geometry-and-trigonometry

[sudden nexus]

yeah

but you also can find it out in some cases

[cold cargo]

No exercise, i just gotta learn it for my second year high

[mossy crow]

that makes it a lot harder then

[cold cargo]

Its ez tho, give some tasks, even ones from google

[sinful iron]

Yup, hence why I was surprised

[cold cargo]

And we analyze one

Brake down

Explain

And it will be done

[mossy crow]

do you not have a textbook?

[cold cargo]

Ehh.... ( ＾ω＾ ) Dad burns them after second year

[mossy crow]

im sorry WHAT

[cold cargo]

Fire starting material

[sudden nexus]

I see

[sudden nexus]

thats kinda crazy

[cold cargo]

( ＾ω＾ )

ω<

Ik

But its not like i can learn anything from them

[sudden nexus]

.<

[cold cargo]

My writing is ancient egyptian

And

[sudden nexus]

lol, same

[cold cargo]

My drawings are more of an suggestion

[cold cargo]

Lol my bro

So

[sudden nexus]

do you study for an exam?

[cold cargo]

Dad js burns em

[sudden nexus]

well, in worst case you have google

xD

[cold cargo]

Rn i study for all exams upcoming in second year

[cold cargo]

Google sucks

I tried

[mossy crow]

hmm

[sudden nexus]

Thats kinda.... too much. At least for me

[mossy crow]

do you not have a teacher

[sudden nexus]

it also depends on what do you search and how

[mossy crow]

if you have exams you gotta have a teacher

[sudden nexus]

fair

[cold cargo]

Its summer brake >ω<

[mossy crow]

also, tell your dad to stop burning textbooks

[sudden nexus]

why do you study then 😭

[mossy crow]

you never know when you need to look at them again, from my experience

[cold cargo]

To be ready

[sudden nexus]

may be a good idea

[sudden nexus]

Do you know the topics?

[cold cargo]

Look up and see why

[mossy crow]

as long as you have a good teacher, you'll be fine!

a responsible instructor can explain things from bottom to top

well

[cold cargo]

Not really, i just recently learned pemdas and Bodmas

[cold cargo]

Nah

It doesnt work like that here

[mossy crow]

that's prealg, not geo

[cold cargo]

Kimda

[sudden nexus]

okayyy

[cold cargo]

Most teachers here js say, you should have learned it in elementry

[sudden nexus]

true

[mossy crow]

what geo topics youve learned?

[cold cargo]

I know, js showcasing how utterly far behind i lack

[cold cargo]

None

[mossy crow]

oh uh

[cold cargo]

Oh uh indeed

They left 🥀

[mossy crow]

best advice i can give you is to get your hands on some practice probs

and do them

[sudden nexus]

🥀

[cold cargo]

Idk how to thats the problem 🥀

[sudden nexus]

agree. Once you do them you will know where you have difficulties and what you need to study Alex

[cold cargo]

Its giving my teacher vibes

[sudden nexus]

Looked online?

[cold cargo]

For sum reason

[cold cargo]

Yeah

Not really helpful ig

[sudden nexus]

Why?

[mossy crow]

where have you looked?

[cold cargo]

Cuz i dont have internet always

🥀

[cold cargo]

Wiki

[sudden nexus]

you can download things

[mossy crow]

https://www.khanacademy.org/math/geometry/hs-geo-circles

[sudden nexus]

outside of it?

[cold cargo]

I have 11mb of storage left

[cold cargo]

Some fishy site

[sudden nexus]

its actually good link, though never heard about this website xD

[cold cargo]

I have the app and idk where to go for what 4/10

Imo

[sudden nexus]

There are some good sites where you can find useful info. Or you can always look for good, old books in your local library

[mossy crow]

:0 youve never heard of khan academy?

[sudden nexus]

ermmmm..... no 👉 👈

[mossy crow]

are you american?

its everywhere here in the us

[sudden nexus]

I am not, so ig thats why

[cold cargo]

Local library quiet literally and phisically doesn't exist in my town, if it did i wouldn't be here (🌧️ ω🌧️ )

[sudden nexus]

😭

[mossy crow]

icic

[sudden nexus]

lol

[cold cargo]

European here that was neglected by corruption btw

[cold cargo]

I know

😭

What can i do?

I tried

Yt

I tried

Chan academy

[mossy crow]

uhm

what happened?

[cold cargo]

I tried those math apps that suck

[cold cargo]

I got lost in Khan academy menus and options cuz i dunno what means what and where it takes

[sudden nexus]

you must live in a very small town

[sudden nexus]

How long have you tried to figure it out ? It took me some time and was confusing ngl. Though it wasn't super hard

[cold cargo]

By population it has 3k ppl, so no, phsically yes, most land is taken by private investors due to good real estate position which allows for lower fuel costs to transport materials and faster delivery to capital

[sudden nexus]

Mhmmm. I see

[cold cargo]

You just said you didnt know it existed 🥀

[sudden nexus]

Yes. So I just clicked on that link and went there

[cold cargo]

Let me guess, 2 monitor build?

[sudden nexus]

One monitor

[cold cargo]

Still better then me

Im using

Nokia 3310

Jk

Some 2019 Smasnug

[sudden nexus]

Bruh. I had that as my first phone 💀

[cold cargo]

Lol

[sudden nexus]

No joke

[cold cargo]

Respect

I legit used it

As hunting tool

When it died

[sudden nexus]

💀

[cold cargo]

Imma add you as bro if you dont mind

[sudden nexus]

Sure

Though I gtg

[cold cargo]

Ok

Bye

[halcyon matrix]

X times 5 equals 25 find k

I meant x

I know it's wrong

[mossy crow]

i believe this belongs in #prealg-and-algebra ?

[summer cradle]

by adjacent you mean share a common side?

[upper karma]

Is there someone here?? I am in the mood to solve geometry problem

[summer cradle]

I have one

[upper karma]

Let mee see

[summer cradle]

http://youtube.com/post/Ugkxs9xSyYjDi_bGL2SyA6V8aGldcG_k4e4d?si=0CRAsVGXAoS51nEA

[upper karma]

O_o

[summer cradle]

try to solve it

[upper karma]

6

[summer cradle]

correct

[upper karma]

I let EH=X and GH=Y

So....

Using similiar triangle theorem

Thank you, that was fun

[summer cradle]

its EF actually

[summer cradle]

np glad you liked it

[upper karma]

oohh yeah

It is fun

Math somehow is fun lol

Do you do that a lot too?

[summer cradle]

do math?

[upper karma]

Yes

[summer cradle]

yes

[upper karma]

I wish i could teach someone in real life

[summer cradle]

I like math too

[upper karma]

I wonder, what if we only know AE length and FG length

Could we try it?

[summer cradle]

sure

[upper karma]

I see this impossible.

[summer cradle]

it's even easier cuz you know one side

[upper karma]

And... how to know the other side?

oohh yeah, trigonometry

[summer cradle]

wait nvm

[upper karma]

Yeah

It seems impossible lol

[summer cradle]

yea

[upper karma]

I am at hospital

And the only healthy thing i could do when playing is solving matu

What the fuck lol

[exotic birch]

Hello

[chrome stump]

yo i need help with homework can anybody help me?

[simple vigil]

Which class

[dark sparrow]

just post your hw question(s)

!da2a

[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! See https://dontasktoask.com/

[wooden rain]

When it comes to basic trigonometry, valuing values of π/3, π/6, etc. would it be advisable to NOT use a calculator to help with conceptual understanding?

I use calculators to utilize trigonometric functions every time, but I'm afraid it's hindering my understanding

[warm shuttle]

Honestly the trick is to realize it's symmetric and the only quadrant that matters is I

then forget about tan, cuz you can figure out tan as sin/cos

Then realize that 0° and 90° is utterly trivial

do the only interesting angles are 30°, 45°, and 60°

Then remember euler formula with e^{ix} = cis(x) = cos(x) + i sin(x)

and i is the vertical axis

So cos is the projection onto the x axis, and sin is the projection onto the y axis

and the three values trig functions take for angles of 30°, 45°, and 60°

are √1/2, √2/2 and √3/2

and then you can just eyeball what the correct values are

[wooden rain]

I see a lot of math papers suggesting to memorize the unit circle, but I'm wondering how understanding these concepts can differ how I look at trig completely

[warm shuttle]

e.g. cos(30°) = √3/2, sin(30°) = 1/2

[wooden rain]

I can see the π/n is used for its elegance

I'm revising trignometry for pre-calculus by the way

[warm shuttle]

Well pi/n is just = 2pi/2n = (1/2n) 2pi, so it's 1 2n-th of a full turn

see tau manifesto lol

[warm shuttle]

Trig is necessary for calculus

annoyingly

You have to be pretty good at it

[wooden rain]

Yeah, I totally see, my fear is how I always use calculators for trigonometric functions, that I might have knowledge gaps or problems with differentiation rules and their interworking

[warm shuttle]

It's mostly trig identities specifically

and in the US, you should also learn identities involving csc, sec, cot

[wooden rain]

Looking at this table, it seems deceptively straight-forward, I can't say for certain until I give it a hand https://en.wikipedia.org/wiki/Differentiation_of_trigonometric_functions

[warm shuttle]

But thing to note here is that there will be more complicated integrals

where you would have to manipulate trig function sums/products into a form that's actually easy to integrate

[wooden rain]

In hindsight, I think it's quite interesting how this expression these make visual sense

$$ \frac{d}{dx}(sin(x)) = cos(x) $$

[somber coyote]

Kolo

[wanton moss]

how do i do this question?

[dire jackal]

are you prepping for jee by any chance

since this has been asked before

also the way to solve this is like, from 0 to pi/4, tan is from 0 to 1 and cot is from 1 to infinity

try to use this

[wanton moss]

no lol js trying to improve my trig which i feel im highk weak at

[dire jackal]

ah okay

[dire jackal]

the simplest way to solve is to just assume an angle between 0 and pi/4 and compare the 4 expressions

[wanton moss]

that would be considered a valid solution right? or would u have to somehow prove for all theta within that domain

[dire jackal]

if you just want the answer, assuming the values will work, if you want to show how, you have to use what i said earlier, this thing to be precise i think

[wanton moss]

Yeah this is what I did so far

[dire jackal]

yes, now see t1, t2, t3, t4

lets say tan is small and cot is large

t1 is small^small

t2 is small^large

t3 is large^small

t4 is large^large

you can see that t4 is the largest

[wanton moss]

yeah

would it be t4, t3, t1, t2 then?

[dire jackal]

yep!

[wanton moss]

thanks for the help

[dire jackal]

np!

[lost lichen]

t1 smollest and t4 greatest t3 is larger then t2

cause however smol tn(theta) is the value of t3 will never go below 0 cause if u keep taking roots ur min is 1 while t2 is like fraction to the power of a real number greater than 1 which jut results in a smaller fraction

OH RIGHT AM DUMB

[lost lichen]

ye ok its this

i checked smol and big for those but not t1 and t4

[upper karma]

Hey, I was just looking at this ques, am just learning intro. to trigno. but if you mind than can you tell that how t1 was greater than t2

[lost lichen]

tan theta is a fraction

so fraction power fraction is basically like a root of the fraction

why increases the value of the farction

fraction*

but cot theta is a number greater than 1

[upper karma]

ohh ye i got it

thanks

[lost lichen]

so fraction power a number greater than 1 is a lesser fraction

[lost lichen]

ite

np

[warm shuttle]

I have a slightly silly trig question

What are all the trig identities?

[lost lichen]

like their definition or what they are?

[warm shuttle]

No, literally, i didn't quite come up with a way to mathematically formulate this question

[upper karma]

Sin
cos
Tan
Csc
Sec
cot

[warm shuttle]

But there has to be some way to classify all of them

surely

Like i think we can limit ourselves to sin, because we can construct the rest from it

[upper karma]

Okay

[dire jackal]

yes that can be done

[warm shuttle]

It should be something like "what are all the expressions in sin(x) that are always true, that wouldn't hold with a general f(x)"

[upper karma]

Do you know what the value of sin come from?

[warm shuttle]

Sure

[lost lichen]

what do u mean by a "general" f(x)

[upper karma]

Now that you know, how about cos?

[dire jackal]

cos is just square root of (1-sin^2(x) ), for example

[lost lichen]

this is one of those shower thoughts isnt it?

like ur own shower thought

[upper karma]

I always think about triangle when it comes to know the trig identity

[lost lichen]

but it isnt quite the square root func

cause the square root func only has positive values

[warm shuttle]

Like e.g. "sin(x)² + sin(π/2 - x)² = 1" holds for all x (is that correct expression for cos???), but it wouldn't for "f(x)² + f(π/2 - x)² = 1" for some f

[warm shuttle]

yep

[warm shuttle]

not literally shower, but yes

[upper karma]

Good, now remember this:
cos(90-a)=sin(a)

Just that one and you will know other trig identities

[warm shuttle]

So the question then is, find some way to classify all such identities

[warm shuttle]

Yeah cuz complement angle in a right triangle, i get it

[warm shuttle]

So how do i derive e.g. chebyshev polynomials from that?

This does not seem straightforward

[upper karma]

so you want to be smart.

[warm shuttle]

I'm just saying that they are a family of trig identities

So surely there is some classification?

[upper karma]

I don't think so, i am still learning

What is that app name that can make regenerate graph?

[warm shuttle]

regenerate?

[upper karma]

Maybe that can help you derive things.

[upper karma]

I am sorry, english is not my first language

[warm shuttle]

like desmos?

[upper karma]

yes

Maybe that can help you.

[warm shuttle]

Maybe

[upper karma]

good luck

[warm shuttle]

Tbh i have no clue if this is trivial and i'm just too dum to see it, or if that's one of those too hard to even consider problems.

Or if there is some known result i couldn't find by googling

[sly patrol]

Hi

I accidentally cancelled discussion for myself how do I get it back

[dire jackal]

click on mathematics here and click on show all channels

that could work

[sly patrol]

Basically

I did some random command

Just cause everyone else was doing it

😭

[grave pond]

"Show all channels" won't be enough.

You need to go to the "channels & roles" screen (top of the channel list) and uncheck "hide the social channels".

Alternayively .iamnot studying in #bots

[sly patrol]

Damn I was not study

[solemn verge]

Can someone help me understand pythagorean identity?

[stiff spoke]

Vros

[grave pond]

Can you say somwthing about which problem you have with it?

[stiff spoke]

sin^2x + cos^2x = 1 you have sin so you can solve for cosine

[exotic yarrow]

Reposted here: #help-43 message

[azure rivet]

,tex \Tau

[somber coyote]

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

[warm obsidian]

hey guys

is there a reason that the equations of an ellipse/ellipsoid are always equal to 1?

for a circle/sphere, for example the equations are equal to different numbers (the radius of the circle/sphere squared)

but what does the "1" mean in an ellipse equation?

[cunning lion]

you could rewrite the circle equation to read
[ \frac{x^2}{r^2} + \frac{y^2}{r^2} = 1 ]

[somber coyote]

cloud

[cunning lion]

the point is to have a standard form which you can easily read relevant information (i.e. semi-major and semi-minor axes) from

[warm obsidian]

why = 1 tho?

is it just convention?

[grave pond]

It's a convention that means the denominators on the left are exactly the semi-major and semi-minor axes.

[warm obsidian]

so then what exactly does the "= 1" mean?

or does it not really have a specific meaning?

[grave pond]

Huh?

[warm obsidian]

like in the ellipse equation

it equals 1

[grave pond]

The ellipse consists of those points (x,y) where you get 1 when you add (x/a)² and (y/b)².

[warm obsidian]

?

[grave pond]

That's what =1 means.

[warm obsidian]

is 1 related to any of the semidiameters of the ellipse?

[grave pond]

1 is the number one.

[warm obsidian]

yeah ik but why 1

like what would change if instead u had it equal to 5?

[grave pond]

Then the denominators a and b would not be the semi-(major/minor) axes of the ellipse you get out of it.

[warm obsidian]

by that do u mean they are the lengths of the semi-major/minor axes?

[grave pond]

You can think of it as "take the unit circle x²+y²=1 and scale it separately by a and b in the x and y direction, which you do by writing x/a instead of x, and y/b instead of y".

[grave pond]

One of them is the semimajor axis, the other is the semiminor axis (depending on which is larger).

[warm obsidian]

so if it's equal to 1

then the denominators a and b are the lengths of the semi-major and semi-minor axes?

[grave pond]

Yes.

[warm obsidian]

ok, thanks

[rose arch]

i just realized this is where the ellipse formula comes from

i am mildly stupid

[pale lance]

Well what's the catch though?

[pale lance]

Ya exactly, like infact an ellipse with eccentricity = 0 is nothing but a circle

Eccentricity is like the measure of deviation of any conic section from being a perfect circle

[pale lance]

There's nothing wrong

[quaint otter]

i like more the analytic derivation of the trig function derivatives. I mean Limits are the backbone of calculus.

[mortal raft]

does anyone know anything about ALEKS placement test?

[knotty quiver]

Under what condition on $y_1$, $y_2$, $y_3$ do the points $(0,y_1)$, $(1,y_2)$, $(2,y_3)$ lie on a straight line\?

[maiden brook]

the slope between any pair of points must be fixed

[somber coyote]

glass

[knotty quiver]

-# I know

[maiden brook]

then it should be clear…

[knotty quiver]

-# yes. Thanks

-# I wasn't asking for help but appreciate it anyway

[maiden brook]

?

[knotty quiver]

-# I assumed your response was you trying to help me on the problem

-# I guess I just want to gain some response for posting it

-# because it turns out to be a recommended problem in an intro chapter in linear algebra

-# because the problem is easy

-# but the problem is too simple for me to see its significance in linear algebra. Like maybe I'm missing a key takeaway

-# ig I'll cross post it on #linear-algebra

[quiet drift]

hello guys new here but is there a good video that sumarise geometry?

[grave pond]

There's no royal road to geometry.

[obsidian harness]

Khan Academy

Evan Chen has EGMO notes for Olympiad geometry

but you can't expect to have a video that summarises all of geometry

even if it were 10 hours long

and you'd still need to practice anyways

[crimson jolt]

Nah i got 84/85, is it right?

[obsidian harness]

correct!

[dawn iris]

Where's hyperbolic functions

[dark sparrow]

probably #precalculus. or if you're doing calculus shit w them then #calculus.

[dawn iris]

Ok

[lusty dirge]

Hi im 3rd grade highschool and I have problem with geometry do you have any tips learning?

[hazy tree]

Please explain your difficulties in more details.

Which topics in geometry you're not understanding?

[fossil heron]

can somone help with copying an angle in geometry

[hazy tree]

sure

[untold oracle]

[pearl hinge]

Apply pythagoras

[untold oracle]

I'm way too stupid to know what this means

[pearl hinge]

Do u know what's a right angled triangle?

[untold oracle]

Kinda

[pearl hinge]

Can u state it

[untold oracle]

90°?

I'm so lost sorry

[pearl hinge]

The triangle which has 90° as one of its angles ,yea

[pearl hinge]

Nw ,js try to understand

[pearl hinge]

Did u understand this

[untold oracle]

Thats obtuse right

If it's over 90°

Under 90° is acute

[pearl hinge]

I didnt mention anything abt the angles being over 90°

[untold oracle]

Oh

Okay

Well

[pearl hinge]

In a right angles triangle
One of the 3 angles is 90° and the rest 2 are acute angles

[untold oracle]

Yes

[pearl hinge]

Can u identify the right angle in this figure?

[untold oracle]

6?

[rapid comet]

I need help plotting this one

I need help pls

[pearl hinge]

6 is a length of a side

[untold oracle]

X?

[pearl hinge]

Can u plot the coordinates in the graph

[pearl hinge]

That is also a length

[rapid comet]

Yes

[untold oracle]

idk then

[pearl hinge]

These 3 r angles @untold oracle

[pearl hinge]

Can u try plotting them then?

[rapid comet]

The thing is idk how 😢 I plotted this

[untold oracle]

Okayokay

[pearl hinge]

The coordinate is not right ,ent we doing the previous question,this looks like a different question

[pearl hinge]

So can u identify the right angle amoung these 3

[rapid comet]

Ye it is a dif I just need help idk how to plot it

[untold oracle]

No

[pearl hinge]

Alr so
For the 1st questions, one of the coordinates is (-6,2) the first one represents x axis and 2nd one represents y axis

[pearl hinge]

Lemme find a vid

[rapid comet]

Ok

Then how do I graph it now

[pearl hinge]

https://youtu.be/D5lZ3thuEeA?si=Z9gYKPuiDRX-dDJa

https://youtu.be/AA6RfgP-AHU?si=T7ym_Rm3l6NFT5Jp

These 2 should be helpful @untold oracle

[pearl hinge]

Do u k which one is x axis and which one is y axis

[rapid comet]

-6,2 is x axis

Then 2,5 is y

Axis

[pearl hinge]

Ah no,give me a moment

[rapid comet]

Okay

[pearl hinge]

https://youtu.be/cL3f0EhKRn0?si=u8Szd23t3MlXi5KR

Check this out

[fiery lotus]

Do you know the midpoint formula

[pearl hinge]

See red line is y axis and blue is x axis

[rapid comet]

Nope

[untold oracle]

Thank youuu

[rapid comet]

Yes

There’s a example

I can show

[pearl hinge]

Try watching the vid,ig u xan plot from there
As for finding the midpoint,we will need formula

[rapid comet]

That’s the example

[pearl hinge]

I get it

[rapid comet]

So basically I just graph it then I plot the midpoint

?

Then create that square

[pearl hinge]

But a vid can explain how to plot coordinates

Better than texts

[pearl hinge]

U dont need to

[rapid comet]

oh ok

[fiery lotus]

Basically you take the average/mean of the xs and the ys.
For example, if we want to find the midpoint of (3,5) and (6,8), then we find the average of 3 and 6 (the x points) and the average of the y points. The average/mean of 3 and 6 is (3+6)/2 = 9/2 which means the x coordinate of the midpoint is 9/2.
The average of the y coordinates are (5+8)/2, which is 13/2. therefore the midpoint of these two points is (9/2, 13/2)

[rapid comet]

Okay got it

[pearl hinge]

For 1st coordinate,
When u plot -6 ,u need to go 6 steps from origin towards negative x axis

[rapid comet]

Ok

How about the 2nd

[pearl hinge]

When u plot 2 along y axis u need to move 2 steps from origin towards positive y axis

[rapid comet]

Ok

[pearl hinge]

Well this is the 1st coordinate

[rapid comet]

Okay

[pearl hinge]

The intersecting point of these 2 coordinates is (-6,2)

So here red is (-6,2)

[pearl hinge]

But watching the video would make it clear,so maybe do that?

[pearl hinge]

U did (-6,-2)

Here

[rapid comet]

Ok I’ll watch it

I watched the video but I’m still kinda confused

Can u just show me how to graph that one it’s so that I can understand the next problem

[pearl hinge]

This is the 1st coordinate

For second one (2,5)

We need to move 2 steps along positive side of x axis and 5 steps along positive side of y axis
And then the intersecting point is (2,5)

@rapid comet

[rapid comet]

Ok at

Okay*

Is the red one the first

Corrdinate

?

[pearl hinge]

Yes

[pearl hinge]

(-6,2)

[rapid comet]

Ok

[pearl hinge]

The green one is the 2nd coordinate

[rapid comet]

Ok

[pearl hinge]

And joining red and green gives u the line

[pearl hinge]

Yes

[rapid comet]

Ok now what

[pearl hinge]

Finding the midpoint requires the fomula

[pearl hinge]

He explained it well here

[rapid comet]

Ok

I need to create the rectangle it says

Then idk where the yellow dot goes

[pearl hinge]

Alr

[pearl hinge]

Use formula for it,did u understand what A GAME THEORY said

[rapid comet]

Ok I plotted it

The yellow dot

[pearl hinge]

What is it

[rapid comet]

(-2,3.5)

[pearl hinge]

Yes,i hope u understood how u got that

[rapid comet]

So is that the right answer?

It says incorrect

When I checked

Answer

[pearl hinge]

Yes it is

[pearl hinge]

How tho

[rapid comet]

Idk it says incorrect

[pearl hinge]

Xan u show me how u plotted it

[rapid comet]

Ok

I can’t show it anymore I already clicked next problem

[pearl hinge]

Sorry i did a mistake

[rapid comet]

It’s alright I can show next one

[pearl hinge]

But understand where i did wrong

[pearl hinge]

This is supposed to be the 2nd coordinate

I mistook x as y and y as x

[pearl hinge]

It's similar try plotting the points

[rapid comet]

K

[pearl hinge]

It's wrong

U mistook y for x and x for y as well

[rapid comet]

Oh

[pearl hinge]

That's one coordinate

this is 2nd

[rapid comet]

Yo I got it bro thanks

I got all of them correct

Except for the one wrong question

[lusty dirge]

2 things first I cant remember formulas and second like I don’t see simple solutions like I write 2 pages of equations and then I realize you could just just Pythagoras theorem

[simple vigil]

Which class you are in I feel like geometry is one of the most practical topic to learn

[gloomy dagger]

i dont remember the formula prperly but i gues it is ----> x coordinate = (x1 +x2)/2 , y coordinate= (y1 +y2)/2

[lusty dirge]

3rd class high school (Poland)

[vague sonnet]

What this website is?

[pearl hinge]

@rapid comet , he might know

[obsidian harness]

it's Canvas, an online learning platform schools and unis use

[vague sonnet]

Thank you!

[vague sonnet]

Excuse me, could you tell me how to use this?

[obsidian harness]

your school needs to sign up to it

you can't access it freely

[vague sonnet]

Oh, I see. Thank you for telling me that

[rapid comet]

I can send u link

To it

It’s delta math

[light eagle]

How do you translate a logarithmic spiral on x-axis in polar form?

[light eagle]

out of curiosity more than necessity lol

[knotty quiver]

ig you transform the coordinates first to rectangular

then add to x

[knotty quiver]

is logarithmic spiral something in the form of
[
r=e^\theta \quad ?
]

[somber coyote]

glass

[light eagle]

that makes sense.

[light eagle]

yes

[dark sparrow]

r = e^(k theta)

[light eagle]

oh yeah with k sorry

you can also have a constant out front

and there are various other forms for different spirals

for example the golden spiral can be given in polar form as
$\phi^{\frac{2\theta}{\pi}}$

[somber coyote]

Stick

[light eagle]

oof

$\phi^{\frac{2\theta}{\pi}}$

[somber coyote]

Stick

[knotty quiver]

oh

,w plot r=phi^(2 theta/pi)

[light eagle]

note that the constant in front is the initial radius

where it's just 1 here, but could be anything

[knotty quiver]

i see. graph shows it

[simple vigil]

@obsidian harness Where are you bro?

I got 30/35 in trigonometry test

[dreamy blade]

I'm working with this problem I created.

The square has dimensions of r by r
The semicircle has radius of r
And the tangent line intersects at pi/4 radians
I've worked out the problem for radius of 1
But the generalized form is a little harder than that

[plush quail]

what u solving for

[dreamy blade]

Red shaded area

[grave pond]

Are of the square, minus area of quarter circle, minus area of triangle.
The triangle is half of a smaller square with side length r·2·(1-sqrt(1/2)).

[dreamy blade]

Found the area in generalized

[grave pond]

This should work out to r² times the answer you got for r=1.

How sure are you about the 1/2 factor in front of (sqrt2-1)²?

[dreamy blade]

because triangle

when using x and y to represent the horizontal and vertical positions

we see that the slope of the tangent is -1

and our intersect is (r/sqrt2,r/sqrt2)

gives us two equations for the same line which allow for the two intersects to be found

y=sqrt2r-x
x=sqrt2r-y

[grave pond]

The red altitude is sqrt(2)-1, but the blue base is twice as long as that, so when you multiply them and divide by 2 you should just get (sqrt(2)-1)².

[dreamy blade]

so the intersect points are x=r(sqrt2-1) and y=r(sqrt2-1)

Are you sure the red length is sqrt2-1?
Since the blue and red intersect at (1/sqrt2,1/sqrt2)

[grave pond]

Yes -- the diagonal in the big square is sqrt(2)r, and then we subtract the radius of r, leaving sqrt(2)r-r = (sqrt(2)-1)r.

[crimson rose]

Can u guys show me how to do this kind of exercise?

Btw,im at grade 9

[crimson rose]

---

Problem 26.
Given a right triangle with the right angle at , let be the altitude. The feet of on are , respectively. Let be the circle with diameter , and the circle with diameter . Prove that:

a) Point lies on circle and point lies on circle ;

b) The two circles and are externally tangent;

c) is the common tangent of circles

d) AH=DE

e) The area of quadrilateral DEOO' is equal to half the area of triangle .
Help me do the e exercise

[summer cradle]

Area(DHF) = Area(FAD)
Area(DOH) = Area(DOB)
Area(FO'H) = Area(FO'C)

[pearl mantle]

"Given acute triangle ABC, circle with center (O') and diameter BC intersects AB, AC at D, E respectively. H is the intersection point of BE and CD, F is the intersection point of AH and BC. Through D draw a line perpendicular to BC intersecting (O') at D'. Draw BB' perpendicular to DE and CC' perpendicular to DE. Prove that FD + FE = B'C'." Can someone help me with this problem?

[simple heath]

What are some techniques to calculate arctan limits

[dark sparrow]

wdym by "arctan limits"

also maybe you should be asking this in #precalculus or #calculus @simple heath

[simple heath]

Uh limits that has the arctangente function but thanks anndreas ill do

[dark sparrow]

"anndreas"?

[deep bison]

Hello

Quite beautiful

[obsidian hornet]

I don’t understand why we have information about D’ coz its useless, you just need to use excircle properties for orthotriangle (DEF) and then you will easily proof the statement

[hallow niche]

Hello Everyone I need help with Geomotry

INTRODUCTION

How DO I do this

[obsidian harness]

do you know what perpendicular means?

if so, you should be able to find the symbol on the diagram that means two lines are perpendicular

[dark sparrow]

OP up and left @obsidian harness lmao

[simple vigil]

Hey what is difference between orthocenter and centroid

[dark sparrow]

centroid is the meeting point of the medians

orthocenter is the meeting point of the altitudes

@simple vigil

[upper karma]

centroid is boring and orthocentre is goated

[grave pond]

There's lots more to choose between if you want to be hipster about it: See https://en.wikipedia.org/wiki/Encyclopedia_of_Triangle_Centers

[glossy pike]

Hey there, I was going through some trigno worksheets and I found this question. Anyone know how to solve?

[lost lichen]

my solution if u didn't get it nd am very late (can someone verify this):
||length of tangent=2x
u can pretty easily prove that the triangle is right isosceles and sides=r-x and 2x
4x^2=2(r-x)^2
4x^2=2(r^2-2rx+x^2)
2x^2+2rx-2r^2=0
x^2+xr-r^2=0
x=-r+-root(5r^2)/2
length can't be negative
x=(-r+rroot(5))/2=r/2(root(5)-1)
then r-x=r/2(3-root(5))
so area of red shaded part= r^2-pi r^2/4-1/2*r^2/4(14-6root(5))=r^2(1-pi/4-(7-3root(5))/4)
FINAL ANSWR = r^2(1-pi/4-(7-3root(5))/4)||

oh wai am stooped

i think that works?

[upper karma]

why is X56 not a notable point in Wikipedia smh

[upper karma]

i would help you
but your nickname 😠 😠 😠

[obsidian harness]

try finding $a^2 + b^2 = \sin^2 A + \sin^2 B + 2 \sin A \sin B + \cos^2 A + \cos^2 B + 2 \cos A \cos B$ and immediately some identities should jump out

[somber coyote]

south

[obsidian harness]

the question should be cos(A - B) = (a^2 + b^2 - 2)/2 instead

[lost lichen]

@obsidian harness cn u verify?

[obsidian harness]

no thx

[lost lichen]

*cos(A-B)

[lost lichen]

ite ite

[worthy eagle]

x+2xcos(45°) = r
x(1+2/√2) = r
x(1+√2) = r
x = r/(1+√2)

||easier way to do that||

[obsidian harness]

@fading jasper finally figured out a neat solution

construct the reflection across AE

then since CA = CB, 2x = 2y or x = y

[worthy eagle]

2x^2+2rx-2r^2=0
it should be 4rx

[queen venture]

Let f (x) = sin6x + cos6x + k(sin4x + cos4x) for some real number k. Determine
(a) all real numbers k for which f (x) is constant for all values of x.
(b) all real numbers k for which there exists a real number 'c' such that f (c) = 0.
(c) If k = – 0.7, determine all solutions to the equation f (x) = 0.

[dark sparrow]

is that sin(6x) + cos(6x) etc?

or sin^6(x)

actually do you even still need help with this

[golden knot]

just let me fail bro. 🙁

i've already fouind one solution

but i cannot find another for the LIFE of me.

[upper karma]

What solution did you find

[golden knot]

heron's for area, split into two triangles using a partition down from C and connect the tangent points to the partition from C and then just 1/2 b. h from there

[upper karma]

Ok

[upper karma]

you can notice that ||the center X of the semicircle is the foot of the angle bisector of C on AB||

actually can you use trig? I forgot to ask

[golden knot]

Not really

It’s a non calc thing

So

[upper karma]

Wdym

[golden knot]

well i've done cosine rule and trig identities for my first solution

but if ur tlaking about something like sin17 for example dats a no

[golden knot]

and also i'm sorry but i don't really understand what "the foot of" means

[upper karma]

the intersection of that line with AB

[upper karma]

Uh you don't need trig for your approach

[golden knot]

oh yeah sorry i realized that it's easier to do heron's after i did that

so i just said that

[upper karma]

oh ok

my approach would involve finding sinB

Thats the only trig part

Because you know how to calculate BX so the radius is then BXsinB

[golden knot]

so if i understand you right

i label the center of the semi circle as X

then

draw a triangle connecting B and the top of the circle? 😭

[upper karma]

wdym top of the circle

No just say the tangency point of the semicircle with BC is P, then XPB is a right triangle and XP is a radius so XP=BXsinB

[golden knot]

omg

THANK YOu!!

Is XB = 10?

[upper karma]

No

X is not the midpoint of AB

[golden knot]

oh ok

hi sorry

how do i get the value for B?

[upper karma]

For sinB?

[golden knot]

mhm

[upper karma]

lol what

[golden knot]

no please don't say that 😭

[upper karma]

i asked you a question and you said mhm

[golden knot]

oh

sorry

i meant yes by that

[upper karma]

oh ok

Lol

[golden knot]

lmao 😭

[upper karma]

basically you can find cosB

From the law of cosines

[golden knot]

and then trig identity ?

[upper karma]

yes

[golden knot]

ooohhh right

thhank you again

[summer cradle]

I found 2 ways without Trigonometry

[rare cargo]

Hey guys can someone help me with this math question?

[hoary totem]

hint: there are two semicircles, combine them

[rare cargo]

Ohhh I see Thanks

[upper delta]

can anyone help me with this question please?

[summer cradle]

draw the height of BAC

[upper delta]

What after that?

[summer cradle]

let the height be BG

BGC is right triangle

[upper delta]

Correct

[summer cradle]

and since BAC is isosceles then
AG = CG = AC/2

so AC = 8/2 = 4

[upper delta]

If BC is the same as EF, can't I use EF and the angle at C, to find EC, with EF / Sin40

which = 5.56cm for EC

[summer cradle]

yes

[upper delta]

so I got 5.56 from all that right? But MathGPT says it's 4.56cm

[summer cradle]

should be 9.68

what do you get for BC?

[upper delta]

4.14

I'll retry it

give me a minute, maybe I did something wrong

[summer cradle]

yea should be 6.22

[upper delta]

yeah I got 4.14

hm

maybe I forgot something

if GC = 4cm, and the angle is 50, we have the adjacent, and are trying to find the hypotenuse, in which we do adjacent / Cos50, right?

if i do 4 / 0.6428, in which 0.6429 is Cos 50, it equals 6.22, but if I do 4 / Cos50, it equals 4.14, am i doing something wrong in google using it as a calculator?

[simple vigil]

How to find angles like 20 50 70 degrees for trigonometric ratios?

[summer cradle]

you shouldn't round it off

[upper delta]

identify the side of the right angle triangle you have the value of, and the side you are trying to find. For example, you dont know the value of the opposite side, but you know the value for the adjacent side, and you are trying to find the side of hypotenuse using the adjacent side and the angle. This is where you use Sin Cos and Tan, in this case we are using Cos. So in this example, you would put in your calculator, adjacent / Cos (x)

[summer cradle]

avoid rounding off in any case aside from the final answer

[upper delta]

okay, thank you so much bro

[summer cradle]

np

[dire jackal]

technically we can find stuff like sin70, sin50, sin20 without any constructions. we know sin30, from which we can break it down to get sin15, which we can break down to get sin5, from sin5 get sin10 and get the rest

[golden knot]

REALLY?

how without trig? that seems impossible

[summer cradle]

yes

[golden knot]

could you tell me

[summer cradle]

first method

[golden knot]

i had this same thing but my teacher told me it was wrong :(

[summer cradle]

y = 20-x

ODA:
r² + (11-r)² = x²

OEB:
r² + (13-r)² = (20-x)²

2 equations, 2 variables
sove for r and x, and diameter = 2r

[summer cradle]

why?

[golden knot]

what i had :

apparently

my assumption for assuming the top z's are the same as the radius

[summer cradle]

oh wait

[summer cradle]

yes mb it's wrong

[golden knot]

no worries

[summer cradle]

I mean yes the top 2 are the same length but not equal to r

[golden knot]

is there anything we could do with that possibly

[summer cradle]

lemme think

[obsidian harness]

You can compare areas!

Herons formula for the area of the triangle

[summer cradle]

unfortunately I think there are no other workarounds for this method

[golden knot]

then split the triangle from C right

[summer cradle]

but I have another method

[obsidian harness]

Then you have two smaller triangles with bases 11, 13 and heights r

[obsidian harness]

Indeed

[golden knot]

yep this was my first method!!

[summer cradle]

using areas

[golden knot]

unfortunately the question's hard part comes from asking for anothe rmethod to solving it

that's what i'm realllyyy struggling with

[obsidian harness]

Interesting

[summer cradle]

I thought you already have a trig method

[golden knot]

yes it's the area comparison

only method i could find

[obsidian harness]

Oh isn't it just Pythagoras then

[summer cradle]

yes this is also exactly my method

[obsidian harness]

11 - z and z
z and 13 - z

These make up two right triangles

[summer cradle]

it doesn't require trig

[golden knot]

the way i did it required trig but yeah if u just heron's formula then it is much faster

[summer cradle]

I see

[golden knot]

if you want to try any of ur own methods for this issue the general answer is ||diameter = 11||

i think i'll just take the L on this question

[slender bluff]

guys

is this claim valid

if a right angled triangle can be put inside a semicircle then its also possible to put the triangle in the same semicircle in such a way that the largest side lies on the diameter

[summer cradle]

yes

[slender bluff]

can someone help me with the proof

[summer cradle]

the largest right triangles have their longest side lie on the diameter (circle theorem)

for each of those possible right triangle you can create smaller similar right triangles

it's obvious that all of those smaller right triangles will fit in cuz the largest of them fits in

hence proved

[slender bluff]

thanks bro

[summer cradle]

you're welcome

[deep bison]

To Whoever has birthday today.

[queen venture]

yeah mb this was the question

just the a, and b, part

[foggy swallow]

can someone just clear a doubt of mine quite dumb ik but is 1/-3 same as -1/3?

[elfin saddle]

yes

[foggy swallow]

oh ok

[eternal bane]

Yes

1/-3 = -1/3 = - 1/3

[foggy swallow]

thanks

[eternal bane]

There's also -1/-3 = 1/3

[foggy swallow]

yah i know atleast that much hehe

[obsidian harness]

you can multiply the numerator and denominator by the same number and it doesn't change the number

so $\frac{1}{-3} \cdot \frac{-1}{-1} = \frac{-1}{3}$

[somber coyote]

south

[summer cradle]

sin⁶x+cos⁶x = (sin²x+cos²x)³-3(sin²x)(cos²x)(sin²x+cos²x)
sin⁶x+cos⁶x = 1-3(sin²x)(cos²x)

sin⁴x+cos⁴x = (sin²x+cos²x)²-2(sin²x)(cos²x)
sin⁴x+cos⁴x = 1-2(sin²x)(cos²x)

f(x) = 1-3(sin²x)(cos²x)+k(1-2(sin²x)(cos²x))
f(x) = 1-3(sin²x)(cos²x)+k-2k(sin²x)(cos²x)
f(x) = 1+k-2k(sin²x)(cos²x)-3(sin²x)(cos²x)
f(x) = 1+k-(sin²x)(cos²x)(2k+3)
(a) -(sin²x)(cos²x)(2k+3) = 0
2k+3 = 0
k = -3/2
(b) (sin²x)(cos²x) = ¼(4)(sin²x)(cos²x)
(sin²x)(cos²x) = ¼(2²)(sin²x)(cos²x)
(sin²x)(cos²x) = ¼(2 sin x cos x)²
(sin²x)(cos²x) = ¼sin²2x
f(x) = 1+k-¼(sin²2x)(2k+3)
f(c) = 0
1+k-¼(sin²2c)(2k+3) = 0
¼(sin²2c)(2k+3) = 1+k
(sin²2c)(2k+3) = 4+4k
sin²2c = (4+4k)/(2k+3)
we know that 0<=sin²a<=1
0<=(4+4k)/(2k+3)<=1
solve the inequality to find the range of k

[golden knot]

could someone tell me how the angle bisector theorem works ?

[lost lichen]

oh yeeeeee

ty

me failing to do basic math

[obsidian harness]

here's the theorem

and here's a straightforward worked example

[mellow quest]

||C' is the reflection of C from the line AB
areaABCC' = 2areaABC = pr
r = 2areaABC/p||

[upper karma]

that's basically the same sol

[obsidian harness]

@teal warren https://bestexamhelp.com/exam/cambridge-igcse/mathematics-0580/2018/0580-s18-ms-43.php

question 6b, ii)

the way I showed you was correct and your one actually wasn't correct

wait

[teal warren]

Huh

Good thing I didn’t submit the assignment yet

[teal warren]

Wow I just went through what you showed and it’s spot on

Thanks dude

Really

Just got an easy 4 marks

[teal warren]

@obsidian harness Stellar

Yeah it makes sense why they ask for the diagonal across the plane now

[obsidian harness]

https://cdn.discordapp.com/attachments/576461701332074517/1411953183143493763/image.png?ex=68b6875f&is=68b535df&hm=b8cda5b9c03a076c792d8eacaa9157fb010d78881703bd93eff76866eaeabf46&

Indeed!

[whole ridge]

u guys are in what grade?

[glossy pike]

Ty

[reef elm]

geometry is hard

[sly urchin]

True

[short dawn]

Gonna start freshman year and do Honors Algebra 2

[upper karma]

geo is love

[torpid current]

just a side question, is cos^2theta-1=-sin^2theta?

[knotty quiver]

nvm. i didn't see the minus sign. sorry

[knotty quiver]

-# therefore love is hard?

[willow granite]

Man can anyone help me with this one
(tan theta)/(1 - cot theta) + (cot theta)/(1 - cot theta) = 1 + csc theta sec theta
Prove LHS = RHS

[restive gazelle]

can anyone work with me on coordinate geometry?

[obsidian harness]

!da2a

[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! See https://dontasktoask.com/

[obsidian harness]

Have you tried converting the LHS to sin, cos?

[reef elm]

true

[obsidian harness]

challenge problem: find the exact area of the green circle

[upper karma]

is there a synth sol

[dark sparrow]

he did say exact...

[obsidian harness]

not sure; found it on BlueSky

[solemn nova]

In triangle ABC, D is mid point of BC. M is mid point of AD.
Ray BM intersects AC at F. Prove that
(i) 2AF = CF. (ii) 3MF = BM.

pls help

tools mid point theorem and median concurrency

[dark sparrow]

do you have a diagram?

[simple vigil]

Is this even solvable by me as a class 10 student

[obsidian harness]

idk

[upper karma]

i wanted to come up with a funny sol

but there's still algebra so it's bad

||invert at the origin with radius 1 and you only get 1 parameter to solve for (x coordinate of the inverted circle's center)||

the only issue is ||you get a quartic that must have a double root. You can just compute gcd of the quartic and its derivative but it's long and boring||

But at least I get that the radius is ||r=1/4|| so since its not an ugly number maybe there's a nice sol

[obsidian harness]

damn you figured it out

I actually got stuck

[upper karma]

i have the synth 😍

idk I didn't notice this stuff at the start

[obsidian harness]

I would never have thought of circle inversion

[upper karma]

I just didn't want to set 3 tangencies and die

but it's not like it's that nice anyway

I'll write the synth soon™

[upper echo]

||Denote a=tan(x) and b=tan(3x), then your equation is equivalent to (a + b)(ab + a - b + 1)=0 and ab≠1. Then just apply b=a(3-a²)/(1-3a²).||

[solemn nova]

no

[dark sparrow]

then make a diagram.

[upper karma]

ok tbh I don't wanna spend too much time making it look nicer, so I'll just keep it like this and write a sketch

||Let the focus of the parabola be F. Let the centers of the 2 circles on the axis of the parabola be A and B and name them w_a and w_b.
A key property is that if a circle centered at Z on the axis of the parabola is tangent to it at points X and Y then FZ=FX=FY (this follows from the optical properties of parabolas).
Using that or in whatever way you want, prove that the radius of the big circle is 3/2.
Now let T be on the parabola with FT=1 and let K on AB be the center of a circle tangent to the parabola at T. Then prove KFT is equilateral.
Let the line parallel to FT through A intersect KT and w_a at D, E. Let G=BDintw_b. It's easy to see that DE=DT=1/4.
Compute BD with the law of cosines or whatever you want, and check that GD=1/4.
So (EGT) is has radius 1/4, and its center lies on BG,KT,AE so it's tangent to the parabola, w_a, w_b. So it's the circle we wanted and has radius 1/4.||

it feels a bit dumb but whatever

[solemn nova]

bruh

[simple vigil]

What is this looks hard !?

[tired oracle]

i think sl loney is good for 12

[upper karma]

idk probably just skill issue, if you find a better synth lemme know

[upper karma]

but I found out that in general (when the tangency point of w_a with the parabola is not necessarily at its vertex) then KT=(BO+AO)/2 and K=(A+B +2O)/4.
Oh and also, r_b=r_a+4FP (P is the vertex of the parabola)

[shadow dome]

could someone please refer me to some sort of youtube video that will teach me how to do questions like this?

[wide flume]

desmos?

[weak elbow]

Whoever needs any kind of support or help be sure to reach out to me.

Thank you

[light eagle]

https://youtu.be/j7J_Gid7nKg?si=viZ1chjEDfIbmOpl

https://youtu.be/8oEjxGjAeQI?si=1APFoQMv2qbyspHN

[strong root]

Can someone help me with this problem? I've been trying to solve it for a few hours and I don't know how :c.
In triangle ABC, interior bisectors AD and BE are drawn, which intersect at P, such that AD=AC and PC=CE. Find m∠PAE.

[inland tendon]

Hello is someone here I can talk to about geomtry

[rich jasper]

hello

i can be at your service

[inland tendon]

Oh u 😆😆

[inland tendon]

Uh! It was just the same question from earlier 😅

[rich jasper]

do you have an image

[inland tendon]

Mhmmm

Do you need my notes or??

[rich jasper]

hmmm

i could take it for extra info

[inland tendon]

Okay I working on homework right now

[rich jasper]

alright

i can help if i must

[inland tendon]

And remembering what to do

That all

[rich jasper]

you can mainly just revise

look over stuff and it can end up catching back onto you

[inland tendon]

Is so confusing

Is like

In class

We keep moving to new topic of geomtry

[fickle anvil]

What grade are you

I’m trying to learn grade 12 math

I’m in 9 rn

I kinda have it down

Like Ik how to solve some stuff but I’d like to learn more

[quiet vigil]

Lol same here I started studying 11th and 12th stuff from 9th

And before that like at 7th and 8th i used to watch vids of 11th and 12th stuff tho i understood or not

TRIGONOMETRY HAS SO MUCH FREAKING HARD MANIPULATIONS

[quiet vigil]

Is it just me or is it fr?

[fickle anvil]

It is

[quiet vigil]

Phew!

[fickle anvil]

Haha

[quiet vigil]

How much 11th and 12th did ya cover?

[fickle anvil]

Not sure haha

Just learning WAM levels rn

Idek what they are but that’s why my teacher said

[swift patio]

would plane tilings be best put in here or another channel?
i just created this tile and i think it is aperiodic (not sure tho)
https://cdn.discordapp.com/attachments/1375085745890398333/1412688497495769108/IMG_4871.jpg?ex=68b93430&is=68b7e2b0&hm=91a78d86e7ad2e5f8cf67b8143b3af6eaf8f8c7caeced3b99b3783c8176f5a86&

[next python]

Guys I am new and want to understand all of trig is there any thing I should study because I have only covered basic trig and I think I might need it for pre calc if I even get there

[dark sparrow]

organic chemistry tutor, and/or khanacademy

the latter also has practice questions for trig

[next python]

Thanks a lot I will try khan academy

[dark sparrow]

ochem tutor has a lot of videos breaking down all kinds of math questions so you can usually find one if you're stuck on a practice thing

[inland tendon]

I in 11

[proper fable]

how do i do trigonometry man

[spiral lodge]

By studying the theory, understanding the formulas and the rest is only practising a lot.
And, very important, knowing all the basics math (such as how to solve equations, inequalities, and I mean both first and second degree at least, computing things with powers/exponents) very well

[wide flume]

Khan academy helps

[gloomy spear]

how tf does trig even work bro they tell me oh SOHCAHTOA like wtf even is sin

wow it does this magically

or its Opposite/Hypotenuse

like why

and what even is it

[visual skiff]

skill issue

[gloomy spear]

why is sin theta opp/hyp for instance

[visual skiff]

honestly idk but you can look it up

[gloomy spear]

i've been dying bro i genuienly hate this

no one teaches it right

oh bro js SOHCAHTOA

like stfu

teach it properly

[mystic wedge]

It's defined as the ratio of opposite to hypothenuse of a right angle triangle

[gloomy spear]

so its just a ratio between 2 numbers how does that help me

or what does it tell you

[mystic wedge]

That's how sine function is defined for a right angle triangle

[gloomy spear]

so if sin theta is 0.5 that means that opp is half of hyp

that kind of self implies

[mystic wedge]

Exactly

[gloomy spear]

but why is this useful?

[mystic wedge]

It's used in many places especially engineering if you are asking about real world applications

[gloomy spear]

ok i get what it tells u

what is sin

isnt it a function

[lost lichen]

yes

[mystic wedge]

Yes it is

[gloomy spear]

i've seen it graphed

[mystic wedge]

Yeah?

[gloomy spear]

so whats the function

[lost lichen]

its just sine

[mystic wedge]

The sine function?

[gloomy spear]

yeah

[lost lichen]

and it has its own definition

[gloomy spear]

like g(x) = 2x + 4

sin(x) = ?

[mystic wedge]

It really depends on what x is

[lost lichen]

ion think u can represent it like that maybe?

[mystic wedge]

Yeah you can't

[gloomy spear]

im js trying to understand sin, coz and tan

[lost lichen]

yk how to write functions in ordered pair system?

[mystic wedge]

Those are just ratios of different sides of a right angled triangle

[gloomy spear]

ok so they are ratios but if they are also functions they have to be defined by something somehow

especially i saw their graphs

apparently sine is symmetrical

on y

so it has a value right

[gloomy spear]

no sry

[lost lichen]

ok

[mystic wedge]

Alright kid you are not getting it

Analytic Definition (Power Series)

Sine can be defined as the following infinite Taylor series, convergent for all real (and complex) numbers :

\sin z = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} z^{2n+1}
= z - \frac{z^3}{3!} + \frac{z^5}{5!} - \frac{z^7}{7!} + \cdots

This definition is independent of geometry and works directly in analysis and complex variables.

[lost lichen]

sine is like u when construct a triangle with the given one angle as x and another as 90 and take the ratio of the opp and hypotenuse side of the triangle ig??

[lost lichen]

ye that is tru

[lost lichen]

kind of literally the definition

[mystic wedge]

Sine of an angle has a limt
-1 ≤ sinx ≤ 1

[lost lichen]

u accept this scout?

[gloomy spear]

why? is it a percentage?

[mystic wedge]

Yes

[mystic wedge]

It's not?

[lost lichen]

where percentage?

[gloomy spear]

cuz why would it be limited between -1 and 1

oh wait

its cuz

hyp is biggest

always

[lost lichen]

ye

[mystic wedge]

Yeah genius

[gloomy spear]

how can you have -1

negative side length??

[lost lichen]

cause u plot it in a graph

u use the cartesian system

[mystic wedge]

Ion think you'll get it now but the angle can be Greater than 90° tbf it could be any angle

Do you know the ASTC rule?

[lost lichen]

every question in geometry u have done is actually using the modulus of the disp

[gloomy spear]

no i js started learning trig

[mystic wedge]

No not really

[lost lichen]

whats that

oh all silver tea cups