geometry-and-trigonometry

[dark sparrow]

did you find the area of the shape?

[chrome ruin]

The point A(-8,4) is plotted on a grid. A line segment joins A to the origin, forming the obtuse angle P Find the sin cos and tan ratios for triangleP.
how would i start this question?

[glossy minnow]

Well

What do you know that can help solve the problem

[foggy oxide]

for u @chrome ruin

better one

[dark sparrow]

obtuse

what you've drawn is not an obtuse angle

[foggy oxide]

wait

i didn't read it

here :

but

isn't he supposed to calculate this angle and subtract from 180 ?

[foggy oxide]

omg

that's why i have a inner affliction with analytic geometry

even knowing that it's so useful

and just saying

it's easier to solve if you work with the angle i drawn

[dark sparrow]

i mean, it's rather trivial regardless

@chrome ruin since you're online: are you able to figure out how far A is from the origin?

...welp

[slim gorge]

lol

[minor quest]

Can someone explain to me the circle formula?

[dark sparrow]

what do you mean by "circle formula"?

the one that gives you the area of a circle?

area = π * radius^2?

[minor quest]

(X-h)^2 (X-k)^2=r^2

[dark sparrow]

surely you meant

(x-h)^2 + (y-k)^2 = r^2?

[minor quest]

Yes

[dark sparrow]

okay, so

[minor quest]

What's the use?

[dark sparrow]

hm?

[minor quest]

For the formula

[dark sparrow]

i... wouldn't really call it a "formula" per se?

[minor quest]

Alright

Equation?

[dark sparrow]

yeah, this is the equation for a circle in the xy plane

centered at (h,k) with radius r

[minor quest]

What's the purpose of it?

I mean I can call something the equation for a circle or square etc...

[dark sparrow]

you know how lines in the plane can be defined by equations that look like ax + by + c = 0?

[minor quest]

By?

[dark sparrow]

...do you know what it means for an equation to define a set of points in the plane?

(idk what exactly you're looking for when you say "purpose", admittedly)

[minor quest]

Like what we use it for

To find the radius? To find the center point?

[dark sparrow]

...honestly? the only "usage" i can think of right now is if you want to find where a circle intersects some other shape that's defined by another equation in x and y

and such implicit equations, as they're called, are useful for checking if a given point lies on whatever shapes they define or not

[minor quest]

Alright

[dark sparrow]

does the equation (x-h)^2 + (y-k)^2 = r^2 itself need explaining? as in, why exactly this defines a circle, and not some weird sort of shape?

[minor quest]

Well I know where h and k comes from but what about the X and y?

[dark sparrow]

x and y refer to the coordinates of an arbitrary point on the circle in question

as in, the circle is the set of all points (x,y) that make the equation true

[minor quest]

And h and k is more specific points then X and y?

[dark sparrow]

h and k are fixed

they're the coordinates of the center of the circle

[naive scarab]

@minor quest have you seen the equation derived? It's pretty easy to explain if you haven't.

[dark sparrow]

it's trivial if you know how to find the distance between two points on a plane

[naive scarab]

yeah, but they don't usually explain that

[dark sparrow]

?!

[naive scarab]

or at least, they didn't to me when I first learned it

[dark sparrow]

they did to me

lol

[naive scarab]

and this is usually the first thing I explain when I tutor people lol

[dark sparrow]

ayy

[naive scarab]

it's pretty bad, usually I have to back up and explain why the distance formula works...

[dark sparrow]

pythagorean theorem

it's literally a restatement of that

[naive scarab]

yeah but people don't know if you don't tell them :p

to give the explanation incase @minor quest wants it:

the distance between the points (x_1,y_2) and (x_2,y_2) is sqrt((x_1-x_2)^2 + (y_1-y_2)^2)

this reason is pythagorean theorem, see this picture:

[dark sparrow]

=tex \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}

[naive scarab]

so you'd like to describe a circle with center at (h,k) of radius r

this is the set of all points (x,y) with distance r from (h,k)

that is to say, x and y need to satisfy this equality

=tex \sqrt{(x-h)^2 + (y-k)^2} = r

[naive scarab]

and if you square both sides of that, you get the formula you were taught

[minor quest]

Oh I see

But why the square root?

[dark sparrow]

hypotenuse**^2** = leg1^2 + leg2^2

that's what the pythagorean theorem states

[minor quest]

Well I usually only squareroot the hypotenuse

[dark sparrow]

what do you mean?

[minor quest]

Oh nvm I just realized what you're talking about lol

[cedar prawn]

I have a problem I asked about a few days ago and didn't get an answer I could really understand... maybe someone on this channel can help

Given three points in the plane, A, B, and C - if you imagine A is "looking" at B and C, which one will it see on its left, and which on its right? How do you calculate that using only their coordinates?

Only thing I can think of is to translate all three so that A becomes the origin, rotate them until B is above A on the Y axis, and check whether C's X coordinate is positive (on A's right), negative (on A's left), or zero (in front)

but that seems like a lot to do for a seemingly simple problem

[haughty prawn]

I don't know what you mean by left and right, those have no meaning on a plane

[cedar prawn]

I'm trying to draw a picture to describe it

Somehow nobody I ask this manages to understand what I mean until they see it -_-

[azure storm]

Rotation Matrix?

Wait

You say A looking at B,C that doesnt make sense

at the middle?

[azure storm]

i.e A in the perpendicular bisector?

Cause the problem doesnt make sense without defining A orientation

What if B A and C are aligned in that order?

[cedar prawn]

How could I know using only the values, whether it is B or C that is on the right, or if they are in the same line of sight?

Look at the picture

[azure storm]

You drew a particular case

idc about a particular case, i need a general definition of what "looking at" means x)

[cedar prawn]

I drew the general idea

You can derive what I'm trying to say by actually LOOKING AT THE PICTURE

and thinking about it

please!! nobody has managed to help me with this yet!

[haughty prawn]

Wut?

[cedar prawn]

I need to solve this problem for a program I am making and I just can't figure it out on my own

[haughty prawn]

Yeah, what in the world does looking at mean?

[azure storm]

As i told you, define what looking at means with word or math...

[cedar prawn]

A is a person. It is looking at people B and C

[azure storm]

A drawing isnt a definition

[cedar prawn]

Who is on its left?

[vapid kettle]

this is bbest done using complex numberse

[cedar prawn]

Who is on its right?

[vapid kettle]

woo my typing skills!

[cedar prawn]

How do you calculate THAT given only coordinates?

[azure storm]

Only thing i can say in general is to use A orientation and the Rotation Matrix

[haughty prawn]

You mean that a vector denoting the direction it's "looking" from A is pointed between the vectors from A to B and A to C?

[cedar prawn]

I don't know all these big words. I know how to do basic transformations and I know how to program Python.

I guess???

[vapid kettle]

well here is a way to do it

[cedar prawn]

I think like a human thinks. If I were A, and I were looking at people B and C, one is on my left, one is on my right, unless one is standing behind the other

[vapid kettle]

I didn't read full buy you didn't define rigorously "looking at B and C"

[azure storm]

So A facing the perpendicular bisector of [AB]?

[vapid kettle]

so I'll give you some formalizations

if you draw a directed line from A to B

goes through both

left/right can be well defined

and if you draw a directed line from A to C

same thing can be well defined

so think of all the points as complex numbers ~

[cedar prawn]

So as I thought, you just have to rotate the whole system until that line is vertical?

[azure storm]

You could maybe just get OAB and OAC and compare too

That'd be applying the matrix rotation, but that means you define first what A is facing with accuracy

[vapid kettle]

and a directed line as a function from R to complex numbers given by z = a + bt

the points for which Im (z-a)/b < 0 is either the right or the left of the line (I forget)

but it's easy to get that line equation between two points in C

and then a little arithmetic with complex numbers

[cedar prawn]

I don't know much about complex numbers

[vapid kettle]

and you have left and right

[cedar prawn]

I mean, I understand the general idea but what you are saying doesn't mean much to me

All I can think of it as is a rotation

but it seems like that's too complicated, there should be an equation

[vapid kettle]

that is the equation :p

[haughty prawn]

Why not just like... Find the angles from the vector along the direction A is looking to the vectors AB and AC?

[cedar prawn]

I know but like... I don't understand it very well

[vapid kettle]

well the way I presented it

I wouldn't expect you to now

but you can work it out ~

here is some motivation:

if you have the complex plane

[cedar prawn]

Everyone always tells me to work out things I have no experience with

[vapid kettle]

work out Im z > 0 and Im z < 0

[cedar prawn]

lm??

what does that mean?

[vapid kettle]

imaginary part

you'll get left/right of the line of reals

[cedar prawn]

Okay........

[vapid kettle]

and then you can shift the reals/rotate them

[haughty prawn]

That seems ridiculously over complicated......

[vapid kettle]

well if you think about what a computer would do

it's much simpler ~

[azure storm]

Checking if OAC<OAB should be enough, wouldnt it?

[vapid kettle]

computers like their short sweet linear equations/inequalities

[cedar prawn]

Ugh you people are only confusing me even more

just like I expected

I understand algebra (up to quadratic equations) very well. I know basic trigonometry, as in, what sines and cosines and tangents are and how to use them. I understand the basic idea of what vectors are. I don't really know anything beyond that.

So far I have never needed anything more to create the programs I have wanted to create

I'm sorry to be rude but I get confused easily when everyone is telling me different things at the same time and none of it means anything to me

[haughty prawn]

You must have something to denote which direction A is looking, like a scalar for an angle or some sort of vector right?

[cedar prawn]

Well I was thinking more of the general idea and I wasn't thinking about how to be precise about it

But let's say you've got the line of sight as a directed line from A to B

and you want to know which side of that line C is on

that may be the better method

[haughty prawn]

As a directed line? Make a vector from the line and find the angle between that vector and the other vector. Should be simple from there unless I'm missing something.

Oh, and do that with dot product.

[cedar prawn]

Okay so everyone talks about dot product and I'm like, what is that. I looked at it on Wikipedia but my eyes glazed over...

How do I do it, and how do I apply it to solving that particular problem of finding the angle?

[haughty prawn]

So you have your point A at (x1, x2) and C at (y1, y2) right?

Unsurprisingly, I fucked up. Use cross product instead.

I dunno really how you'll use a line to denote the direction your character is looking tbh. But let's just say you have your vector from that <i1, i2>. Cross product <i1, i2> with <y1-x1, y2-x2> then find the magnitude of that vector, and then divide by the magnitudes of the 2 original vectors. This should net you sin(whatever the angle is). Arcsin that to get [whatever the angle is] or 180-[whatever the angle is]. Both give you the info you need.

[cedar prawn]

head spinning

I can sort of follow that if I read it over and over

sort of

so you're doing better than anyone has yet, congrats

Btw the vector you mention, <i1,i2>, is that like just point B minus point A?

And then you're doing cross product of that with <y1-x1,y2-x2>?

[haughty prawn]

Huh, the first vector is the direction A is looking. Since you said you were doing it with a line.

Even this might be overcomplicated now that I think about it...

[cedar prawn]

Well my initial idea was even worse: translate all three points, A, B (which it's looking at), and C in such a way that A is at the origin. Then rotate them all so that B is above A on the Y axis. Then look at C's X coordinate - if positive, it's to the right; if negative, to the left

That's a lot of transformation to go through

[haughty prawn]

Oh God Yeh lol

One idea I didn't think of cuz I'm dumb was, find the angle from the vector <i1, i2> to horizontal with like arctangent. Then find the angle from <y1-x1, y2-x2> with arctangent too. Then subtract the first from the second. That might work, and it'd be easy to implement.

[cedar prawn]

I may need a picture :3

I think about math visually tbh

[haughty prawn]

I kinda don't have pen and paper rn, do if you're gonna get an image, its gonna be the shittiest image you've ever seen.

I don't even have pen, it'd basically be in Ms paint, but worse.

Still want it?

[cedar prawn]

Yeah sure

Sorry I am not replying quickly

I am tutoring a guy in math at the same time

Yeah I make pictures in Photoshop all the time, my computer-handwriting is just the devil's laugh lines let me tell you

That is a stupid af metaphor

[haughty prawn]

It's on a phone, even worse

[cedar prawn]

omg that is so ugly that it's beautiful ❤

[haughty prawn]

It's better than this one

[cedar prawn]

lord and lady wtf

XD

paranormal distribution, that's great

[haughty prawn]

Anyway, you can get theta with arctan(i2/i1)

[cedar prawn]

so what is <i1,i2> exactly?

is that the vector to point C or what?

[haughty prawn]

A has to be looking somewhere right? That's what I'm using to denote where it's looking.

[cedar prawn]

Ohhh

Okay. Sorta...

So that's like, the exact point it's looking at?

[haughty prawn]

That's a vector

[cedar prawn]

I mean

Point it's looking at minus its own location

[haughty prawn]

Yeh, basically

[cedar prawn]

<x1-x0,y1-y0> if A is <x0,y0>

basically

[haughty prawn]

Yeh

In the case you just store the angle info, you get to skip a step. 🤷

[cedar prawn]

So. The angle is supposed to be arctan(i2/i1)

I can see that I think

but couldn't that be like a bunch of different angles? like, there's at least two angles that would have the same arctangent inside a circle, right?

or am I crazy

tangent excuse me

[haughty prawn]

Yeh there are, that's what the first method works against by using sin. But this one might work with a little messing around, and be easier to implement too.

[cedar prawn]

So this means it could be either the correct angle or π minus the correct angle, right? Or plus...

Plus I think

[haughty prawn]

Plus iirc

[cedar prawn]

Yeah yeah

So the issue is which one is right

[haughty prawn]

That's why you might wanna use the first thingy I said, despite the fact that it'll be more annoying.

[cedar prawn]

I feel like it should be possible to tell... but I dunno how

I may have to test this with known angles at some point, see if I can figure out on my own what to do...

[haughty prawn]

You can check if the object has an x value of greater than or less than A, that could do something. Maybe.

[cedar prawn]

Hmm... blargh

What was the first thing then? With the sine?

[haughty prawn]

Cross product

[cedar prawn]

This stuff is so confusing

Math makes sense to me until I'm trying to talk about it with someone else

WHICH IS WHY WE NEED MATHLANG but maybe you weren't in that conversation I can't remember

and even then I would still be confused XD

[haughty prawn]

Wut dat?

[cedar prawn]

I have been working on a way to speak math, like verbally, with one syllable per symbol

An artificial language

I love constructed languages

[haughty prawn]

Ack, conlangs

[cedar prawn]

Everyone says there's no point but like

it's cool though

[haughty prawn]

I hate conlangs lol

[cedar prawn]

Whyyy

[haughty prawn]

They make no sense to me 😵

[cedar prawn]

Well they're languages

made by one person

or one group of people

on purpose

there you go

[haughty prawn]

Nah, I know that much. Just the other stuff is weird, linguistics really isn't my thing.

[cedar prawn]

I have several conlangs of my own

one of which I am highly fluent-ish in

some things make more sense in my conlangs than they do in English

Anyway though. I can't remember what you wanted me to do with the cross product, please for give me, I am an airhead

[haughty prawn]

Take you direction vector and cross it with the vector from A to B.

Take the magnitude of that, and divide by the magnitude of the vector from A to B, (the direction vector already has magnitude one). You're left with sin of the angle between the vectors

[cedar prawn]

Uhh

I will have to draw that

but I will save your explanation and draw it tomorrow because it's beddy-bye time for me

I think I see what you mean but visual = good, so tomorrow I'll try it all out and tell you how it goes

thanks so much for your help

[umbral rivet]

👀

Vectors 👀

[foggy oxide]

.-

[cedar prawn]

@haughty prawn It's actually even simpler than you thought. I don't even need to know the angle. My preliminary findings, after reading up on vector products, suggest that, given B and C which are being "looked at" by A, all I need is to know the sign of the vector product of (B-A) times (C-A). If negative, C is to the left of B according to A's perspective; if positive it's on the right; if 0 they are in the same line of sight from A.

Which, now that I think of it, is what someone else told me to do to begin with - I just didn't understand at the time

It took further research into vector products for me to understand it

[haughty prawn]

Oh, neat. 👍

[cedar prawn]

Now I just have to see if I can clean up the bug in my program by using this. You see, this entire venture was in order to deal with a bug I couldn't figure out how to get rid of in a program I was making that was drawing tessellations of the plane XD

It started from a single point placed in the center and then used angles to find the locations of other points to be arrayed around, in rings around that center point, in order to produce a regular tiling - but it only works if they are always being placed in a sort of spiral either clockwise or counterclockwise - so I had to be able to tell the difference between the two

so now MAYBE I can solve that. Or maybe the problem is something else :3

[sacred vessel]

if someone could help i would apeciate it

i dont understand at all

[dark sparrow]

what do you want to find

[sacred vessel]

the area

[dark sparrow]

of the parallelogram?

[sacred vessel]

yeah

that stick is to show how tall it is

[dark sparrow]

what length is the 4.2 cm marking meant to represent?

oh wait

this isn't a parallelogram

it's a trapezoid

do you know how to find the area of a trapezoid?

[sacred vessel]

so thats what it is

no could you give me the formula

so i can figure it out

thanks

[dark sparrow]

area = 1/2 * (base1 + base2) * height

[sacred vessel]

i got 5.06

as the area

how close am i

?

[dark sparrow]

== 0.5 * (4.2 + 5) * 1.1

[charred spear]

5.06

[sacred vessel]

right on target 😄

[dark sparrow]

yeah, 5.06 cm^2

[sacred vessel]

thanks a thousand

i thought it was a parallelogram but i couldnt apply the formula so i came here thanks again

how do i calculate for b

i know A and i know h

in a triangle

[dark sparrow]

you know the formula for the area of a triangle, right?

A = bh/2

[sacred vessel]

yeah

i know it

[dark sparrow]

multiply both sides by 2/h

what do you get

[sacred vessel]

so wait A divide in two and then times h?

[dark sparrow]

no

=tex A = \frac{1}{2}bh \ 2A = 2 \cdot \frac{1}{2}bh \ 2A = bh \\frac{2A}{h} = \frac{bh}{h} \ b = \frac{2A}{h}

[sacred vessel]

oh i get it now

thanks

i catch on quick but its just so confusing

its 22

[sacred vessel]

um how do i get height in a Parallelogram

?

[foggy oxide]

@sacred vessel do you have a angle ?

[glossy minnow]

How can you show that light rays emitted from one focus of an ellipse will reflect back to the other focus?

Wait I can ask questions right

Lol

[dark sparrow]

pretty sure that boils down to showing that lines drawn to a point on the ellipse from its foci are at the same angle to the tangent there

[fickle birch]

can u explain about how to sketch some graphs too @dark sparrow ?

[dark sparrow]

what graphs

as in, of what functions?

[dark sparrow]

okay, so, this is a sinusoidal function

[fickle birch]

i found maximum =120

min 80

[dark sparrow]

yes

what is its midline?

[fickle birch]

amplitude =120-80/2 = 20

[dark sparrow]

yes

[fickle birch]

then i dont what should i do next?

[dark sparrow]

what is the position halfway between its max and min?

[fickle birch]

what is the keyword for it?

[dark sparrow]

midline

or average value

what number is halfway between 80 and 120?

[fickle birch]

so 100

?

[dark sparrow]

yes

okay, so

[fickle birch]

its simple so i can calculate

[dark sparrow]

since this function starts at its minimum, it looks like 100 - 20cos(kt)

for some constant k which we'll find out in just a moment

is that clear?

[fickle birch]

yeah

[dark sparrow]

okay

[fickle birch]

u put 100 and 20 there

i see

[dark sparrow]

the period of your function is 1

[fickle birch]

how do we calculate the period?

i couldnt understand this

[dark sparrow]

it's given

[fickle birch]

given?

[dark sparrow]

once every second

one heartbeat takes one second

[fickle birch]

i see

[dark sparrow]

gtg

[fickle birch]

oh ok

can someone please help?!

[dark sparrow]

@fickle birch i'm back

[fickle birch]

thanks

[dark sparrow]

what is the period of the "raw", unmodified function f(x) = cos(x)?

[fickle birch]

Is it pi?

[dark sparrow]

no

how long does it take the cosine and sine functions to repeat themselves?

[fickle birch]

finish in 2pi right?

[foggy oxide]

i'm hear

[dark sparrow]

yes

so now

what is the period of cos(ax), for constant a?

[foggy oxide]

sorry for interrupting

[fickle birch]

so 2pia?

[dark sparrow]

no

[fickle birch]

omg im so bad at this

[dark sparrow]

think about what happens if, say, a = 3

it takes less time for 3x to reach 2π than x if x increases at a constant rate

do you agree?

so when x changes from 0 to 2π/3, cos(3x) must go through one period, while cos(1x) only goes through one-third of a period

does this make sense?

[fickle birch]

hmm not really

[fickle birch]

what is the x now?

[dark sparrow]

...ok i'll probably have to go to sleep now sorry

[foggy oxide]

i can try to help you

but i don't understand some words in this question

[fickle birch]

thanks a lot

[foggy oxide]

and i don't really know what it is asking

just the graph ?

or you have to say some information ?

[fickle birch]

produce a trigonometric function

[foggy oxide]

this is the blood pressure in function of time right

[fickle birch]

yeah

[foggy oxide]

where did u guys ended ?

[fickle birch]

100 - 20cos(kt)

we got this so far

and find k

[foggy oxide]

http://www.wolframalpha.com/input/?i=f(t)+%3D+100+-+20cos(2pit)

is this correct

?

[fickle birch]

so k=2pi?

[foggy oxide]

i think it is

or just 1, idk

wait

wolfram is loading

[fickle birch]

how to find Coecient for t?

[foggy oxide]

i'll try to explain thyat

wait, just checking something on wolfram

do you still here

?

@fickle birch

[fickle birch]

im

[foggy oxide]

i'm sorry

i was talking with my father

there are a lot of ways to find k

think that k has to satisfy this equation 100 - 20cos(2pi*k) = 120

[fickle birch]

why =120?

ahhh

i see

[foggy oxide]

forget everything

and think like this

100 - 20cos(kt)

you know why it's like this right ?

[fickle birch]

yeah

i understood this part

[foggy oxide]

for wich f(t) is supposed to evaluate 120 ?

every heart beat is a wave

Which lasts 1 second

[fickle birch]

hmm

sorry but i really dont get this part

[foggy oxide]

it starts relaxed 80 then go up go up then 120 then down down 80 and it repeats like that

[fickle birch]

can u please explain a bit?

[foggy oxide]

well, if you know wich part of second 120 is supposed to be evaluated

you can set a equation

[fickle birch]

half second?

[foggy oxide]

yes

that is the peak

[fickle birch]

okay

then how to solve k if we know half second is peak?

[foggy oxide]

you have to think a little bit about the unit circle

that's the easiest way i think

[fickle birch]

ok thanks i will try to do it

[foggy oxide]

but you can manipulate the equation too

it'll help you to visualize it

100 - 20cos(k*1/2) = 120

• 20cos(k*1/2) = 20

• cos(k/2) = 1

cos(k/2) = - 1

so think about the unit circle

where cos(x) it's equal -1 ?

@fickle birch

do you still here ? xD

[fickle birch]

yes?

[foggy oxide]

when cos(x) it's equal to -1 ?

[fickle birch]

i just found the formular

to find the coefficient of x

its 2pi/period right?

[foggy oxide]

yes yes

[fickle birch]

so its 2pit

[foggy oxide]

i thought it was 2 (2 π n + π)

[fickle birch]

there is no shift

[foggy oxide]

i firstly thought it was 2pi too

but sqrt2 said it wasn't

[fickle birch]

how to check that?

[foggy oxide]

with the equation

100 - 20cos(k*1/2) = 120

• 20cos(k*1/2) = 20
• cos(k/2) = 1
  cos(k/2) = - 1

cos(k/2n) = cos(pin)

k/2n = pin

[fickle birch]

yeah x=2pi+4pin

but how to plug it into the function?!

[foggy oxide]

here

f(t) = 100 - 20cos(2pi*t)

it's correct xd

[fickle birch]

this is solution?

[foggy oxide]

yes

idk why sqrt 2 said it wasn't

http://www.wolframalpha.com/input/?i=f(3%2F2)+%3D+100+-+20cos(2pi*1%2F2)

look it gives 120

[fickle birch]

thanks

[foggy oxide]

just solve the equation like i did

100 - 20cos(k1/2) = 120

^^

you try to put everything in terms of cosine

[fickle birch]

sorry do u mind to explain how to sketch a graph from a function?

[foggy oxide]

there are a lot of ways to do that

but first you

have to mark the minimum and maximum

and try to mark the middle points between them

[fickle birch]

like i got this

like i got this

i finished a b c

but i dont know how to sketch

[foggy oxide]

well dude

i have to say that this is very boring to do

i would just copy the graph

[fickle birch]

i want to

but they dont have exact number

on x axis

and the graph just touch some where on the x axis

[foggy oxide]

it's hard to know how detailed they want u to draw it

just do that

since

[fickle birch]

mark the point where the of the peak on x axis

[foggy oxide]

your function it's f(t) = 100 - 20cos(2pi*t)

try to plot

f(1/4)

first

then tell me the value of this

remember, f(x) = y ( this means for a x value you set you got a correspondent y point { i think you already know that } )

then, plot f(3/4)

[fickle birch]

i trying to sketch it now

[foggy oxide]

try to do it like that first

that's how i sketch my graphs

[fickle birch]

i can sketch this one

but the other one is so small

for this one

[foggy oxide]

@fickle birch

you just have to draw it to the correct interval [-1 , 1 ]

you can make it bigger

[trim belfry]

i love this gif so much

[foggy oxide]

very nice

[fickle birch]

i remember there is somewhere they make something like this gif too and u can edit the params or inputs live

[cedar prawn]

Wow MrDigger you use the exact same profile pic as a guy I follow on Quora

whose name I forget

facepalm

[eager saddle]

Ive forgotten how to work with trig equations

How do I 8sin^2x-5sinx=2?

Sliiightly panicking.

[dark sparrow]

let y = sin(x)

you get a quadratic in y

you can solve quadratics, right?

8y^2 - 5y - 2 = 0

[eager saddle]

But there's not really any factors out of that that I can see while panicked.

Factors of 16 that add up to -5?

[dark sparrow]

eh

it's easiest to use the qf here honestly

[eager saddle]

I have an assignment due at midnight, and I've hit a block and can't answer the questions.

[dark sparrow]

you know the quadratic formula, right?

[eager saddle]

Yeah.

[dark sparrow]

yeah

use it

=tex y = \frac{5 \pm \sqrt{89}}{16}

[eager saddle]

I don't know what to do with this information to solve over a period from 0 to 360 degrees.

[dark sparrow]

okay, so

each sine value other than 1 and -1 corresponds to precisely two angles from 0 to 360°

for a sine value of S, these are arcsin(S) and 180°-arcsin(S)

does that need explaining?

[eager saddle]

_<

This assignment is failed.

[haughty prawn]

Don't give up! You can do it! ✊

[dark sparrow]

what did you attempt to put in?

is this a message you got from an online thing? or what

[haughty prawn]

What?

[dark sparrow]

was asking @eager saddle

[haughty prawn]

Oh Oop, my b

[eager saddle]

I got a 30%. ._.

[haughty prawn]

Oh, I guess it was already over. That sucks 😬

[dark sparrow]

time limit?

[eager saddle]

Okay I can still attempt to do other things, but still due in 40 minutes and I don't remember stuff that I was doing yesterday.

No, I just gave up. >_<

[haughty prawn]

:(

Don't do that, Anto can help

[eager saddle]

Sorta time limit, cause due at mindnight in 40 minutes

midnight

[haughty prawn]

Just calm down a little. 40 minutes is enough for more than you might think.

[eager saddle]

It's not hard, I just don't know what I'm doing. @_@

I don't know how to manipulate the numbers/I was told not to divide like you do in algebra.

[dark sparrow]

do you understand what i've walked you through?

@eager saddle

[eager saddle]

I opened up the assignment again, I can try it until midnight/35-ish minutes

[dark sparrow]

okay

so

i'll reiterate my question

do you understand what i've walked you through so far?

[eager saddle]

I cannot answer the math problem from what I can read, so no.

Forget about that, I think we should start over from a different question.

[dark sparrow]

we've yet to arrive at an answer.

but okay

can you post the question you want to start over from?

[eager saddle]

I'll find one.

Okay, let's start simple, tehn...

If I have sin(3x)=1, I don't know what to do with the 3 if it's inside the parenthesis.

[dark sparrow]

is there a range you're asked to solve this over?

[eager saddle]

This is what's stopping me. I don't know what to do with that coefficient.

Yes, form 0 to 360

degrees.

[dark sparrow]

okay

so in that case

if it makes things simpler for you, you can let y = 3x

and solve for y, over the range 0 ≤ y ≤ 1080°

and then solve for x once you get that

[eager saddle]

So I multiply the range by the coefficient?

Bleh. >_<

[dark sparrow]

i mean, if 0 ≤ x ≤ 360, then 0 ≤ 3x ≤ 1080, as is hopefully obvious?

[eager saddle]

Okay, so I read that as 'the sine of three times an angle is one'

[dark sparrow]

yes

[eager saddle]

I'm failing at basic things because panic, one sec.

[dark sparrow]

the main idea is that instead of trying to rewrite sin(3x) somehow, you first find what 3x can be

also, calm down. please.

[eager saddle]

Even when it's y=3x I don't quite understand what I'm trying to find.

[dark sparrow]

ok

[eager saddle]

I solve for x, I get x=y/3

[dark sparrow]

so

yes, that's how you're going to obtain x once you've solved for y

anyway

ignoring the range restriction, can you tell me what values of y make the equation sin(y) = 1 true?

[eager saddle]

90?

[dark sparrow]

90° and any angle coterminal to it

90 + 360n degrees, for arbitrary integer n

make sense?

[eager saddle]

Yes.

[dark sparrow]

okay

[eager saddle]

So do I divide 90 +360n by 3?

[dark sparrow]

indeed

that gives you x = 30 + 120n

[eager saddle]

So 30 + 120n, so my angles would be...

30,150,270 within that range

[dark sparrow]

yup! you got it

👍

[eager saddle]

😅

[idle bobcat]

hi

[dark sparrow]

hi

[eager saddle]

Why am I finding this so difficult?

Okay, so 9tanx-3cotx=0

[rare talon]

Struggling is the way to learn math, don't worry

[eager saddle]

I don't know what to do with this.

Turning it into 9x-3y=0, but then I'm forced to divide across the equals with a variable, which I'm not supposed to do

Oh, over 0 to 360 degree range

[dark sparrow]

cot(x) = 1/tan(x)

[eager saddle]

so 9tanx - 1/3tanx = 0?

[dark sparrow]

no

[eager saddle]

Or is it 3/tanx

[dark sparrow]

9 tan(x) - 3/tan(x) = 0, yes

you can divide through by 3 to get 3 tan(x) - 1/tan(x) = 0

you can now multiply both sides by tan(x), noting that it cannot equal zero

[eager saddle]

so 3-1 = 0, no the solution is an empty set

[dark sparrow]

no!

[eager saddle]

so no solution?

[dark sparrow]

tan(x) * tan(x) doesn't equal 1, does it?

=tex 3\tan^2(x) - 1 = 0

[eager saddle]

Okay

and then I use a pythagorean identity?

[dark sparrow]

no

tan^2(x) = 1/3

tan(x) = ±1/sqrt(3)

(does this make sense?)

[eager saddle]

Yeah.

I'm sorry I'm being so frustrating. >_<

[dark sparrow]

you aren't

anyway

[eager saddle]

tanx = sqrt(3)/3

[dark sparrow]

±sqrt(3)/3

[eager saddle]

^

[dark sparrow]

i missed that sign initially even though i had intended to type it >.>

anyway

do you know what angles have sqrt(3)/3 as their tangent?

[eager saddle]

then I use the inverse

30 degrees?

[dark sparrow]

30 + 180n

do you know what angles have **-**sqrt(3)/3 as their tangent?

[eager saddle]

30 degrees, 210 degrees,

[dark sparrow]

do you know what angles have **-**sqrt(3)/3 as their tangent?

[eager saddle]

Er...

_<

Really close to midnight, ah

[dark sparrow]

hint: tan is an odd function

[eager saddle]

30,150,210,330?

[dark sparrow]

yes

those are the solutions to tan(x) = ±sqrt(3)/3

i was expecting -30 + 180n

[eager saddle]

Thank you. ^^

RAn out of time. Failed.

[idle bobcat]

can I have some help with circles

[dark sparrow]

yes

is there a problem you're stuck on?

[idle bobcat]

yeah, so this question says, "What can you conclude about the daigram? State a postulate or theorem that justifies your answer"

hmm can i send an imgur of that circle?

[dark sparrow]

...can you post the diagram?

and yes

[idle bobcat]

https://imgur.com/gallery/y7i3o

[dark sparrow]

which one do you want help with

[idle bobcat]

22 please

[dark sparrow]

ok

so you see here that chords AB and CD are equal

what can you conclude from this?

[idle bobcat]

congruent?

[dark sparrow]

what two things are you claiming to be congruent?

[idle bobcat]

line AB and DC

[dark sparrow]

i mean, duh

that's given

that's not what i meant

i wanted you to conclude that the arcs subtended by them are also equal

equal chords subtend equal arcs and are at equal distances from the center, and that works in every direction

is that clear?

[idle bobcat]

yes

[dark sparrow]

do you need help with anything else in that picture?

[idle bobcat]

not really for now, but I'll let you know if I get stuck

thank you

[foggy oxide]

good evening gentlemen

haha

[surreal bolt]

hahaha

[foggy oxide]

xDDD

[upper karma]

Okay I have a max point at

=tex -\frac{5\pi}{4},6.7

Just wanted to try latex

[dark sparrow]

=tex \left( -\frac{5\pi}{4},6.7 \right)

[upper karma]

Ah yeah nice

[dark sparrow]

ok

[upper karma]

Amplitude is 4.1

All we know

Now they're asking me for midpoint

[dark sparrow]

well, what do you think the midpoint is?

[upper karma]

I know that midpoint = (highesty + lowesty)/2

[dark sparrow]

that's true

[upper karma]

But they want us to do it through the amplitude probably?

[dark sparrow]

the amplitude of a sinusoid is how far its max and min are away from the midpoint

[upper karma]

I noted: Amplitude is how much midline varies from highest to lowest y

yeah

[dark sparrow]

6.7 is 4.1 units away from what?

[upper karma]

Midpoint

[dark sparrow]

no, what number?

[upper karma]

uh

2.6?

[dark sparrow]

💯

[upper karma]

Waitw hat

[dark sparrow]

yes

[upper karma]

I don't get it

http://puu.sh/wSQaG/3d5bd1fc26.png

[dark sparrow]

amplitude = max - mid = mid - min

[upper karma]

I'm algebraically retarded

[dark sparrow]

if you mark off the max, min and mid points on the y axis they're going to be evenly spaced

and the space between them is precisely the amplitude

[upper karma]

So amplitude is always same distance never average?

[dark sparrow]

hm?

[upper karma]

Like you can have 4.1 only and not 4.1 and 3.2 or something

[dark sparrow]

i mean, yes

if your midpoint was at different distances from the max and min it wouldn't be a midpoint, would it?

[upper karma]

yeah ohly fuck

how can i be this dumb

[dark sparrow]

dw

[upper karma]

how can you even remember all this

I will forget this after one exam

all of it

[dark sparrow]

🤷

i just do

[upper karma]

You read your notes every day?

[dark sparrow]

i don't take notes

[upper karma]

WHAT

????????????????????????????????????????????

[dark sparrow]

i mean, i do occasionally but

the best way i can describe it is that all of my mathematical knowledge ends up being very inter-connected and

when i really know a fact i can relate it to other facts i know

leading to better retention

[upper karma]

That's probably true for higher mathematics but not highschool where you do very small chapters of big topics like statistics and complex numbers

Nothing connects you only get a brief overview of many different topics

[dark sparrow]

ah, our high school educations must have differed

mine was very much all about the connections and exactly how everything made sense

we didn't do small chapters

big topics would have weeks if not months dedicated to them

[upper karma]

i'd say its more in higher maths that you lose connxions when everything gets so far appart

[dark sparrow]

but then, i went to a school with a more extensive math curriculum than other schools

also @upper karma 3b1b videos where he reveals connections between seemingly unrelated fields of math come to mind 😄

[upper karma]

WTf you must be ghoing to a god school

3b1b is not really advanced

I mean its not highschool

but its basics

Also sorry im retarded again, How to swap 4.1 = 6.7 - mid without making it - 6.7 + 4.1 = - mid?? :DDDDD

Yeah

[dark sparrow]

-mid = -6.7 + 4.1 = -2.6

😛

[upper karma]

._.

[dark sparrow]

anyway

[upper karma]

is that how you do it?

[dark sparrow]

=tex a = b - c \iff c = b - a \ a = \frac{b}{c} \iff c = \frac{b}{a}

[upper karma]

is that associativity

[dark sparrow]

@upper karma no

associativity is (a $ b) $ c = a $ (b $ c)

where $ is an operation

be it multiplication, addition, or anything else that you know to be associative

[upper karma]

Then what was that you showed me?

[dark sparrow]

a couple of rearrangement shortcuts, really

the first one can be stated as "add c to both sides and subtract a from both sides"

the second one as "multiply both sides by c and divide both sides by a"

[upper karma]

does it have a name (like a law) like associativity, commutiativity, distribution

[dark sparrow]

it... doesn't, really?

not that i heard of anyway

[upper karma]

ahh okay

[dark sparrow]

it's just a specific case of doing the same thing to both sides

lol

[upper karma]

That one is very useful and is like the reason I failed physics

I couldn't translate an equation when a variable was like 5 = g/3

or anything alike

5 = 3/h

[dark sparrow]

come to think of it, physics is where it came up a lot

[upper karma]

Yeah it was like the fundamental

I couldn't wrap my head around it

As soon as I had to convert something in the equation I wanted to die :D

I asked teacher if she could bring out a gun from her drawer and just blow my brain out :D

[dark sparrow]

...

[upper karma]

True story right there

[upper karma]

Math is too damn difficult

101010011001011010100110

How did you go from

Midpoint = highesty + lowesty/2

To your thing

[dark sparrow]

mid = (max + min)/2

2 * mid = max + min

mid + mid = max + min

mid + mid - min = max

mid - min = max - mid

[upper karma]

Holy shit

You are like an algebra god

[dark sparrow]

more like goddess if anything lol

and i mean

algebra is something i've simply had tons of practice in? lol

[upper karma]

mid + mid - min = max
mid - min = max - mid

I don't understand this step

[dark sparrow]

subtract mid from both sides

[foggy oxide]

lol goddess 😄

[dark sparrow]

mid + (mid-min) = max
mid + (mid-min) - mid = max - mid

@upper karma that clearer?

[upper karma]

Yeah

Algebraically

Wait no that made less sence

Shouldn't it be

mid + (mid-mid) - min = max - mid

[dark sparrow]

no

mid + (mid-min) = max
mid + (mid-min) - mid = max - mid

[upper karma]

Nope...
mid + mid - min = max || -mid
mid - min = max - mid
would have made more sense

[dark sparrow]

that's exactly what i did though

[upper karma]

No you did confusing paranthesis on those two

You could've like
mid + (mid-mid) - min = max (-mid)
mid - min = max - mid

I get the idea tho

Nope

All went wrong

We have

Min point -8.2

Amplitude 5.4

With our equation

5.4 = mid - (-8.2)

5.4 - 8.2 = mid ?

[dark sparrow]

yes

[upper karma]

Got wrong answer

[dark sparrow]

-2.8

what were you asked for?

the midpoint? it's definitely -2.8

[upper karma]

It is

Somehow I calculated it to -3.6

[dark sparrow]

that's an issue with arithmetic, then

[upper karma]

Everything is issue with me

Im so bad

Living proof that math isn't for anyone

[dark sparrow]

hey

don't beat yourself up

[upper karma]

If I somehow manage to one day get into uni in physics or math, I will be hard proof that math can be done by everyone

[dark sparrow]

arithmetic errors happen to everyone

i miscalculated 3 * 4 as 6 once in a test

[upper karma]

I don't bleieve that ur too good

Brainlet error not possible

If not, I promise I will hold a TEDx Talk about how math is not for everyone and not make people waste their lives

[dark sparrow]

believe it or not, such mistakes do happen to me sometimes

more than once

[upper karma]

That kind of error won't get you into university

[dark sparrow]

i remember staring at a page of seemingly airtight algebra leading to an absurd answer for like

5 minutes?

before discovering a missed minus sign

in exams, triple-check everything

that's how i 100'd my CS final

(also, offtopic, but mochizuki's name is actually もちづき. source: Japanese Wikipedia)

but yeah anyway

don't beat yourself up for this

occasional slip-ups in arithmetic don't make you bad at math

[foggy oxide]

lol

i agree

sometimes i can't sum numbers when their unity are respectively 8 and 5

i have no idea why

[upper karma]

I don't have time to even double check

Because when I'm done it's literally the last minute

And thanks weeb ill fix

[dark sparrow]

former weeb, if anything

anyway, practice practice practice
eventually you'll pick up speed

time yourself

[upper karma]

If only I enjoyed doing math as much as I do games

Have like 3k hours on LoL and 1k+ on CS:GO

Imagine those 4000+ hours on math

I'd be ascending

But difference is I'm good at games and beat everyone up I have like an intuition for it and I pick up new games pretty easily

But math is just full struggle and failure, it's very de-motivating and makes me not want to do it

[dark sparrow]

mm

i understand where you're coming from

[foggy oxide]

i wish i had spent more time with math in my chillhood

[dark sparrow]

i have to go to sleep now though, so

👋

[foggy oxide]

the brain develops more in the childhood

[upper karma]

same

[foggy oxide]

xD

[upper karma]

we all wish that prelude

this would be fields medalist discord if we all did

[foggy oxide]

i used to like math in my chillhood but at some time of my life all i could think was video games

😭

rip

[upper karma]

cause video games are fun, math isn't

Unless you're a prodigy like those asians

[haughty prawn]

😒

Wow

Rip my emote

[hazy field]

yeah we don't have that enabled

[foggy oxide]

Well, i feel like math is very fun

[hazy field]

but @upper karma don't

[foggy oxide]

usually

[upper karma]

It's fun when you do something right, it's rewarding, you do less mistakes in games which makes it more rewarding = more fun

@hazy field dont what

[foggy oxide]

you're right

[hazy field]

@upper karma don't be edgy and annoying

[upper karma]

I'm having a conversation wtf?

[foggy oxide]

well ... i've been always good with videogames idk how it feels to be bad at this

[upper karma]

If anything you're the one annoying for pointing it out for no reason other than personal things.

@foggy oxide I agree it's weird

[foggy oxide]

like, i've born into this world playing videogames i remember playing super mario at age of 2

[upper karma]

You think that prevented you from approaching math?

[foggy oxide]

a little bit

[upper karma]

Or became an obstacle

[foggy oxide]

that became a obstacle at some part of my life

i think we should change to #discussion

[rustic spire]

Bruh where the géo questions at

[sacred vessel]

can someone help

i need to find height when i only know the perimeter

[rare talon]

What do you mean?

I don't think it's possible to find height ONLY given the perimeter (I'm assuming you're talking about triangles, but let's actually see your problem)

[sacred vessel]

here

i tryed using herems formula but i dont know if thats what i should do

@rare talon

[rare talon]

Let me think

The 1.3 dm is the shorter solid line right?

[sacred vessel]

yeah

i doubled it

so that i get the perimeter

when combined

[rare talon]

No

You don't get the perimeter

Also you didn't know what is the stripped line length?

[sacred vessel]

yeah its unclear

[rare talon]

I don't think it's possible tbh

[sacred vessel]

can you try to help with the same thing but other shape

?

[rare talon]

Let's see

Give the letter of the shape next time

So I can refer it better

I'm lazy

B(

[sacred vessel]

paralesomething

paralesomething

😄

i need to find h

[rare talon]

Well I don't actually mean the name of the shape

And I'm still thinking on the first problem dw

[sacred vessel]

i need to find h with the given information

[rare talon]

Anyway

Before we got to that

Is 3dm the height the of the triangle?

[sacred vessel]

yeah

[rare talon]

You didn't tell before!

I'll help you with that now

[sacred vessel]

sorry

😄

okay