geometry-and-trigonometry

[lost lichen]

lel

[gloomy spear]

aare we talking about sine of sine of theta

[lost lichen]

nope

[gloomy spear]

on the graph

[lost lichen]

we are just talking about sine of theta

[gloomy spear]

sine has to be symetrical (sign independent for input)

[lost lichen]

ye

[lost lichen]

nope

thats cos and sec

[gloomy spear]

if its symmetrical on y

oh is sine not symmetrical

[mystic wedge]

It ise

[visual flume]

help pls

i got √3/24 and its wrong

[lost lichen]

that is not symmetrical

[mystic wedge]

The amplitude of a sine wave is between
-1 and 1

[visual flume]

f(x) = 67sin(x)

[-67, 67]

boi 67 so tuff

[gloomy spear]

why though i get why sin theta has to be between -1 and 1

but why does the sin(x) output

have to be

[visual flume]

help pls

[visual flume]

:3

have to be what

-1 and 1?

[mystic wedge]

The graph and and sinx is telling the same thing

[visual flume]

HELEP9AWOEUIPHP3IUHOI

UWUWUUWUWUW PLS

[lost lichen]

i think maybe u find the central plane find theta and use sin theta maybe

[gloomy spear]

i get it

but why is sin(x) constricted to -1, 1

i get sin theta

because there hyp is bigger

but sin(x) isnt opp/hyp

[mystic wedge]

How can you have a hypotenuse bigger than the opposite ?

[visual flume]

boi what

nah

[mystic wedge]

It literally is

[visual flume]

were not supposed to use trig for this prblem

[gloomy spear]

i thought its some infinite series

[mystic wedge]

For you it isn't

[gloomy spear]

sin(theta) is opp/hyp i thought? is sin(x) = sin(theta)

[mystic wedge]

All that is higher mathematics which you won't get unless you understand the basics

[mystic wedge]

Θ and x are literally variables they could be anything

[lost lichen]

demk

then why send it here?

[gloomy spear]

yes so i dont get why sin(theta) is opp/hyp because i dont get the sin function but are you saying i cant undertand it until later

[mystic wedge]

The sine function is opp/hyp

[gloomy spear]

sin is a ratio yes

[gloomy spear]

so how do u plot opp/hyp

[mystic wedge]

To help you get it
The ratio opp/hyp is coined as sine

[gloomy spear]

ok but isnt the ratio of sine different from the function sine then

i get the ratio of sine

[mystic wedge]

To understand the graphs of you need trig function you need to wait

[mystic wedge]

No

[gloomy spear]

its defined as the raatio between opp and hyp

[mystic wedge]

They are literally the same thing

[mystic wedge]

Yes finally you are getting it

[gloomy spear]

yes but how do you express that as a function

because a functio doesnt care bout triangles

or anything

[mystic wedge]

It is a function you doughnut

[gloomy spear]

ok sin(theta) simplifies to opp/hyp somehow right?

but what is it originally

what does it do to the input

[lost lichen]

and i was right about how u were finna crash out

[lost lichen]

IT DRAWS A TRIANGLE AND FIND THE RATIO

[mystic wedge]

All right you doughnut listen carefully
The ratio opp/hyp is called sine
It isn't derived but rather named

You don't solve 100 pages long equations to get that sine is opp/hyp

[mystic wedge]

I mean this kid thinks that what he's being told isn't right and being presented in the wrong way and had the wrong understanding of everything

[gloomy spear]

yes thats ratio sine i get it ratio sine is just a name we gave to a ratio

[lost lichen]

lel

[mystic wedge]

Yes

You are finally getting it

[gloomy spear]

im trying to understand 😭

[lost lichen]

tbh every function is just named fr

[gloomy spear]

yes i got that a while ago

but i have a question

[mystic wedge]

Yeah

[mystic wedge]

What's that?

It better not be something stupid

[gloomy spear]

on the difference between sin function and sine ratio, sine ratio is just discovered and named right i get it but the function sine isnt a ratio, rather its a function so what is the function sin() y=sin(x) sine the ratio isnt sine(x) sine the ratio just means opp/hyp of a right angle triangle i dot even get why you start talking about some sin(theta) when theta (the input) isnt even in the function

[lost lichen]

ye alr i quit but ye u didnt quite teach him that

[mystic wedge]

That's where you are wrong I'm sure that you don't even know what a function is

[gloomy spear]

a function takes an innput x and produces an output

sin(theta) isnt doing that

[mystic wedge]

What do you think a sine does?

[gloomy spear]

why do you need theta

you can see the opp/hyp from any triangle

anyways

if theta is marked

so it should just be sine of triangle

[mystic wedge]

You can't? A hypothenuse is defined for right triangles

[gloomy spear]

the input theta is doing nothing

[gloomy spear]

.

[mystic wedge]

It literally is what makes triangles different for different angles you'll get different lengths hence different ratios

You have no idea of a basic geometry

[mystic wedge]

No wonder

Tbh he isn't not that level yet

[gloomy spear]

the definition of the stupid function is f(x) take x and make output so then if you take any angle theta as input say 30 tthen you aarent using it in opp/hyp directly (i get that theta defines the triangle but then it shouldnt be sin(theta) as a function it should be sine of the triangle)

[mystic wedge]

A sine function is dependent on θ where θ is the angle

It'll eventually cancel out

[gloomy spear]

yes its dependent on that

but also on the triangles side length

and hypotenuse length

[lost lichen]

tru

[mystic wedge]

Take these triangles for example it's clearly not of same sides but the trig ratios will always be same

[lost lichen]

which grade u in @gloomy spear ?

[gloomy spear]

english year 11

idk american

[mystic wedge]

You from the UK?

[gloomy spear]

u are too i presume ("doughnut")

[lost lichen]

ahhhhh ok ok

[mystic wedge]

OMG in that case I'm done explaining it to you the gcse curriculum is a mistake I've seen it they reach high schoolers middle/elementary school maths/science

[gloomy spear]

what if its opp/hyp and opp is a different size in the 2

[mystic wedge]

I'm not

[gloomy spear]

wdym

the GCSE curriculum doesnt go into depth

[mystic wedge]

Exactly

[gloomy spear]

i personally wanna go into maths heavy fields

[lost lichen]

well the ratio would still be the same

[gloomy spear]

so im going into depth on my own accord

[lost lichen]

since the triangles are similar

[mystic wedge]

That's why you are not getting what a sine is

[gloomy spear]

i care about the math not the grade

[mystic wedge]

Okay ig I was being rude i always encourage learning and curiosity

[mystic wedge]

Yes

Yes

[gloomy spear]

oh similar triangles

like in the pythagoras

[lost lichen]

LEL

[gloomy spear]

proof

?

[lost lichen]

YES

[mystic wedge]

Kinda

[lost lichen]

U LEARNT ABOUT THAT PROOF BUT NOT SINE

wild

[mystic wedge]

Lol

[gloomy spear]

i learned it outside skl

[lost lichen]

ahhh

ok

so sine is actually based on similarity

thats why we say its equal if angle equal even for different right angle triangles

[mystic wedge]

It doesn't matter what the length is as long as the angle right

[gloomy spear]

because angles dictate length

in a triangle

[lost lichen]

yep

[mystic wedge]

Yes!!

[gloomy spear]

anyways

oh

[lost lichen]

basically

like ratios of lengths but whatevr

[gloomy spear]

so then here opp and adj are swapped

to give the same thing

[lost lichen]

yes

[gloomy spear]

if 30 stays theta

or 60

[mystic wedge]

Since the sum of internal angles of a triangle is 180°

[lost lichen]

alr ye i gotta go write my record and sleep

gl

hf

bye bye

[gloomy spear]

ok so that means if a triangle has 2 equal angles (besides the right one) the sin of each angle will be the same

in both triangles

[mystic wedge]

No really but
In that case sinx = cosx
That's why we can say that
Sinx = cos90-x
If the sides are not swapped

[mystic wedge]

That's a 45-45-90° triangle

[gloomy spear]

i mean in the first one if we keep 30 as theta

and then go to the second one

and 30 is still our theta

then the opp and adj are swapped

right?

which is what makes the sin the same

[mystic wedge]

Yea that's kinda what I just explained

[gloomy spear]

i get it now thx

[mystic wedge]

Byee

[gloomy spear]

oh here i meant if 2 triangles

[mystic wedge]

These are from allied angles you'll learn it in the future

[gloomy spear]

sry wrong wording

like if 1 triangle has 30-60-90 and the other 60-30-90

yeah

i get it why sin is based on angles

[mystic wedge]

They are similar triangle m

[gloomy spear]

and why it stays for similar triangles

[mystic wedge]

Good!!!

Yessss

You finally got it

[gloomy spear]

yes thank you

[mystic wedge]

Welcome

[gloomy spear]

what schooling system r u in?

[mystic wedge]

India

And it's hell

[gloomy spear]

what grade?

[mystic wedge]

11

[gloomy spear]

i can imagine

is that 15-16 for u asw

ye india, china, korea

are hardcore

[mystic wedge]

Yeah there's an engineering entrance exam called jee it tests you on maths, physics and chemistry on level just below university

[mystic wedge]

Yep

[mystic wedge]

True

[gloomy spear]

btw GCSE isnt our final exams before uni

its an intermediary

[mystic wedge]

Ohh

[gloomy spear]

before a levels

which are way harder

and deeper

[mystic wedge]

Ohhh

[gloomy spear]

GCSEs are easy lol compared to a level

[mystic wedge]

Makes sense not everyone can get into a uni as good as cambridge right 😂

[gloomy spear]

did u think UK skl was that dumb lol

[mystic wedge]

I really did

[gloomy spear]

a levels in comp sci are too easy though tbh

comp sci in the west

is too easy

not rigorous

[mystic wedge]

Is that so?

[gloomy spear]

yes at a level they dont even expect u to be able to build an API or even describe how a fullstack app could work

[mystic wedge]

Idk about comp sci with regard to academics but I do also learn coding

[gloomy spear]

what lang?

[mystic wedge]

That's crazy no way

[gloomy spear]

but im already doing ts privately

so its all good

ik python and js

[mystic wedge]

Right I'm not doing it as I'm focusing on jee but before I was learning JavaScript and python

[mystic wedge]

That's nice

[gloomy spear]

yeah i am proficient (can never say i "know" a lang)

at them

no one actually knows a whole language

[mystic wedge]

Yeah

That's true😂

[gloomy spear]

unless they have like 15+ years of experience

maybe

they might be close

[mystic wedge]

I don't think even they know

[gloomy spear]

yes they could know one niche

but never all the modules and libraries lol

[mystic wedge]

Alright I'm going to sleep it's like 1:40am over here

[gloomy spear]

gn

[mystic wedge]

True

[mystic wedge]

Thanks

Byee

[gloomy spear]

bye

[graceful jacinth]

Just started geometry and algebra (dont mind the smart score...)

[wide flume]

cool! Do you need help on this question?

[queen venture]

how did you come up with this?

[upper echo]

Just substitute tan(4x)=(a+b)/(1-ab) into equation and factor out a+b and 1-ab in the numerator.

[queen venture]

ooh thats clever

[lost lichen]

ik am very late but i dont like the way that this looks

[light eagle]

Does anyone know a site or git or anything that compiles all known solutions for polygonal and polyhedral packings?

Or somewhere that has the vast majority of them in one place?

[mystic umbra]

This is mostly like highschool-middleschool geometry yo 🥀

you're not going to find many people discussing that here

[light eagle]

you never know!

[graceful jacinth]

am in 8th grade and was not expecting to encounter so many theorms in geometry , fun.

[neat oracle]

Hi guys, can you guys help me with this simple question, where did they get the 1+1 on the second to last step? This equation is being used in a unit circle, b1 and b2 are the coordinates of point B, a1 and a2 are the coordinates for point A

Not really sure if i ahve to open a channel for this since its not that hard of a question, i just got a bit confused and wanna get clarification

[knotty quiver]

1 is the hypotenuse of A and B

in a unit circle, the coordinates are also the length of the sides of a right triangle formed by a coordinate

[neat oracle]

Yes i understand that but how did that apear in the place of a1^2+a2^2+b1^2+b2^2

[knotty quiver]

pythagoras theorem

\begin{tikzpicture}[scale=5]
\draw (0,0) circle[radius=1];
\coordinate (A) at (0.87,0.5);
\draw (0,0) -- node[above left] {$1$} (A) node {$A=(a_1,a_2)$};
\draw[dashed] (0,0) -- node[below] {$a_1$} (0.87,0);
\draw[dashed] (0.87,0) --  node[right] {$a_2$} (0.87,0.5);
\node at (0,-0.5) {$a_1^2+a_2^2=1$};
\end{tikzpicture}

[somber coyote]

glass

[knotty quiver]

here's one for A

[neat oracle]

Ah of course

I just forgot about pythagoras lmao

[knotty quiver]

ah i see

[neat oracle]

Thanks tho 😁

[azure rivet]

my god

idk how to write this shi

no idea

[knotty quiver]

if you want to learn, you can read the first part of the TikZ documentation (check #latex-help pins)

then you can search it for anything specific you want. (Hopefully your pdf reader has search functionality)

[azure rivet]

[queen venture]

what are you facing difficulty in? struggling to choose which theorem to use or remembering them?

[covert tinsel]

I do not understand this question for the sake of me I tried solving it twice only got so far as in the height gets doubled but I don’t know where to go from here

[proven estuary]

Can someone help me? How do you solve these? 🥹 my teacher doesn't explain well

[exotic yarrow]

The surface area of the two prisms combined without overlap is 2K. However, the surface area of the resulting prisms is 92/47 K. This loss is due to not counting one of the square bases in the surface area for each prism. So, the area of one square base is 1/2 (2K - 92/47 K) = 1/47 K. See where you can get from here.

Also, google: https://youtu.be/UHWKPglwxpo?si=0kF9hAvSgz8mBea6

[exotic yarrow]

Recall that the distance from a point to the center of rotation is equal to the distance from the image of said point to the center of rotation

So for two pairs of points (image and preimage), you can find the locus of points equidistant from each point in the pair by constructing

Whichever point is in both pairs is your centre of rotation

https://mammothmemory.net/maths/geometry/manipulating-shapes/how-do-you-find-the-centre-of-rotation.html

That’s the logic that they’re using here

[proven estuary]

I am not trying to find points. Idk how to explain it but here's my teacher's work

I am not sure where to start or what to do

[exotic yarrow]

What? The centre of rotation is a point.

It looks like your teacher is just doing trial and error on each marked point to see which matches the given angle of rotation though

[proven estuary]

idk how to do this bruh

I regret taking accel geometry im dumb

[exotic yarrow]

In 35, you have points D, E, F being marked that aren’t part of the image or preimage

The implicit message is that one of those is the centre of rotation

So you test all of them, as I said earlier

Ex. Your teacher testing F by seeing what the angle of rotation would be about F from A to A’

This angle is acute

So that can’t possibly be it

They did a similar thin for E between B and B’

But that angle is way too big, so no

Then they tested D between A and A’

Which gives a reasonably 100 degree angle

Combined with the fact F and E are eliminated, that leaves D as the answer

[proven estuary]

oh I also noticed E and F are CW, only D is CCW 🫠 welp thank you

how r u so smart

[devout pollen]

help how do we get fdg

[red smelt]

FDE = 180 - CDE = 180 - 138 = 42, FDG = 90 - 42 = 48

[devout pollen]

TYYYY we had figured this out by now tysm

Do you know how to get x

[red smelt]

x is correct

[devout pollen]

Ii ment to say W I'm so sorry

😭🙏

[red smelt]

w = 8

JCB = HCD (opposite angle) = 50

50 = 6w + 2, solve for w, w = 8

[final frost]

Measure of fdg is 48 degreees

Mb I didn’t see it was answered

[hoary totem]

draw arcs connecting the points

what is the center of these partial circles

looks like D for 35 and F for 36

[frozen pilot]

can someone slide the rest of the identities this is all i have rn 🥺

[dire jackal]

there might be a few extras on here

[dark sparrow]

heres a thing i wrote several years ago (section 6 is a bit of a showoff thing and you dont rly need to memorize those at all)

[knotty quiver]

pythagoras identities

also it's already pinned

[graceful jacinth]

Man I started studying geometry personally three days ago. so much learning to do 🥹

[upper karma]

you didn't mentioned the "tan(90-x) = cot(x)" one's if am right :<

[dark sparrow]

yeah i suppose i didnt

[upper karma]

:>

[lost lichen]

did u actually prove section 6 when u were that day years old?

[dark sparrow]

yes

[vagrant kestrel]

im currently using khan to learn. would someone be able to vc and explain what im missing?

[obsidian harness]

we don't do VCs on this server

[vagrant kestrel]

right, but this is discord.

[obsidian harness]

and I'm telling you that the chance that someone wants to VC you instead of just type on this server is low

[hasty token]

Sorry for pinging, can you share it as a screenshot? I don't know but i can't download the PDF.

[dark sparrow]

here you go

[hasty token]

Thank you so much 🙏

[lost lichen]

wild

i assume u were like 14 or 15?

[dark sparrow]

this was like 7 years ago, and i am 26 now. so nope, you underestimated my age back then.

[lost lichen]

ohh

cool

[lost lichen]

oh ye where do u ask combinatorics questions again?

[dark sparrow]

#discrete-math or i guess #precalculus

(irrelevant reply-ping...)

[lost lichen]

ite

[graceful jacinth]

wait help pls?

I was think additive popery of length or midpoint for the first one.

For the second I was thinking SSS

Or side-side-side therom

wait

yeah its prob additive property of length

OK

google ai says midpoint

and SSS and be deduced because al three sides are congruent to each other

I GOT IT

[brazen vessel]

$\frac{\tan \left(θ\right)}{\sec \left(θ\right)-\cos \left(θ\right)}=\sin \left(\pi \right)$

[somber coyote]

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

[fresh forge]

qn6 what step i wrong the ans is tanx

[worthy eagle]

send a better photo, nobody can see what's written

[fresh forge]

sorry

[worthy eagle]

in 2nd step how did you go from
1-tan²A / 2tanA → 1-sin²A / 2sinA*cosA ?

you are wrong here

maybe try again

[simple vigil]

Bro the writing 💀

[proper fjord]

I don't even know where to start.

I'm genuinely so cooked.

This is supposed to be a review and I have a test on this tomorrow 💔

[light ingot]

I would start by writing down the geometry words that you don't know the definitions of and try to understand what they mean

For example: complementary angles, angle bisector, etc.

It will be helpful to draw some examples of what each of these words mean in pictures. Then to solve the problems where you need to find measures of angles, practice understanding angle addition/subtraction as well as the algebra needed to find x in each of those.

If any of the notation is confusing as well you could write down what they mean from a youtube video or Khan Academy and practice using them.

[barren flint]

hello quick question! i got the answer 3.134 and rounded it down to 3.1 and it said i was wrong lol. if anyone could point out what i did wrong that’d be great

[light ingot]

You will have to use tan instead of sin because tangent results in the opposite/adjacent side lengths of the angle input. sin results in the opposite/hypotenus side lengths of the angle input, you don't know the length of the hypotenuse

Well actually nevermind lol

[barren flint]

wait i just realized i put 9.4 instead of 9.7 😭

[light ingot]

oh I gotcha

[barren flint]

thank you though lol!!

[light ingot]

do you feel comfortable setting up the trig ratios to find the side lengths?

[barren flint]

by using sin i can

im not too sure abt tan tho cuz none of the example problems listed use it

[brazen cloud]

just remember SOHCAHTOA

[fading folio]

hey guys

Im in precalc right now, but in class we're reviewing trig/algebra 2 rn bc its the start of school and I lowkey forgot everything

[silent plank]

review using Khan academy

[fading folio]

ok

[wheat elm]

how is #8 negative cause i’m confused

[cunning lion]

,calc 840 - 2*360

[somber coyote]

Result:

120

[cunning lion]

it's in the second quadrant, where sin is positive and cos is negative

[karmic pulsar]

hey guys, i really need help bc i dont understand, so to find the distance between 2 point, we use the pythagorean theorem, but also everything is squared, so can we cancel the square by rooting everything???

[cunning lion]

square roots don't distribute over addition, so in general
[ \sqrt{a^2 + b^2} \ne \sqrt{a^2} + \sqrt{b^2} ]

[somber coyote]

cloud

[karmic pulsar]

no but i mean, if the equation is c² = a² + b², and we use this equation to find the distance between 2 points, cant we just root c², and since we are doing that on 1 side we have to do it on the other

you know what i mean?

[cunning lion]

well you can always use the rearranged formula
[ c = \sqrt{a^2 + b^2} ]
if that's what you mean

[somber coyote]

cloud

[karmic pulsar]

thats my point, why are we cancelling the square of D, but not the rest of the equation?

[cunning lion]

are you asking why we aren't doing
[ \sqrt{a^2 + b^2} = a + b? ]

[somber coyote]

cloud

[karmic pulsar]

no, what i mean is if to find the distance we use c²=a²+b², why do we square both sides and destroy the roots. so use this equation instead c=a+b

[cunning lion]

but that equation isn't true

like using
[ c^2 = a^2 + b^2 ]
and
[ c = a + b ]
will give you different values for $c$ (try it for some sample values and see)

[somber coyote]

cloud

[plucky crag]

the square root of a^2+b^2 is not a+b. for example a = 3 b =5 (sqrt(9+25) therefore sqrt 34, whilst 3+5 is 8 )

[hoary totem]

because if it was $c = a + b$, that implies $c^2 = a^2 + 2ab + b^2$ which is evidently not the pythagorean theorem, which says $c^2 = a^2 + b^2$

[somber coyote]

shiru

[obsidian harness]

you can't unsee this

[simple vigil]

Yo @obsidian harness Are you good in linear equation in two variables?

[spiral lodge]

https://dontasktoask.com

[dire jackal]

dont post the same thing on multiple channels

[hollow field]

can someone help me

[pearl hinge]

!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/

[karmic pulsar]

if we take 100-3x=20+2x, is X= 24 or x=16???

[austere silo]

#prealg-and-algebra

[austere silo]

But add 3x-20 to both sides

[karmic pulsar]

Ok but why isnt it -100-2x

[austere silo]

Do that, you get -5x=-80

[silent plank]

you could do that,
most people prefer to avoid excessive negative signs

if you're adding the same amount to both sides, it'll still be valid

and lead to the same end result

[opal dawn]

hi guys

can someone help me understand geometry

like point, line, line segment, ray, plane, colinar, coplanar, and congruent segments

[devout hull]

Study geometry positional for a better understanding of what they are, depending what geometry are you studying, they change.

[opal dawn]

can u look at my post

im really confused about that because i thought f was in both planes

[devout hull]

You are studying plane geometry, right?

[opal dawn]

yes

plz help me :(

[devout hull]

A geometric plane is a two-dimensional location

[opal dawn]

how is f not in both planes

i dont get it

:(

[devout hull]

I find it confusing, but it depends on where F is.

[opal dawn]

its in the middle

do u understand

[devout hull]

You can see that like a function

[devout hull]

Ok, to be in both planes it has to be in the middle of the planes

[opal dawn]

wdym

[devout hull]

Since these planes are perpendicular, for a point to belong to both it needs to be part of the intersection of these planes.

[opal dawn]

so f and d are both?

i dont get it

:(

[devout hull]

I see two letters F, am I correct?

[opal dawn]

f and d go through the lines

i thin te one on the left is an e

[devout hull]

ED It is on both planes

[opal dawn]

so ED is the only one that involves both?

[devout hull]

Yes

[opal dawn]

so only ED

so A, C, m, and n are on x right?

[devout hull]

ED, EB and BD

[opal dawn]

wdym

i thought it was only ED

oh same line

[devout hull]

You can track the other points as well

[opal dawn]

ok

those are all that one line right?

[devout hull]

Yes

[opal dawn]

ok

[devout hull]

See

[opal dawn]

thank you

[devout hull]

The Cartesian plane explains it better

[subtle oak]

can anyone help me with geometry proofs?

[trail mural]

sure

and

https://dontasktoask.com

[patent wigeon]

Can someone show me an example of a proof

[trail mural]

what proof exactly

[patent wigeon]

Direct

[trail mural]

no like what topic

what type

thats so vague

[patent wigeon]

Rigid transformations if that’s possible

[trail mural]

im so sorry but thats still very vague for me to show an example of proof of

what do u want exactly

[patent wigeon]

Nvm

[hoary totem]

proof that c × 0 = 0 (where c is any number)

we know that 0 is the "additive identity", i.e. c + 0 = c for any c
therefore 0 + 0 = 0
therefore c × 0 = c × (0 + 0)
now distribute the c to get
c × 0 = (c × 0) + (c × 0)
now subtract (c × 0) from both sides to get
0 = c × 0
■

(the square means end of proof)

[wooden swan]

Does anyone know what I’m doing wrong? I keep getting decibel

[dire jackal]

you didnt subract 2 from RHS i think

in the second step its still 11

[wooden swan]

I thought you have to add it

Had*

[dire jackal]

you always do the same operation on both sides of the equation

if you subtract 2 from LHS, you need to do the same from the RHS to keep both sides equal

[wooden swan]

Ohhhhhh

Ok

I see

Thank you

[dire jackal]

yw!

[obsidian harness]

this analogy should help

[wooden swan]

Thank you!

[obsidian harness]

np!

[dire jackal]

thats honestly a great analogy, saving it!

[vocal mantle]

how do we prove section formula with trigo?

[obsidian harness]

you don't need trigo at all....

it's 100% algebra

[vocal mantle]

ah

[lost lichen]

not 4th?

oh nvm tht was a decade old msg

[hidden fossil]

guys is there any good book on geometry

[queen venture]

coordinate geometry or like triangles?

[hidden fossil]

just geometry in general honestly

[queen venture]

coordinate by sl loney is good

[hidden fossil]

okokokok ill look it up thanks!!!

[upper karma]

he asked for geo and you give him bash? Who hurt you

[hidden fossil]

CRYING

is the book scary???

[cunning lion]

120 degrees is in the 2nd quadrant last i checked

[pastel bone]

broooo I miss this shit sm 💔

it was fun ash

[mint iris]

Chill guys

[queen venture]

is it not relevant?

[keen hollow]

mersenne jumpscare

[digital wigeon]

so my teacher gave this to my class and asked us to prove it..

hooray!! :D (my brian hurts)

[winter sail]

I need someone who can teach me trignometry in zoom

[signal bridge]

I can

[trail mural]

😱

[winter sail]

how?

[proper hornet]

Dash

[signal bridge]

How do you want?

[winter sail]

By zoom

[winter sail]

Meeting

[viscid nimbus]

i'd focus on expanding the right side

[upper karma]

yo chat, i was just peeking at this and couldn't get how the coordinates (sqrt3/2,1/2) were found in 30 degree sector

like you can find the arc length and then find the coords but you should know atleast one point (x or y) or am i just losing it

and moreover how do you find pie related identities through it :/ like if you draw up a perpendicular on radius then it will something like 1 - x (where x is something which is left out by the perpendicular)

[obsidian harness]

it all comes from the 30-60-90 (and the 45-45-90) triangle

take an equilateral triangle with sides 2, 2, 2

now split it in half, so the side 2 gets split into halves 1, 1

then we look at one half of the equilateral triangle only

[upper karma]

ohh yeahh i get itt

forgot abt the interior angles

[obsidian harness]

no worries!

[upper karma]

..

[obsidian harness]

yeah then by Pythagoras we have b^2 + 1^2 = 2^2 or b = sqrt(3)

[obsidian harness]

I'm not sure what you mean

[upper karma]

i mean there are identities like cos(pie/2) etc., they came up through the unit circle right?

[obsidian harness]

yes

so the unit circle definition says that cos(theta) is the x-coordinate

[upper karma]

nvm

i am dumb fr

got it

thnx

[obsidian harness]

no worries!

[upper karma]

ae yo bro, sorry for disturbing again but sin30 = 1/2 but if i take the ratio of sides for sin it comes out as (sqrt3)/2

[obsidian harness]

the side 1 is opposite to the 30 degree angle

[upper karma]

oh yeah, finally i geddit :/

[fresh shell]

This question is confusing me. It says angle 4 is 48 degrees but angle 3 and angle 4 have to be 180 total and 4 is clearly a wider angle than 3?

[exotic yarrow]

Diagrams can be deceiving

What's more likely is they copied the general diagram and picked a random number for one of the angles

[fresh shell]

Yeah, I looked at the other questions and that’s exactly what they did 😭

Thank you

[tame jasper]

i once had a test that had one of these angles come out to 114 degrees when it was clearly an acute angle…

ref do something ?

[fresh shell]

Its so confusing when that happens 😭

[blissful sandal]

i had an agle go to 200+ degrees... its j about the numbers not the angle math part

[analog ether]

E

[regal kraken]

just covert sec and tan into sin and cos

in numerator it would be 1+sinx

=

write 1 as sin ^2 x/2 + cos ^ 2 x/2

[upper karma]

Use half angle identity and simplify both sides (lhs into cos and rhs into tan) then show them equal (if you still haven't done it)

[tropic cairn]

I think ts the ans

[frigid meteor]

how to get good at math

[cunning lion]

practice

[fresh shell]

Nah I got it already dw, the teacher said he just forgot to change the diagrams 😭

[tropic cairn]

Can any1 help me in like the thinking process u should have while doing trigonometry and dealing with identities, like which identify is right to apply in the question and stuff, its a bit confusing for me

[brave karma]

same doubt for me too

[sterile kayak]

ur 4's look like 9s

also the diagram on that problem pmo sm

[river schooner]

<@&286206848099549185>

<@&286206848099549185>

HELPP PLS

[native pier]

hi, welcome to the mathcord! please only ping helpers once!

[river schooner]

Oh sorry

[native pier]

also, how can we help you if we don't know your question

[river schooner]

Need help on arc length

[native pier]

can you post your question here?

[river schooner]

Yes

[native pier]

picture on the way?

[native pier]

okay! what have you tried

[river schooner]

He attempts to explain it but I dont understand what you do with the radians

What do I minus from 360

[native pier]

the arc length for <AOB

or wait

[river schooner]

Is there a circumference formula

[native pier]

let me rephrase that

that wasn't a good explanation

[river schooner]

Ok

[native pier]

do you agree that arc AOB is (2pi/3)/2pi of the whole circumference?

[river schooner]

Yes

[native pier]

okay. so that simplifies to: AOB is 1/3 of the whole circumference

do you also agree that arc ACB is [whole circumeference]-AOB?

(please let me know if you have questions by the way!)

[river schooner]

OH

Yes I understand now ty

[native pier]

no problem! let us know if you have any more questions

[sterile kayak]

i would personally do it by seeing ACB as 2/3 of the circle so I just multiply the circumference by 2/3

[native pier]

that works too!

[little viper]

yo im looking for a tutor for the sat im willing to pay if reasonable and you have credentials dm if interested

[brave karma]

it could also be solved using (pirtheta)/180

[paper rose]

Wow you are all genius

[tropic cairn]

Thats jus how i write

[next gale]

Try to find complex solution for z
$$
sin(z) = 2
$$

[somber coyote]

TheSup_3912

[teal patio]

can someone help with this

oh nvm i figured it out lol

[hoary totem]

if $\sin(z) = 2$, then one cosine value corresponding to $z$ is $\cos(z) = \sqrt{1 - 2^2} = 3i$. therefore one of the solutions for $z$ satisfies $e^{iz} = \sqrt3 i + 2i = i(2 + \sqrt3)$. therefore one of the solutuins is $z = \ln(i(2 + \sqrt3))/i = -i\ln(i(2 + \sqrt3))$

[somber coyote]

shiru

[sudden ice]

does anybody have resources for conics questions

[lost lichen]

oh u were doing the 840 one i was doing like te one before that

[cold veldt]

2 pi

right ?

[lost lichen]

ye it is

[lime crown]

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

[ocean canopy]

i think the cos²+sin²=1 identity would be useful here, just a lil hint

[spring cedar]

not sure if this is the ans you r looking for, but ig this could be one of the way to approach it

[ocean canopy]

can you explain your answer, I dont really understand it

[spring cedar]

sure, which part

[ocean canopy]

top right

[spring cedar]

i just took sinx common from sinx + sin^2 x

[ocean canopy]

shi i forgor 😔

[spring cedar]

happens👍

[night cape]

hi, im terms of an angle and the sine/cosine/tan functions
is it correct to say that a trig function is eqivalent to the angle theta?
and vise versa?
as in its a fractinoal representation of the angle
like .5 is sin(30) degrees

[cunning lion]

there is only a 1-1 correspondence between angles and trigonometric ratios if you restrict the angles to be acute

[night cape]

ohhh

[azure helm]

if I know cos^2(t) = 0 and that sen^2(t) = 1, how do I find t?

[upper echo]

And of course sin(x)=(sqrt(5)-1)/2 from (2)

[chilly fiber]

Wait guys

I forgot how to find the diameter

So I used one end point and another endpoint, using a straight line

And found the distance

Is that still considered the diameter?

(Yes I went through the middle of the circle)

[cunning lion]

any straight line passing through the center of a circle is a diameter of that circle

and its length is 'the diameter'

[upper karma]

after getting sin^2x (1+sinx) i took sin^2x as 1-sinx (rearranging the given) which leads to 1 - sin^2 or cos^2

[upper karma]

maybe there are multiple answers, you know in what terms you have to show the answer in?

[storm ridge]

how to change to degree mode

[cunning lion]

wrench button in top right corner

[storm ridge]

I see🙏

[queen folio]

this stupid math question that took me several hours before I gave up, I was seriously considering wrighting y=ix+i

[visual flume]

help pls

i got 1/6 but its wrong

[silver basalt]

Hi, I’m struggling on #6 and #7 can anyone help me?

[civic rock]

,rotate

[upper karma]

any1 can help me prove shors equation?

[azure rivet]

$V_f = (1/3) \pi h(R^{2} + r^{2} + Rr)=\frac{7}{12}\pi R^{2}h $

[azure rivet]

now simplify

[frozen mason]

Im struggling on this section

[sterile kayak]

the answer is 5

[frozen mason]

For which question?

[brave bison]

In 7th question, your answer is missing 0, 1, 10
8th is 2, 3, 5, 7
9th is nothing (∅) Clear contradiction there.

[frozen mason]

Thank you

[brave bison]

You're welcome.

[sterile kayak]

idk

[frozen mason]

Are you sure 9 is correct?

[brave bison]

For x to be satisfied, it would have to be a prime number and and not a prime number at the same time, that's just not possible.

[small spade]

who can explain me this problem?

Calculate the value of k for the lines:
x + 2y - 3 = 0, x - ky + 4 = 0
so that
a) they are parallel
b) they are perpendicular

[silver basalt]

Can anyone help me with problems 6 and 7 please?

[jovial compass]

U should convert both the equation to a slope intercept form:y=mx+ c
Where m is the slope

[glad tendon]

i made a python program to do trig for me lol

[sterile kayak]

https://tenor.com/view/shaquille-o'neal-shaq-67-6-7-gif-5184942973847308913

[sterile kayak]

when the triangle isnt a right triangle:

https://tenor.com/view/kermit-kermit-the-frog-kermit-the-frog-meme-worried-kermit-sad-kermit-gif-2908512056935738589

[cunning lion]

law of sines/ law of cosines :)

[sterile kayak]

yeah but his code isnt made to do those

[glad tendon]

yea ok lol

GET OU-

[half heron]

gonna be honest i rlly dont get the law of sines

[worthy eagle]

Law of sines says that the ratio of any side of a triangle to the sine of the angle opposite to the chosen side is constant.
So in triangle of sides a,b,c and their corresponding opposite angles A,B,C, we have a/sinA = b/sinB = c/sinC

[sterile kayak]

is bro Shakespeare

https://tenor.com/view/fire-writing-gif-24533171

[dark sparrow]

???

[long yarrow]

That’s poetry to me.

[hoary totem]

because a/sinA = b/sinB = c/sinC = diameter of circumcircle

[sterile kayak]

also because because a/sinA = b/sinB = c/sinC = diameter of circumcircle

[upper karma]

i might be wrong but i think thats for the same reason as why a/sinA = b/sinB = c/sinC = diameter of circumcircle

[cinder ledge]

Is c²=a²+b²-2ab.cos(c) valid for right-triangles?

[exotic yarrow]

Yes - the law of cosines holds for all triangles.

[cinder ledge]

Alright 👍

Humm, cosines just cancels itself in case of 90 degrees angle?

[silent plank]

cos(90°) = 0 which leaves you with pythag

also use capital C for the angle when using lowercase c for the side

[livid vector]

Here's a Voronoi diagram with the metric varying between Manhattan and Euclidean distance, and you can see how each boundary has at least one fixed point (spoilered because ugly lol). Does every diagram like this have fixed points?

Here's the code I used to make it. It's really unoptimized so don't be scared if the tab freezes lol

If you add ≡₁(×Weights) between lines 8 and 9, it adds random multiplicative weighting

Also, is there a way to predict how many fixed points there'll be in a given boundary?

[manic swan]

lol

[livid vector]

if it works, it works 💀

it will work eventually it just takes like 30s

Interesting counterexample bc I think the boundary doesn't even exist in Manhattan distance

I think all 2-region diagrams have at least one (usually at least 2) fixed points though

(change [4 2] to [2 2] in line 3 to see those)

[upper karma]

Does anyone understand how to get better understand soh-cah-toa with angels, decimal numbers, and solving expressions with them. Im self taught so far and cant find anywhere on youtube that explain kind of what Im looking for. For example If cos(236 degrees) = -0.559, find the value of sin(236 degrees).

[opal dawn]

hi

would plane ABC work for plane W as well?

Also, with an intersection, how would I know if it's a line, ray, etc . . . if it doesn't mention anything?

[exotic yarrow]

No because there are points in plane ABC that are not part of the intersection

ex. B is in plane ABC

but is not in plane ADE

[opal dawn]

i dont get it

[exotic yarrow]

two planes either don't intersect at all or have a line as their intersection

[opal dawn]

can u explain for plane DAC

doesnt ABC make a plane?

[exotic yarrow]

planes DAC and ABC are the same plane

namely plane W

[opal dawn]

so am i right?

u said i was wrong earlier

[exotic yarrow]

hold on I was talking about 7 for some reason????

yes plane ABC also works for question 6

[exotic yarrow]

this commentary still holds

disregard the rest

[opal dawn]

how about ABC being the same as plane W

does that work

[exotic yarrow]

yes plane ABC also works for question 6

[opal dawn]

same with DAB

[exotic yarrow]

yes

[opal dawn]

ok

ty

[exotic yarrow]

any selection of three (or four) letters from {A,B,C,D} works

[opal dawn]

so the intersection of ABE and ABD, AB would intersect and would be a line?

for example to the other question

[exotic yarrow]

yes the intersection of plane ABE and ABD is line AB

[opal dawn]

so <-->AB

ok

ty ty

line segment ----

ray --->

line <--->

oh yeah

when do I use commas?

for example is it plane A,B,C or ABC

and for mentioning points/planes/etc . . . would it be with commas?

[exotic yarrow]

don't use commas for any of those

line segment, ray, line, or plane

[opal dawn]

like which 2 planes are collinear, i would put A,B

the question prob doesnt make sense

[exotic yarrow]

in that case you're using the comma as a grammatical object, so that's fine (assuming you used actually correct plane names)

to separate two things

[opal dawn]

just an example

so for mentioning a list i use them in geometry?

like points A,B,E are collinear

not ABE?

[exotic yarrow]

yes because that's the function of a comma in language in general

[opal dawn]

ok

does it matter whether I use the comma or not?

[exotic yarrow]

you should

because again you're separating the points in a list

they are not coagulating into a single entity

anyway I gtg

[opal dawn]

noo

bye

guys

[opal dawn]

can someone explain C:

[opal dawn]

also why is not included in the intersection of ZVY and R

is it because its higher up than the other 4?

[upper karma]

Think of it like this
When two lines meet their intersection is a point
Scaling up from 2D to 3D
When 2 planes meet what would their intersection be?

[upper karma]

They said intersection which means it should lie on both the planes
Is Z in both plane R and ZVY?

[opal dawn]

i dont think so

but idk :(

[opal dawn]

im not sure

z can be apart of the planes but idk if its the same plane as R

[upper karma]

Let's go step by step
Is Z on the plane ZVY?

[opal dawn]

yes

[upper karma]

Good

Is Z on the plane R?

[opal dawn]

idk

[upper karma]

Just check if Z lies on the plane or not

[opal dawn]

how do i do that

[upper karma]

Well from the image you can say that Z is above the plane R
Agreed?

[opal dawn]

yes

so plane ZXW is not the same as R

[upper karma]

So that obviously means Z is not on the plane R

[opal dawn]

ok sorry

[upper karma]

No no

I mean yes they both are not the same

[opal dawn]

ok

so only the bottom four can be equal to r

bottom four points

[upper karma]

I think you are having trouble imagining the planes

Think of the plane R as a sheet of cloth
And then we stick 4 small circles on this piece of cloth
These points are V W X Y
And now place a ball above this piece of cloth(plane R) with some distance between them
This ball is the point R

[upper karma]

Yes the bottom four points are on the plane R

[opal dawn]

ok thanks

how do i know whether or not a point is on two points or one point

when they intersect (planes)

[upper karma]

Point is on two points?

[opal dawn]

point is on two planes

not points mb

[upper karma]

Not sure how to explain that but let's try an example

[upper karma]

From here say
Let's say we are considering the plane ZVW

Is point X on the plane ZVW?

[opal dawn]

thats the same question

[whole flame]

Let's get this easy for u

Every face is a plane

How do u recognise this

So

Imagine the face of triangle any one

With paper stuck from behind

So the 3 points r on the plane of paper

So we can say the face of triangle is a plane as well

This is same for all faces

So we can say point z

It is the intersection of all planes except the plane R

Which is bottom

[hollow flower]

idk

[silent plank]

look up adventitious triangle

[weary minnow]

@here can anyone help me

[wintry coral]

shure

[wintry coral]

can i help u

[azure rivet]

Can anybody try to prove it? I did it like a few days ago- but wasn't quite satisfied by my method of proving this result at all, because it was quite tiresome and lengthy.

Do try it.

[proper ember]

is there a range or interval for A,B and C ?

because ifA=B=C=pi , then LHS is 1 and RHS is 3 , so they arent equal

[obsidian harness]

there's a lot of videos online on this exact problem

[proper ember]

https://tenor.com/view/sad-face-saddog-really-no-gif-19397106

[whole flame]

120° ??

[whole flame]

Relation between A B C??

[azure rivet]

Nah. It ain't like that. These are just the angles in a triangle, no other information given in the offending question

[azure rivet]

Oh mb forgot about it, these all are the angles in a triangle

My bad that I didn't inform this tidbit of info correctly

My very bad.

[proper ember]

alr

[whole flame]

A+B+C=π that's the only relation?

[whole flame]

Did u try -2sin(c+d/2)sin(c-d/2) = cosC-cosD

[azure rivet]

yea

nothing else

i had used sin (a+b+c)'s identity btw @whole flame

[tropic cairn]

HELP ME PLSPSLSLSLSL

[azure rivet]

grade X.

[tropic cairn]

Yes.

[azure rivet]

@whole flame @proper ember did you two do it or....??

[whole flame]

pi - A = B + C, \quad \pi - B = A + C, \quad \pi - C = A + B.

1 + 4 \sin\left(\frac{B+C}{4}\right)\sin\left(\frac{A+C}{4}\right)\sin\left(\frac{A+B}{4}\right).

Let
\frac{A}{2} = x, \quad \frac{B}{2} = y, \quad \frac{C}{2} = z,

x + y + z = \frac{\pi}{2}.

Rhs

\sin x + \sin y + \sin z.

\frac{B+C}{4} = \frac{y+z}{2}, \quad \frac{A+C}{4} = \frac{x+z}{2}, \quad \frac{A+B}{4} = \frac{x+y}{2}.
So
1 + 4 \sin!\left(\tfrac{y+z}{2}\right)\sin!\left(\tfrac{x+z}{2}\right)\sin!\left(\tfrac{x+y}{2}\right).

\frac{y+z}{2} = \frac{\pi}{4} - \frac{x}{2}, \quad
\frac{x+z}{2} = \frac{\pi}{4} - \frac{y}{2}, \quad
\frac{x+y}{2} = \frac{\pi}{4} - \frac{z}{2}.

So
LHS = 1 + 4 \sin!\Big(\frac{\pi}{4} - \tfrac{x}{2}\Big)\sin!\Big(\frac{\pi}{4} - \tfrac{y}{2}\Big)\sin!\Big(\frac{\pi}{4} - \tfrac{z}{2}\Big).

I used identity

\sin!\Big(\frac{\pi}{4} - \tfrac{t}{2}\Big) = \frac{1}{\sqrt{2}}(\cos\tfrac{t}{2} - \sin\tfrac{t}{2}).

So
\sin!\Big(\frac{\pi}{4} - \tfrac{x}{2}\Big)\sin!\Big(\frac{\pi}{4} - \tfrac{y}{2}\Big)\sin!\Big(\frac{\pi}{4} - \tfrac{z}{2}\Big) = \frac{1}{( \sqrt{2})^3} \prod_{cyc} \big(\cos\tfrac{x}{2} - \sin\tfrac{x}{2}\big)

LHS = 1 + \frac{4}{2\sqrt{2}} \prod_{cyc} \big(\cos\tfrac{x}{2} - \sin\tfrac{x}{2}\big).

LHS = 1 + \sqrt{2} \prod_{cyc} \big(\cos\tfrac{x}{2} - \sin\tfrac{x}{2}\big).

It is known that:

1 + 4 \sin!\Big(\frac{\pi}{4} - \frac{x}{2}\Big)\sin!\Big(\frac{\pi}{4} - \frac{y}{2}\Big)\sin!\Big(\frac{\pi}{4} - \frac{z}{2}\Big) = \sin x + \sin y + \sin z,

Fkkkkkkk

[whole flame]

Brooo

I typed the whole shit

What the hell

[whole flame]

Someone fix this

[proper ember]

lmaooo

[proper ember]

trying

[whole flame]

😂

How does that bot convert text

I did all right i guess

But bot ditched me

[obsidian harness]

$\sin x$

[somber coyote]

south

[proper ember]

🤣

[proper ember]

loll put dollar

[azure rivet]

$\pi - A = B + C, \quad \pi - B = A + C, \quad \pi - C = A + B$

$1 + 4 \sin\left(\frac{B+C}{4}\right)\sin\left(\frac{A+C}{4}\right)\sin\left(\frac{A+B}{4}\right)$

Let
\frac{A}{2} = x, \quad \frac{B}{2} = y, \quad \frac{C}{2} = z,

x + y + z = \frac{\pi}{2}.

Rhs

$\sin x + \sin y + \sin z$.

$\frac{B+C}{4} = \frac{y+z}{2}, \quad \frac{A+C}{4} = \frac{x+z}{2}, \quad \frac{A+B}{4}=
\frac{x+y}{2}$.
So
$1 + 4 \sin\left(\tfrac{y+z}{2}\right)\sin!\left(\tfrac{x+z}{2}\right)\sin!\left(\tfrac{x+y}{2}\right)$

$\frac{y+z}{2} = \frac{\pi}{4} - \frac{x}{2}, \quad $
$\frac{x+z}{2} = \frac{\pi}{4} - \frac{y}{2}, \quad $
$\frac{x+y}{2} = \frac{\pi}{4} - \frac{z}{2}$

So
$LHS = 1 + 4 \sin!\Big(\frac{\pi}{4} - \tfrac{x}{2}\Big)\sin!\Big(\frac{\pi}{4} - \tfrac{y}{2}\Big)\sin!\Big(\frac{\pi}{4} - \tfrac{z}{2}\Big)$

I used identity

$\sin!\Big(\frac{\pi}{4} - \tfrac{t}{2}\Big) = \frac{1}{\sqrt{2}}(\cos\tfrac{t}{2} - \sin\tfrac{t}{2})$

So
$\sin\Big(\frac{\pi}{4} - \tfrac{x}{2}\Big)\sin\Big(\frac{\pi}{4} - \tfrac{y}{2}\Big)\sin\Big(\frac{\pi}{4} - \tfrac{z}{2}\Big) = \frac{1}{( \sqrt{2})^3} \prod{cyc} \big(\cos\tfrac{x}{2} - \sin\tfrac{x}{2}\big)$

$LHS = 1 + \frac{4}{2\sqrt{2}} \prod{cyc} \big(\cos\tfrac{x}{2} - \sin\tfrac{x}{2}\big)$

$LHS = 1 + \sqrt{2} \prod_{cyc} \big(\cos\tfrac{x}{2} - \sin\tfrac{x}{2}\big)$

It is known that:

$1 + 4 \sin\Big(\frac{\pi}{4} - \frac{x}{2}\Big)\sin!\Big(\frac{\pi}{4} - \frac{y}$
${2}\Big)\sin\Big(\frac{\pi}{4} - \frac{z}{2}\Big) = \sin x + \sin y + \sin z$

[somber coyote]

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

[azure rivet]

wasnt my fault.

better you couldve done on paper via pen

[azure rivet]

diy then.

no

not a typo

search on the web

and you'll end up with results

[tropic cairn]

I alr solved it..

[proper ember]

x+y+z = pi/2 right ? how is it pi

[obsidian harness]

pi radians is 180 degrees

[proper ember]

ye

[proper ember]

but x y and z are half of A B and C

[obsidian harness]

yeah it's assumed that A, B, C are the angles in a triangle

oh let me read

[obsidian harness]

they formatted it wrong lol; they did actually write pi/2

[proper ember]

huh

ok

yeee mb

[azure rivet]

the question was based on A,B AND C being the three interior angles of a triangle.

dont mess it up.

[proper ember]

[azure rivet]

well mb i pinged you for no reason lol

[whole flame]

Dang

[whole flame]

I know bro

I just tagged to ask

Bcz i am in bus

N there was this dude standing next to my seat

Peeping in my phone

[azure rivet]

BMTC, aint it?

[azure rivet]

whore ass fool

[whole flame]

What's that

[azure rivet]

mb i thought you were south asian

mb

[proper ember]

lol

[whole flame]

Did u do the same way

[azure rivet]

it is a local means of transport in my country uk

[azure rivet]

nah entirely differently

[whole flame]

Ahhh yeah

[azure rivet]

well first clean up the code to make it comprehendable

[whole flame]

Bus

I am from India

[whole flame]

Did u understand

What i did

[azure rivet]

then how dont you know BMTC?

[azure rivet]

i didnt look into it much though

give a hint atleast what was done

[whole flame]

We call it brt

Different names

Depends on region

Pmpl

[azure rivet]

yea

leave it

[whole flame]

In my region

[azure rivet]

@whole flame maybe you proved something incorrigibly wrong ig

you obtained this:

well we were meant to prove it

[whole flame]

I assumed x as A/2

[azure rivet]

and first clean up the LaTeX coding behind it

[whole flame]

y as B/2

And z as C/2

[azure rivet]

oh ok

[whole flame]

Nothing wrong

😅

[azure rivet]

didnt look into it deep enough

mb

[whole flame]

It's okk

I used simple identity6

[azure rivet]

my method was huge and maybe, to say, a bit long

do you need it @proper ember @whole flame ?

or shall i abstain from sending it here?

[whole flame]

Ahh go ahead

I would like seeing that identity

Of sina+b+c

[azure rivet]

i didnt derive it but used it heavily for the starting steps

[azure rivet]

see the time/name of the files

was geeting too late that night

[whole flame]

Damm daamm

U were confused in beginning

Like pro tip from be

Where ever u see sin together

Use cosC-cosD

[tropic cairn]

Is this the solution of ONE question?

[whole flame]

Yeah but his approach was length

Took him time to figure out

Check my solution

[tropic cairn]

Oh damn i had a heart attack lookin at that

[whole flame]

This

[whole flame]

..

[tropic cairn]

Its still really long, for my level-

[whole flame]

Ohh 😅

It's okk

U can learn

[tropic cairn]

Im in 10th grade so yea :’)

[whole flame]

Ohhh

I had this in 11th

But trigo is abt mastering

[azure rivet]

yea. it went too long tbh

[azure rivet]

nah man idts

i simplified it radically

but was too lengthy tbh

[azure rivet]

why do you think like that? i wasnt confused/anything

@whole flame can u clean up the latex a bit? i was planning to save it for further thoughts

[whole flame]

U started of by rearranging LHS and RHS

Then again after 3 steps u went to RHS

It's clean proof if u start without changing LHS and RHS terms of question

So maybe

[whole flame]

Bro yk what

I just tried those commands for first time

N the results were 😅

It didn't work

So i don't know exactly how to type and fix this shit yet I am a c++ coder

U can ask me if u get problem understanding any step I'll tell yaa

But sorry u will have to edit it urself

Bcz for me tbh

I solved this with reference formula for u

If i was to write this question i would have done it in 6 steps minimum

[azure rivet]

it aint so hard to tell man

i modified it too much but tbh i find it nice

a little harsh on those who dont know trigo at all but yea

idts i was confused while writing the solution to this at all, maybe confusions here and there might've occured

but that is my MO

to prove things

[whole flame]

👍

U did great

[azure rivet]

nah

was lengthy asf

i was quite dissapointed and wanted to do it in a shorter way, perhaps