay if you can't figure out, thank you for trying

I don't wanna just connect with segments cuz the professor will see it

I'll send an example he showed us

https://www.geogebra.org/m/fvbnn9r7

[sly urchin]

Are you crocheting with math?

[atomic flame]

Atp idk what bro wants us to do ✊
The subject is called geometrical representations

[sly urchin]

Usually in problems where variables are triangle side lengths, the Ravi subsitution can be used. What about with quadrilateral side lengths, are there any similar subsitutions?

[atomic flame]

My bro, I really wish I could understand you rn

I barely knew what I was doing when making this

I think I'll just let it be, even if it's not closed

I have no clue how to close it

[sly urchin]

That wasn't directed to you, i was asking my own question

[atomic flame]

Alr alr

[echo storm]

hello can someone like help me with geometry im failing and i like cannot understand proofs at all

[silent plank]

show specific questions you're stuck on
for uninterrupted help its recommended you claim your own channel #❓how-to-get-help @echo storm

[obsidian hornet]

In terms of this picture i think its a good way to start from proving that JHI is an orthotriangle of LMG or AJ is an external angle bisector of HJI

[obsidian hornet]

.

[sly urchin]

Uhmm... I'll send a picture with the original points

[sly urchin]

Well that's basically proving AJ ⊥ JG (or in the original problem: AT ⊥ AD)

If that could be proven then T is the A-Dumpty point of ∆AEF. Note that D is the circumcenter of (AEF)

I got 2 solutions for the first part of this problem btw (1 by calling a few more points and using similar triangles to angle chase, and 1 by complex coordinates), i just wanted to use the Dumpty point

[obsidian hornet]

I got solution by inversion in G

[obsidian hornet]

Can u share complex solution?

[sly urchin]

Let o=0, a=1 => e=2b-1, f=2c-1. Let X,Y be the reflection of T through AB, AC then x=1+b-bt̅, y=1+c-ct̅.
We also have EX//AM//FY so (x-e)/(m-a) and (y-f)/(m-a) ∈ ℝ
=> (2-b-bt̅)/(m-1)=(2-b̅-b̅t)/(m̅-1), (2-c-ct̅)/(m-1)=(2-c̅-c̅t)/(m̅-1)
=> (c(2-b-bt̅)-b(2-c-ct̅))/(m-1)=(c(2-b̅-b̅t)-b(2-c̅-c̅t))/(m̅-1)
Some simple calculations give t= -m(m̅-1)/m̅ (m̅-1)
=> |t|=1 => T ∈ (O)

[sly urchin]

I'm terrible at inversion, ugh

[sly urchin]

Radius GA ?

[obsidian hornet]

GP*GA

[shut thorn]

if sinA+cosA = sqrt(2). Prove that tanA+cotA = 2.
This just feels right but i literally cant prove it

do i multiply both sides by something or what

if i uhh square both sides. You get 1+2sinAcosA = 2

[orchid rampart]

Express tanA + cotA in sin and cos

[shut thorn]

sinA/cosA + cosA/sinA
-> sin^2(A)+cos^2(A)/cosAsinA
-> 1/sinAcosA

oh oh wait

[obsidian harness]

sin A + cos A = sqrt2 sin(A + pi/4)
There's a way without this but that's easier if you know

[shut thorn]

what if i

1 = 2sinAcosA

god damnit

well uh

thanks i guess

[orchid rampart]

This too

[shut thorn]

i do NOT wanna mess with radians yet

[sly urchin]

Well, you should though

You can't learn trig and choose not to learn about radians

[shut thorn]

theres not a single question on radians in my book. For trigonometry that is

[shut thorn]

yet 😭

all i know till know is how to convert radians to deg

[sly urchin]

Man i'm starting to like inversion

I know there are some people obsessed with inversion out there and I wish to be one of them

[obsidian hornet]

I'm not fan of it but its very useful in some moments, additionally if u use inversion and symmetry

[olive kindle]

Is anything going wrong, i am a bit stuck here

[sly urchin]

The problem i sent has a different solution using inversion too btw

Around A, "ratio" AB.AC (idk the word in English)

[obsidian hornet]

Like sqrt(AB*AC)? Its logical due to inversion+symmetry translate isogonal lines such as median and symmedian

[sly urchin]

Eh... It's the inversion around A that turns (O) into BC

[obsidian hornet]

Any inversion in A turn O into BC

[sly urchin]

Turns B into C?

Turns F into the midpoint of AB

[obsidian hornet]

Yes

Its inversion and symmetry with radius sqrt(AB*AC)

[craggy relic]

What identies do you know about trig functions and squares

Personally I would manipulate that fraction a little bit before I plug anything in

(but you could also just continue from what you have once you work out how sin and cos are related)

[pearl vigil]

am I correct?

[shut thorn]

yeah that works

[pearl vigil]

😀

[karmic spindle]

im using khan academy to prem myself for trig within 4 weeks, should i use something else or am i taking the right path?

[pearl oxide]

sin(x) = opposite/hypotenuse
from that, sin(x) * hypotenuse = opposite

let me name the sides of your triangle rq

sin(a) = xy/xz
cos(b)= xz/1

so sin(a) * cos(b) = xy/xz * xz which is xy which is... that

what do you mean "assume"?

you cant assume. you can only conclude it from given information (unless its already given...)

if we knew that xz and xk are cos(b) and sin(b) then from pytha's, we can conclude that the hypotenuse^2 = sin(b)^2 + cos(b)^2

which is = 1.

[buoyant grove]

im trying to derive trig sum and differences from scratch using the idea that the sum of 2 angles can be illustrated as drawing an initial angle, and then another angle ontop of it. so would the initial triangle, A have the hypotenuse 1?

[pearl oxide]

can I see what you started with?

like the first two triangles you drew

[buoyant grove]

lol

[pearl oxide]

and then you went on and assumed the hypotenuse of the triangle with angle B = 1

[buoyant grove]

And then I did this

oops

bottom should be cosA

[pearl oxide]

did you... assume that?

[buoyant grove]

yea

is that troll

[pearl oxide]

I haven't tried proving the trig angle difference/addition

so I dont know.

but yeah you practically dont assume that a right triangle has a hypotenuse of 1...

but if you're tryna build smth up from there

then again, I never tried this before.

perhaps wait until another person sees this.

best of wishes.

[buoyant grove]

okok thanks for ur time and patience!

[atomic python]

does anyone have a sheet which I can use to remember n practice trig identities

[hollow lily]

Check the pinned messages on this channel

[jovial forge]

hey anyone here can help me p with this question

i mean i cant even think an approach

[floral stirrup]

what the fuck

where do i even start

[sick wraith]

think first

what do we know?

[jovial forge]

thats what i am asking

[sick wraith]

so you are clueless?

no idea how to start

was this in a test?

@jovial forge
what is the value of bc?

[jovial forge]

yes its a sample que

[jovial forge]

the value BC is given

[sick wraith]

yes what is it?

if i just solve it then u wont understand

[jovial forge]

14

[sick wraith]

yes

[jovial forge]

bro listen i have tried it once

but i was stuck in a loop

[sick wraith]

what is the incircle property?

[jovial forge]

i used tanget theorems

and this question will be solved with some topics which is not in our syllabus but its in sample paper so ig i need to do this

[atomic python]

i wanted like question sheets

if u have

[orchid rampart]

two tangents drawn from a point to a circle are equal

use this

[hollow lily]

Well I do, but it isn't in English, have you tried find one on google?

[timber kraken]

Good luck I’m still learning proofs

There’s SO MANY

[unique roost]

What is intercept theorem of quadrilaterals

[sly urchin]

Nice

[atomic flame]

I figured it out

[pearl vigil]

can anyone give me a hint on how to make an approach for this problem

[sudden zinc]

1+1 = 2

[hollow lily]

I would start by making a rough sketch

[pearl vigil]

what do u mean

[sudden zinc]

1+1 = 2

[silent plank]

Don't troll.

[upper karma]

Man ita easy

I have that on my textbook

Make the first eqn by the area of traingle in terms of 1/2 ×bh and make the second as the area of triangle which was like sqrt[(s-a)(s-b] .... i forgot the formulae name. And since there is only one variable (due to tangent theroem) you can solve it

[plucky notch]

Yeah that formula is √{s(s-a)(s-b)(s-c) where s is the half perimeter of the triangle and a b c are it's sides

[upper karma]

[plucky notch]

It's called heron's formula aka brahmagupta's formula

[upper karma]

Yeah exactly

My mind was saying euclids formula god knows why

[plucky notch]

Ohh

[harsh gale]

Use the property of auxiliary circle and write the coordinates in paramentric form

I will send you how

[timber kraken]

More proofs 💔💔

This one was fairly easy

[sharp inlet]

I js taught myself trigonometry

But only like

[winged oyster]

since the unit circle's radius is just 1 that means you can always find what y is based on what x is usibg pythagorean theorem

since itd just be x²+y²=1

[storm zephyr]

you can't use $\frac{1}{2}bh$ because h is not given

[somber coyote]

Flax.

[upper karma]

you can create the circles radius by construction

and take it as h

[storm zephyr]

wdym create the circle's radius? the inradius is given to be 4cm you have to use A = rs

[upper karma]

join BO,OE,OF and try to do it

sorry i mean OB,OD,OC,OE,OA,OF

do you want me to do it? If you are having real trouble doing it

[storm zephyr]

ok do it

btw AD, BE and CF are NOT altitudes

[storm zephyr]

brahmagupta's formula is for cyclic quadrilaterals, heron's formula is just a special case.

[storm zephyr]

ohh you mean using pythagorean theorem to find BO, AO and CO, I didn't get that b4

eitherway point O does not lie on AD or BE or CF

[orchid rampart]

we can use area of triangle = semiperimeter * inradius

equate that with heron

[storm zephyr]

The points D, E and F are touchpoints of the incircle inside of the triangle. Point D can therefore be denoted as $T_A$ because it's opposite to the vertex A, and Point E will be $T_B$ and Point F will be $T_C$. Therefore the lines $\overline{AD}$, $\overline{BE}$ and $\overline{CF}$ become $\overline{AT_A}$, $\overline{BT_B}$ and $\overline{CT_C}$ and I define these lines to be "Gergonne Cevians" (not Gergonne Line). Gergonne Cevians can pretty much never be Altitudes if the triangle is non-equilateral.

[storm zephyr]

yeah

[somber coyote]

Flax.

[upper karma]

no not pythogoras, maybe you can but that wasnt my way

i will just do it and send you nvm

am too lazy to type

[steep mango]

Can anyone tell me the shortest formula to find prime numbers?

[harsh gale]

just square sinA+cosA, you will get
(sinA+cosA)² = (√2)²

sin²(A)+cos²(A)+2sinAcosA= 2

{we know sin²(A)+cos²(A)=1}
So, 1+2sinAcosA=2
2sinAcosA=1
sin2A=1
sin2A=sin(π/2)
2A=(π/2)
A=(π/4)

tan(π/4)+cot(π/4)= 1+1=2 (Proved!)

[timber kraken]

All primes greater than 3 are in the form 6k (plus or minus) 1

So for example plug in 1 for k it’s either 7 or 5

If you plug in 2 it’s either 11 or 13

and so on

Does this help @steep mango

$6k \pm 1$

[somber coyote]

varq

[tranquil iron]

Does ts bot generate latex?

Pretty cool

Totally stealing it to my server(dead)

[exotic yarrow]

yes

the texit bot renders latex

and also runs wolfram

amongst other things

[ripe flame]

yall help me

im getting so caught up on this easy ass question its only 2 marks i dont know what im missing

c)

[exotic yarrow]

you should be able to just equate coefficients, no?

(since a and c are non-parallel)

[ripe flame]

im overcomplicating things but its confusing me because i dont have 2 expressions for op

[exotic yarrow]

what

[ripe flame]

and c is stating that i should

[exotic yarrow]

don't you have one from part a and one from part b

[ripe flame]

wait

oh yeah thats where stupid mistake

i answered for AP

😭

[exotic yarrow]

rip

[ripe flame]

so then

you have OP = wavelength(a+1/2c) and weirdu(-a-c)

is it do with the c being half less in the second part?

[exotic yarrow]

I assume you're referring to lambda and mu

um

[ripe flame]

i thought that symbol meant wavelength

anyway thats not important

just call them k and z

or something

[exotic yarrow]

how did you get $\overrightarrow{OP}=\mu(-\mathbf{a}-\mathbf{c})$?

[somber coyote]

Civil Service Pigeon

[ripe flame]

oh god mightve done too much math today

horrible mistakes

should have -a + c right?

[exotic yarrow]

read carefully

[ripe flame]

yeah...

[ripe flame]

yes

so op = u(-a +c)

and y(a+c/2

right? 😭

[exotic yarrow]

What?

\begin{align*}
\overrightarrow{AC} &= -\mathbf{a}+\mathbf{c} \
\overrightarrow{AP} &= \mu(-\mathbf{a}+\mathbf{c}) \
\overrightarrow{OP} &= \overrightarrow{OA}+\overrightarrow{AP}
\end{align*}

[somber coyote]

Civil Service Pigeon

[ripe flame]

oh i forgot the second part holy crap

right i think i have it now..

how do i make one of them images?

[exotic yarrow]

what is "images" supposed to mean

hello?

[ripe flame]

sorry man i was trying to figure it out , i mean the math you are uploading

is it code into a discord bot or something?

texits the name

[exotic yarrow]

https://top.gg/bot/510789298321096704

[ripe flame]

thanks i already got this figured out though

was worth the time i spent looking

sorry, my brain isnt super in gear, is it like this ?

$$\overrightarrow{OP} = \lambda(a + \frac{1}{2}c)$$

$$\overrightarrow{OP} = \mu(-a + c) + a$$

[somber coyote]

danny ✭🌈

[ripe flame]

can it delete that

not really needed

sorry i meant, $\overrightarrow{OP} = \lambda(a + \frac{1}{2}c)$

$\overrightarrow{OP} = \mu(-a + c) + a$

right

.

this pmo

i dont need it to keep repasting what ive already wrote

how do i disable it for me

[exotic yarrow]

bot auto-activates whenever it sees latex

you can click the trash can emoji to delete it tho

until it times out

then you're just kinda stuck with it there

[cunning lion]

you can disable the latex auto detection if you want

[exotic yarrow]

oh

did not know that

[lime crown]

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

[untold trout]

1

How do I utilize the ratio of lengths here

[hollow lily]

The typesetting looks like a JEE problem

The only way I can think of is using trig

[upper karma]

Forgive my bad handwriting

[untold trout]

Hmm

[hollow lily]

I mean it's pretty obvious with unit circle

The only thing to worry about is computing cos(pi/12)

[strong sierra]

class 10 cbse ch cirlces

[upper karma]

Ye exactly lol

[red raven]

You write like you're in 10th grade

[slate needle]

m

[rustic mauve]

Chill bro

[nocturne breach]

As someone in 10th grade I can confirm that it’s good handwriting 👍🏾

[wind iris]

Fr better than my shi ahh writing.

All the specific notations and arrows makes it look like he is a 10th grade CBSE student

[nocturne breach]

I don’t know what cbse is but I agree

[cunning nebula]

hey guys is this proof good

[craggy relic]

The calculation of D_qr is overkill

It's correct but there is a much simpler way to see it

And technically the should be an absolute value for the length of D_qr which goes away when you square it

[slim plinth]

I love it

What about proof of m1=m2

[cunning nebula]

For parallel lines?

[slim plinth]

Yes

[cunning nebula]

If you have the two lines in y-intercept form, just compare their respective slopes and see if they are equal

If you don’t have their slopes, you can make the two equations for y equal to another

[slim plinth]

Ik

$y=m_1x+b$ and $y=m_2x+b$

[somber coyote]

Jake

[slim plinth]

Then
$m_1x+b=m_2x+b$

[somber coyote]

Jake

[cunning nebula]

If $x \nexists$ for $m_1x+b_1=m_2x+b_2$, then $m_1 = m_2 \land b_1 \neq b_2$

[somber coyote]

Koja Mori

[cunning nebula]

@slim plinth

[slim plinth]

Yes

[stiff spire]

y need prove tho

[tacit hawk]

State board
Basically in india when u give 10&12 exam
The paper are checked by government officials and exam is held at different different school across the city

Also permanent record is kept in case u lost ur result

[nocturne breach]

Oh cool

[vestal dirge]

yeh same in Pakitan too

[spring lion]

Help

I got 3 right

Now I gotta solve this one.

10/10√3

I'm not sure with this one.

√3/3 right?

Now I find arctan of that.

In radians!

0.52.

Is that right?

Now...

Is that the right answer?

[spring lion]

Since this is in 4th quadrant...

Hmmm.

[obsidian harness]

nope

[spring lion]

Oh

[obsidian harness]

you need the exact form

[spring lion]

Third

Third third

Lmao my mind slipped for a sec

My bad

[obsidian harness]

so as a fraction of pi, this is?

[spring lion]

No I know I know I meant if the radian was right at least.

[spring lion]

Ahhh. The question did indeed have a pi.

π/6?

[obsidian harness]

yep!

[spring lion]

Quick maths lol

So

Cos and sin will be π/6?

Now I just need to find r.

[spring lion]

Is r 20, by chance?

[obsidian harness]

yeah, so without thinking you just do (10 sqrt3)^2 + (10)^2 then sqrt

if you recognise the triangle though, the 1 - sqrt3 - 2 one

[distant umbra]

The arguement of this one is in the 3th quadrant since the angle is arctan(pi/6) so the principal argument is pi/6 - pi so you get

-5pi/6

[obsidian harness]

you just scale that up by 10

so 10 - 10sqrt3 - 20

20 is the hypotenuse and that's r

[distant umbra]

The modulus is sqrt(400)

20

[spring lion]

Ohhhh

30 60 90?

[obsidian harness]

yep!!

[spring lion]

Yeah I did mention the third quadrant, almost forgot about that man

[gray light]

What grade math is this specifically I'm not asking for help but I do just want to know?

[distant umbra]

Its basic complex numbers

10th or 11th

[gray light]

Thanks Elfie!

[spring lion]

Why did we subtract pi?

That part I didn't get about tan honestly

[obsidian harness]

okay, the period of tan is 180 degrees

if you think about tan t = (sin t)/(cos t)
= y/x by the unit circle definition

what this means is that tan represents the slope

[distant umbra]

The arguement in the 3rd quadrant is π + θ

Since the principal argument is between -pi and pi we can subtract 2π from 7π/6 or just write it as θ-π

[obsidian harness]

30 degrees and 210 degrees have the exact same slope

it's just the arrows are pointing in opposite directions (from the origin)

so that's why you need to check the quadrants

[spring lion]

Ohhhh

[obsidian harness]

it's the 3rd quadrant, so you don't want 30 deg

you want 210 deg (=7pi/6 rad)

[spring lion]

You go with a 210

[distant umbra]

Yeah you just remember the extreme values and add/subtract acxordingly

Similar to how trig equations are solved

[obsidian harness]

the value of tan gives you the slope, but every slope has 2 equally valid angles

if that makes sense

[spring lion]

Yeahh

[distant umbra]

Yes but principal arguement is between -180 and 180 so you subtract 360 from it

[spring lion]

Wait let me understand this

[obsidian harness]

it's not hard to draw it if you want to have a proper think

[spring lion]

For tan?

[obsidian harness]

draw 30 degree and 210 degree vectors

they make a straight line

same for any angles 180 degrees apart

[spring lion]

Yeah I can visualize that!

[distant umbra]

For complex numbers

A number can have infinite arguements which is why we define the principal arguement between -180 and 180

[spring lion]

Ohhh

Okay

So since that's 2π

But why did we subtract it 🙏

[spring lion]

.

[spring lion]

Wait so let me think

We found a π/6, but that's the first quadrant.

[distant umbra]

Adding/Subbing 2pi is basically doing a full rotation

i.e you have the same angle

Since 7pi/6 was greater than the 180 we subtracted 2pi to get it between -180 and 180

[spring lion]

So we just go to the same angle

Which is 180 degrees apart

So we subtract π

Or nah

[spring lion]

Hmmm

[distant umbra]

We take the +ve x axis as a 0 degree angle

Our theta was pi/6

But we need the theta to be in the third quadrant

So we start at the third quadrant(-ve x axis) which is an angle of pi and add a further pi/6 to it to get our required angle

Ehich is how we get 7pi/6

But 7pi/6 > pi

Generally we consider the principal arguement to be between -π and π

So to get the principal argument we rotate the angle by 360 backwards

i.e subtract 2π

[spring lion]

OHHHHH

[distant umbra]

So we get 7π/6 - 2π

= -5π/6

[spring lion]

So this is basically

Same thing as π/6?

[distant umbra]

Texhnically in the question they said the angle can be between 0 and 2π so 7π/6 is actually correct

But the majority of complex numbers usually asks for the principal arguement

[distant umbra]

Yes

[spring lion]

Okay

[distant umbra]

Its the equivalent angle of pi/6 in the third quadrant

[spring lion]

Okay thank God I get it lol

[spring lion]

Yeah

[distant umbra]

For example if you ever want to find where sinx is -1/2

We know sin is negative in the third quadrant

And sinx = 1/2 at pi/6

[spring lion]

Wait let me see

[distant umbra]

So you just find the equivalent of pi/6 in 3rd quadrant

Which is 7pi/6

[spring lion]

Yeah

[distant umbra]

Another valid answer would be 2pi - pi/6

Which is the equivalent in the 4th quadrant

11pi/6

[spring lion]

Wait I didn't get that

Or wait

Adding pi

Ohhh wait adding or subtracting 2pi is the same thing anyway

[distant umbra]

Yes

[spring lion]

And to find the negative equivalent you can just add one pi

[distant umbra]

We use these logics in physics often in the form of phasar diagrams

[spring lion]

Sounds exciting lol

[distant umbra]

Depends on the quadrant

For example for the seconr quadrant

You can either do π-θ or π/2 + θ

[spring lion]

Yeah

So the answer is 20(cos(7π/6)+i sin(7π/6)?

[distant umbra]

Yeppp

[spring lion]

LET'S GOOOO

I spent 40 minutes getting one right btw before this

😭

And this is BASIC apparently

[distant umbra]

As far as complex numbers go yeah

Beautiful thing to study but it can get difficult

There's a portion of questions that require a lot of calculation

And later on it merges into conic sections and geometry

[spring lion]

I see.

I'm gonna take a 20 minute break man, thanks for the help.

[distant umbra]

Np

Feel free to ask this is also the topic im currently teaching my class

[spring lion]

Damnnn that's awesome LOL

[distant umbra]

Yeah we're doing the triangle inequality stuff

[spring lion]

Chat...we might be a little cooked.

I have GENUINELY no idea what the hell to do here.

I'm gonna rewatch the explanation then come back.

Nope. I don't get it. 🙏

[craggy relic]

Try putting it on polar form

Sqrt3 should remind you of 30-60-90 triangles

(or the hint literally lets you skip figuring out the polar form and skip straight to the value)

[spring lion]

Hmmm.

[spring lion]

Oh

2*(cos30)+i sin(30)?

Correct?

But we multiply that by 5

[craggy relic]

sin 30 is certainly not 1

But the form you want is e to the power of something

[spring lion]

Polar form is this though no?

[spring lion]

Where did I make a mistake?

[craggy relic]

Do you know what e^(ix) is?

[spring lion]

Yes

[craggy relic]

You have mismatched parents so it's hard to tell what u meant lol

Oh wait

No it's just wrong

Your 2 needs to multiply the whole thing

[spring lion]

Is that all?

[craggy relic]

Anyway from there you can convert to re^ix

[spring lion]

Okay

[craggy relic]

Which makes taking the power trivial

Then convert back

[spring lion]

Though so far I've only dealt with polar and rectangular how do I write it like that?

So let me write it again

2*((cos30)+i sin(30))?

[craggy relic]

Yes

[spring lion]

And well

To the 5th power

With powers you just add the argument that many times

[craggy relic]

Do it that way then

[spring lion]

And the modulus is just multiplied normally

[spring lion]

32*((cos150)+i(sin150))

This feels wrong tho

I feel like I'm missing something?

No?