geometry-and-trigonometry

[dire copper]

2x - 3(a + 1)y + a - 1 = 0
the slope is 2, what is a?

can anyone help?

[dark sparrow]

@dire copper do you still need help with this?

[dire copper]

yes:(

[dark sparrow]

do you know how to find the slope of a straight line from its equation?

[dire copper]

I guess no and that's why I couldn't:(

[silent plank]

would you be able to determine the slope of something like
$$2x + y = 5$$

@dire copper

[somber coyote]

ℝamonov

[dark sparrow]

they DM'd me with their answer

[nova tiger]

I got the equation arctan(100w)+arctan(w) = 90 degrees
From what I remember arctan(a)+ arctan(b) =arctan(a+b/(1-ab))
so what I did was to tan both sides and got a+b/(1-ab) = tan(90) which is undefined
but when I look at solutions what they did was: arctan(a)+ arctan(b) = sign(a) x 90 degrees, if ab = 1 then they set 100w^2 = 1 and solved

so my question would be, why was the first method I was using wrong? and whats the theory behind the solution? ive never heard about it

[wary veldt]

Hello all.

TLDR: My question is - how do I convert a decimal (e.g. 1.414213562 is root2) to the root form?

---

I'm working on a pre-calc course and getting some basic trig. It's an online class with no professor. I'm inputing sin/cos/tan etc problems in my calculator and getting decimal results where the text book is providing roots.

For example, the problem is sec(pi/4)
I input 1/cos(pi/4)
answer = 1.414213562
Book answer = root2, which comes out to 1.414213562

I think I SHOULD be correct if I answer the decimal form on the quiz / tests, however I'd rather avoid having to justify it if I can give the root version of the answer.

Sorry this is so long.

My question is - how do I convert this decimal to the root form?

[old fable]

Why does the radial component of the weight have a minus sign? The force is in the r hat direction so why would it have a minus sign?

[dim stirrup]

they are both in the opposite direction to r hat (i.e. towards the center of the circle)

[old fable]

But the radial direction is towards r hat or not?

[dim stirrup]

r hat is pointing away from the centre of the circle

the acceleration is towards the center of the circle

[old fable]

So radial acceleration is always to the Center of the circle

[dim stirrup]

in this case, circular motion

T is always towards the center

the weight isn't

[old fable]

Where can I find resources on this radial acceleration

As it’s so confusing

[dim stirrup]

search for centripetal force

[old fable]

Yes but that’s not always the radial acceleration, I heard some dude say

[dim stirrup]

im not sure what he means by that haha

[old fable]

Wait I’ll send the vid

https://youtu.be/JId4DO95Vfg

[dim stirrup]

oh okay

[old fable]

Min 3:31

[dim stirrup]

the tangential force in your case is given by Wcos(theta)

so the weight always points down

as the mass moves around the circle

you need to split the components of that vector into radial and tangential

think about it physically

if the mass is at the bottom of the circle

the weight won't make it move

however, if the mass is at 90 degrees, it will fall down

[old fable]

Oh ok

But how do you explain what he says in the video

[dim stirrup]

oh okay, so the difference here is how you're defining r hat in your model

it doesn't actually matter, just pick whichever direction makes your calculation convenient

whichever one you choose will only add or remove some minus signs

it's like deciding xhat points like <- instead of ->

doesnt matter

[old fable]

Ohh ok just like he says in the vid

Thanks a lot

[dim stirrup]

no worries

[old fable]

I’ve been on this for like 2 hours cuz no replied in the physics server

[dim stirrup]

feel free to ping me with any physics questions, i have 2 masters degrees in it haha

[old fable]

Oh damn

I’ll sure do

Thanks again

[dim stirrup]

might be

yes

what do ya need lmao?

[grizzled lantern]

cap

what class is it?

[dim stirrup]

high school?

[grizzled lantern]

Lol

fuck trig

[dim stirrup]

show me the questions lol

[grizzled lantern]

1/2bh

?

[dim stirrup]

whatever, im bored and can give you a hand with a few

[grizzled lantern]

that a radius

[dim stirrup]

i dont need your money

[grizzled lantern]

18 is half your height

and all the sides are the same

[dim stirrup]

i failed maths in school bro i feel your pain

alrite bro dm me the qs

[plush cloud]

Help?

[simple bear]

<@&286206848099549185>

[dark sparrow]

@plush cloud do you still need help with this?

[nocturne remnant]

E is a point on the interior of quadrilateral ABCD s.t.
[AEB][CED]=[BEC][DEA]
(where [] denotes the area of a triangle. )
Question: Is it possible that E lies on neither AC nor BD?

[dark sparrow]

i can say for certain that it's impossible if ABCD is a parallelogram

not sure about the general case tho

[nova tiger]

why is arctan(a)+ arctan(b)=sign(a) * 90 degrees if ab = 1?

[nova tiger]

<@&286206848099549185>

[nocturne remnant]

let a=tan x, b=tan y
( -90<x,y<90 by range of arctan)

tan x tan y = 1
tan x tan 90-x = 1 (identity)
So x+y=90 or -90

If x+y=90, x and y both positive, so a is positive, done
x+y=-90 just the same

[nocturne remnant]

I think I have an epiphany

(Refer to diagram)
The area condition reduces to

sin a sin c = sin b sin d
(Other than trivial sols with a+b=c+d=180 etc.)

Suppose a non trivial sol exists.
Then let T be a point in the region with smallest angle in (a,b,c,d) in this case c
Then reflect Ray EA over T, it should meet EC again (since this solution is not trivial) at P, then by def P’ would be back on Ray EA
(Same for Q AND Q’)
Then PQP’Q’ is parallelogram with a nontrivial sol. Which we know cannot happen.
Done

[old fable]

,w simplify (-csc x)(cot^2x) + csc^3x

[somber coyote]

Results provided by WolframAlpha

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

Requested by Thalapathy#3541

[old fable]

does this actually simplify to cscx?

looks kinda weird to me cuz the second part is way more complicated

[plush bloom]

please

help

https://cdn.discordapp.com/attachments/750369692157345925/883905654773932082/unknown.png

[wanton silo]

Any tips to remember trig identity

[upper karma]

(3/4, 3.5) (5.25, 10.5)

[mossy geyser]

From the dimensions of the black rectangle and the green line length, I want to calculate the length of the blue line. So, I thought maybe I should determine, first, the value of the violet angle. But how can I do that? I'm confused.

[orchid quartz]

$$\frac{5230 \text{cm}}{1} \cdot \frac{1 \text{m}}{100 \text{cm}} \cdot \frac{1 \text{hm}}{100 \text{m}}$$

$$\frac{5230 \text{cm}}{1} \cdot \frac{1 \text{m}}{100 \text{cm}} \cdot \frac{1 \text{hm}}{100 \text{m}}$$

[somber coyote]

Mr Syndicate7734

[silk venture]

let a=tan x, b=tan y
( -90<x,y<90 by range of arctan)

tan x tan y = 1
tan x tan 90-x = 1 (identity)
So x+y=90 or -90

If x+y=90, x and y both positive, so a is positive, done
x+y=-90 just the fuck

[unreal ivy]

oh wrong channel

[shell raven]

so i have a question

if I had a circle with a radius of 1cm

and i had a piece of string 1cm and put it around the circle is a radian

wait no

that's stupid

i don't get radians

[nocturne remnant]

What seems to be the problem

[shell raven]

so

uh

how do radians even work

like

wait

nvm

it clicked

[upper karma]

hey can someone just tell if this is solvable, or the info is sufficient or not

[crimson dust]

I got 3 for the big circle

what is the small one

[wheat osprey]

@crimson dust how did you get 3?

If the question had an equilateral triangle then we could have done it by finding the angle OCF... But out here there's no symmetry... So how did you go about it?

[fallen girder]

https://mathschallenge.net/full/inscribed_circle_in_isosceles_triangle

2 different ways of finding the area of the triangle allows an easy way to solve for the inscribed radius

[wicked crag]

how am i supposed to solve this

"Find the value of the following, put it in its simplest form.
cos(pi/12)

[wheat osprey]

Cos(2x) = 2cos^2(x) - 1

We know cos(pi/6) = root(3)/2

[marsh bobcat]

hey, can anyone help me with a trig question? i forgot trigonometry lol

[wanton silo]

can anoyne please help me with geometry

[wanton silo]

sure

[marsh bobcat]

can i dm you?

[wanton silo]

k

[torn sluice]

hey guys

I have a trigonometry question

what do I need to do here?

[silk patio]

What’s the question

[torn sluice]

find coordinates? @silk patio

[silk patio]

Of?

[torn sluice]

idk :((

[silk patio]

Then go away

[torn sluice]

my teacher didnt explain anything :I

[valid harbor]

sine and cosine law

that's how

[novel mango]

hi, $\sin^2(x)=\frac{1-\cos(2x)}{2}$ youll get a telescoping sum

[somber coyote]

God V2

[upper karma]

,

[boreal niche]

can someone pleas help me with some bearings and scale drawings?

1. A taxi travels 8 km south, then 3 km west.
   a) Draw a sketch to illustrate this information.
   b) Using a scale of 1 cm represents 1 km draw an accurate drawing to represent the above information.
   c) From your accurate diagram, determine;
   i. How far the taxi is from the starting point.
   ii. The bearing of the taxi’s finishing point from the starting point.

[lone arch]

i think this geometry since its a shape but anyways can somebody help i forgot how to solve it and the answers and explinations i getting online arent helping+ i have to find area of these

[vague bloom]

Uhh

You know the formula for area of a circle?

The formula is pi x r^2

so in the first circle the radius is 12, so substitute it into the formula

it would be pi x 12^2

which is 144pi

so for the first one the answer would be 144pi km^2

and just repeat the process for the second circle

@lone arch got it?

[lone arch]

Tysm

[high estuary]

Am I doing this right? Basically need to solve for c if a is 15

[upper karma]

correct

[upper karma]

Hey, currently learning analytic geometry and was wondering how can you find the rectangle points when all you know is the x and y values of only 2 points?

[upper karma]

If only two points are given, then these are of opposite coner of rectangle. Find the difference of abscissa and ordinates of both the points and you will get both the sides, then add or subtract the sides in the given points to get other two points.

[upper karma]

It's a rectangle where each vector length of 2 points equals to the others. But it's not grid aligned.

[upper karma]

What are these two vectors? value

A (0,0), B (4,3)

@upper karma

I know the answers for the 2 possible options for both point C and D but I don't know how to get to those values nor I understand how to start solving this kind of problem.

What's that

[upper karma]

Tell me the answer, I will try.

Oh it's fine 😅

[upper karma]

One sec

(7, - 1), (3, -4) or (1,7), (-3,4)

Thanks

[upper karma]

Its a square not rectangle according to these other points. You need to find the slope of the lines perpendicular to the line joining the two given points and passing through these two points. Then add and subtract the side(distance between the two given points) in the direction of these two lines.

Thanks will try it out, and yeah I guess that's the right term for what I was trying to say ^^

[lean ferry]

Can someone please explain double identities I have no idea what to do

I have the formulas but idk how to do anything

[flat tartan]

Is this the correct way to write where tan3x is continuous ?

[fleet marsh]

what

whyd i get pinged

[hard geyser]

literal ghost ping

[fleet marsh]

im not a helper

[night crown]

what's good brother

[fleet marsh]

[upper karma]

y da ping?

[night crown]

applied math gang

LOL

[fleet marsh]

no

[hard geyser]

just read the rules dude

[tacit whale]

420 degrees

[night crown]

6

[tacit whale]

qed

[upper karma]

Lol

[night crown]

diagram says 6 degrees bro

[upper karma]

Bro. I just got ghost pinged

[fleet marsh]

the answer is probably 1

[hard geyser]

[fleet marsh]

[hard geyser]

stop pinging the roles

[night crown]

who tf

[upper karma]

STOP PINGING

[fleet marsh]

how are they pinging

[upper karma]

IDK

[night crown]

@ ADMIN

[upper karma]

Uh

I got pinged

[upper karma]

same

[stuck torrent]

[spark stag]

user has been banned

move on

apologies for the disruption.

[hard geyser]

thks nami

[upper karma]

Oh ok

[upper karma]

thx

[hard geyser]

[remote hill]

2 pings nice

[gloomy trail]

How would I do this?

<@&286206848099549185>

[valid harbor]

if the lenght is x, then what is the lenght of its supplement

@gloomy trail

[gloomy trail]

2x+40?

[valid harbor]

no

try again

[gloomy trail]

I have no idea

[valid harbor]

when do two angles are supplementary

[gloomy trail]

when they add to 180

[valid harbor]

yes

then can you deduce the measue of its supplement from that

[gloomy trail]

oh 2x+40+x=180

is the formula?

[valid harbor]

no

i'm asking the measure of its supplement

[gloomy trail]

Tbh idk how the supplement thing works

[valid harbor]

if <DCB = x then what is <ACD

[gloomy trail]

x+ACD needs to equal 180

[valid harbor]

then <ACD is ....

[gloomy trail]

IDK how to set it up to get the measure or equation for it tho

[valid harbor]

substract <ACD from both sides....

what do you get

[gloomy trail]

180-ACD

=x

[valid harbor]

i rather meant substract x from both sides

my bad

[gloomy trail]

180-x=acd

[valid harbor]

yes

what is the measure of its complentary

[gloomy trail]

90

[valid harbor]

no

[gloomy trail]

90-acd?

[valid harbor]

the complementary to x...

[gloomy trail]

is there a youtube video you could send me, I feel like I'm not fully grasping it

[valid harbor]

no

(do you still need my help)

[pearl tendon]

Ok this might be a weird question

How would we put this measurement into deceimal form

11 7/8 “

[valid harbor]

i really hope that your lack of responses is due to a lack of connection and not because of the fact that you are blatantly ignoring me

[verbal night]

Math GRE question

My current work is:

Note that A, B, C sit on the same circle centered at O so OA=OB=OC. Note also that O, B, C sit on the circle centered at A so OA=AO=AB=AC. That is, the lines drawn between the centers of either of the circles to any points are all of equal length. 

Then the strategy I imagine would be to construct equilateral triangles and use some kind of angle complement argument to get it

[upper karma]

hey

who pinged me?

[weary hearth]

anyone tagged me?

[primal inlet]

ping?

okay last straw im removing all my roles

[upper karma]

Did someone ping me?

[jolly hollow]

Idk who ping me 2 times?

[surreal pike]

Anyone knows how to eliminate the x in this

[gritty fractal]

whos good with yr 10 maths
i need help lol

[vague bloom]

ok

Post the question here.

[worn sparrow]

wtf is sen lol

[lone copper]

What is the problem?

[lone copper]

For question 1 find a word to describe OA and OC (remember O is the centre of the circle). For question 2 just compare the length ratio of the sides and whether it increases or decreases

[storm heath]

Q1) Option C because both the triangles are congruent.

[storm heath]

Q2) Answer is +3 since dividing the sides of the bigger triangle by the smaller triangle give that.

[undone totem]

sin he misspelled

[versed river]

sen is how sin is written in some spanish adjacent languages lol

its not a typo

[undone totem]

wait is this actully yr 10 maths

[undone totem]

o

oops

[sleek prawn]

the factor is 3

and ab=cd i think

and both triangle

[full dome]

im a bit confused

[quick cedar]

do you know what it means for two angles to be supplementary?

[full dome]

i solved it, nevermind

[left lake]

4 hours later it means they add up to 180˚

[humble pulsar]

it also... wasnt a question

it was a prompt for Viper

[left lake]

nvm I'm dumb occasionally

thanks for clearing my confusion @humble pulsar

I didn't realise

[upper karma]

@surreal pike still looking for help?

[upper karma]

sen is sin in spanish.

[storm heath]

Lmao.

[storm heath]

Well sure.

Yee.

I have studied that better.

[potent orbit]

Hey guys! I'm in a kinda odd situation

My professor gave me this question and everything that has been given to me doesn't work, even what the professor gave me doesn't work and i don't know what to do

[river vortex]

I mean it looks right. Not sure why it's counting it wrong

[potent orbit]

this is pretty odd tbh

[river vortex]

Is there any kind of weird order you have to put them in

I don't see why

But worth a try

[potent orbit]

That's an old pic btw, I fixed (b), but c is messed up

All I have for c is AB, AC, BC

[undone totem]

helppp

nvm i solved it

[nocturne remnant]

I think he meant the possible segments which are not already on the diagram?

Such as BE?

[potent orbit]

Oh shoot, maybe you're right

In the meantime, I gotta figure this out and I think I'm going the wrong direction

[silent plank]

ABC and ACD aren't lines

[potent orbit]

Yeah, realized that a little late

I've been brute forcing some of it for a while

[silent plank]

can you identify two lines that are perpendicular to each other?

[potent orbit]

I'll be honest, I've forgotten so much

All I remember is that it sums up to intersecting lines that result in a right angle

Like a plus sign

[silent plank]

perpendicular lines meet at a 90° angle

can you identify pairs that meet at 90°?

[potent orbit]

So like AB, BC or AC, CD

[silent plank]

yes

[potent orbit]

wait

god damn

[silent plank]

its unclear how they want you to enter the answers

try
AB||BC
AC||CD

[potent orbit]

jeeeeezzzzzz, why is this so weird

[potent orbit]

@silent plankHey, so I fixed it, but I'm on one last question

[surreal pike]

@surreal pike still looking for help?
@upper karma YES PLZ

[full dome]

can someone please draw thsese instructions with a straightedge and compass

Open the compass more than half the length of line
Then keep at the two ends of line and draw an arc from both sides on both above and below the line
The arcs drawn will intersect at a point above and below the line
Join the two points, this is the perpendicular bisector of the line

[silent plank]

@full domehave you tried doing it yourself

[full dome]

i dont have a compass so i cant

[silent plank]

use a cd

fk sry for random ping

<@&268886789983436800>

[full dome]

what is a cd

[silent plank]

compact disc lol

blu-ray disc? but way cheaper

or a decent sized round object will suffice

[full dome]

something like this?

[silent plank]

yes

very good

[slender wasp]

yes, and remember cos(x+pi) = - cos(x)

Actually finding an expression for cos(2pi/5) is a slight pain :(

[slender wasp]

Oh, that's easier then

I would basically just say it's 27pi/5 shifted by multiples of 2pi until its in (0, 2pi)

Oh sorry silly mistake

Ignore that please I made a mistake cri

OK so we already had that cos(27/5 pi) = cos(7pi/5) from what you said

the trick I'd use next is the symmetry of cos x

(about pi)

cos(2pi-x) = cos(x)

This is because we need to get something in [0,pi]

Limitation on cos^-1's range ye

A number x in the interval [0,pi] such that cos(x) = cos(7pi/5),yes

Ye

Well with sine it's slightly easier as the range of arcsin is [-pi/2,pi/2]

cos(x) = cos(2pi-x), sin(x) =sin(pi-x)

Ah okay

Oh I mean those are correct

But that's not what you were quoting / what we were using

Well we were using cos(2pi-x) = cos(x)

The identities there use + signs

Different thing

And sorry I can't vc

We used cos(2pi-x) = cos(x) to say cos(7pi/5) = cos(3pi/5)

Not the 2pi+x identity or anything

Although yes, you used the identity with 2pi+x earlier too to get to 7pi/5 in the first place

Yes

yeah

so here I'd use the sin(pi-x) = sin(x) identity

It's not the same thing, no, it's just another identity

Npp

a lot can be derived from just staring at the graph, like lines of symmetry

For example cos(2pi-x) = cos(x) because of symmetry about pi

but you could also find another list of identities ofc if need be

But yeah np

[lusty plover]

Quick question, what are a 3d vector's version of magnitude and direction and the equations to convert them into their components?

[silk patio]

What would you guess

[dark sparrow]

@slender wasp did someone delete all their messages after talking to you or did they get b&

[thorn finch]

I have maybe a little Advanced question

I have some part of a circle, which is controlled by the Korde, is there a formel, which Can determine if i have moved 1cm on the “korde” from the begining i have then moved x degrees, but if i move 1cm on the “korde” from the middle i have then move Y degrees. Y > x

[nocturne remnant]

You mean, move the Korde without changing its direction? …So it gets shorter near the edge and longer near the center?

[thorn finch]

Yes. Start point is the left side of the “korde”. And the middle of the circle is static point, if i am walking on the “korde” and i start from the begining, if i move 1cm i have example made 1 degree, but if i am in the middle and move 1 cm from the same “korde” i have now by moving 1 cm made 5 degree change

Hope that made sense 🙂

[dark sparrow]

the english word for korde is chord

[thorn finch]

Thanks 🙂

[nocturne remnant]

That sounds interesting

I’ll try

[thorn finch]

I also try haha, it is a little brain twister

[dark sparrow]

it's going to take some calculus

maybe? idk

[nocturne remnant]

I think I did it without calc

The A represents the area of obtuse triangle

[thorn finch]

I’ll try it out

I have a question, if You calculate the area of the obtuse triangle, what is the degrees then

[nocturne remnant]

?

[thorn finch]

I need to calculate the length of the arc, if i move 1cm on the chord i have moved example 1,5cm on the arc from the start, and also the degrees, if i move 1cm from the start i have moved 2 degrees but if i move 1cm from the middle of the chord i have moved 8 degrees

But Your illustration helped me alot

[nocturne remnant]

y is the angle you’re looking for

[thorn finch]

Is there a Way You Can show that in a formal, where i get Y and not A, and im super grateful for Your help

[slender wasp]

I'm honestly not sure, I didn't see them delete anything though

[thorn finch]

@nocturne remnant i solved it only thanks to you mate :)! thank you

[old fable]

2/L sin^2(pi nx/L) where can i find the graph of this function

if i put it in wolfram it doesnt give the graph

[silent plank]

what's n and L

[old fable]

constants

but nvm

i found it already

[peak mesa]

So idk how to explain this but the black body and the orange arm are connected so if the position of the black body changes, the orange changes too, and when the rotation of the orange thing changes, is around of the right side of the black body according the red circle (the rotation) but i want to achieve a state when the orange thing is all the time poiting to the green point independ the rotation (the black body can move or something), so i need to know what i can do to achieve this (i dont know what math i need to use on the rotation of the arm & position of the black body)

i tried to use sin cos for the rotation of the arm and then the black position add sin/cos but inverted but did not work as expected

[peak mesa]

lol i just added 180º on the arm and thats it...

[upper karma]

In 2nd question, slope (m) tends to infinity at point (1,2). Can't use point slope form to derive the equation of tangent... Any other way?? <@&286206848099549185>

[wanton kestrel]

please don't simultaneously post the same question in different channels

but #help-0 is occupied atm

so you may continue that here instead

[upper karma]

it's a geometry question so it's better to ask here ig

Should I delete that from #help-0

[wanton kestrel]

i would say it's a calculus question, actually

[upper karma]

Yeah a bit of differentiation is also used

[wanton kestrel]

your choice

[upper karma]

I figured out the solution. Nvm

[visual girder]

where do i begin to solve this triangle?

[steel trout]

Solve the whole thing and then solve the top triangle.

Whole triangle - top triangle = bottom quadrilateral

[flat schooner]

can someone plz explain what this is and how it works? |

[visual girder]

how do I find angles ACB and DEC

[upper karma]

So korde is the actual word from where it was stolen?

[dark sparrow]

no, i think it was "stolen" from a greek word

[surreal pike]

Help i need to find the area of the colored area in the unit circle

And also say true or false and justify

[potent orbit]

dis sucks, any thoughts?

[storm heath]

Any way to know the reason why their graphs is such without knowing radians and stuff?

[dark sparrow]

@storm heath suck?

what do you mean

[storm heath]

Edited.

I meant "such" instead of "suck".

[nocturne remnant]

How such?
They look normal to me

[dark sparrow]

yeah these do not look weird to me

but desmos does measure angles in radians by default

there is a setting to make it use degrees instead

[storm heath]

I mean to say.

What is the reasoning behind them.

[nocturne remnant]

.?

Behind what

[storm heath]

The graphs.

What else.

[nocturne remnant]

Hmm

Do you understand the sine one

[storm heath]

Npe.

Nthing.

Must I learn Radians?

[dark sparrow]

you're just starting trig right

[storm heath]

T understand these?

[storm heath]

Yup.

[dark sparrow]

okay then you'll learn everything in due time

radians are just a unit of angle measurement nothing more

[storm heath]

Damn inverse trigonometry is cool but hard.

[storm heath]

I see I see.

[dark sparrow]

get comfortable with right triangle trig and circle trig will follow

[potent orbit]

I hate this math

[storm heath]

Oooo.

[potent orbit]

It's due in 30 minutes and I have an english paper due in 30 minutes too

[dark sparrow]

@potent orbit and you want us to do what about that, exactly?

[potent orbit]

My questions are slightly easier than what you're dealing with rn

[storm heath]

GiVe DiReCt AnSwErS.

[potent orbit]

whatever, I have 7/9 total

Look, I need a, c and d. I can't really figure this out

[storm heath]

What the duck is this inverse trignmetry.

Fun but hard.

[potent orbit]

my stuff or your stuff?

[storm heath]

My stuff.

[dark sparrow]

...your part (a) looks weird to me

no clue what conclusion they expect you to draw from a=e

[potent orbit]

Yeah...

this is why I've been stresdsing

[nocturne remnant]

Then e=a?
Seems like the only reasonable thing

[potent orbit]

Tried

[dark sparrow]

at least c and d are somewhat more transparent since the angles in question are well-positioned

[storm heath]

Yeah exactly.
What is angle of s's?

[potent orbit]

I've even done b=e/e=b

just.. c/d is confusing

[dark sparrow]

(c) you can look at angle 3

(d) you can look at the sums of two pairs of angles

thats as much as can be said without just giving away the intended answer\

[potent orbit]

I'm so confused, I'll get to you if I fail the question or get it

Oh yeah, I also have this too

[storm heath]

D nt yu think #❓how-to-get-help wuld be better since that place is mre fr help than discuss?

[potent orbit]

I got it partially corrrrect

[storm heath]

Ayo wtf.

[potent orbit]

i was thinking of only one pair instead of actually using multiples

[dark sparrow]

D nt yu think ❓how-to-get-help wuld be better since that place is mre fr help than discuss?
is your O key broken?

[storm heath]

How is the 2nd one the opposite of the first?

[storm heath]

Yup.

[nocturne remnant]

It kinda looks likes the line is flipped across the x=y diagonal

[storm heath]

I need t use this but it is tedius.

[storm heath]

Frm what angle?

[nocturne remnant]

The 45 degree diagonal line

[storm heath]

I see.

But-

[nocturne remnant]

Actually, for any function f, the inverse of that function should look like it’s reflection over the diagonal line
(Disregarding the domain stuff)

[storm heath]

This yu mean?

[verbal beacon]

no ,the line y=x

[storm heath]

Duck I need t re-learn Algebra 1 which I started and get inverse trignmetry well.

[storm heath]

That is what I sent.

[verbal beacon]

you've sent a diagonal of a square

[storm heath]

Cnsider the square the graph ff.

But tan 35/65 is 0.0093 smething.

[storm heath]

Wait so here the 2nd function has the inverse of itself in the graph?

[woven yacht]

.

[desert pewter]

i just asked a triangle question in #help-6 if anyone can help

[potent orbit]

I can try

[lone cave]

#help-9 is for you geometry people

[lone cave]

,help

[somber coyote]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[verbal night]

Does anyone have hints on how to do this?

math GRE question for context

[valid harbor]

@verbal night impossible

[verbal night]

This is how I feel every single time I like at a MGRE geometry question

[sour folio]

can someone help me with transformations

[spark stag]

https://dontasktoask.com/

[humble pulsar]

even/odd ness of a function is a property of the function

so doesnt matter

[junior veldt]

Idk how to help without giving answers, but I’d suggest to think of how midpoints work? (Given can be a thing you put in reasons)

Look at 3

[upper karma]

how do i evaluate this

Sin(arccot(1/2csc(2pi/3)))

[verbal night]

Not impossible. <RPT=<QPS as P,Q,R sit on the same line as well as P,T,S do. This means that RT/PR=QS/PS. Solve for PS to get PS~10.29

[clever sluice]

Can anyone teach me how to evaluate the volume of a spherical hexagon given its angles and sphere radius?

[lone cave]

Been struggling on this one for days now

Given: ray OP (pointing right) and ray OR (pointing right) trisect ∠NOS, m∠NOP = 3x - 4y, m∠POR = x - 6, and m∠ROS = y - 10. Find the m∠ROS by creating and solving a system of equations.

[silk patio]

What’s a spherical hexagon

[molten delta]

Do you know how to solve system of equations?

[lone cave]

yes but i do not know how to set it up

[lone cave]

because we have to create one

[ocean socket]

Oum Hi

[wary veldt]

Hello all,

My text book has not explained how to complete these problems. I was wondering if someone would be willing to walk me through the process with problem 38

Or any of the even numbered problems really, as the answer key in the book only covers the odd numbered

Disregard - I've figured out to find the x/y intercepts

[valid harbor]

The slope of the line represents the tangent of the angle made by the x-axis and the lane

@wary veldt

[solid mantle]

Can anyone explain how it’s rad3 from tan60?

[lone copper]

Special right-angled triangles in trigonometry, there are mainly 2 to remember, for this one just draw an equilateral, draw a line from the top vertex to nicely ÷ the triangle into two smaller triangles, ratio of hypotenuse: smallest length=2:1
Now use pythagoras theorem to find the opposite length and tangent ratio is opposite/adjacent=√3/1=√3

[upper blaze]

Hy guys, does anyone know any good book that has stareometry and properties of pyramids and stuff?

[upper karma]

Doubt

[lone copper]

Seems correct

I think it cannot be factorised

[hallow tulip]

If it only gives you that information you could theoretically say that:

$\tan{\theta}=\frac{\sin{\theta}}{\cos{\theta}} then \sin{\theta}=2x(x+1) and \cos{\theta}=2x+1$

[somber coyote]

DR

[dark sparrow]

no you couldn't say that

who says 2x(x+1) is between -1 and 1

or that 2x+1 is between -1 and 1

[solid mantle]

Oh! Thank you!

[solid mantle]

Can anyone help explain how they got that answer? Plugging it into the calculator also seems to be wrong

[upper karma]

,w approx 1/(sin(25 degrees))

[somber coyote]

Results provided by WolframAlpha

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

Requested by Al3dium#4084

[upper karma]

can you type exactly what you are plugging into your calculator?

[dark sparrow]

👻

[kind rain]

I need help with this

[kind rain]

<@&286206848099549185>

[valid harbor]

what did you try sir

[kind rain]

I need help with this one now

[valid harbor]

very good reply tho

also is that a test

[kind rain]

No

It’s homework

[raw tiger]

a segment bisector is just a line/ray that divides a segement in half

[split nexus]

can anyone help me on my hw

[left lake]

Don't ask to ask, just ask! ( #❓how-to-get-help )

go ahead and post the question

someone can help

[nocturne tusk]

for geometry, to have an opposite ray, do the points have to be on the same line?

[storm portal]

If (in 2D), a ray with a starting point has an opposite ray, then I'm pretty sure they lie on the same line

The opposite ray may or may not be able to have its starting point elsewhere.

[lone copper]

Split the top circle into a semicircle

Then the bottom part of the top circle not overlapping with other circles=area of square with sides 1cm- area of quadrant with radius 1cm

Each side of the blue square is 1cm

So u can find area of black portion by taking 1²-area of quadrant

[lone copper]

Send ur workings u r probably doing something incorrect

Yes

That is an even better diagram than mine

So it’s (2+area of one circle)cm²

It’s ok we are always careless

As long as u understand where u went wrong, ur good

Np

[lone cave]

#help-3

[smoky talon]

tell me if ur good at 2 column proofs

[junior veldt]

do you mean like statemnet reason ones?

[smoky talon]

yeah

[junior veldt]

if you need helpw iththem, i might be able tohelp wtih simpler ones? i could try

[smoky talon]

k

i need help with hard ones, ez ones are ez

[junior veldt]

which ones are the hard ones? i could try to help with them?

[fluid wasp]

quadrant VI is the same as quadrant II?

[dark sparrow]

i think they meant quadrant IV but typo'd

[pure void]

hi

i need help with math xdd

[timid pier]

a question specifically or a concept?

[pure void]

yes

like conditional

and stuff

@timid pier pls help me understand this better

[timid pier]

aaa sorry but i have a limited understanding on geometry stuff, idk this stuff

sorry

[pure void]

oh i forgot to add false

[pure void]

oh ok

[upper karma]

I had a question

Do you know what cosec is

Yes

Do you know what cot is

Yes

Trigonometric ratios

Do you know that sin^2 t + cos^2 t = 1

Yes it's an identity

Did you try expressing cosec and cot in terms of sin and cos and using this identity to simplify your expression?

...lemme do again. It all messed up

Oh. I see

Like this?

Yes, exactly

Then should I square them

Yes

Did it

Alright.

Yeah. Now plug it in and simplify

Oo lemme try

But how

🤔🤔

First, multiply nominator and denominator by sin^2 A

Okei

Now

[upper karma]

Use this formula

The denominator :- sin^4 A

Should I solve numerator part by quadratic equations formula?

[upper karma]

Huh

Show me your calculations

Okei

It's not this that you were supposed to multiply

I meant (q^2-1)/(q^2+1)

Plug in q^2 and multiply nominator and denominator by sin^2 A

@upper karma and how is it going?

[upper karma]

Oh sorry.. 😅I slept on my table

[upper karma]

Oo

[upper karma]

I'll do it again today

[south onyx]

help please https://cdn.discordapp.com/attachments/887133609239142481/889313912338677780/image0.jpg

[upper karma]

Okei

1st one or second

[upper karma]

BTW 1st one is correct

[winged torrent]

How the frick did i forget dis property

can someone help

[junior veldt]

I think (this is my guess) that if both segments are the same length then the other 2 angles should be the same? So like 180-100 then take whatever you get and divide by 2?(this is a guess)
@winged torrent

[winged torrent]

yyah i just forgot the property btw the answer is 40

can someone help on this

<@&286206848099549185>

[junior veldt]

There’s a supplementary angle in there

Use it

[junior veldt]

Use supplementary angle theorem thing to find angle ABC, then go on from there

[winged torrent]

ok

thnx

[bitter nimbus]

<@&286206848099549185>

someone please help

[raw saffron]

can I help?

[bitter nimbus]

yea

[raw saffron]

ok

so angle B is 55 degrees due to corresponding angles

and angle A is 125 degrees

@bitter nimbus

[bitter nimbus]

ohhhhh

okay thank you so much

[upper karma]

Hello

Anyone free?

I wanted help

[dark sparrow]

@upper karma do you still need help with this?

[upper karma]

Still?

It's a new one

[dark sparrow]

maybe you could have had an idea in the 5 minutes before me noticing it

in any case, what's giving you trouble here?

it appears all you'd need to do here is substitute the values of x, y and z into x^2+y^2+z^2 and simplify.

[slender wasp]

because it's multiple choice just consider A = 0 bad advice

[nocturne remnant]

Obviously the answer is proportional to r^2
So that leaves A or B
Subbing r=1, a=b=0 gives the answer is B

[dark sparrow]

or you could also not engage in any of that bullshit

it is not hard to simplify r^2 sin^2(A) cos^2(B) + r^2 sin^2(A) sin^2(B) + r^2 cos^2(A) using the sin^2 + cos^2 identity twice

[nocturne remnant]

Or…
You can prove with actual geometry

[elder pebble]

pls help

[winter rover]

Hey could yall help me with some geometry problems? They're pretty basic

[lyric topaz]

hi guys i was wonding i my question is right

My classmate said it is incorrect

[junior veldt]

What do you now know how to do?

For first 8magem do you know what alt int and alt ext angles are?

[winter rover]

I know what they are, but I don't know how to explain them.

[junior veldt]

For sending image, use the supplementary angles

[junior veldt]

The thing is asking to give a pair of alt interior and alt exterior

[storm heath]

Anyone here who has studied AoPs Introduction to Geometry?

[noble thunder]

Introduction to Geometry?

didn't saw that before on aops

is that a pdf or a post

oh it's a book from aops...

it says for grade 7 to 10... I don't think it could be very advanced though I never read it before

[upper karma]

please help

[quiet blade]

The solution is quite long, and it's been solved on other websites such as: https://www.toppr.com/ask/question/in-a-triangle-abcrarbrc-are-the-radii-of-the-circles-which-touch-the-incircle-and/

If you need a hint, start with the fact that the center of the incircle and (one of the*) smaller circles must lie on one of the angle bisectors

[upper karma]

centre of $\Gamma_A$ lies on angle bisector of $\angle A$
\this applies to every of the smaller circles (with their corresponding angle) right? why only one of them?

[somber coyote]

OvercomplicatedHexagonAAAAAAAA

[quiet blade]

yes the result generalises to the other circles as well

but only one small circle* would lie on the angle bisector of A

[upper karma]

yeah ok fair enough

how to do this?

https://www.youtube.com/watch?v=baD79KCWcsQ
@quiet blade btw what the hell is this lmao
i assume this is wrong?

@upper karma could you perhaps help?

sure

ty

as $MQN$ is similar to $MPO$
\$\frac{MQ}{MP} = \frac{MN}{MO} = \frac{QN}{PO}$

[somber coyote]

OvercomplicatedHexagonAAAAAAAA

[upper karma]

just substitute values now

ah okay ic ic, i was just confused as to what is similar to what etc

[upper karma]

ohshi I just realised this is in Hindi and you might not know it xD
he's saying r = r_a + r_b + r_c

[upper karma]

5/8 = 3/5 = x/x+9/4

is this what I should get after substituting the values?

@upper karma ^

[upper karma]

MP = 5+3 not 3

edited

still seems wrong

[upper karma]

what is wrong

$MO = 6 + \frac{18}{5} = \frac{48}{5}$

[somber coyote]

OvercomplicatedHexagonAAAAAAAA

[upper karma]

so $\frac{MN}{MO} = \frac{6}{\frac{48}{5}} = \frac{5}{8}$

[somber coyote]

OvercomplicatedHexagonAAAAAAAA

[upper karma]

ohhhh

[upper karma]

you deadass wrote 5/8 = 3/5 lol that gives 25 = 24

[upper karma]

lmao

ok so im getting 5/8 = 5/8 = x/x+9/4 @upper karma

i really appreciate the help btw

nice

[upper karma]

x/(x+9/4) btw brackets can change a lot of stuff

[upper karma]

ic ic, yes

so from now on, what do I have to do?

[upper karma]

solve for x

kk lemme do that

[upper karma]

x = 15/4?

i think so yes

[upper karma]

tysm dude,

:)

[ruby seal]

i need help calculating the circumfrance and the area of a circle

no questions just revison

[sweet linden]

iirc,
Circumfurence = 2 * pi *r
Area = pi * (r^2)

[eternal plover]

Why does the circle have to move when graphing trigonometric ratios?

nvm I think I get it now

[slate steppe]

Could someone help me with this? I’m stuck on 24

[onyx cloud]

can we assume that OZ is parallel to OB?

[slate steppe]

I think so

Most likely yes

[onyx cloud]

actually i don't know if that's helpful

the immediate thing i see is that AOZ looks like a 90 degree angle

but i can't quite show it is with the information they gave you

[slate steppe]

That’s what I think as well but when I try doing it it dosent work

[strange river]

Um hello?

[onyx cloud]

hello

[strange river]

Do you know how to do this?

[onyx cloud]

yes

do you know what the area of the first siren would be?

[strange river]

About 5.3

I think

[onyx cloud]

ypu

yup

1.3^2 * pi, correct?

[strange river]

Yes

[onyx cloud]

so what we did there is
A = pi*r^2

and we solved for the area

now, instead, what we want to do is solve for the radius

if the area is double what we had before, we can set A to be 2 * pi * 1.3^2

so

2 * pi * 1.3^2 = pi * r^2

can you solve for r here?

[strange river]

10.6 is the new area right?

[onyx cloud]

it's best to keep it exact

[onyx cloud]

as i did here. the left side is about 10.6, but it's better to keep it in this way

[strange river]

Ok

[slate steppe]

Could someone help me understand this?

I’m really bad at proofs

I need help solving it but I also want to understand how to do these on my own

[storm portal]

Well, we can say that CT = CA + AT

And there for CT = DO + OG

because CA = DO, and AT = OG

And because CT = DO + OG, and DO + OG = DG

then CT = DG

[upper karma]

i have a question - sec411º to rounded two decimal places

can anyone show me steps for this?

[silent plank]

do you know what sec(x) is?

[upper karma]

i've seen it but i dont know much

i think i heard there is a way to solve it with calculator?

[silent plank]

yes

but in most cases will require you to know the definition of sec

$\sec(x) = \frac{1}{\cos(x)}$

[somber coyote]

ℝamonov

[upper karma]

oh

[silent plank]

which is something that could be entered in a scientific calculator

also make sure it is set to degrees since you're working with degrees

[upper karma]

ty so much ill try that

can i ask one more question?

secØ<0, sinØ<0 sinØ<0, tanØ>0

i have to name the quadrant where the angle lies

[silent plank]

don't use Ø as a variable (assuming you're using it in the place as theta)

the simple letter t is a much more appropriate substitute for theta

have you done much with the unit circle yet?

[upper karma]

no we just started with this chapter

[silent plank]

not much or literally nothing at all?

[upper karma]

i should say nothing at all

i've never seen one of these

[silent plank]

i mean you should at least have some information since,
work with quadrant is dependent on knowledge of the relation between trig and circles

otherwise the teacher hasn't provided you with the information required to do this

[upper karma]

oh this isn't a assignment that's due i just wanted to look ahead what we were gonna do

[silent plank]

try to find an in-depth guide on the unit circle and its properties

[upper karma]

ok

[timber tapir]

how do you know what specific values of a trig function

?

in high school covid kicked me out of school when we covered the unit circle

and idk how ppl just look at sin(pi/2) or sin(0) and know what it is

[weary drift]

you can see how trig function values relate to certain right triangles or cram a values table (https://trigidentities.net/wp-content/uploads/2021/06/Trigonometric-Table-of-All-Angles.png)

[upper blaze]

Guys there is one interesting problem I am working on 😄 If anyone has any ideas it would be apriciated. I got a bit rusty on those stuff 😄 I will post a picture too 🙂
Problem: Equilateral triangles $\triangle A_1BC, \triangle AB_1C, \triangle ABC_1$ are constructed outwards on sides of triangle $\triangle ABC$ and equilateral triangles $\triangle A_2BC, \triangle AB_2C, \triangle ABC_2$ are constructed inwards on sides of triangle $\triangle ABC$. Let $A_0,B_0,C_0$ be midpoints of $BC,AC$ and $AB$ respectivly and let $C_3$ be midpoint of $A_1B_1$. (With $A_3$, and $B_3$ you can mark midpoints of sides$ B_1C_1$ and $A_1C_1$ if needed). Also let $A_4$ and $B_4$ be midpoints of $B_2C_2$ and $A_2C_2$ respectivly (you can set $C_4$ the same way). Prove that triangles $\triangle A_4B_4C$ and $\triangle A_0B_0C_3$ have the same center.

[somber coyote]

MotionMath

[upper blaze]

Two things I know for sure ahha Both of those triangles are equilateral and C C_2 anc C_3 are colinear

and CC_3=C_2C_3

[timber tapir]

so just memorize this?

[upper blaze]

I think you just need to memorize just value of sin of angles 30 45 and 60

all others you can deduce from those 3

[timber tapir]

really? how

[upper blaze]

well 0 and 90 are 0 and 1 always on sine and cosine

cosx= sin(90-x) so you got cos 30 45 60 from there

[upper blaze]

same for 270 and 360

[timber tapir]

wait what im lost

[upper blaze]

HAhaha

[timber tapir]

sorry im not good at trig stuff

if u can't tell

but that's why im here

[upper blaze]

values at 0 90 180 360 are always 0 or 1 in sin and cos

[timber tapir]

ok

[upper blaze]

and you just need to understand where you look cosine values and where you look sin values to know them

so those are easy

you go that?

and 270 same

[timber tapir]

"where you look cosine values"

idk what you mean by this

[upper blaze]

Hmm, when you draw unit circle, you look cosine values on x-axis and sin values on y-axis

you know that?

[timber tapir]

i do now

[timber tapir]

last half of precalc got cancelled

[upper blaze]

nice thats where you start

Now when you got like cos(90) its easy just remember cos is on x axis and you immidiatly know thats 0 😄

[timber tapir]

ok so i know the thing where the denom is 0 and the numerator is 1,2,3 and 3,2,1 then sqrt everything

[upper blaze]

hmm ok I dont get what you want to say on this xD

[timber tapir]

idk it's like a trick to remember something my teacher showed

[upper blaze]

Hmm you in high school?

[timber tapir]

https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&cad=rja&uact=8&ved=2ahUKEwiE5_PTppPzAhXcmmoFHUP3DuAQFnoECAsQAw&url=https%3A%2F%2Fwww.youtube.com%2Fwatch%3Fv%3Dao4EJzNWmK8&usg=AOvVaw3v-8Nj2n2BGj0YX5MkA7zM

no im in college

i passed calc last year in high school and got the college credit but i wanted to take it again in college just to make sure i have a solid foundation before hitting calc II

[upper blaze]

Hmm Idk maybe this works for someone but I wouldnt learn this method honestly haha

[timber tapir]

okay

[upper blaze]

but you can to this too if you see its good for you

[timber tapir]

so teach me the way of the unit circle

pls haha

[upper blaze]

DId you understand what I said about 0 90 180 270 and 360 first?

those should be easy

[timber tapir]

sure but i usually see the values in radians

so it would b easier to convert to that bc i more often end with like sin(pi/2)

than sin(what ever pi/2 is)

[upper blaze]

okay I can do that 😄

so we are done with those values

now all you need to learn

is sin(pi/6) sin(pi/4) and sin(pi/3)

[timber tapir]

so they're always 1 or 0, so i just need to memorize which are 1 and which are 0?

[upper blaze]

you dont need to know which one is which you can deduce that from understanding unit circle 😄

[timber tapir]

wait is pi/2 the angle at (0,1)

ar

[upper blaze]

yes

[timber tapir]

ok so now what

like idk where to even start to understand the unit circle

like i have a circle drawn with x and y axis in front of me

in my notes

[upper blaze]

Uhh seems to me like you miss theory there :/

[timber tapir]

huh?

[upper blaze]

The way I teach trigonometry is that I ask my students to understand sin and cosine functions completly

[timber tapir]

okay well i am willing to learn that

[upper blaze]

and how to connect them from right trigangle to unit circle

[timber tapir]

yes i just have very little understanding of the unit circle bc of aforementioned reasons

[upper blaze]

Well for values that you asked not much is needed actualy

[timber tapir]

yeah but i dont want to just learn for this test

like ideally i'd like to understand the unit circle

[upper blaze]

you know that it is and you know what points determines 0,pi/2,pi,3pi/2 and 2pi points right?

[timber tapir]

no

all i know is the info from the vid i sent

which is the angles and their coordinate values

[upper blaze]

good

now

When you want to find values of sin function

you always look at y-axes

for example

wait I will draw you picture

[timber tapir]

oh so (x,y) is cos, sin

[upper blaze]

yes 😄

[timber tapir]

ok nice

good to know

[upper blaze]

okay and now

when you have sin0

you need to see what values is determined on y axis by its 2nd line

[timber tapir]

wait so what are the coordinates of pi/2

[upper blaze]

pi/2 is at x=0 y=1 so (0,1)

1st line of angle is always on positive part of x axis

Look at this picture

https://prnt.sc/1t9cqq9

orange part is value of sin and pink one is cos value of angle that I draw there

[timber tapir]

ohh yeah duh

so do i only need to know the angles for that one quadrant?

and then i can just make x and/or y negative based on whichever quadrant im in?

[upper blaze]

Basicly yes

you just need to be very careful there and thats it

And you dont need to know both sin and cos

you just need one

I memorize sin in first quadrant

and rest I deduce from formulas and using this fact that you look sin on y-axis and cos on x-axis

Thats how I do at least

when you give me an angle it takes max 2s to deduce but thats becouse of practice though at begining it may take you 5-10s to calculate in your head or its even good to draw a picture for yourself

dont be afraid of doing it on the little picture on the side of your paper

its better to understand how it works than to memorize that table honestly

its more valuable for you

[timber tapir]

yeah i think for now i will def draw the picture when i need to

thank you

[upper blaze]

I really dont know what book is good for it though

I have never learn that from book

[timber tapir]

so then i find tangent with trig conversions

[upper blaze]

tan=sin/cos

nothing else you need honestly

those sqrt(3) stuff I have never learned in my life xD

[timber tapir]

like if i need tan(pi/2), i do cos(pi/2) [0]/ sin(pi/2) [1]

[upper blaze]

yes

basicly

[timber tapir]

lol ok cool

well thanks i hope that works

wait

[upper blaze]

the one I know though is arctan 😄

but you probably dont need those

[timber tapir]

what's the angle at (1,0)

[upper blaze]

its 0

[timber tapir]

oh ok

[upper blaze]

both lines coincide

[timber tapir]

so the pi/2 angle is the 90 degrees angle

[upper blaze]

yes

those radians is best to learn just pi=180 😄

[timber tapir]

wait so when would i ever use (-1,0)

[upper blaze]

thats pi

[timber tapir]

oh so 1,0 is 0 and -1,0 is pi

[upper blaze]

Thats when one line of angle is positive part of x axis and 2nd on negative part

yes

[timber tapir]

what about 0,-1

-pi/2

?

[upper blaze]

3pi/2 or -pi/2

yes

[timber tapir]

ok this makes sense

thank you again

[upper blaze]

np anytime 😄

[timber tapir]

im glad i finally understand the unit circle

its been haunting me in calc haha

every time i see a lim and then a trig function

[upper blaze]

Hahahah just practice it every time you need it

and after 10 uses you will do it blindfolded 😄

[prime vigil]

hey i need help

i was sick and missed my lesson

so i need help with this geo review