geometry-and-trigonometry

[halcyon wave]

Can someone explain the ambiguous caseto me

@silent plank

I know everything about this unit BUT the ambiguous case 😭

i was out the day it was taught and i watched my teacher's video lesson and it made no sense

[silent plank]

You shouldn't ping specific people without their permission

$\textbf{ambiguity of the sine rule / law of sines:}$ \
there are two solutions to
$$\sin(k) = c$$
where $0<c<1$ \
for $0 < k < 180\deg$ or $0 < k < \pi$

[somber coyote]

ραμOmeganato5

[halcyon wave]

Oh really? Sorry

[halcyon wave]

I can't really understand this explanation

I'll show the way it was taught to me one sec I'll send

The only thing im struggling with is how to know if there is a second triangle

and the way my teacher explained it as shown here kind of makes no sense because when I tried that way on another question I got it wrong

so I don't really get it

[silent plank]

firstly do you understand that there are two angles
that when sin is applied will give you the ratio

[halcyon wave]

yes I understand that

[silent plank]

ok then the part after that is checking whether its possible for the angles of the triangle to sum to 180°

in that above example, the sum exceeds 180° (the angle sum of a triangle) so that isn't possible

so B can't be 148.5°

[halcyon wave]

yeah but then on other questions i do that strategy and it says it's wrong

and then a different strategy is shown

like for this one my teacher for some reason did 180-C+A instead of 180-B+A like they did before

[silent plank]

use whatever angle(s) you have

what you know about a,b,c,A,B,C are different in every question.
they're just variables.
some questions won't even use those letters

180-B was in the previous question because B was the unknown angle when applying sine rule

180-C was used in this question because C is the unknown angle when applying the sine rule

same strat

whos saying its wrong

[halcyon wave]

oh

ok thanks

dang that actually helped a lot i was tryna figure this out for 3 days 😭

thank you deadass bro

[stoic field]

Yo I have a small terminological question regarding geometry. If I have a hypershape that consists of the cross product of a hypercube and a simplex of some dimension, would this be considered a "hyper prism"? Since a classical 3d prism is the crossproduct of a 1D cube and a 2D simplex so to say, but I've found some definition of "hyper prism" online that doesn't seem to match my shape

[silver jolt]

no i wont call it a hyper prism at all

i would call it a polytope

a duoprism most likely

[stoic field]

I see, thanks :)

[quaint chasm]

this is probably obvious but are there equivalents to pythagorean triples in higher dimensions?

like a sum a²+b²+c²…. = z²

such that all a,b,c… z are integers

i would also assume its not possible for an exponent above 2 since fermat’s last theorem, but maybe it works differently in higher dimensions?

oh wait ig there’s the trivial case of 1 for all axes

[exotic yarrow]

You can extend it to an arbitrary number of integer dimensions.

[quaint chasm]

thanks

[exotic yarrow]

[slim plinth]

I did this

[dusky jacinth]

@agile saddle

[full light]

If sin(-x)= -sinx , what is tan(-x)=????

[agile saddle]

You scared me

[dusky jacinth]

mb

[agile saddle]

HAHA

[tiny acorn]

Try

Like

[modest mulch]

geometry dash

[tiny acorn]

Using tan x = sin x/cos x

And the even or odd properties

[trim stump]

Probably -tan(x)

[hard ore]

am i on the right way and how do i solve 1-64/49

[exotic yarrow]

$(-8/7)^2=64/49$, not $-64/49$.

[somber coyote]

Civil Service Pigeon

[exotic yarrow]

negative times negative is positive

I stopped reading there

[hard ore]

thanks

atleast something

[honest pumice]

anyone familiar with geogebra? i need to graph (rcosO,rsinO) = (x,y) for 2 <= r <= 3 and pi/4 <= O <= 3pi/8 in the xy place but the most i was able to get was something like

i THINK it looks like a rectangle but 1% of my head thinks it can be a parallelogram

[exotic yarrow]

This is the parameterisation of a circle

[honest pumice]

yeah, it's a portion of it, thank you

whoa

<@&268886789983436800>

[spare ibex]

I graduated chat

Still a grade level ahead (still in honors i didn't get demoted ⭐ )

[marsh marlin]

orz

[next helm]

why dont we use tau instead of pi

[grave pond]

Tradition.

[spare ibex]

Wa?

[cerulean blade]

@cerulean merlin

[cerulean merlin]

@rigid crescent

[light pumice]

Can someone help me with a trig question?

[lime crown]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[tiny acorn]

!da2a

[lime crown]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[covert sun]

Hey

Any short trick for trigo proof?

[silent plank]

what sort of trig proof

[covert sun]

For these kinda questions

[silent plank]

,rcw

[covert sun]

I asked short tricks to know what to do to make these easy

[silent plank]

maybe express stuff in terms of sine and cos first

[covert sun]

Yeah

[exotic chasm]

btw should we make trig id short notations\
unclear values:\
$\sin(x) = s\
\cos(x) = c\
(\sin(mx))^n=s_m^n\
(\cos(mx))^n=c_m^n\$
clear values \
$\sin(k)=S_k\$
you can use t for tangents

[somber coyote]

tsuitachi (tuitati)

[exotic chasm]

only for proofing trigonometric identities! not for geometry!

[tiny acorn]

What about cot,sec,csc

[exotic chasm]

1/t, 1/c, 1/s

unfortunately this only caters sin cos tan

oh, special for reciprocals: write as power form, but for -1 power, you can just write -

$\csc(x) = s^-$

[somber coyote]

tsuitachi (tuitati)

[exotic chasm]

no. 30 can be written as the\ nonformal trigonometric identity notation\
$\frac{t}{1-t^-}+ \frac{t^-}{1-t} =1+ t+ t^-$

[somber coyote]

tsuitachi (tuitati)

[gritty tree]

sin cos tan

[vague plover]

<@&268886789983436800>

it’s in every channel with image perms

[tiny acorn]

Mrbeast scams

[exotic yarrow]

In the future, please show what you've done so far when asking for help along with any other relevant context — it gives us more to work with and saves time from explaining things unnecessarily. \

I think it would help to substitute $u=x+y$. This gives $y=u-x$ and $2x+y=x+u$. Then,
$$\sin(x+u)-2\sin(u-x)=0.$$
What do you get when you expand this using the addition/subtraction identities?

[somber coyote]

Civil Service Pigeon

[dawn creek]

oh sorry it's just i solved it liked the solution and thought everyone here would appreciate it too

[trim stump]

I WANNA learn how to use this bot, I always write mine in paper💔

[tiny acorn]

#latex-help the pins should help

[wheat wing]

How can i draw here the shape i want to discuss about?

[tiny acorn]

I would recommend geogebra

If you just want to sketch then any online whiteboard/paint clone is fine tho

[wheat wing]

its like i draw there later take screenshoot of it then post here?

[tiny acorn]

Yea

[wheat wing]

okay

you know what i didnt like this geogebra stuff

let me use paint

the shape is this

AB being the diameter and the small circle inside is tangent to bigger circle

the given equation is in the image: 1/r = 1/a + 1/c

does anyone know this theorems name?

How to browse for stuff like this by the way

@tiny acorn any idea?

[normal rapids]

i think ive seen this somewhere...in french its called moyenne harmonique..maybe its a literal translation too idk?

so like forgive me

[wheat wing]

moyenne harmonique huh

[normal rapids]

i didnt see the pic

[wheat wing]

no problem at all

[normal rapids]

thats so tension and stuff not this

its not it

[wheat wing]

well okay

[normal rapids]

math text books

i mean

you can find

names

of

stuff

there

catch my drift?

[wheat wing]

yeah i understood but

i mean how can i find them easily

looking in math textbooks would take hours

[normal rapids]

ur right

[wheat wing]

i also saw this in a math textbook but there is no name given for this theorem

i have found a video regarding to this theorem

https://youtu.be/ENo8_ZKNEgw

@normal rapids How do they teach mathematics in france?

with the proof and reasoning backed or just formulas?

[normal rapids]

um

[normal rapids]

oh

[normal rapids]

well
ig its like: problem--> solution-->then formula-->properties

[wheat wing]

hmm

in turkey we get properties -> formula -> problem -> solution

[normal rapids]

ohhh

i think most of the time we get an exemple then we must find ourselves the theoreme(if one) then they give us a the formulas then how we could use those formulas

not to be detailed...

um yeah and then they give how to get to that formula...and then what you could get from it

theres prob words for what im stating but cant think of any...

[wheat wing]

you meant proof ?

[normal rapids]

ugh.....

mmh

ig

[wheat wing]

well okay

[lone yew]

a building lies at (3, 8)
a transformer lies at (0, 4)
each unit is 10 meters
the building is elevated 15 meters above the transformer

what is the minimum length of a cable connecting the two? (round)

this was from an extra worksheet, friend and i both answered 33.5 but im pretty sure we're wrong but he says we're correct (i think its 52)

we both made right triangles thinking the representation was like a picture and from 2d facing forward.
both intially thought "hey thats a 3-4-5 right triangle HOW are we gonna put the 1.5 in there" but we substituted the vertical (4) and did some pythagorean theorem.
i'm now reconsidering if we were supposed to first take the 5 from the hypotenuse and treat the graph like a top-down map or something and then apply pythagorean theorem with 5 and 1.5 to get around 5.2 as if it was now a 3d projection

[exotic yarrow]

seems like your friend interpreted the graph as a 2D side view

but this completely ignores the y coordinates

I'm with you that it should be around 52

[lone yew]

exactly

hes on some "it doesnt matter now we turned it in"

sacrificed a portion of my grade for this

[exotic chasm]

"blue area" means the semicircle with unknown radius MINUS the red circle?

[slim plinth]

Ik 30-60-90

[rustic crater]

the theorem has no given name as it's a common property for two tangent circles

[exotic chasm]

btw is the double angle property only happens at a circle and not other conic sections?

[exotic yarrow]

I'd start by finding this

[rustic crater]

it only applies to a circle, other conic sections also have unique properties tho

[minor heron]

am i tweaking i swear none of the answer options are right 🥀

[exotic chasm]

||18tan(30deg)||

[exotic yarrow]

Should be ||D||.

||The blue is 24/2 = 12. So the bottom leg of the left triangle is 18. Hence x=18/sqrt3 = 6sqrt3.||

[slim plinth]

Problem is is there a triple that has hypotenuse of 24?

[minor heron]

oh fack i mixed up the sides i knew it was 12 but then while tryna headsolve it i mixed it up

oops

nvmnvm

[slim plinth]

I doubt it

It had to be a sqrt ans

[minor heron]

not that... have you found the blue leg yet?

[exotic chasm]

i think this one doesnt use the so called pythagorean triple where everything is an integer

[minor heron]

oh the pythag triple

i only know like 3 4 5 and that sabout it for triples who cares

[slim plinth]

I’ll do it

[slim plinth]

Me too and beyond

[minor heron]

never bothered learning more

[exotic chasm]

this one uses the sin cos tan of right angles

[minor heron]

well i know 5 12 and 13 but ye

solvable thru trig

[slim plinth]

Ik SOHCAHTOA

[minor heron]

ye can use those to figure out

[slim plinth]

That was my idea

[minor heron]

uhh

you can't figure out the angles of the other triangle

[slim plinth]

10-24-26

[minor heron]

dont need it

[slim plinth]

After trying to use that

X=5

That’s not the ans choice

[pale coyote]

You don't know it's a 10-24-26 triangle

[slim plinth]

I do

10-24-26 is from 5-12-13

It’s double of its original

[rustic crater]

yo interesting

[pale coyote]

It could be for example a 24-24-24√2 triangle

[rustic crater]

such a shame that the other triangle isn't a right one as well

[rustic crater]

but at least it makes it harder

[rustic crater]

how did u evaluate the 26?

[slim plinth]

But there’s no 45-45-90

[pale coyote]

It can't physically be 26

[slim plinth]

Pythagorean theorem

[rustic crater]

wait no

how did you get the 10

[pale coyote]

The hypotenuse of that triangle must be larger than 30

[rustic crater]

i mean that side is equal to 2x via sine of 30

but how would you be able to calculate that it's 10 with the given information

[pale coyote]

It's not 10 👀

[minor heron]

you can't assume a pythagorean triple if thats what ur trying to do

its not needed

[rustic crater]

true

[pale coyote]

I think its 10√3 but that's not an option so I got no clue

[slim plinth]

Then only way is what u said do SOHCAHTOA

[exotic chasm]

btw we can make more of pythagorean triples\
select 2 integers, $x$ and $y$, with $x<y$\
the triple will consist of\
$a=\sqrt{c^2-b^2}\
b=y^2-x^2\
c=y^2+x^2$

[minor heron]

oh icic

[somber coyote]

tsuitachi (tuitati)

[pale coyote]

x=6√3

[exotic chasm]

||btw for more expert mathematicians: generalize to n dimensions 🤣||

R^n moment

[exotic chasm]

||(imagine they putting factorials, e, or pi to geometry)||

||-# wait pi is already in geometry||

[slim plinth]

What about do pythagorean theorem?

With that 2x and 24

(2x)^2+(24)^2=y^2

[rustic crater]

it would've made things easier if they told you that the second triangle really is right

[slim plinth]

Yes

[rustic crater]

I think you can assume the right angle and even somewhat prove it

[exotic chasm]

should use unknown angle and answer will be generalized from that angle

also btw for a 30 60 90 triangle, pythagorean triple is irrational
1:sqrt(3):2

[trim stump]

I feel like I could solve this

[trim stump]

.

[rustic crater]

now I feel like it truly makes no sense

[slim plinth]

Btw that one with “y^2” should be the hypotenuse

[trim stump]

I feel like this is wrong

[rustic crater]

it looks wrong and is wrong on many levels

[slim plinth]

I got that one 30

Now it’s rectangle

[rustic crater]

look at the angle between 2x and y

how can it be 30 degrees

[slim plinth]

30-60-90

Also used

[rustic crater]

the angle creates between x and y cannot be a right angle as y is not parallel to 30

which proves why that angle is not 30

degrees

[slim plinth]

This

[rustic crater]

you need parallel lines

and then you intersect them with a new segment

[slim plinth]

Ohh

[rustic crater]

that's when you receive the alpha angles

the 60 degree angle up top is wrong as well, that's just assumed by you and not calculated

[slim plinth]

Forgot to erase

[rustic crater]

anyways, there is indeed a way to prove the second triangle is a right one so

from that point on you should be good on your own

[slim plinth]

Btw

Good news is

Proved it

It’s another 30-60-90

[trim stump]

How would u know if the right triangle is a right triangle

[slim plinth]

By definition

Rt triangle has to be 90 degrees

And sum of these has to be 180

It is 6sqrt(3)

[trim stump]

I still don't get the proof for it being 90°

[slim plinth]

Think it as a rectangle

[trim stump]

Yeah but I'd think of parallelogram

[slim plinth]

Ok

[trim stump]

Wait I'm trying to prove it

[slim plinth]

Ok

[rustic crater]

because the angle formed between two triangles (where the first one is a right triangle) is a right angle, and that the longer leg of the right triangle as well as the base of the second triangle both lie on the same line, we can therefore deduce that a side of the second triangle MUST be perpendicular to the line, making it so that the angle formed between the "base" of the triangle and it's side is right. qed

it's kind of self-explanatory I can't lie

[trim stump]

Ohhh alr

I think it's not asier to solve that nowzwait

[rustic crater]

now it is way easier

because we can use the little tool known as trigonometry for the second triangle

[torn gulch]

magic
||right triangle is similar to left triangle so the base opposite to 30 would be 24/2=12 30-12=18 now x:18=1:sqrt3=> x=18/sqrt3=6sqrt3=D.||

[torn gulch]

why is this what u did dawg-OHHHHH the right triangle isn't a right triangle

is it possible
I mean

30 kinda indicates it is a line

[slim plinth]

Is this an interesting question

[exotic chasm]

https://cdn.discordapp.com/attachments/1397581939501961357/1504325881706053642/588143980747685889.png?ex=6a069427&is=6a0542a7&hm=019475c607361ac4b5ff3b9234e29887881be508fabf57e27faf81ed58c37172&

[tiny acorn]

What shape is this

[tiny acorn]

Oh

[exotic chasm]

problem is, i thought it was really obvious to imagine the shape from the formula

as if the formula speaks what the shape is

[tiny acorn]

Is this like a square but you place 4 circles centered at the 4 vertices and subtract them from the square

[exotic chasm]

wanted to say "quarter star" as the shape name but it isn't descriptive enough

and there are many curvy 4 star variations can be made with other functions, but their area formula belongs more to #calculus

[tiny acorn]

Ye

[slim plinth]

Is it just A_square - A_circle

[exotic chasm]

yeah

[tiny acorn]

Yup

[slim plinth]

Just take the quarter circle

Merge it back in

Then take that subtract the square

That’s ur area of that shadow

[exotic chasm]

now can we generalize for all regular polygons though

[slim plinth]

So it’s A_shadow = s^2-2pir

[exotic chasm]

stupid question; we can

[tiny acorn]

How about irregular polygons

[exotic chasm]

actually better question

[exotic chasm]

btw its cousin, the ginkgo, is actually better\

$A=\frac 12 s^2\ P= \pi s$

[somber coyote]

tsuitachi (tuitati)

[exotic chasm]

or axehead

[exotic chasm]

only possible for triangles

[tiny acorn]

Oh

[static lotus]

hi i am new

[trail patio]

W larps

[exotic chasm]

should we classify pentagons like how we classify quadrilaterals?

like by existence of right angles, same lengths, same angles... etc

oh, also equiangular, equilateral, etc

[exotic chasm]

oh, equiangular has constraint: sidelengths will either be 2 pairs of same and 1 different

[winged oyster]

uh

people prolly wouldnt care

[exotic chasm]

eh true

it's 4 sides max that matters

[shy flower]

Can anyone help me?

[exotic chasm]

should we add a "secret point" O that is the center of the circle

this.... adds nothing to actually work the solution out but now we have OBD triangle that is a right triangle

and ODA

[shy flower]

Yeah but how far does it take us after that, can we use Menelaus theorem?

[grave pond]

Is there really enough information in that problem? Feels almost magical.

[grave totem]

i want to complete geometory and trigno can anyone suggest me something

[civic rock]

Hint: BF is a symmedian of ∆DEF, so PD/PE = (FD/FE)^2. also notice ∆TFE ~ ∆TDF

[tiny acorn]

<@&268886789983436800> nword

[tender hull]

To know what’s the adjacent and opposite sides we take the perspective of the angle right?

[tiny acorn]

erm what do you mean by perspective

[tender hull]

Like imagine your the angle

[tiny acorn]

in case you dont know there are formal definitions of adjacent and opposite sides that does not require perspective

[tiny acorn]

this works usually

[tender hull]

Bet

[grave pond]

Yes, "adjacent" means the leg that touches the angle you're measuring; "opposite" means the one that doesn't.

[tender hull]

Thank you

Could someone help me understand how we get the angles here? I feel like this is rage baiting me

[slender crag]

For q1
Tan(x) = opp/adj
Plug in opp and adj and put it in solver, you will get the answer

[tender hull]

The other ones be crazy

[slender crag]

For q2, I have no idea why north is positive x since it should be east
Q3 is pretty simple, it holds the ground plane and the x axis plane
Q4, north is positive x
Q5, what did they even do
Q6, water surface is x axis plain

If it was me, I would probably do them saying east as positive x axis, and during exams, if you say that east is positive x, there shouldn't be any problem

[tender hull]

Thank you

[grave pond]

Guess that the triangle is probably meant to be equilateral, and check that in that case the numbers match.

[trim stump]

It's probably not scaled correctly, but yeah if it is equi, 60° should be = logx^30 right?

[grave pond]

Actually even without guessing that, the repeated x means it is at least isosceles, so the lower left corner must be 60°, and then the lower right one must be too.

[tiny acorn]

,w 60 = log_10(x^30)

[somber coyote]

Results provided by WolframAlpha

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

Requested by hexagon69_

[tiny acorn]

Welp solving for y should be trivial now

[grave pond]

So here's a better question:

[trim stump]

Is 60=logx³⁰ because they have the same opposite side length

[trim stump]

Ahhh

But wouldn't that mean y=x

[tiny acorn]

Log10 since it gives a nicer answer (100)

[grave pond]

Any other base would get a wrong answer, since we get x=b² and y=x because equilateral, and we need 6sqrt(y) to be 60.

[trim stump]

I just know y=100 so x must be 100 too

[tender hull]

How do we know in general what to use? Sin cos or tan ?

I always have issues knowing which one to use

[grave pond]

When in doubt always draw a diagram.

[trim stump]

Id u have the adjacent given, with use cosine if it's trying to look for hypotenuse, or use tan if it's looking for the opposite

Just think of soh cah toa

And look at what's given and what they're looking for

[trim stump]

In no. 19, adjacent is given and yours looking for opposite

So I suggest using tan

[tender hull]

So when I want the opposite it’s tan

In 20 I know I should use sin I think but idk why

[grave pond]

Did you draw a diagram yet?

[trim stump]

No judging ok

Here adjacent is given

And yours looking for the length of the tree which is the opposite

[trim stump]

Nooo, it depends

Look at the given length and the missing

[grave pond]

I hoped I could get Kiwi to draw the diagram herself, which is more instructive.

[trim stump]

Wait does this mean I have no future at being a painter or😕

[grave pond]

(Sorry if your diagram was not a response to my calling for it).

[slender crag]

Remember
Soh Cah Toa
Sin is opposite over hypotenuse
Cos is adjacent over hypotenuse
Tan is opposite over adjacent

[tiny acorn]

We might need a !sohcahtoa at this point

[oak tapir]

Since when did trees grow from walls

[potent hollow]

in australia

[exotic chasm]

suppose lunar radius of 2 Mm, terrestrial radius 6 Mm, and center-center distance 380 Mm. how many steradian the moon covers the sky?

would this formula work\
$\textup{coverage}=\frac{a^2+h^2}{4}$

wait wrong formula

but basically using spherical cap, then over 2π

thought about this

[trim stump]

I live in Y axis dude

[unkempt ember]

tan function where 5 pointed star is used as unit "circle"

pi=10 ☺️

[slim plinth]

I’m doing this

Think sin/cos

sin x: 0,1,0,-1,0
cos x: 1,0,-1,0,1

[minor heron]

interesting... i just remember that it goes thru -1 0 and 1 and where the asymptootes are...

[slim plinth]

Or the pole

[slim plinth]

Can u check mine on #precalculus

[minor heron]

the pole is just the x axis so ye..

alr checking rn

[limber dagger]

could somebody help me with this?

and this

[tiny acorn]

Err

Just find the surface area of all the faces and add them up

[limber dagger]

uhm

okay

[manic lynx]

What he means is: use a well-known formula about right triangles to do that

[tiny acorn]

...

Dude

[limber dagger]

where do i even start im so confused

[tiny acorn]

Nvm.

[tiny acorn]

Err lets try the triangular faces first

Do you know the formula for the area of a triangle

[limber dagger]

ok

i got

yes

i got it right

it says

[tiny acorn]

Says what?

[limber dagger]

i did something right

well i used

formula sheet thing

my teacher gave me

idk i figured it out

eventually.

[tiny acorn]

Great. Can you solve the second one now or do you need help

[limber dagger]

let me see

i got it

thxs

i just ggot it wrong the firs ttime

idk what i did

thanks tho

[exotic chasm]

||tadc discord server|| 👀

[exotic chasm]

btw the bottom side is a regular hexagon right

thus makes it easier to calculate

[exotic chasm]

as in, no trigono bs needed

-# cos you can get height of equilateral triangle without even knowing what sin(60°) or cos(60°) is
-# and regular hexagon is 6 equilateral triangles together

-# pentagon is kinda tricky but to get the pentagon diagonal is using similarity/congruency that later the result is has smth to do with golden ratio

[old root]

:() hi

[winged oyster]

8×10×6 + (3sqrt(3)*8²)/2 i think

how do i even derive the formula for area of a hexagon

i only remembered it because i googled it so many times

[tiny acorn]

sum the 6 triangles up i think

[winged oyster]

oh yeah i forgot the hexagon is made up of triangles 😭

[glass spire]

What is the area that is in color black?

[grave pond]

By symmetry, and if we ignore the blue ostrich (?), it seems like it ought to be half the area of the circle.

[wanton snow]

i need someone to help me in geometry have an exam in a few days need a tutor to help me with my friend

[exotic chasm]

1stly we dont know the tested level of geometry so perhaps instead of giving vague requests you should tell which part of geometry that is tested in the exam

cos when one of us tutored in prism/pyramid/cylinder/cone surface area but the tested subjects are angles, similarity, and congruence, the studying becomes vain

[wanton snow]

The exam covers Modules 5 to 7 in Geometry

Angles of Triangles
Congruent Triangles
Proving Triangles Congruent (SSS, SAS, ASA, AAS)
Right Triangle Congruence
Isosceles and Equilateral Triangles
Perpendicular Bisectors
Angle Bisectors
Medians and Altitudes
Angles of Polygons
Parallelograms
Rectangles
Rhombi and Squares
Trapezoids and Kites

[exotic chasm]

....rectangles?

isnt that just, A=ab and P=2(a+b)

or do you have to find rectangle's diagonal or other weird stuffs like ratioes, similarity and congruence of rectangles, special ratioes like 1:sqrt(2) and 1:φ,

also angle bisector as in, finding the special properties of triangles given a line that divides the angle by two (l=2ab/(a+b). cos(α/2))?

[wanton snow]

Yeah diagonals congruence and basic angle bisector stuff not golden ratio or those advanced formulas though

[slow smelt]

guys

[slim plinth]

So do I need two triangles

[vestal basin]

great

wait what

not sure if you applied pythagoras correctly...

oh you crossed out 8^2...

what the fuk

ok please write every step on a separate line

also it's 4*20 = 80

[slim plinth]

It’s 4sqrt(5(

And it’s (56+16sqrt(10))/108

Then it’s A

[marsh marlin]

why do u always find these random ahh problems

[slim plinth]

Yes

[exotic chasm]

i mean you can even make one on the spot

[exotic chasm]

now let's reconstruct laws of...

1. triangle altitude
2. triangle median
3. traingle angular bisector

progress for awhile

[exotic chasm]

i think altitude doesnt need trig funx

[jade ether]

is anyone actually good at geometry

[raw mountain]

Who knows

[exotic chasm]

i'm not, so that's 1 less person to ask

[small marlin]

I'm not to sure, have you tried using a 2D truss analysis software like Ftool or SkyCiv or smth?

[tiny acorn]

wow you can just use a tool to presumably solve an exercise that was meant to be solved by hand

[golden island]

Umm it's kinda ez , isn't it?

[barren shoal]

does anyone have like trig worksheets or maybe exponentaisl 💔 /

[karmic frigate]

i got exponentails

[jade ether]

well im failing rn

[rare wave]

at what level? I'd like to learn about new things, so please give me the question.

[jade ether]

highschool

[rare wave]

I may be able to help

[rare wave]

what's the question?

[spiral lodge]

Yeah send the question directly

[jade ether]

its on logs

[spiral lodge]

And what about them specifically?

Send it, don't waste time (for us helpers and for you too)

[jade ether]

5×log (125) = 15
x

it wouldnt copy the sub (x)

[spiral lodge]

$$5\log_x(125) = 15$$

[somber coyote]

Alberto Z.

[spiral lodge]

This one?

[jade ether]

yeah

[rare wave]

base is 5

[spiral lodge]

First thing first, can you divide both sides by 5?

[velvet pond]

yeah

[spiral lodge]

So that we isolate the log and then maybe you'll see what to do next

[jade ether]

aigh

[subtle glacier]

You can just do 5^3 = 125, then take 3 out.

Then divide all by 15 to clear all 15.

Which means logx(5)=1.

And only x can do this is that it must equal to the term inside that log.

But it is just the same i guess.

[spiral lodge]

Yup

[rare wave]

why is this in spanish lol

[queen juniper]

this does not belong here

[civic rock]

likely an advertisement

[muted snow]

Where can i upload my work

[muted snow]

cause i am from argentina

[queen juniper]

in one of the advanced channels

[muted snow]

Can you pass me one?

[queen juniper]

wdym

[rare wave]

We're all preuni bro

[queen juniper]

u can assign urself access to the advanced channels im pretty sure

[rare wave]

how?

[queen juniper]

#info i think

[muted snow]

So where can I upload it?

[rare wave]

in the of the higher channels.
Why did you choose pre uni when joining the server?

[muted snow]

i don't know

sorry

[rare wave]

it's alright, but you're saying you solved a millenium problem?

are you sure the logic is correct?

[muted snow]

yes,i sure

Hodge's conjecture really caught my attention and I worked on itcture

[queen juniper]

you solved hodge's conjecture?

i mean i've seen a lot of people here say that they've solved a millenium prize problem

[muted snow]

Read it first, then tell me what you think

[queen juniper]

well if you actually did solve it then there's basically no chance i'll understand it

[rare wave]

i genuinely don't know anything, but I guess it's good reference material to learn things

[muted snow]

okey

[rare wave]

But no matter what, even if it's not justified, it's more than I can comprehend right now

[muted snow]

thanks

So where can I upload it?

[potent hollow]

im sorry if im mistaken but the formatting does look like chatgpt, especially the underendered latex

[rare wave]

That's automatically done in google docs

[muted snow]

The layout is formatted, yes, but the ideas are mine. Every single tool, logarithm, and theorem in that text was selected by me to build the inference chain. Focus on the mathematical logic, not the template

[rare wave]

go to #info, click manage roles

[potent hollow]

well congrats go claim your 1 million dollars

[tiny acorn]

Sure

[rare wave]

could you like produce a logical, formal maths style proof?

[tiny acorn]

Prove it to the mathematical journal

[potent hollow]

i wonder how much sarcasm were allowed

[tiny acorn]

Oh wait they got muted cuz they spammed the paper and i reported them

[queen juniper]

yeah there have been multiple ai "proofs" sent in these channels

[tiny acorn]

And crank shit in general

[potent hollow]

infinite monkey theorem!!!

surely one will actually prove it eventually

[exotic chasm]

https://media.discordapp.net/attachments/1397581939501961357/1505235045294538853/588143980747685889.png?ex=6a09e2e0&is=6a089160&hm=1dd390bf13083b3532dcb6b17d2c474b2202db260679abcd5f3b2d20254f7ffe&

need to check for term misusage

[tiny acorn]

Bottom is only true if scaled evenly

[exotic chasm]

wait, dilation can just scale in 1 axis only?

[tiny acorn]

If you dilate along x axis and y axis by different amounts it wouldnt be similar

[tiny acorn]

Err idk

Like dilation is synonymous with scaling i think

[exotic chasm]

cos iirc dilation, at least with matrix, it's
k 0
0 k

[tiny acorn]

Err maybe

Im just like

You can scale shit unevenly

[exotic chasm]

a 0
0 b
matrix for starter

[tiny acorn]

Idk matrices 😔

For me its just multipling the x and y coords with a scale factor k

So (x;y) -> (kx;ky)

[exotic chasm]

eh it does the same

[tiny acorn]

(I assume the matrix does the same thing)

(In fact i think it does)

Uneven scaling has like numerous applications tho

When you select shit and scale it unevenly in cad, paint n stuff

[slim plinth]

I wonder if we can use A=1/2 bh to proof a polygons

With A=1/2 a P

[tiny acorn]

Prove what about a polygon

[rare wave]

it's really easy bro.

[slim plinth]

Using area of a rt triangle

[tiny acorn]

Then?

[slim plinth]

Take a polygon and strength it

[tiny acorn]

Err im not getting what your saying

[slim plinth]

How can I use Area of a triangle to proof the area of polygon

With is 1/2 a P

[tiny acorn]

Err sum the area of all of them up

[slim plinth]

a is the apothem and P is the perimeter

[tiny acorn]

P could be trivially converted to edge length tbh

But ye

[slim plinth]

My idea was the h= the apothem and b = the base

Then replace them to get
A=1/2 a P

If I set them equal

Then it’s bh=aP

[rare wave]

what sort of a polygon?

[slim plinth]

This kind

[rare wave]

regular pentagon?

then pa/2

[ sigma x_i y_{i+1} - sigma x_{i+1} y_i]/2

this can be used for all polygons

[muted snow]

Hey Cealum, thanks for asking and for your respectful attitude earlier. To answer your question: yes, I can provide a formal proof. The ideas and the logic are 100% mine, developed by hand. I only used formatting tools to write down the LaTeX formulas neatly, which unfortunately made some people here paranoid about AI. If you ask any AI to solve Hodge, it will fail because they can't create new logic. I’m leaving this server now because the community rules aren't being respected and I'm not interested in a place where people rely on memes and insults instead of actual math. Wish you the best in your studies

[tiny acorn]

"community rules" meanwhile bluds paper isnt even in english

Like

If you seriously wanted ppl to read it

[muted snow]

Did you actually analyze the logic or do you only come here to post memes because reading actual math gives you a headache? Tell me exactly which step of the inference fails or just stay quiet

[tiny acorn]

Have it in english

[tiny acorn]

No

In english

You paper isnt in english

[muted snow]

translate it

[tiny acorn]

Most people can not review it

[tiny acorn]

Im afraid there would be inaccuracies

So its best if you provide your own translation

[muted snow]

That's what translators are for

Are you hurt or what?

[tiny acorn]

Shifting the workload towards the reader isnt a good idea when you publish stuff

[muted snow]

It hurts your pride

[tiny acorn]

As an author you should strive to make it accessible for everyone if you want ppl to read it

[muted snow]

It hurts your pride

Are you hurt or what?

Did you actually analyze the logic or do you only come here to post memes because reading actual math gives you a headache? Tell me exactly which step of the inference fails or just stay quiet

use the translator

[tiny acorn]

Rule 1

No fallacies

[muted snow]

nope

You don't even respect your own rules and you come here to teach me about them.

[tiny acorn]

Wow it hurts my pride that i can not understand Portuguese or whatever that was written in

[muted snow]

If you're so smart, you should know what Spanish is and you could use a translator, common sense, and it hurts your pride to look for errors even in the language

[tiny acorn]

Translators are commonly known to struggle with technical stuff

So i wouldnt use it

Also you cant pull any other argument besides "it hurts your pride"

[muted snow]

You made fun of me when I tried to send something serious that you didn't even bother to look at, so don't talk to me like a professional.

[tiny acorn]

Why should i talk to you like one when you dont act like one yourself?

[muted snow]

And if it's supposed to be AI according to you, you can use it to understand what I wrote, or if you want I can explain it to you here

[tiny acorn]

Okay sure

Summarize it

[muted snow]

Here is the core logic:
​Green's Current & Residues: I define a Green's current G for the cycle Z. Applying Stokes' Theorem on the singularity log |s|^2 forces the Hodge class coefficients to be strictly rational (\mathbb{Q}) via logarithmic residues.
​Siu's Theorem: The Lelong number (local density) of this closed positive current is strictly positive, which mathematically guarantees that the support Z forms a complex analytic subvariety.
​Chow's Theorem: Since X is a smooth complex projective variety, Chow’s theorem ensures that this analytic subvariety is necessarily algebraic (defined by polynomials).
​The bridge from Analysis to Algebra is closed. If you have an actual mathematical counterargument to any of these three steps, I'm listening

[last flume]

https://cdn.discordapp.com/attachments/497227343337881640/1494384382910267453/200.gif

[tiny acorn]

Oh hi ndeltoid

[raw mountain]

Where is the question I need to disprove

[exotic chasm]

by using these formulæ, no apothem needed
https://cdn.discordapp.com/attachments/1397581939501961357/1505247469338890441/588143980747685889.png?ex=6a09ee72&is=6a089cf2&hm=ef092b4452310f8f2dbde59bd3a84c3190d2a741385afa7b0f58ca384649f6c4&

disclaimer: only for regular polygons sadly

[tiny acorn]

Is there any formula for irregular polygons or we have to pull out splitting

Or integrals

Or whatever

[muted snow]

Do you understand it or not?

[exotic chasm]

sadly no so happy splitting

[raw mountain]

Siu's theorem only makes positive-Lelong level sets analytic, it doesn't produce a cycle representing an arbitrary Hodge class

You assumed the missing step?

[muted snow]

That’s exactly where the Green's current construction and regularized analysis come into play. We aren't starting from an arbitrary non-positive class; a Hodge class \omega of type (p,p) on a projective manifold allows the construction of a log potentials singularity system.
​By using the dd^c-lemma and the global green's function G = \log|s|^2, the singularity structure precisely mirrors the topological data of the class. The Lelong number \nu(T, x) isn't just 'positive' by chance; its values are strictly locked to the discrete residue multiplicities forced by Stokes' theorem on the logarithmic singularity. That guarantees the existence and integrality of the dense support, which Siu bridges to analyticity and Chow to algebraicity. The representation isn't arbitrary; it's structurally forced by the dd^c equation.

[raw mountain]

Yeah you can't type this fast

[tiny acorn]

Thjs

[raw mountain]

That was copy pasted

I can tell lmao

[tiny acorn]

Prolly from chatgpt

[raw mountain]

Straight from AI

[raw mountain]

You are assuming the algebraic representative you claim to prove

Tell that to your AI

[muted snow]

I didn't assume it, it's structurally forced by the geometry of smooth complex projective varieties. A Hodge class is not just any random topological class; by Hodge theory, it contains a unique harmonic (p,p)-form.
​Because X is projective, we can use the dd^c-lemma globally. The Green's current G is constructed precisely as the solution to the equation dd^c [G] = \Delta(Z) - \Omega, where the singularity behavior is strictly controlled by the local log potentials. The positivity of the current and the integrality of the Lelong numbers aren't arbitrary assumptions; they are the direct topological consequence of applying Stokes' theorem to the logarithmic residues of that global green function. Every step of the inference is locked by the differential architecture of the variety

[raw mountain]

Alright you guys have seen that right

[tiny acorn]

Firstly, is N a ring?

[raw mountain]

!noai

[lime crown]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[tiny acorn]

!noai

[lime crown]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[tiny acorn]

Yea

[raw mountain]

You're still assuming Z

[tiny acorn]

(This is a trivial question if you know the definition of a ring, any 13 yr old could do this if they looked it up)

[raw mountain]

AI Slop

A bad one

[tiny acorn]

Frfr

At least it isnt collatz

[raw mountain]

Probably using Chatgpt Free Tier

[tiny acorn]

Or rh

[tiny acorn]

Frfr

[raw mountain]

Instant mode 💀

That's why it was so fast

[tiny acorn]

Hmmm math slop

Pretty sure finding y is trivial

,calc (180-48)/2
180-68

[somber coyote]

Results:

66
112

[tiny acorn]

,calc 180-(360-243-66)-112

[somber coyote]

Result:

17

[tiny acorn]

y=17°

[slim plinth]

I got y=20 degree

[exotic yarrow]

I agree

[exotic yarrow]

You're not given that AB and DE are parallel

[slim plinth]

Two sides are same then two angles are the same

[nocturne breach]

In isosceles ABD the angles must add to 180, so 2ADB+48=180, ADB = 66, and the interior ADC should be 360-243=117, and DCB supplementary to BCE so BCD is 112, in the end u get y = 17

[nocturne breach]

No only if they’re parallel, then alternate angle formed by the transversals are the same

[slim plinth]

Okay

[slim plinth]

So Angle D is the same as angle B

[nocturne breach]

No, they are not parallel as civil service pigeon mentioned earlier

[nocturne breach]

^

[slim plinth]

Okay

I wonder what would the tick mark looks like if there parallel

Is it two arrows

[nocturne breach]

It would be like an arrow going through it

Yes I believe so

The tick marks mark they are the same, so the base angles in the isosceles triangle are the same

[slim plinth]

I got currently 66 degrees with angle A and half of angle D

[silent plank]

Wdym half of angle D

[slim plinth]

So it’s

[silent plank]

if it's not exactly half, don't say half

[slim plinth]

66+117+112+48+y=360

Is that right

[silent plank]

use three points to describe your angle

[slim plinth]

So it’s

Angle ADB and Angle DAB = 66 degrees then it’s Angle ADB=Angle DAB

[slim plinth]

66+117+112+48+y= 360
y=360-(66+117+112+48)
y=17

[silent plank]

Yes

[slim plinth]

How would u solve this

[nocturne breach]

Drop an altitude

[slim plinth]

When I see this I almost used pythagorean theorem, but then I thought there is a rt angle

So I can’t use it

[silent plank]

there are trig rules for non-right triangle

[slim plinth]

Can we use trig

[nocturne breach]

Dyk law of cosines?

[slim plinth]

Now that’s what it looks like

[nocturne breach]

Now u can use pythag and standard trig

[slim plinth]

30-60-90

[nocturne breach]

Yes

[slim plinth]

90 is 2x=3

x=3/2

[nocturne breach]

Whoa where

First try to find the height of the altitude from the small 30-60-90, and then use pythag

[slim plinth]

I should of used different variable

[nocturne breach]

Yea that’s right

[slim plinth]

Yay, Pythagorean theorem time

After that I got x=sqrt(283)/2

Can I simplify sqrt(283)

[silent plank]

how'd you get that?

[nocturne breach]

It’s not 8^2, it’s (8-3/2)^2 cuz ur taking the leg of the right triangle not the big triangle

[slim plinth]

Ohhh

[slim plinth]

It’s 13/2

[nocturne breach]

I don’t think so… (8-3/2)^2+(3sqrt3/2)^2=196/4 right?

[slim plinth]

49

[nocturne breach]

Sqrt that and you’d get 14/2=7

[slim plinth]

Yes it’s 7

[slim plinth]

Can I use parallel lines cut by a transversal

Then x=y

[exotic yarrow]

Yes there's an implicit assumption of parallelism here

don't love that but it is what it is

[slim plinth]

That’s what I got

[exotic yarrow]

good

[slim plinth]

yay

x=y

[snow isle]

Clever way of finding the right angle

Props to you

[slim plinth]

Thanks

[plucky crag]

unsure if this is in your course, but a nifty trick anyway is to just use cosine law x^2 = a^+b^2 -2abcos(c) (in this case a= 3, b=8, and C=60 degrees)

it works on any triangle if you have two sides and the included angle, you can find the last side. pretty neat

if its not in your course i recommend learning it and using it to check your answers

just takes less working out from your way

[slim plinth]

Ik cos(60)=sqrt(3)/2

I think

[plucky crag]

close, its 1/2

[slim plinth]

Ohh

So it’s 3^2 + 8^2 -2(1/2)

9+64-2=71

[plucky crag]

no its 3^2 +8^2 -2 x 8x 3 x (1/2) =73-24=49

sqrt 49 = 9

so a^2 +b^2 -2 x ab x cos (C) dont forget the 2x ab part

[slim plinth]

9?

You mean 7?

[plucky crag]

:/.

[plucky crag]

yes

i swear im good at maths 😭

but you see how it works now?

[minor heron]

🥀 son im crine

[plucky crag]

i go from spending hours writing up proofs about logistic functions and their closed derivatives. and then hop on discord and write sqrt 49 = 9

[minor heron]

damn

[plucky crag]

the duality of a math student

[marsh marlin]

https://klipy.com/gifs/brain-iq--k01KRSVN8W4GBGJH6JWVG458PG9

[minor heron]

i go from taking my ap exams and then walk into bio class where it's like "this is an exponential graph!"

[vocal sentinel]

AP?

[exotic yarrow]

"Advanced Placement".

[slim plinth]

Take AP sleep

[minor heron]

Ap is a American education system thing. “College level” stuff with exams in may that give u college credit if u pass and a gpa boost

[marsh marlin]

?

[slim plinth]

Refer to evelyn

[minor heron]

Um

Why sleep when you can math

[marsh marlin]

evelyn what does jake mean

[marsh marlin]

you cant math without sleep

[minor heron]

Yes I can watch me

(Dreams about solving a frq)

[marsh marlin]

but you can sleep without math

[marsh marlin]

ok your actually a geek

[minor heron]

/jk

[winged oyster]

Well, sqrt((4)9)) is 6, which is almost 9

[marsh marlin]

https://klipy.com/gifs/ash-baby-3--k01KRSW5MRJF836BGS81F2T09GM

[winged oyster]

advanced placement sleep is the pro strat

[vocal sentinel]

Antagonising People.

[vocal sentinel]

Locking in bro

[somber coyote]

tsuitachi (tuitati)

[vocal sentinel]

Try draw it out

[exotic chasm]

the proof that i can make actually uses the fact that d2=sqrt(h^2+b^2), d1=sqrt(h^2+a^2)

thus applying A=(1/2)h(a+b)

while getting h=(b-a)

but this requires the parallel lines' length

[exotic chasm]

wait.... this is applicable to all right trapezoids

just realized 🤦‍♂️

[plucky crag]

thanks dude

also if you go sqrt(49) and just like get rid of the sqrt and 4 then 9=9

sooooo

[sly jackal]

Yoo guys I am having some problems regarding trigonometry

I can't even learn the formulas I am doomed

@somber coyote

[lime crown]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[exotic yarrow]

(also texit is a bot)

[exotic chasm]

bros pinging the texit

[tiny acorn]

Do you need help on any problems or techniques?

[sly jackal]

I am doomed bro and can't remember the formulas

[exotic chasm]

start from here:

1. unit circle concept, right triangle, pythagorean theorem, basic sin, basic cos, basic tan, slope, and usage in architecture, navigation
2. sine law (ratioes), cos law (spicy pythagorean theorem), sin of addition, cos of addition, trig identities, etc

[sly jackal]

Alr gonna note it down

[exotic chasm]

oh, you habe difficulty in remembering the formula?

[sly jackal]

Yess dude I can't remember it

@exotic chasm

[exotic chasm]

tbf there's no mnemonics on how trigs work so 🤷‍♂️

yeah i guess one can only go the hard way, by actually using the hard trig concepts irl

[tiny acorn]

Or do practice problems

[exotic chasm]

oh, there's actually a """""clutch""""" that makes trigonometric additions """""easier""""". namely, complex numbers

[plucky crag]

literally just spam revision questions, and when writing questions do full exam style working outs, so write out the formulae you use, then sub in values and then solve. doing this will increase your familiarity with the forumulae

[sly jackal]

Alr brothers thanks for helping

[exotic chasm]

which, is entirely different demon, but once you know the oiler formula

e^(zi) = cos(z) + sin(z) i

trigonometry stops being tricky

because while sin cos tan disallows you to do easy calculations once you insert a+b or smth, exponents, in other hand, allows

[plucky crag]

wait can you give like an example question of where this would simplify some big or annoying trig question. because i learnt this for my complex numbers topic, and then like never used it and just spammed polar form instead. so ive never actually applied it 😭

[exotic chasm]

$\sin(a+b) =\frac{e^ae^b-e^ae^{-b}}{2i}$

[somber coyote]

tsuitachi (tuitati)

[exotic chasm]

oh wait

[exotic chasm]

wait they dont make it easier?

cos i oftenly told that way 😭

like, i genuinely thought that all memorization can be skipped by just counting them in the terms of complex form

[exotic chasm]

oh also i got some details wrong, mb

[plucky crag]

ohhh i see

so is there like euler form formulas for most trig identities like sin (a+b) or is this like a specific one

[exotic chasm]

can be made

[exotic chasm]

then with regroupings and stuff you can get the non complex forms again

[exotic chasm]

should be
$\sin(a+b)=\frac{e^{ai}e^{bi}-e^{-ai}e^{-bi}}{2i}$

[somber coyote]

tsuitachi (tuitati)

[exotic chasm]

oh then when sin(a)sin(b)sin(c) or other weird forms can use complex form

[plucky crag]

how can sin (a+b) (assuming a+b are real numbers) = a complex number but if you use the normal formula it equals a real number (sin(a+b)= sina cos b +sin b cos a )

[tiny acorn]

The is cancel out or sth

[plucky crag]

oh wait sorry i didnt see the 'i' in the exponents do they cancel out

oh yeah

whoops

[tiny acorn]

But im not entirely sure how

[exotic chasm]

now i just remembered, that cos(2π/7) even involve the imaginary unit, YET, a real number

[tiny acorn]

Since top will simplify to x+yi, so the fraction is x+yi/2i

But

How

Do you cancel it

[exotic chasm]

top wont simplify to mere x+yi sadly

[tiny acorn]

Is there any special property that makes the resulting numerator divisible by i

[exotic chasm]

wait, this stops being trigonometry and starts being complex numbers

[tiny acorn]

(Keep going on, its related)

Where would complex numbers go, like #real-complex-analysis but thats like too advanced

[plucky crag]

as soon as you brought up eulers form is stopped being trig 😭

its the same as trig tho so its fien

[plucky crag]

well you can simplify the top part back to 2isin(a+b) through inverse of the proof to get sin(a+b) to equal that but i cannot see any other way to do so

[dense fossil]

It starts from the e^x infinite series which is sum n=0 to infinite of (x^n)/n!

When you plug in ix (meaning e^(ix))

You get 1+ix-x^2/2!-ix^3/3!+x^4/4!..... which when we group the real and the complex we get one part being 1-x^2/2!+x^4/4!..... and the other being ix-ix^3/3!+ix^5/5!..... which is same as i(x-x^3/3!+x^5/5!....)

The first part which is the real gives cos(x) because thats the infinite series for cos(x) while the imaginary part it gives you i*(infinite series for sin(x)) which is isin(x) .

The e^(-ix) its the same idea, but this when grouped gives cos(x)-isin(x).

So by doing e^ix -e^(-ix) gives cos(x)+isin(x)-(cos(x)-isin(x)) whcih is 2isin(x) then to get sin(x) we divide by 2i.

[exotic chasm]

also should we normalize short trig notations in sake of proving trig idents

$\sin(a)=S_a$

[somber coyote]

tsuitachi (tuitati)

[exotic chasm]

like that

[tiny acorn]

Surface area of the shape a

[exotic chasm]

😧

[exotic chasm]

cos this feels so verbose

[dense fossil]

It could potentially get into other function territories such as would sec(a) also be $\sec(a)=S_a$

[somber coyote]

ロリリヴサ

[exotic chasm]

make the notation exclusive into sin and cos

[tiny acorn]

sec =1/cos

[exotic chasm]

and tan

[tiny acorn]

so sec is just 1/cos

[exotic chasm]

for sec csc cot
$C_a^{-1},S_a^{-1},T_a^{-1}$

[somber coyote]

tsuitachi (tuitati)

[dense fossil]

but that would look too much like inverse functions

[exotic chasm]

oh also we can reduce the character count again by defining m=-1

[tiny acorn]

Or just like, 1/C_a, 1/S_a, 1/T_a

[exotic chasm]

nah make the convention of the shorthand notation makes sure that it's 1/smth instead of function inverse

that said this is the exact reason i actually like using arcsin instead of sin^-1

and when i want to use 1/sin, i write
(sin(x))^-1

[dense fossil]

arcsine/arcsin feels so right while sin^-1 just doesnt feel right xD.

[exotic chasm]

or just straight up csc

[tiny acorn]

Or just write 1/sin

[exotic chasm]

oh right, m stands for minus one

[tiny acorn]

Standarđize the sin notation, sin(x)....n times .... sin(x) = sin(x)^n
arcsin(x) = sin^-1(x)

It kinda isnt that hard

Why do we keep having to use sin^n(x) for (sin(x))^n

?????

[exotic chasm]

tbf arcsin vs sin^-1 thing reminds me of the
f^-1 being inverse of f instead of 1/f

[tiny acorn]

exactly

[exotic chasm]

but f^2 (x) means (f(x))^2 somehow

🤣

[tiny acorn]

Ah yes

Why are trig functions so special

Just like

Be every other function

[exotic chasm]

oh, trigonometric solution: use arcf

for inverse of f

[tiny acorn]

And follow f^n(x) = f(f.n times.f(x).)

[tiny acorn]

Alright im clipping this

This like opencry worthy

[slim plinth]

What about f^-1(x) = 1/f(x)

[tiny acorn]

Thats what the arcf is for (i mean for functional inverse of f)

[slim plinth]

What about
P^-1(larp) = 1/P(larp)

[tiny acorn]

Lol

[exotic chasm]

yeah enough of talking abt this
now return to calculating area of a kite

[slim plinth]

I love that

[tiny acorn]

Okay

Wait a minute

[slim plinth]

It’s the product of two diagonals divide by two

[tiny acorn]

Just sum the halves up

[tiny acorn]

Or you can like rearrange it into a rectangle with half the width

[slim plinth]

Guess the formula
A=((d1+d2)(h))/2

[tiny acorn]

@exotic chasm

[slim plinth]

Is it it A=bh and A=s^2

Or h^2

[tiny acorn]

What

[slim plinth]

The rectangle in red

Is bh

Square in blue is a square

[tiny acorn]

It isnt

[slim plinth]

Or b^2

[tiny acorn]

Thats a rectangle

It just so happens to look similar to a square in this case

[slim plinth]

So it’s bh

But it’s b2*h2

[tiny acorn]

Im not sure what do you mean by bh so imma label shi rq

[slim plinth]

Do I use A=1/2 bh

[slim plinth]

base * height

Or length*width

[tiny acorn]

@slim plinth i was thinking this

It was supposed to be 1/2*d_1 there not 1/(2d_1)

[slim plinth]

Then u use 1/2 bh

[tiny acorn]

Im not sure what is the b and the h here

Can you draw a diagram

[exotic chasm]

no h

if with h it becomes kite prism

[tiny acorn]

Oh hi your back

[exotic chasm]

yeah my back

it's kinda itchy

[tiny acorn]

🫂

[exotic chasm]

wait, would line marked with s and t be perpendicular to each other

[exotic chasm]

btw can i give more point marks for this img so we can say which is which

like this

[exotic chasm]

also are ADB actually a right angle?

because if it is, then s and t should be the same, and not different

[crimson glade]

yes it is

[exotic chasm]

hmm

[vocal sentinel]

Show AED = AFD?

Two triangles

[exotic chasm]

both are right ang

[vocal sentinel]

If can't, try to show triangle AED similar to triangle AFD

[exotic chasm]

and if the angle markers are all correct then β can only 45°

[exotic chasm]

oh the triangle not the angle

[vocal sentinel]

Eh that's wrong

You want to find angle DAB

[exotic chasm]

well, for any quadrilateral, since the inner angle sum is 360°, when angle AEF AFD AED are 90°, EDF must be 90° too

oh wait

[vocal sentinel]

So can you show those 2 triangles similar?

Using Angle - Angle

[exotic chasm]

via AAA

as in angle angle angle

[vocal sentinel]

Yes, how?