geometry-and-trigonometry

[wheat granite]

The transversal angle theorems
since they prove eachother it doesnt matter which

[hoary prism]

That if two parallel lines are cut by a transversal then the opposite angles formed are congruent?

[limber dagger]

can somebody explain this too me cause I don't really understand what I'm supposed to do here. This is the first time ive done a problem like this for awhile

[wheat granite]

Yes that

[wheat granite]

perpendicular lines slope is -1/the slope of the original

[limber dagger]

yeah but idk how to find the slope or the perpendicular slope so thats where I am at. also i have no idea what that means

[hoary prism]

Yeah you would need to assume the parallel postulate

[wheat granite]

the slope is the difference in y divided by the difference in x, and then you can use the perp slope to find y and x difference in the perp slope

[hoary prism]

Its an axiomatic thing so i don’t really consider it self referential

[wheat granite]

right

so Im wondering why I seem to have gotten to a conclusion that doesnt support that

[hoary prism]

you’re asking whether you can prove that all angles in a triangle sum to 180 without the parallel postulate?

[wheat granite]

in a right triangle

[limber dagger]

BUT I DONT HAVE ANY SLOPES???????????

im so cooked bro.

[wheat granite]

I just said what a slope was. Find two points, then divide the difference in y between them with difference in x

[limber dagger]

did you just complicated rise over run.

[hoary prism]

In any case you need it including for just a right triangle

[wheat granite]

If you know that then youre good, I just gave those definitions for rise and run

[hoary prism]

For example if we drew a triangle with a right angle on a sphere the sum of internal angles is still strictly greater than 180

[wheat granite]

so why doesnt this work

[limber dagger]

bro ok cus you had me thinking I was brain dead for a second. thank you 🙏 but like idk how im supposed to graph that

[wheat granite]

you just graph a point that is rise up and run to the right of whatever other point you want

[hoary prism]

Cause saying 180-A=B+C is already presuming A+B+C=180

[limber dagger]

Huh

[hoary prism]

I’m saying if the parallel postulate doesn’t hold then you can have A+B+C=T for T not 180

[wheat granite]

Well yes thats what Im moving to

but B and C seem to add to the 90 degree angle when you transfer them over throught he congruent triangle

[wheat granite]

lets say you have a slope of 3/2. Graph some point (x,y) then graph another point (x+3, y+2). This line has a slope of 3/2

[limber dagger]

Oh yeah ok ty

[hoary prism]

I’m not sure wym by this cause even if you break the triangle down into other triangles those internal triangles don’t have the property of internal angles summing to 180 either

[wheat granite]

Okay, are the triangles AEF and BEF congruent? (As well as DAF and FDC)

[hoary prism]

But idk why congruence prescribes the value of the sum of the angles as 180, itd just say that the sum is the same and corresponding angles are the same

[wheat granite]

Because that would mean EAF is equal to EBF and DAF is equal to CDF

Ah I believe I found my error

From my research I thought the midpoint theorem didnt depend on the 5th postulate

[hoary prism]

Oh wait yeah scratch what I said lmfao that’s definitely not the case

Cause not only do triangles not sum to 180 if you don’t assume the parallel postulate they won’t even sum to a fixed angle to begin with it’ll depend on the triangle

So you don’t have all the congruences automatically that you can infer from the internal triangles

[wheat granite]

oh

[wheat granite]

so SAS congruency doesnt work without the paralell postulate?

[hoary prism]

Mmm I don’t think that’s necessarily the case

Yeah this is actually kind of an involved question to answer in general lmfao cause you’d be using a bunch of things that work in some geometries without the parallel postulate but not in others

I gtg but I’ll @ you if I have more thoughts on it I guess

[wheat granite]

alr

[hoary prism]

you would probably get a better answer in #math-discussion fyi

[wheat granite]

I only read first half of midpoint theorem so thats why I got it wrogn

alr

[naive shale]

"THE POLAR OF A GIVEN POINT P W.R.T. ANY CONIC IS THE LOCUS OF THE HARMONIC CONJUGATE OF P W.R.T. THE TWO POINTS IN WHICH ANY LINE THROUGH P CUTS THE CONIC"
the polar of a point wrt to a conic is the point of intersection of the tangents drawn from the two points of intersection (say Q and R) of any line drawn through P intersecting the conic at the points Q and R

can someone tell what the first statement means

[glacial crypt]

why does my science teacher tell me that sin(2x) is not always equal to 2sinxcosx

he is wrong right?

[wise pawn]

ask him for what values of x is it not equal

well ok maybe no reason to be confrontational about it lol. He's probably just mixing it up with some other trig things like sin(x)=tan(x)cos(x) which isn't true for all x, something like that.

[faint hollow]

yep

maybe not in complex number

but always true for all real x

[obtuse steeple]

Hi I need help

Can use help me with this question c

[flint marlin]

We should make a separate channel for pure geometry

[upper karma]

I agree

that way my question won’t get aired again

[runic pasture]

what is your question

[upper karma]

you know the formula for trapezium rule

like

How does it work

[runic pasture]

for integration?

[upper karma]

wait I’ll find a pic

oh my god that quality is horrendous

but that one

[runic pasture]

so for integration

so yk people are individual

[upper karma]

I don’t understand how this gives the area

[runic pasture]

instead of rectangles under the curve

you draw trapeziums

[upper karma]

yeah

[runic pasture]

Area Trapezium = ½ × (a + b) × h

[upper karma]

Oh yeah

What does y0 represent though

And where does the two come from

[runic pasture]

you know the height differ

it's skewed

let me find a pic

[upper karma]

Ok

[upper karma]

yes

[runic pasture]

y_0 and y_1 slightly differ

it's like f(x_0) and f(x_1)

now the thing

[upper karma]

When we refer to y_0 or y_2 etc is this the height of the trapezium

[runic pasture]

y_0 appears once

y_1 is being used twice

hence factor 2

let me draw it for you rosy

[upper karma]

Okay thanks

[runic pasture]

the end points y_0 till here y_4 (or generally y_n) are being used once

y_1 to y_3 twice

for example

for the first trapezium we need y_1

but also for the second

do you understand

[upper karma]

I think so

And for the third you use y_1?

[runic pasture]

no

[upper karma]

oh nvm

[runic pasture]

the third you need y_2 and y_3

and the fourth y_3 and y_4

do you notice how i used needed y_3 twice

[upper karma]

For like

The last two trapeziums?

yes

[upper karma]

im so sorry my brain is really slow today

[runic pasture]

it's fine

mine is slower

[upper karma]

lol

[runic pasture]

Let's do it specifically here

[upper karma]

So we use y_3 for red and purple

[runic pasture]

yes!

[upper karma]

yay

[runic pasture]

same with y_2

orange red

same with y_1

yellow orange

[upper karma]

I see

[runic pasture]

only the endpoints so here y_0 and y_4 we need once

[upper karma]

oh ok

[runic pasture]

So mathematically

[upper karma]

So that’s why there’s a

2

[runic pasture]

yes let me write it

h is this the horizontal width basically here

like the distance between x_0 and x_1 and x_1 and x_2 etc

Now we can group the terms

[upper karma]

Isn’t h like 1

[runic pasture]

h can be anything

[upper karma]

I see

[runic pasture]

we can decide how we cut our interval

h will be of the sort (b-a)/n

but ignore

[upper karma]

thanks

[runic pasture]

no

[upper karma]

Oh

[runic pasture]

I want you to see

the first line is how we would calculate each trapezium

the second we factorize the common term h/2

then the third we add everything up

and you notice the second and pre last term and everything between appears twice

[upper karma]

I see

[runic pasture]

then we can factorize 2

and then we can generalize the idea

y_0 the very first and y_n the very last once

the rest twice

[upper karma]

you know when you say y_1 and y_2 and y_3 etc what number is this actually? Like what would you insert into the calculator

[runic pasture]

it's the function value

let me rewrite

[somber coyote]

bacc the sigma😔🤞

bacc the sigma😔🤞

[runic pasture]

at x_0 we have the height y_0 = f(x_0)

I hope it's understandble if not say it please

[upper karma]

wait one second I have to process

[runic pasture]

ok sorry

[upper karma]

[runic pasture]

dont ping mods

[upper karma]

how did you make this function

[runic pasture]

i didnt make anything

[upper karma]

oh

[runic pasture]

that's like

[upper karma]

my bad

[runic pasture]

my bad

i should explain better

you know y = f(x)

[upper karma]

yes

[runic pasture]

since we have many x values like x_0, x_1 etc

we get different y values

y_0 y_1

f(x_0) = y_0

[upper karma]

so it’s like an equation

[runic pasture]

f(x_1) = y_1

yea it is

its a differnt notation meaning the same thing

for example

y = f(x) = x²

x_0 = 1
x_1 = 2

then y_0 = f(1) = 1

and y_1 = f(2) = 4

simply

[upper karma]

yes

[runic pasture]

is this understandble

[upper karma]

yes

[runic pasture]

do you want a mini exercise

[upper karma]

sure

[runic pasture]

using the rule we discussed I would like you to estimate the area in the interval [1,3]

[upper karma]

is this like

[runic pasture]

yea kinda

how many trapeziums will you calculate?

given the interval

[upper karma]

two

[runic pasture]

yes

now try it brb

[upper karma]

ok I’ll try

wait is the formula: 1/2h [y_0 + y_3 + 2(y_2)]

[runic pasture]

almost you messed the indices

h/2 [y_0 + y_2 + 2(y_1)]

[upper karma]

oh

[runic pasture]

yea

[upper karma]

Oh I completely missed y_1

Is H 1 in this case

[runic pasture]

back

yes

h = 1

(3-1)/2 = 2/2 = 1

[upper karma]

So 1/2 [1 + 3 + 2 (2)]

[runic pasture]

are you sure

[upper karma]

idk

[runic pasture]

ok

peep this

you did 1/2 (x_0+x_2+2x_1) instead

not y but x

[upper karma]

Oh

im dumb

[runic pasture]

you are trying it's okay

[upper karma]

I got 9

1/2 [1 + 9 + 2(4)]

[runic pasture]

,calc 1/2 (1 + 9 + 2(4))

[somber coyote]

Result:

9

[runic pasture]

good

[upper karma]

yay

[runic pasture]

are you interested how the actual area is

,w Integrate[x^2,{x,1,3}]

[somber coyote]

Results provided by WolframAlpha

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

Requested by adoniszgks

[upper karma]

oh

[runic pasture]

good approximation

you see

[upper karma]

nice

[runic pasture]

8.666 is near 9

[runic pasture]

if you look at it closely you will see the trapezium is slightly greater

[upper karma]

yes

it’s an overestimate

[runic pasture]

and the more trapeziums you would use the more accurate

[runic pasture]

yes!

say we used 4 trapeziums

,calc 0.5 * (3-1)/4 * (1^2 + 2*(1.5)^2 + 2 * (2)^2 + 2 * (2.5)^2 + 3^2)

ehmm

[upper karma]

oh

maybe it’s an anomaly

lol

[somber coyote]

Result:

8.75

[runic pasture]

aha

i had a double 2

see

8.75 is way more accurate as we used 4 trapeziums now

and now you can imagine

if we used infinite many thin ones

we would get the exact

[runic pasture]

my math skills yes

[upper karma]

Oh you got it

[upper karma]

yes

Thank you soo much

ur a great teacher

[runic pasture]

no you are a great student

[upper karma]

I got it wrong like four times 😭

[runic pasture]

what did you expect

not everything works out with the first try

[upper karma]

ig

[maiden brook]

isn’t this calculus not geometry

[runic pasture]

you need to try and gain som intuition

[runic pasture]

will you ping and report us to the mods

[upper karma]

thanks so much for ur time

[runic pasture]

in a sense it was geometry

we calculated the area of trapeziums

[upper karma]

I mean I hardly realised icl

[upper karma]

So true

at least nobody was talking in chat so we weren’t disrupting anyone

[junior shore]

im very confused

is f an angle

and if f is an angle would it be

not congruent?

[uneven ember]

they are congruent cuz the angle UFG = GIH
and FUG = IHG = 90 degrees (right angles)
and angles UGF = HGI (vertically opposite angles are equal)

btw F is just a point, the angle at F is UFG

[uneven ember]

this is the same for all the other points

[junior shore]

well i know they are congruent

but with the rules

aas sas HL etc

i need to find the rule

but if f is just a point how i detrmine the angle

[uneven ember]

should be aas

[uneven ember]

the angle is drawn with two lines

they show that the angle at the point F and at the point I are equal

[junior shore]

why is G not an angle

dose it not connect the two points?

[uneven ember]

G does connect two sides, but its just the point at which they connect
im pretty sure the angle is written as the sides that meet to make up the angle

but either way, it works

[junior shore]

okay so then

the line

that has the point q and s

do i consider them angles or sides

im confused on when i should consider it a side or an angle

i know the tick marks symbolize sides

and the little curve symbolizes angle

but sometimes they dont

so for this one would it be

congruent by angle side angle (ASA)

and then PQS is congruent to RQS?

wait

im cooking

[warm hornet]

I'm so not cooking I'm cooked

I forgot what they said in class

[uneven ember]

depends on how you write them

QS is a side
PQR is an angle
PSR is an angle
Q is a point
S is a point

[junior shore]

dose the amount of tick marks have any signicane

significance

[uneven ember]

the tick marks represent the lines being the same length

theres different amounts of tick marks to differentiate between pairs of sides

[tawdry dune]

yo guys

[ember inlet]

Wtf

[tawdry dune]

?

[ember inlet]

It’s stark

So scary

[tawdry dune]

yeah thats me

yeah you can get an autograph

get yourself something nice

[ember inlet]

Thanks!!

The overlords don’t allowing me to save that

But I will take a screenshot

With.. my mind

👀

[ember inlet]

Okay here’s a hard question

Two parallel lines L1 and L2 intersect respectively, the x axis at p > 0, and the y axis at q > 0.

To be more precise, L1 goes through (p, 0) and L2 goes through (0, q) [condition 1]

What is the maximum possible slope of L1?

Where maximum means over all possible parallel lines satisfying [condition 1]

[trail tendon]

IS there a maximum slope of L1?

[ember inlet]

Ye

depends on p and q

question, what is the minimum?

You can find this too

[trail tendon]

is there a minimum?

[ember inlet]

Yes

wait

[trail tendon]

it has to be greater than 0 but it can't be 0...

[ember inlet]

There’s a minimum and it’s

negative. I think lol

[trail tendon]

oh wait

[ember inlet]

But clearly

The horizontal and vertical lines solution have slope 0

[trail tendon]

still it would be like -infinity 💀

[ember inlet]

And infty

[trail tendon]

vertical lines don't have a slope

theoretically they have infinite slope

[ember inlet]

I mean I guess by convention I take that to be infty

[trail tendon]

right but there isn't any real slope

so the "biggest" slope would be "infinity"

[ember inlet]

Okay follow up question -

[trail tendon]

but the maximum slope doesn't exist...

cuz infinity isn't a slope xD

its just the concept of one 💀

[ember inlet]

Let G be the perpendicular bisector of L1 and L2

[trail tendon]

but does this question even have an answer-...

[ember inlet]

What is the maximal distance of G(t) from the origin, assuming L1 and L2 still satisfy condition 1?

[ember inlet]

Yes it’s +infty

The vertical line solution

[trail tendon]

that's not a number 😭

[ember inlet]

Bruh

Use the extended

Reals

[trail tendon]

its not a "maximum" though

[ember inlet]

Why not

[trail tendon]

its like whats the maximum money you can have

infinite

[ember inlet]

bruh

[trail tendon]

👀

[ember inlet]

That’s what the extendedcreals are for

[trail tendon]

thats the same thing as saying theres no maximum

[trail tendon]

[ember inlet]

So that you have a maximum and a minimum for every quantity

[trail tendon]

nobody works in the extended reals 😭

[ember inlet]

The extendedcreals are [-infty, +infty]

[trail tendon]

and by nobody i mean realistically cmon br

☠️

[ember inlet]

Okay then let’s work with actual real numbers

Again, the real question is

[trail tendon]

there are many lines perpendicular to L1 and L2?

i guess you could say the answer is again infinity in the extended reals ☠️

[ember inlet]

Two parallel lines L1 and L2 intersect respectively, the x axis at p > 0, and the y axis at q > 0.

To be more precise, L1 goes through (p, 0) and L2 goes through (0, q) [condition 1]

Let G be the perpendicular bisector of L1, L2 with G(0) on the x axis by convention.

Fix some large t. What is the maximal value of dist (G(t), 0)?

[trail tendon]

infinity?

._.

[ember inlet]

No

For fixed t this is a real number

[trail tendon]

not in the extended reals?

[ember inlet]

Nope

turns out there is a unique solution that

[trail tendon]

then...there is no maximum? ☠️

[trail tendon]

how?

maybe i dont understand the question or smthn

[ember inlet]

Maximises limsup_t dist (G(t), 0)/t

And the limsup is a lim

Maybe easier to use numbers

[trail tendon]

why dividing by t?

[ember inlet]

Cause we want the slope

rate of growth of line

[trail tendon]

i thought we were looking for the distance from G(t) to the origin? no?

[ember inlet]

Yes we are

But we want basically the

Rate of change of

Distance with time

so divide by t

[trail tendon]

the slope and distance is not the same tho?

[ember inlet]

and tak3 limsup to measure long term growth

ofc it’s not

Distance is distance

Slope is slope

With slope u divide by t

[trail tendon]

but we aint finding slope so we aint diving by t

[ember inlet]

Two parallel lines L1 and L2 intersect respectively, the x axis at p > 0, and the y axis at q > 0.

To be more precise, L1 goes through (p, 0) and L2 goes through (0, q) [condition 1]

Let G be the perpendicular bisector of L1, L2 with G(0) on the x axis by convention.

What is the maximal value of limsup_t dist (G(t), 0)/t?

This is the REAL FULL QUESTION I PROMISE

[trail tendon]

oh

💀

[ember inlet]

it’s easy

[trail tendon]

are you trying to make a question that makes sense 😭

[ember inlet]

This makes perfect sense

[trail tendon]

it was very sketchy before 💀💀

[ember inlet]

I was introducing the problem …

I am pretty sure the maximum limsup is achieved by unique configuration

[trail tendon]

pretending infinity is a maximum because you were using the extended reals like nobody ever unless they specified 😮‍💨

😭

[ember inlet]

yeah ok I will be more precise nex5 time

But now what’s not clear

You just wanna max limsup_t dist (G(t), 0)

Maximise long term growth rate of the perpendicular bisector

[trail tendon]

the maximum limit of the distance between G(t) to the origin divided by t...?

[ember inlet]

Yes?

the maximal long term slope of the perpendicular bisector

[trail tendon]

uh

G(t) grows faster than t...

so {G(t)/t} is increasing...

[ember inlet]

I mean there’s nothing stopping the slope of G from being less than 1

[trail tendon]

so the maximum distance is still...infinity...

[ember inlet]

So G(t) does not grow faster than t

[trail tendon]

oh wait im dumb

[ember inlet]

The slop3 of G is always finite

[trail tendon]

i forgot G(t) was the slope of the line

wait its the slope?

[ember inlet]

G(t) is just the valu3 of the line

[trail tendon]

what the heck is a value of a line 😭

[ember inlet]

G(t)/t is the slope

I mean thinking of the line as a function

[trail tendon]

the change in y?

[ember inlet]

with G(0) starting on the x axis

Then G(t) is just the value of the function at t

[trail tendon]

the value of...what function?

[ember inlet]

I suppose we have to require that the line be parametrised by arc length

uh

[trail tendon]

which line 💀

[ember inlet]

Okay so a line can be viewed as a function right

Take an arbitrarily line

[trail tendon]

sure

[ember inlet]

The only difference

[trail tendon]

how arbitrary?

are we talking literally any line in R^2?

[ember inlet]

Yes

[trail tendon]

ok

[ember inlet]

Each parametrisation of the line

Gives rise to a function t -> L(t)

Parametrisation means

Strictly increasing

[trail tendon]

[ember inlet]

Bruh

[trail tendon]

that's....not what parameterization means....?

[ember inlet]

what is not clear 😓

[trail tendon]

unless you're talking about

smthn else

[ember inlet]

It’s what it means to me

I mean

A parametrisariom of a curve c

Is just an increasing function t -> c(t)

[trail tendon]

so what if we say L(t) = (3t,-4t)

[ember inlet]

sure that’s a line

With slope -4/3

[trail tendon]

exactly

its not increasing

[ember inlet]

Oops

Xd

[trail tendon]

😭

[ember inlet]

This my bad

So you want a homeomorphism onto the image

Or a smooth parametrisarioj

[trail tendon]

homeowhatism?

[ember inlet]

Topology

Point set

[trail tendon]

Topology? 😭

[ember inlet]

You don’t know topology?

Shit lol

[trail tendon]

no 😭

also this is a geometry chat 🤣

[ember inlet]

But ur flair says undergrad math

[trail tendon]

takin multi and diffeq

they're under the

[ember inlet]

Point set topology is undergrad

[trail tendon]

undergrad chanel

[ember inlet]

Ah I see

[trail tendon]

thingy

[cunning lion]

if you're only partway through undergrad then the role still applies

[ember inlet]

I see

Well

Blame @lofty spear for giving a too difficult geometry problem

😭

I spent 2 hours last night on it

“Pre university math”

[trail tendon]

what was the problem? :l

[trail tendon]

💀

[ember inlet]

Two parallel lines L1 and L2 intersect respectively, the x axis at p > 0, and the y axis at q > 0.

To be more precise, L1 goes through (p, 0) and L2 goes through (0, q) [condition 1]

Let G be the perpendicular bisector of L1, L2 with G(0) on the x axis by convention.

What is the maximal value of limsup_t dist (G(t), 0)/t?

Note: Here the maximum is taken over all parallel lines L1 and L2 satisfying [condition 1], and by convention we take G to be parametrised by arc length.

Literally this

[trail tendon]

bruh

[ember inlet]

But they didn’t even state the problem in full

fucking troll

[trail tendon]

can't get much higher ahh

[ember inlet]

Highe and higher

[trail tendon]

naw jk 🤣

[lofty spear]

I did not give you this problem

what

I never gave you any problem

[ember inlet]

Sure u didn’t

[trail tendon]

what is G?

[ember inlet]

THE PERPENDICULAR BISECTOR

OF L1 L2

[trail tendon]

ohh you mean like

[ember inlet]

Yeah like

Two points admit a unique

Perpendicular bisector

geometry

Two parallel lines also admit one

[trail tendon]

G(t) is theoretically along the x axis ish

[trail tendon]

they admit...many tho?

like an infinite many

💀

[ember inlet]

There is only one bisector?

given two parallel lines

[trail tendon]

huh?

[ember inlet]

What is the bisector

Uh

[trail tendon]

heres...3...?

[ember inlet]

Wow I can’t share pictures

Uh

Dang it what is the standard terminology for this

It is a line that

[trail tendon]

we're talking perpendicular bisector right?

[ember inlet]

Does not intersect

Either of the parallel lines

[trail tendon]

so...another parallel line is what im hearing 💀

[ember inlet]

This is uniquely solvable

[trail tendon]

supposed to be*

😭

[ember inlet]

yes but also it should bisect

Hmm

How to formulate this

[trail tendon]

so its a bisector that doesn't intersect

[ember inlet]

Well

Once you fix

Two points for reference …

[trail tendon]

we're still in R^2?

😭

[ember inlet]

yeah

God I hate this highers guts

[trail tendon]

we not talkin about a perpendicular or angle bisector ig

[trail tendon]

bro has no idea what ur talking about 😭

also out of curiosity what is a homeomorphism or whatever you said b4 💀

[ember inlet]

:momke,m

Sorry I need a break this is rough 😭

[trail tendon]

😭

[ember inlet]

I have been given a hard task

And problem

[trail tendon]

are you sure the problem makes sense/exists...

😭

[ember inlet]

YES

It’s obvious

[trail tendon]

to you mebbe ;-;

[ember inlet]

[obtuse steeple]

Help me with this

[weary plume]

What resources do you guys recommend for geometry

[silent plank]

pretty much anything you can get your hands on

khan is easily accessible and free
a generic textbook will be sufficient but probs not worth the $

[obsidian harness]

which part?

[obtuse steeple]

b

and c

[obsidian harness]

draw the graph of y = |tan 2x| by reflecting the parts below the x-axis above

then add 1

should look like a V-shape with mininum value 1

[obtuse steeple]

How about c

[obsidian harness]

ok 1 sec

[obtuse steeple]

What’s means the k

[obsidian harness]

k is just any real number, but it's the slope of the black line below

[obsidian harness]

I don't understand what c, i) is asking too

however for c, ii)

the limiting case is when the line passes through (0, -1) and (pi, 0)

below that gradient then there are no solutions

oh wait c, i) is easy

using everything you have |tan 2x| + 1 = kx

so the correct straight line is just y = kx

ahhhh yeah

[obsidian harness]

so you would move everything up in this graph 1 unit up

[zealous pike]

Why is it that C for sine is $-\frac{2\pi}{3}$ and C for cosine is $\frac{\pi}{3}$

I'm having trouble to understand it

[somber coyote]

Pi, a future fluent jp speaker

[obsidian harness]

(1, 1) is on the graph

so you would sub in, 1 = 3 sin(pi/3 * 1 - C) - 2

or 1 = 3 cos(pi/3 * 1 - C) - 2

solving gives those values

[obsidian harness]

that's weird WA says this is wrong

[zealous pike]

How to isolate C

[obsidian harness]

for exmaple, $\sin(\pi/3 - C) = 1$

[somber coyote]

higher's secret twin brother

[obsidian harness]

so $\pi/3 - C = \pi/2$ gives $C = -pi/6$ as one possible value

there are others

[somber coyote]

higher's secret twin brother

[obsidian harness]

oh wait the question never mentioned the C for sin

their C for cos is correct

[zealous pike]

It is to define the general form given the graph

[obsidian harness]

I know

[zealous pike]

Im confused for C

[obsidian harness]

well the better form is $3 \sin(\pi/3 (x - d)) - 2$

[somber coyote]

higher's secret twin brother

[obsidian harness]

so that $d$ directly represents the horizontal shift

[somber coyote]

higher's secret twin brother

[obsidian harness]

that means C/(pi/3) is the horizontal shift

$\frac{\pi}{3} (x - d) = \frac{\pi}{3} x - C \implies \frac{\pi}{3} d = C$

[somber coyote]

higher's secret twin brother

[zealous pike]

I can see that but why is C for sine and cosine different

d = 1 for cosine and d = -2 for sine

[obsidian harness]

cause sin and cos start from different points

[zealous pike]

That considering d = 1?

[obsidian harness]

yes
you know how cos starts at (0, 1) right

so we've moved the graph 1 unit right to get the blue graph

so that's why C = pi/3 * 1 = pi/3 for cos

[bold haven]

Hello guys
I need references for which book I can use for Maths for class 12th boards?

[deft sail]

Sorry 😔 no recommendations

[covert granite]

l need help w geometry

[trail tendon]

lol

[wary fulcrum]

Hi i have a question for trig. its a really basic question like rlly basic

i got everything

except for d

and i know how to determine it except that i dont know why you have to use cosine to find it

[wary fulcrum]

<@&286206848099549185>

[celest berry]

0 is the last angle remaining

[wary fulcrum]

my answer key says 117

because you have to go θ = cos-1 = (-1/ √5)

i just dont know why you have to do cos

[winged mountain]

I need to find "x" FOR FIELD 1

or is there some weird geometry rule i need to take into consideration

(I am helping a friend and I haven't taken Geometry yet)

but I have taken Algebra

I and am currently in II Honors

[round trail]

anybody know how to solve this? I am on circle theorems

[winged mountain]

So whenever you see something that looks like this that is always how you would handle it?

How about this one

would we do the same thing and set y = 180

and this one two

too

@outer reef

so what would you do

could you show an example?

what equation would i get

[winged mountain]

but why couldn't I do that for this?

What would help in determining if or not it is a vertically opposite angle?

so if they are on opposite sides of the line its

vertically opposite then

but if they are on the same i have to set evreything to 180

?

so when vertically opposite you set them equal to eacother

and when its not set evreything to 180

alright

[winged mountain]

so y would be 180 then?

2x+25+y=180 ?

yeah thats what I realized I probably should do

[round oak]

when describing a square, is there a word to describe the "size" of the square? I thought it might be area at first but it isn't as far as I understand. I feel like there should be a simple term that im missing

[silent plank]

using area is fine

[ancient kindle]

i think you generally want to explain something in the highest dimension quantity it can describe. im not sure what that would be called, but for a square that's just area

[spice pendant]

Any jee aspirant here?

[thick bane]

how solve sin 22.5 without calculator?

sin 2x = 2sinxcosx
sin x =?

[obsidian harness]

easier to find cos 22.5 first actually

$2 \cos^2 x - 1 = \cos 2x$

$2 \cos^2 (22.5) - 1 = \cos(45)$ then solve for $\cos(22.5)$, noting that it will be positive cause 1st quadrant angle

[somber coyote]

higher's secret twin brother

[thick bane]

Bro wth thx 😱

Ahhhh I get it now

[obsidian harness]

no worries!!!!

[buoyant crest]

hii, im new here and i kinda need help on smth, it's about finding the exterior of a polygon

i need the formula to get like each exterior angle of a polygon

[worn stag]

Isn't it just 360-interior angle?

[buoyant crest]

oh really??

[worn stag]

For convex regular polygons

[buoyant crest]

ohh okayyy

i understand now

thxxxx

[worn stag]

Take this poorly made square for example

[buoyant crest]

ohhh i get it now

thxxx smmm

[formal agate]

need help with that

[graceful talon]

no hablo turko

[formal agate]

ABCD parallelogram |DE|=3 |AE|, BFG=35 AEF=2

ABF=?

[graceful talon]

what is Alan, area?

[formal agate]

Alan means area in turkish

[graceful talon]

what's the formula for area triangle

[formal agate]

which formula

[graceful talon]

area of a triangle

[formal agate]

1/2 a h

[graceful talon]

idk

[formal agate]

[graceful talon]

do u know the formula for area of parallelogram

,w area of parallelogram

[somber coyote]

Results provided by WolframAlpha

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Requested by lkjsfkjsiu

[charred herald]

cough

[oblique venture]

Hi everyone. Stuck on number 29. Struggling to solve this type of problems. How to do it?

[faint pasture]

Well cot^2(theta) +1 = cosec^2(theta)

Just use that

Also we know theta is in pi/2 to pi

So cot theta is negative

[storm zephyr]

I saw this on Wikipedia:

"If the vertices of the cyclic quadrilateral are A, B, C, and D in order, then the theorem states that:

$\displaystyle AC\cdot BD=AB\cdot CD+BC\cdot AD$

Can I use "if and only if" over here, since they are be biconditional/equivalent? The converse of this statement is true.

Like

(Quadrilateral ABCD is a cyclic quadrilateral) ↔ ($\displaystyle AC\cdot BD=AB\cdot CD+BC\cdot AD$)

[somber coyote]

Flamango

[obsidian harness]

yes, you can use if and only if because the converse is also true

[storm zephyr]

ok

[restive heron]

Say I'm trying to find a sum to product trig identity for cos(x) + sin(x).

I'm finding this one $\sin(x) + \cos(x) = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right)$ but it's from chatgpt and khanmingo. I don't see it in my book.

[somber coyote]

TheJhonny

[cunning lion]

you can use the fact that cos(x) = sin(pi/2 - x), or alternatively that sin(x) = cos(pi/2 - x), then apply that to the standard sum-to-product identities

[drifting zealot]

Im Looking for a self study Geometry course like the BYU one. But my school requires it to be Honors.

[dusk sorrel]

Check out UC Scout. It’s pretty expensive, but it gives you credit for the class. I wasn’t a big fan of their IM2 course, but it gave me what I wanted ig

[drifting zealot]

How much is it?

[dusk sorrel]

Super sorry. I guess they only have the regular class (just checked). It’s like, $350 per semester

Unfortunately, there arent many programs that offer affordable classes for which provide credits

[drifting zealot]

Is jt an Honors Geo course?

[dusk sorrel]

Search up “UC scout” and go through that

[unkempt gulch]

Need help finding h

[unique estuary]

H is 8 root 2 I think

[lyric sonnet]

Is that a quadrant

[unkempt gulch]

It’s quadrant 1

[lyric sonnet]

I mean the white part bounded by the arc

Is that the quadrant of a circle

[unkempt gulch]

No

[lyric sonnet]

Oh ok

[restive heron]

I'm completely stumped on how to solve this. I changed the coefficients of this question.

Consider f(x)=cos(5x)+ sin(5x) find a formula for f(x).

Form of C(x)=Acos(wx+o) + B for w > 0
And
S(x)=Asin(wx+o) +B for w> 0

Forgive me I don't really know latex

[lyric sonnet]

https://en.wikipedia.org/wiki/Inverse_Pythagorean_theorem

@unkempt gulch

[unkempt gulch]

Thanks 🙏

[lyric sonnet]

No problem 😃

[lavish halo]

If A=2Sinx-1 how do i find the potential full number values A^2 can be?Came up on a question

For A i found (4sin^2x)-(4sinx)+1 but not sure how to find the values -(4sinx) can be

Full question was A=sin2x-1 B=(-3cosx+17)/2 How many values can A^2+B^2 be

[faint pasture]

are you completely sure that the que is correct

if you could attach like a picture or something itll be helpful

,w simplify (sin2x-1)^2 + (17-3cosx)^2/4

[lavish halo]

Sec

[somber coyote]

Results provided by WolframAlpha

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Requested by enderdoesntmind

[faint pasture]

yeaa see as you stated it

it has infinite solutions

[lavish halo]

Forgot a (

[lavish halo]

This was where i got while trying to solce it from (x-y)^2

Its not on the question itself

[faint pasture]

,w simplify (2sinx -1)^2 + (17-3cosx)^2/4

[somber coyote]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by enderdoesntmind

[faint pasture]

yea this has infinite possible values too

[lavish halo]

Sec

Question translated is

[lavish halo]

How many different integer values can A^2+B^2 take based on their equality?

[faint pasture]

oh

do you know differentiation?

[lavish halo]

I might but not sure since different languages

,w differentiation

[somber coyote]

Results provided by WolframAlpha

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Requested by hiya0461

[lavish halo]

This?

[faint pasture]

ye

[lavish halo]

I dont

[faint pasture]

uuuuuuh then i dont think that question is possible

[lavish halo]

What is differentiation btw

[lavish halo]

Gotcha

[faint pasture]

like if you draw the graph of any function

and draw the tangent line at any point

differentiation tells you the slope of that tangent

[lavish halo]

Ic how can i use that here?

[faint pasture]

uh i cant really explain all of differentiation and how it works over text

[lavish halo]

To youtube i go then

[faint pasture]

and i dont really know any good sources for you to learn cuz of the language barrier

[lavish halo]

Gotcha btw is differentiation used here aswell?
A=(3Sin2x+5)/2 Integrr values A can be

[faint pasture]

nah this one can be done without that

see sin(2x) is always between -1 to 1

so 3 * sin(2x) is always between -3 to 3

5 + 3*sin(2x) is between 2 to 8

im sure you can do the rest

[lavish halo]

Is the order of + - * / same as normal when doing that btw

Or can i do those in any order

[faint pasture]

doing what?

[lavish halo]

Addition subatraction division stuff like that

[faint pasture]

do that in any order BUT the final expression should be what you need

[lavish halo]

Gotcha ty

[faint pasture]

so like (5 + sin(2x) )*3 wouldnt work

[dusk sorrel]

You don’t have to explain this if it’s too difficult/you don’t want to, but how would taking the derivative of that be helpful?

[faint pasture]

if you know the basics it isnt tough, so in short the derivative would tell you the minimum and maximum values of that function

[dusk sorrel]

Ohhh yeah oops

[faint pasture]

and then you can just count the integers between them

[dusk sorrel]

Ah okay

[faint pasture]

without that i dont really see a solution

please do tell me if you can think of one

[dusk sorrel]

I’ll look at the equation a bit; i don’t think I’ll be able to find one

[faint pasture]

generally when sins and cosines are intermingled like that and the arguement is different too

there isnt really another method

but maybe the question makers did something special so that there would be

[dusk sorrel]

What’s the problem from? I doubt a regular trig class would require a student to use differentiation

[faint pasture]

i mean why not? my class covered it

[dusk sorrel]

Oh fr??

Mine never did 😭

[lavish halo]

Name of the test is trigonometry-3 Can always be an slightly outdated test since the curriculum changes often

[dusk sorrel]

Ah okay

[faint pasture]

probably different countries

[dusk sorrel]

That’s probably the case

[lavish halo]

How do i simplify Cosx/(1+sinx)+Cosx/(1-Sinx) got 2Cosx/(1-sin(x)^2)

[silent plank]

pythag trig identity for the denominator

[lavish halo]

Is that the 1-sinx=cosx?

Anyways got 2/1+cosx

,w pythag trig identity

[somber coyote]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by hiya0461

[obsidian harness]

$\sin^2 x + \cos^2 x = 1$

[somber coyote]

higher's secret twin brother

[lavish halo]

Ah gotcha

[obsidian harness]

so this is not correct

[lavish halo]

Where do i find cos(x)^2 from here?

[obsidian harness]

yes, you get $\frac{2 \cos x}{\cos^2 x}$ as a consequence

[somber coyote]

higher's secret twin brother

[lavish halo]

Ah ic

1-sinx^2 is cosx^2 gotcha

Ty

[obsidian harness]

np!

[fallen fox]

I think there is way to solve without differentiation. The sinx takes the values [-1,1] so 2sinx takes the values [-2,2] then 2sinx -1 takes the values [-3,1] we can same thing for the cosx too. Again cosx takes the values [-1,1] , -cosx takes the values [-1,1] too, -3cosx takes the values [-3,3] , -3cosx +17 takes the values [14,20] fand finally (-3cosx +17)/2 takes the values [7,10]. Now we need to find minimum and maximum values of A^2 +B^2 where A ∈[-3,1] and B∈[7,10] so the maximum value for A^2 +B^2 is (-3)^2 + (10)^2 and the minimum is 0^2 + 7^2. Then the values for A^2 +B^2 is [49,109] and there are 61 integers in this interval

[fallen fox]

I forgot to reply the question

[faint pasture]

but it isnt neccessary that when B = 7, then A = -3

and the same for the other bound

ie

these values happen at different values of x

if it were x in A and y in B sure this woulda been the way to go

but if the variable is same in both

this doesnt work

uuuuh lemme give you an example

consider cos(x) + sin(x)
since both of them have the minimum value -1, the total minimum value of the function should be -2 by your logic right?

but we know the minimum is -sqrt(2)

so your logic fails

[fallen fox]

Well if I take the( -3)^2 + 7^2 would it be correct

[faint pasture]

did you even read my message?

[fallen fox]

Yes

Ooo

[faint pasture]

it wont be correct

at all

[fallen fox]

Okay I got it now

[scenic rapids]

Can sb solve this problem?? It’s about Thales theorem. I apologize for it being in Spanish but I’m sure you’ll get it anyway.

[upbeat compass]

Since CA does not equal BC

Because of the definition of the Pythagorean Theorem

and if A'<CB

[lavish halo]

Mükemmel isim

61 is in one of the answers

[lavish halo]

Oh makes sense why that doesnt work

[lavish halo]

If tan18=a what is (tan162+cot198)/(tan198+tan45) how do i solve this doesnt the top become 0 from -cot18+cot18

[obsidian harness]

no, the top doesn't become 0

the numerator becomes -tan(18) + 1/tan(18)

so you're done

[lavish halo]

Why 1/tan(18)? What rule is that

Oh wait the cotx=1/tanx one? Makes sense

[lavish halo]

Btw how is tan162=-tan18?

Nvm just checked on an unit circle makes sense

[jagged venture]

The regular hexagonal prism ABCDEFA'B'C'D'E'F' has AB=1 m and CC'=2 m. Show that the lines AF' and BC' are coplanar and calculate the sine of the angle formed by these lines.

Ive solved the first part

But idk ab the second

[obsidian harness]

Yes tan is periodic and tan is also odd

Nice you could check yourself

Unit circle W up top

[upbeat glacier]

hello

is

$\frac{2}{\tan^2{x}+1} = \frac{2}{1-\tan^2{x}} $

is this true?

[faint pasture]

no

there are solutions yes but it isnt an identity

[lilac onyx]

tan²x+1=sec²x

[upbeat glacier]

ah ok, i was asking because my answer was the left but the example answer was the left

range is between -pi/2 to pi/2

[faint pasture]

not even needed, just reciprocal both sides, cancel out stuff you get 1 = -1

[lilac onyx]

yep

[faint pasture]

uh do you want the solutions or just if its an identity or not

nd range is not that

[upbeat glacier]

one sec

[faint pasture]

domain maybe

[lilac onyx]

just a info

[faint pasture]

but not range

[upbeat glacier]

can i send my working so far

this is a part of the example

where is my mistake, example has it tan^2 + 1, i have 1 - tan^2

found it

sin^2 = 1 - cos^2 not cos^2 - 1

[wintry loom]

assuming you've solved the first half already*, this is what i got.
line L bisects θ and line C'F'. some pythagoras is left out of working because i'm tired :p, sry
(nb: the long diagonal of a regular hexagon with side lengths 1 is equal to 2)
(nb: line BC' = PB by similar triangles)
that stray alpha label is an accident, sry for not removing that

*easily demonstrable by noticing AB || C'F'

[somber owl]

CaN anybody help me with constructing

i desperatly need help

its not even funny

[obsidian harness]

constructing what

[unkempt gulch]

How do I find the height of the building?

[pseudo nebula]

notice the triangles

[unkempt gulch]

I noticed the triangles but I'm not sure what to extract from them

[warm shuttle]

they are similar

and right angled

also this building has some low ceilings

she'd have to crouch to even get inside

[upbeat compass]

So theres a hella lot a triangles

find the areas of them

for example

the girl

[faint pasture]

what if she doesnt wanna get in

[wintry loom]

if the bottom side of the smallest triangle here is 7m then this problem is easily solved with some extra algebra. i say if since it looks ambiguous but that may just be my amazing /s drawing skills

[upbeat compass]

Bro how do you draw so good

when i do math my paper looks like chris chan monkey scratchings

[wintry loom]

dedicated drawing program :p that and being an artist i guess, but im derailing

[somber owl]

squares and stuff

[pseudo nebula]

think about the properties of a square

[turbid reef]

Does anyone know the answers for these?

[jagged venture]

thanks a lot! I'm just beginning with spatial geometry and I coudln't really figure out this one

[fossil laurel]

Well first of all in order to know what the value of the other legs is we will be using a^2+b^2=c^2 since we know the hypotenuse or the c we can plug in 410 squared. And we know the other lenegth to be 400 ft or A. Knowing this we can substitute the values and get that b^2 equals to 8100 which the sqaure root is 90. To determine the area of the parking lot we use the formula length times width. Since only right perfect 90 degree triangles can be a pythagorean therom. And a square or recantangle area formula is Length multiplied by length or length times width. So since we know that the lengths of the triangle which is 90 and 400 you can multiply those by each other and then by two giving you the area. I think after that you can solve question three

Feel free to dm me for more questions

[wintry loom]

happy to be of help :3

[sacred estuary]

hi

[graceful talon]

hi

[radiant vortex]

if N + M is 60degree what is A ?

[nocturne magnet]

N and M are the same so each is 30

[radiant vortex]

nvm

[nocturne magnet]

so i think that

[radiant vortex]

and about this?what is main idea?

[nocturne magnet]

180 - A will be the tip of an isoceles triangle

with side angles 30

so 180 - A = 120

so A is 60

[radiant vortex]

ohhhh, got ittt.

[nocturne magnet]

this is beyond me

[radiant vortex]

thats wrong.

[nocturne magnet]

why wrong

[radiant vortex]

160, 70, 140, 80 are option.

[nocturne magnet]

clearly i am no geometry

[faint pasture]

No human is geometry

[zealous pike]

Is the distance between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ equals $\pi$?

[somber coyote]

Pi, a future fluent jp speaker

[graceful talon]

yes

distance = |a-b|

[zealous pike]

If the period is $\pi$

[somber coyote]

Pi, a future fluent jp speaker

[zealous pike]

Then |B| is 1

[zealous pike]

This is the graph, correct?

[unkempt gulch]

How do I calculate the perimeter of the triangle AFE knowing that the perimeter of the pentagon is 30?

[obsidian harness]

try showing that CFD and FAE are congruent and that they are both isosceles

[unkempt gulch]

But what about FE?

[obsidian harness]

golden ratio pentagon yields a lot of search results btw

[unkempt gulch]

This helps a lot

[obsidian harness]

yeah so PSR = 36 is what you would find first

and then if you call the centre O, I bet that you can tell that OR is perpendicular to PS

[obsidian harness]

oh yeah then the shaded right triangle has sides x, x, 1 and you know all the angles cause it's isosceles

[unkempt gulch]

But how do I exactly prove this?

[craggy bay]

@timber canopy

<@&286206848099549185>

[silent crystal]

I’m sorry I’m just dumb asf, but what is the relationship between all these, I needa know this to solve it and they screwed up how it was written

[ancient kindle]

so you notice how those two lines are parallel right

these two

because they're parallel, that means they're just shifted versions of each other

which means that the angles it makes with the line are basically the same when they're in the same 'spot' if that makes sense

that means the ones with matching colors are equal angles

also notice how 117 + 3y is equal to 180, because they form the angle of a straight line

[quaint oasis]

Hello

Is there someone here that could help me w geometry-9th grade?

[wintry loom]

what specifically?

[quaint oasis]

I just want some recources that could help me study because im very good at algebra but I am starting to get B's and c's on my test

And I was to get As or B+

Im on 3.5 rn I think (Perpiniducular measurements and stuff like that)

[wintry loom]

khan academy, corbettmaths, ...