geometry-and-trigonometry

[upper karma]

But now I think it’s 200pi but can’t remember if that was an answer.

[weak shoal]

Draw a picture of the scenario

[upper karma]

200pi seems too much.

[weak shoal]

And highlight the distance that needs to be calculated

[upper karma]

Ya I will once I get home sorry

Could not find anything online.

[weak shoal]

Sure

[upper karma]

Something kinda like that and get from A to B going all about the semicircles.

From A to B diameter is 100

[weak shoal]

Well okay, what's the radius of each semicircle?

Have you attempted to calculate that?

Also, draw a bigger and nicer diagram

That can make the difference

[upper karma]

That was all that was given.

I tried to do L = 180/360 * 2(pi)50

[weak shoal]

No, like I said, draw bigger and better picture first

[upper karma]

Ok

Is better?

[burnt plinth]

@upper karma that picture is awful but do u mean linesegment AB is 100 units long?

[upper karma]

Bruh.

Yes thought.

And we where suppose to find the path using the semi circles.

[burnt plinth]

ok well draw line segment AB

[upper karma]

I got 50 pi

[burnt plinth]

🤔 doesnt quite sound right to me

consider drawing line segment AB first of all

you know the 4 semicircles are congruent, so you really only need to find the length of traveling along one of them and multiply by 4 if that makes sense

https://media.discordapp.net/attachments/326138757474680852/657628854650929190/image0.jpg?width=507&height=676

ill redraw the problem in ms paint and u tell me if i drew it properly

[upper karma]

Does the location of the semi circles really matter? As long as they all touch like that but ok.

Well I did I tried to do L = 180/360 * 2(pi)50

And got 50pi but is that just one

[burnt plinth]

ok im struggling with paint

well firstly i would label the mid point between A and B

you want to find a relation between that line segment and the circles

well, you know you can find the length of the semicircle as 1/2 * the circumference, so that's 1/2 * 2 pi * r which is pi * r

so now you just need to relate the radius to the length of the segment AB

well, you know segment AB so if you label its midpoint C

consider the top left semi circle

you know AC is length 50

so how do you relate AC to some important property of the circle?

hint: what do you know about the triangles formed by the diameters of the circle and segment AC

looking at ur formula, it looks like you didnt get the radius of the circle correctly

[upper karma]

Oh ya I just used the whole diameter

So you need the radius of one of the semi circles?

Is the diameter of one of them 25?

[burnt plinth]

why do you say that?

@upper karma

[upper karma]

Because there are 4 semicircles and the total is 100. So 100/4?

[burnt plinth]

no you cant just divide like that because the diameters dont add up to 100

[upper karma]

Oh ok.

[burnt plinth]

you need to consider the triangles that are formed

there are very nice properties with right triangles you can use

draw and label some numbers on the triangle

[upper karma]

Ok

[upper karma]

so, I'm looking for the packing efficiency in 9-dimensions

I know I know its not proven yet

I don't need the EXACT perfect percentage

I'd like a close approximation

I was thinking of trying to do this the hard way, of tyring to squeeze spheres of the same radius into a finite volume randomly

What branch of mathematics does the topic of packing efficiency in higher dimensions fall under?

[summer spire]

Wikipedia lists it as discrete geometry

[tribal rose]

On Paul's Math notes, I dont see a Trig section, only a review in Calculus 1 ?

What should I be looking at? The algebra section doesnt have much trig

[devout harbor]

you can try examsolutions

[dark sparrow]

precalc?

[tribal rose]

I just finished algebra, taking trig in spring, then its calc 1 in fall 2020

Idk what I should be looking at. I dont think Paul covers much trig?

[tribal rose]

Yeah its precalc ?

[dark sparrow]

precalc should have trig

[upper karma]

How many lines of symmetry does a rectangle that is not a square have? https://i.imgur.com/ikH7pAM.png

[white cradle]

@upper karma how many do you think?

[upper karma]

@white cradle I don't know. I am just stupid.

[white cradle]

no, am genuinely asking you how many you think

that way we can fix whatever doubt you have

[upper karma]

Either 2, or 4.

Which one is correct, I don't know.

I can't reason it.

[white cradle]

why not

what's your logical reasoning to choose 2 or 4

[upper karma]

I don't even know. That's just what intuitively feels right.

[white cradle]

alright

here's an easy way

if the line is a line of symmetry

it divides the figure into two halves

and both halves would look identical

this is just another way of seeing that reflection results in the same figure

so now try to find out which of the four lines you've drawn are lines of symmetry

[dark sparrow]

it divides the figure into two halves
and both halves would look identical
not quite

[white cradle]

well ig identical is a bit vague

i honestly don't know how else to explain this

after all that we've already tried the other day

[upper karma]

So how many are there for a rectangle?

That is not a square?

[white cradle]

we can't just tell you the answer

what's the point if you won't be able to figure it on your own

what happens when you have to find for some other figure and we're not there to tell you

[upper karma]

Then tell me how to figure it out...

I'm going to make a drawing..

[white cradle]

I mean you still don't know what a line of symmetry is

how about watching some videos on it?

visualizing might help much better than explaining could

atleast for this case

[upper karma]

https://i.imgur.com/0VLWZOa.png

Is this a line of symmetry?

In my opinion, it is.

[white cradle]

is it?

[upper karma]

Why not?

[white cradle]

reason

[upper karma]

If you reflect it, the figure is the same.

Is it not?

[white cradle]

well, yes it is A rectangle still

but is it the SAME rectangle?

after reflection is it still oriented the same?

can you draw on the same image

the reflected rectangle

that might help us out

[upper karma]

No, can you?

If I could I would, but clearly I don't get it

[white cradle]

alright sure

ABCD is the original rectangle

BD is the reflection axis

DECF is roughly the reflected rectangle

[upper karma]

Wait, what just happened?

I'm reflecting the "right triangle"

But you kind of rotated it instead?

[white cradle]

you reflect every point on the rectangle

about that line

essentially ig you could see reflection as a 180 deg rotation

[upper karma]

But if you have a line going through the diagonal like that on a piece of paper. You could fold it in like the diagonal, and unfold it

It would be the same on both sides, no?

[white cradle]

would it perfectly overlap?

[upper karma]

I don't know

Let me get a piece of paper one sec..

I never tried this with paper

[white cradle]

yes try it

the perfect demonstration and you thought of it

👍

[dark sparrow]

essentially ig you could see reflection as a 180 deg rotation

no

[white cradle]

why not

[dark sparrow]

well

[white cradle]

atleast in 2D

[dark sparrow]

no

definitely not

[white cradle]

:(

[dark sparrow]

if you talk about 3D space

[white cradle]

I mean

yes 3D

[upper karma]

Wait, you are right...

[white cradle]

fuck

[dark sparrow]

then you could see it as rotating the entire plane 180° around that axis

[upper karma]

What is going on?

[dark sparrow]

but that's unnecessarily overcomplicated

[white cradle]

ig

that's what I meant

didn't phrase it right

[upper karma]

@white cradle In your picture, what is E and F supposed to be?

Like what were they before?

[white cradle]

they weren't there before the reflection

they're the new vertices formed after reflection

[upper karma]

Hmm, ok

You suggested a video

Do you have a video for this?

[white cradle]

well not particularly

[upper karma]

Damn I should've done this with a piece of paper earlier..

[white cradle]

mm hm

well atleast now you have some idea as to how it works

[upper karma]

Well ok, but how do I know for which shapes it works and where it works and when it doesn't?

Like when I draw the hypotenuse in that rectangle, both pieces look even

[white cradle]

yes but, say you reflect one of those right triangles about that diagonal

you wouldn't get the reflected image to superimpose on the other right triangle

like reflecting figures is easy

for each point you just draw a point on the other side of the line, at the same distance as the original point was from that line

do that for a few points of the figure and you'll have an idea whether it's symmetric or not

[upper karma]

I don't think it's working for me

One sec

@white cradle https://i.imgur.com/UE9vVFq.png

I'm messing it up..

How should I plot the points @white cradle

[white cradle]

you're reflecting on the wrong line

don't plot exactly

just rough

[upper karma]

Yeah how?

[white cradle]

draw a line from the point to the diagonal

[upper karma]

Middle of the diagonal?

Or where?

[white cradle]

extend that line to the other side till it's roughly twice the length it was

shortest distance @upper karma

drop a perpendicular

[upper karma]

From where?

[white cradle]

from whatever point you want to reflect on the diagonal

[upper karma]

Say I want to reflect (5,-3) in that drawing

Where do I draw the perpendicular to?

[white cradle]

draw a line from that point to the diagonal such that it is perpendicular to the diagonal

just roughly, you don't need the exact points

draw on paper by hand

[upper karma]

Ok

So I draw the perpendicular to the diagonal from the corner or some other point

And then I extend the perpendicular straight so it's twice the length of one perpendicular?

[white cradle]

yeah

and mark the end of it

that's the reflection of the point

[upper karma]

What's a reflection point sorry?

[white cradle]

??

I'm just telling you how to find the reflection of a point about some line

[upper karma]

Yeah ok

Ok

So I did that

[white cradle]

you can do it for any point

[upper karma]

And it's past the "other" side of the rectangle

[white cradle]

yup

[upper karma]

Does that mean it isn't a line of symmetry?

[white cradle]

yes!

[upper karma]

Nice

Ok, so how do I finish the reflection now? Like I have this: https://i.imgur.com/bkFNWtL.png

[white cradle]

on the other hand, if all of the points after reflection fall on the "other" side of the figure, it's a line of symm

find the reflection of the vertex of the other half rectangle also

[upper karma]

Do I keep drawing lines like this?

[white cradle]

well it's not the best way but it's the only one I've managed to make you understand

[upper karma]

https://i.imgur.com/F4zIBuK.png

(I know I could connect it earlier but I mean just to be "proper")

[white cradle]

oh ok

reflect the other half of the triangle too

[upper karma]

Do I have to?

[white cradle]

rectangle **

[upper karma]

Doesn't this guarantee it's not a line of symmetry

[white cradle]

yes it does

[upper karma]

Because the other side is past the left side of the rectangle

[white cradle]

yup

[upper karma]

Like the reflected side goes above

Ok thanks

You really helped me a lot. Holy.

[white cradle]

👍

[upper karma]

Just so you know, if someone else asks this and they don't instantly get it, tell them to get a piece of paper and do it. And afterwards you can tell them to reflect it for any figure, they can simply draw the perpendiculars and extend them onwards

I think it's a very good explanation

[white cradle]

mm hm it didn't strike to me earlier

but indeed, it's the best possible explanation

[upper karma]

Probably because it's obvious to you

[white cradle]

yeah..

[upper karma]

@white cradle It's because I am fucking stupid

My book says it as well

[white cradle]

Ohh lmao

might've skipped it when you read

meh happens to the best of us

[upper karma]

Mathematically speaking, the image when we reflect point A over line m is the point A' such that m is the perpendicular bisector of AA'. In other words, if we folded our paper along line m, A and A' would coincide.

A over line m is the point A' such that m is the perpendicular bisector of AA'

[white cradle]

yeas

[upper karma]

😦

To be honest I feel like a graphic would've been nice there

I read it but I think it didn't "absorb"

[weak shoal]

You're not 'fucking stupid'

Sometimes, it just takes time

[white cradle]

you were arrogant tho, ngl

[upper karma]

Nah, I was just frustrated.. Sorry..

[white cradle]

but that may have simply been the frustration

yeah it's all good

[upper karma]

Yeah, I understand it now. For example, https://i.imgur.com/oLeS5Wq.png

Right?

[white cradle]

yeah

[upper karma]

Nice

[tribal rose]

"How many lines of symmetry does a rectangle that is not a square have? https://i.imgur.com/ikH7pAM.png"

So back to this, I dont understand how to answer thequestion? I see how you drew up a reflection

The answer is 2 ?

What is this called? I want to look it up

[polar sable]

6

[dark sparrow]

2

this is called symmetry

geometric symmetry maybe

[tribal rose]

Thanks

[restive nymph]

so I need to find the remaining trigonometric ratios for this, I believe the hypotenuse is the square root of 10, and the sin and csc are both negative but what about the rest?

[thick bloom]

if you draw a triangle with angle beta that had 1 opposite beta and 3 adjacent to beta you should be able to get all the ratios just by looking at the triangle, using sohcahtoa, and using the CAST rule for the unit circle

[restive nymph]

this is what I think
sin is 3 divided by the square root of 10
cos is 1 divided by the square root of 10
tan is 1/3
cot is 3 of course
sec is the square root of 10
csc is the square root of 10 devided by 3

is that true?

I am not really sure how to know whether they are positive or negative is the problem

[silent plank]

unit circle.
you are told that pi < B < 2pi
cot B is also positive,
which quadrant is B in?

[restive nymph]

oh it's the 4th one!

thank you so much

because cot b would be negative if it was in the third quadrant

[silent plank]

uh, are you sure?

[restive nymph]

wait hold up

sin and cos both need to be negative

so it's the third one I think

[silent plank]

yes

[restive nymph]

since tan there is positive

alright, just messed up a little bit but yeah

so only tan and cot are positive and the rest is negative

[silent plank]

yeh

[tardy junco]

I didn't follow how they did this step.

(it's a trig simplification step).

What trig identitity did they use?

Please ping me if you answer (I'm going offline)

[weak shoal]

Urm there’s a general formula for these

Hold on let me see if I can find them

[tardy junco]

Hmm product of sin and sin?

[tawdry pivot]

$\sin a\sin b=\frac12(\cos(a-b)-\cos(a+b))$

[tardy junco]

Oh gosh you're right. That makes sense why there is a half

[somber coyote]

EpicGuy4227:

[tardy junco]

Thx epicguy

[weak shoal]

Some fuck shit like that

[tardy junco]

K thx both of you.

[weak shoal]

You can easily prove it from basic ideas but you never remember stuff like this lol

[tardy junco]

That's so true

[weak shoal]

I mean, it’s good to prove it once in your life i guess

[tawdry pivot]

the only ones I remember are sin(a+b), sin(2a), cos(a+b), cos(2a) cuz I use them the most. I derive all the other ones when needed

[weak shoal]

Yea

[gritty sail]

those other formulas are hard to remember

[weak shoal]

That's why you never remember them

[gritty sail]

just big brain it

[weak shoal]

You prove them at least once in your life and then you forget about them till you gotta derive them

[white cradle]

yeah

[silk crown]

lol I have like 50 formulas in trig memorised on a huge chart I have hanging around

[quiet mason]

wth

[weak shoal]

?????

What's the point of memorizing all of that?

[sly marlin]

Let me see how many trig formulas are reasonable to memorise (Not counting hyperbolic sine/cosine...)
sin^2+cos^2=1, and we can divide that by sin^2 and cos^2 to get 2 more identities (3)
sin(a+b), cos(a+b), tan(a+b) + changing to a-b + deriving csc, sec, cot from there (3 to 12)
double angle/half angle (4?)
product to sum / sum to product (5?)
R-formula (not really a formula, more of a concept)
De Moivre's (1?)
That's just 25 to 30.

[weak shoal]

just 25 to 30

[robust socket]

$y=\frac{(x^2+\pi^2)(1-\sqrt{\sin(x)-1}}{(x^2-\pi^2)(1+\sqrt(\cos(x)+1)}$

[somber coyote]

Umma.Gumma:

[robust socket]

I've got to find the domain of this stuff here

I would proceed with a check on the denominator(which looks rather daunting) and the square roots

is that correct or should I consider something else?

[weak shoal]

What do you think?

[robust socket]

well I' not sure that's why I'm asking

:D

tbh I sense there is something going on related to those two parentheses before the sq roots

[weak shoal]

Well, there are a bunch of square roots as well. Might wanna consider those

[robust socket]

yep, I considered those above, second sentence

[robust socket]

I'd appreciate if someone could evaluate that domain, I'd be interesting in see how you proceed

[idle bloom]

No one is gonna do it for you but people could help you

[umbral snow]

You really only need sin(a ± b) and cos(a ± b). If your teacher is really mean enough to force you to memorize tan or sec sum identities, you can get them from sin and cos

Double or half can be easily obtained by doing sin(x + x)

[robust socket]

Thanks Kaynex, I'm not absolutely sure what do you mean with a and b here

thing is, I'm not seeing which identities we are talking about here

[umbral snow]

a and b just being labels for variables. Could replace them with x and y

Oh sry I was talking about the earlier convo

Oh IC you want that domain

Anything under a square root has to be positive, and anything in the denominator can't be 0

[robust socket]

yeah that's pretty clear, the thing is, how do I check that denominator

that product is rather nasty

[weak shoal]

Determine all the points of discontinuity. You have x = pi, x = -pi and so on

[umbral snow]

You just have to check both factors:
x² - π² = 0
1 + √[cos(x) + 1] = 0

As if either of them are 0, the denominator is 0

[robust socket]

great, the second factor is an irrational eq but what about the first one?

[idle bloom]

difference of squares

$(a^2 - b^2) = (a - b)(a + b)$

[somber coyote]

Spamakin🎷:

[idle bloom]

that should be one of things you learn to recognize instantly

[robust socket]

right thank you

[sage lodge]

How do I convert between cartesian and spherical coordinates? I wrote these two short methods but they aren't functional https://cdn.discordapp.com/attachments/564666592282148864/657820049599496202/unknown.png (in my case xz plane is pan and y plane is tilt)

or rather I think cartesianToPolar works but not polarToCartesian

I can give some input/output examples

[upper karma]

@sage lodge spherical coords?

do you mean polar?

[sage lodge]

ye

3d polar coords

same thing

[upper karma]

ah, 3d polar

is the formula for cartesian -> spherical not $ r = \sqrt{x^2 + y^2 + z^2}, \phi = \arctan \frac{y}{x}, \theta = \arccos \frac{z}{r}$?

[somber coyote]

Sun:

[upper karma]

Mispoke ^, deleted.

oh, your polartocart doesn't work

not the other one

one sec

polar.mag = ?

I think you typod a cos instead of a sin for the x coord in polartocart

[sage lodge]

thanks ill try to make those changes

[quiet mason]

changs

[sage lodge]

[upper karma]

I think just on the line of double x = ....

i think that's the error

[sage lodge]

might be both lines. Just realized I'm taking the sin of the yangle. If the yangle is 0, it will give me 0 for my x and z components which doesn't make any sense.

[upper karma]

@sage lodge $x = r\sin \theta \cos \psi, y = r \sin \theta \sin \psi, z = r \cos \theta$

[somber coyote]

Sun:

[upper karma]

$\theta \in [0, \pi], \psi \in [0, 2\pi)$

[somber coyote]

Sun:

[upper karma]

r = radius, $\theta$ inclination, $\psi$ azimuth.

[somber coyote]

Sun:

[sage lodge]

Ok I tried that, and it seems like its working now except if i convert a vector to polar, and the polar vector right back to cartesian, the Z value will be negative (with some magnitude)

[upper karma]

@sage lodge make sure you have your variables correct

[upper karma]

Hey, does anyone know how to do proofs involving parallel lines?

[dark sparrow]

i'm gonna say yes even though that's quite a vague criterion

[upper karma]

no need to worry, they're just doing a survey

[onyx basin]

why does sin(-x) flip sin(x), but cos(-x) is the same as cos(x)

[weary drift]

You can try showing this yourself by rewriting sin & cos using complex exponentials

[dark sparrow]

lmao

do you really think someone going in this channel even knows what a complex number is

[weary drift]

Worth a try imo

[onyx basin]

not gonna lie i dont know what a complex exponential is

i know what a complex number is but not a complex exponential

anyways there's gotta be some logic behind it tho right

[dark sparrow]

ever seen the unit circle?

[onyx basin]

yep

[elder surge]

yep

[dark sparrow]

yeah ok so like

[elder surge]

If you're stumped with what complex exponentials are, go watch 3Blue1Brown's video about e^i*pi

[dark sparrow]

going from θ to -θ is flipping across the x-axis

[onyx basin]

for sin yea

[dark sparrow]

no i mean like

[onyx basin]

o my bad

[dark sparrow]

every angle corresponds to a point on the circle itself

[onyx basin]

yeah

[onyx basin]

ok so ive been thinking

cos0 = 1

is that why

because you can't really do cos0 = -1

since cos0 will only have one answer

[dark sparrow]

it'd be a bit obtuse to call that alone the reason

[onyx basin]

yea

but i can't really think of another reason

so im just gonna stick with that one for now to satisfy me

though there's gotta be some deeper logic behind it

[elder surge]

why would you make cos(0) be equal to -1?

cos(x) is the distance between a point on the unit circle with an angle of x radians to the point [0,0] and the y axis. so cos(0) is 1.

[upper karma]

,w calculate cos(0)

[somber coyote]

Display results online and refine query

Requested by Sun#3606

[sage lodge]

@upper karma i tried using a different formula. now the xz is proper sign but the y is negated ;/

[upper karma]

@sage lodge I don't know C#(?)

Put it in pseudocode and I'll help.

[sage lodge]

ok

for Cartesian to Polar I'm using $ r = \sqrt{x^2 + y^2 + z^2}, \phi = \arctan \frac{z}{x}, \theta = \arctan \frac{sqrt{x^2 + z^2}}{y}$?

[somber coyote]

SpeedDart:

[sage lodge]

oh well the sqrt didnt work

In the system I'm using the z and the y is flip flopped. This may be the source of the confusion I'm having but unfortunately a requirement for what I'm working on

[upper karma]

that's not pseduocode 😅

why not $\theta = \arccos \frac{z}{r}$?

[somber coyote]

Sun:

[sage lodge]

i'm pretty sure it's a math issue and not a logic issue so i thought i might as well just put it as a formula ;p

[upper karma]

i'm assuming radius r, inclination θ, azimuth φ

[sage lodge]

previously had that. if i use that, the z and x values will be negated instead

[upper karma]

ok, so

using your formulae

if you can take an input

convert it to spherical

then convert it back to cartesian

then your formulae are correct

and it's logical

so try that by hand, if it doesn't work, type it up on pseudocode. you'll probably find your error while typing it

[sage lodge]

i'lll try it by hand then

[upper karma]

make sure you're using the right tan function

[sage lodge]

i get the same issue by hand. x is negated (although only the x value rather than x and z)

[upper karma]

then fix ur formulae

I've given you correct ones (from wikipedia, so at least 50% correct)

[upper karma]

Hello. I stumbled upon this geometric problem. I've resolved it, but by employing a rather complex approach. Apparently there's an easier method to tackle it. Could anybody give me a hand?

Find the relationship between the square's perimeter and the circle's circumference.
The leftmost intersection cuts the circle in half.

[undone shuttle]

There are no information about right intersection points?

Information about anything related to the circle

[silent plank]

can you tell us your method first?
@upper karma

[rancid verge]

can someone explain this please

[upper karma]

@silent plank

@undone shuttle No. The right points are randomly placed.

[silent plank]

7-8 lines of work isn't too bad

[clear haven]

@rancid verge are u familiar with transformations of sine/ cosine function

[rancid verge]

sorta

[upper karma]

@silent plank Yep, but I feel there's an approach that doesn't involve the law of sines or even the Pythagoras theorem.

[silent plank]

apply sin to the arccos directly instead of evaluating the angle

hmmm

[clear haven]

what do you know so far then @rancid verge

[rancid verge]

know how to get the amplitude

[clear haven]

cool

[rancid verge]

that doing cos (2pi) makes more cycles

[clear haven]

yeah that's ur frequency thing

[rancid verge]

but the question has time on the x-axis

[clear haven]

yes

[rancid verge]

and I'm used to pi on the x-axis

[clear haven]

wdym

[rancid verge]

well after 2 pi we have a cyle

cycle

for a sin or cos graph

and thats usually on the x-axis

[clear haven]

yeah

i mean here it is also on the x axis ?

[rancid verge]

no

it's time

on the x-axis

[clear haven]

i mean

same difference though right

as time progresses

[rancid verge]

hmm i guess

[clear haven]

time can take on the value 2pi and cause a cosine or sine to make a full period

just imagine time as just

the independent variable

just like "x" is

[rancid verge]

ok

[clear haven]

"x" and "t" are just

names

for that sort of thing

anyway

so from wat u know so far

[silent plank]

@upper karma
did someone actually confirm there was an easier way?
also your first line should be b=
and you should use side length s instead of 1 to be more rigourous.

[clear haven]

what's the amplitude of the graph

[rancid verge]

3

[clear haven]

good

then what is the frequency

[rancid verge]

and then shouldnt the period be 0.7?

approx

time for one cycle

[clear haven]

something like that

i cant tell for sure

but

uh

[rancid verge]

so the frequency is 1.5

[clear haven]

hmm

sounds right

but from the answer choices something is weird

[rancid verge]

so answer A

[clear haven]

lemme check

hmmm

i dont think so

[rancid verge]

so B?

[clear haven]

probably

but im being super dumb rn

and cant comput

gimme a sec

[lusty reef]

Yea, it's B

[clear haven]

yus

lol

crippling self doubt

is bad

[rancid verge]

because there is 3 cycles in 2sec?

is that a way to find the answer

[clear haven]

sorta

to get the frequency

you take 2pi / the coefficient of the x thing

well first really ya gotta do a bit of factoring

since the answer choices u have are not in the form of Acos(B(x-C))

[rancid verge]

what is the coefficient of the x thing

[clear haven]

B

[upper karma]

I came up with an easier solution, which does not involve angles.

[silent plank]

I just realised that too

[upper karma]

What does it mean by this?

7:5 ratio applied twice

btw this is the problem

[weak shoal]

7/5 = 35/25

[upper karma]

why is that cb:cf

why not ac:cb

[onyx basin]

@elder surge little late, but i didn't say cos0 = -1, i was saying that cos0 cannot equal -1

[clear haven]

https://cdn.discordapp.com/attachments/488104216678760469/658859020521832449/229743249877762061.png

i dunno what crazy ass identity i need to figure out how to get this to happen

they are equal when i desmos both

so that's ok

but

[weak shoal]

???

[clear haven]

beyond just making the 1-cos(theta) a 2sin^2(theta/2) i dunno wat else i can do

yeah i know right

[weak shoal]

Okay here's thing

Have you tried simplifying the denominator?

$\cos((n+1)x) - \cos(nx) = \cos(nx +x) - \cos(nx) = \cos(nx)\cos(x) - \sin(nx)\sin(x) - \cos(nx) = \cos(nx)( \cos(x) -1) -\sin(nx)\sin(x)$

Then, start simplifying

Wtf

[clear haven]

oh right i can do stuff with the n+1 stuff

[somber coyote]

Abhijeet Vats:

[weak shoal]

DUDE, WHY AM I JUST SUCH AN IDIOT WITH LATEX

Jesus

[clear haven]

it ok

e

i got the gist of wat u want me to do lol

[weak shoal]

Okay yea, then start simplifying using half angle identities

Same thing with the denominator

[clear haven]

ty

[weak shoal]

I am just an idiot with this man

Oof

[golden sail]

Is this correct

Why my answer diffrent from the book

[weak shoal]

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

What are you doing

[golden sail]

Intersection @weak shoal

Polar graph

[weak shoal]

Have there been any conditions placed on what theta can be

[golden sail]

No

[weak shoal]

Show the exact problem

[golden sail]

Question 9b

[weak shoal]

Oof i don't see any restrictions on theta.

[golden sail]

But is my answer correct??

Becoz its so far different from the answer from book

[devout shell]

Convert degrees to radians

You got the (3/2, π/3)

And then 300 degrees is the same as -60 degrees

[golden sail]

Where 3/2 from

[devout shell]

1.5 = 3/2

[golden sail]

Ok

[devout shell]

You have it correct

[golden sail]

J see

[weak shoal]

Well, 300 degrees is not the same as -60 degrees but they do give the same value when used as inputs in cos(theta)

[devout shell]

Should have said coterminal then

[golden sail]

Thanks guys @weak shoal @devout shell

[devout shell]

Nice pusheen cat

[silk crown]

I think the angle in [0, 2π] is what is asked when no range is there for the angle

HS textbooks aren't always very explicit lol

[dark sparrow]

this is polar

[silk crown]

Isn't the problem just to find a point

oh wait

nvm

still the "most likely" way is probably the way

for like most problems

[copper moss]

Question: I know how to calculate a scalar product, but I'm confused as to what it represents. So the product of scalars u and v would be the image below, but what does that mean?

[eternal crag]

the dot product of u and v is the product of their length times the cosine of angle between then

[copper moss]

Yes I know, but what is the dot product for? is it for solving for the angle?

[eternal crag]

well, for one, dot product tells you if u and v are perpendicular to each other

[copper moss]

if uxv is 0

[eternal crag]

cross product is a different thing

if u X v = 0 then u and v are parallel

and yes you can solve for the angle if you want from the formula

[copper moss]

oh ok

Well tyvm 🙂

[eternal crag]

glad i can help

[swift moth]

is a=a

maybe I should've chosen a different name for the other angle

[weak shoal]

What

Lmao

[swift moth]

I know they're equal, I just wanna understand the logic behind it

[weak shoal]

Well, what do you think?

Run me through why they are equal

Your arguments, that is

[swift moth]

hold on

there is another angle a between the blue line and the red line

[weak shoal]

Begin by labelling all the points where two lines intersect in your diagram

Then, use that as a way of referring to your angles

It makes it easy

[swift moth]

like so?

I'm sorry for the shitty caligraphy

[weak shoal]

Okay what the fuck is that thing in the middle lol

[swift moth]

wait, I know why they are equal now

a P

an attempt at a P

then you can use the fact that when two lines intersect opposite angles are equal

so there is another angle a

[weak shoal]

Well yes, CAB = ACD

Mmh go on

[swift moth]

and then, if you have a line between connecting two parallel lines you know the alternate angles will be equal

hence a=a=a=a

[weak shoal]

What

Use the capital letters to describe your angles lmao

[swift moth]

I guess it'll be clearer if I label my angles properly

or letters

CAB = PCD = BQC = PBQ

Q is the point where the dashed red line and the blue line intersect

[weak shoal]

Do you know, for certain, that lines DB and AC intersect at right angles?

[swift moth]

yes

[weak shoal]

Aight then?

[civic nexus]

Hey guys, how do I tell if a negative sin function is in quadrant 3 or 4?

a part of my question has a function sin inverse (-12/13) and im not sure if it is in quadrant 3 or 4

[silent plank]

specifically, what is the question asking for?

[civic nexus]

is it better if i type it out or take a picture?

[silent plank]

picture

[civic nexus]

oops thats a shitty picture

but its the one above "Show that tan"

basically i needed to let both functions be A and B then expand tan(a+B)

but when im doing the special triangle for sin inverse (-12/13) im not sure if the tan value should be positive or negative 12/5

[silent plank]

do you have a better pic?

[civic nexus]

its part G

the question is "Find the exact value of:"

[silent plank]

i derped for a sec,
recall the range of arccos and arcsin

[civic nexus]

does arccos mean cos^-1?

[silent plank]

yes

[civic nexus]

-pi/2 to pi/2

[silent plank]

for arcsin, yes

[civic nexus]

cos is 0 to pi

[silent plank]

which means if the ratio of sin is negative, then the angle from arcsin is in quadrant:

[civic nexus]

._.

sorry but i cant see the connection

[silent plank]

-pi/2 to pi/2 cover which quadrants?

[civic nexus]

1 and 3

[silent plank]

no

[civic nexus]

oh wait its 1 and 4

[silent plank]

yes

[civic nexus]

so the range of arcsin covers quadrant 1 and 4, therfore when sin is negative it is in quad 4?

[silent plank]

yes

well the result of the arcsin will be in quad 4 to be specific (if the ratio from sin was neg)

[civic nexus]

wouldnt it be in quad 1 if it is positive?

[silent plank]

yeh, i mean for this question

[civic nexus]

oh ok, thanks a lot!! ill think about the theory by myself

[civic nexus]

Can someone help me with question 9

I'm not sure how to work with 3 tan functions and how to prove something is pi/2 since there is no solution for tan(pi/2)

[weak shoal]

Well, i suggest taking the tangent of the left-hand side

Then, use identities

But before that, first let:

$\alpha = \tan^{-1} (\frac{1}{2})$

$\beta = \tan^{-1} (\frac{2}{5})$

$\gamma = \tan^{-1} (\frac{8}{9})$

[somber coyote]

Abhijeet Vats:

[weak shoal]

Then, work in terms of the variables & get a formula for the tangent of the left-hand side

[civic nexus]

Yea thats what I thought as well, but I'm not sure how to form a formula with 3 variables

and also if I form a tangent with the left side, i would get tan (pi/2) for the right which is unsolvable

[silent plank]

argh

[weak shoal]

Haha lmao, get some practise with it

[civic nexus]

Im not sure how to solve the part with pi/2

[silent plank]

you can probs apply a different trig function

[weak shoal]

$\tan(\alpha + \beta) = \frac{\tan(\alpha) + \tan(\beta)}{1-\tan(\alpha)\tan(\beta)}$

[somber coyote]

Abhijeet Vats:

[weak shoal]

^Start with that sort of thing. Then, go from there.

[civic nexus]

Yea but i get stuck when i get to tan (pi/2)

[silent plank]

try applying sin or cos instead of tan

[civic nexus]

alright ill try that thanks

i solved it with sin(a+b)=sin(c-d) thanks!

[weak shoal]

Noice

[silent plank]

you may also need to briefly explain how
arctan(1/2) + arctan(2/5) + arctan(8/9) is within (0, pi)

[civic nexus]

i dont understand that part, i assume it has something to do with the range of arccos?

[silent plank]

yeh, eg.
sin(a) = sin(pi/2) doesn't necessarily mean that a = pi/2

arctan could actually be used here.
since (arctan(1/2) + arctan(2/5)) is in [0,pi/2)
(arctan(1/2) + arctan(2/5)) = arctan ( tan (arctan(1/2) + arctan(2/5)) = arctan(9/8)

then your LHS becomes
arctan(9/8) + arctan(8/9)
which is pi/2 from the properties of a right triangle

[civic nexus]

Woah I need a bit of time to digest that lol

I think I'll leave that for the future haha, probably going out of my syllabus rn

[silent plank]

was there any part of that you didn't understand?

[civic nexus]

Yea, way too many LOL

4 parts to be exact

but i dont wanna trouble you with that, I think im good for now

thanks a lot!

[sterile hinge]

How do we find the equation of a parabola?

Using 4PY?

[weak shoal]

What

[sterile hinge]

exactly I dont get it either

[weak shoal]

So how can we answer the question if we don’t know what any of that is

Perhaps google?

[sterile hinge]

alright man sorry to bother you

[weak shoal]

Nah you didn’t

[sterile hinge]

🙂

[fringe crater]

I think you're talking about the focus of a parabola

[weak shoal]

Of course

4py = x^2

It’s, however, better for him to do some research on his own first and figure things out before asking questions here

[upper karma]

Can someone help be with this cosine function transformation question, ive never heard of the letter terminology. A cosine function has been vertically stretched by a dilation factor of 45, reflected by the x-axis, horizontally stretched by a dilation factor of 10 and translated up by 33 units. Determine the values of the parameters a, k, d, and c

[rich wolf]

@upper karma what is the parent function

[upper karma]

f(x)=cosx

[rich wolf]

Now what do you do to that function to carry out those transformations

[upper karma]

it should be f(x)=acos(bx+c)+d

[late frost]

can someone help me find length of blue line in terms of radius R

[rich wolf]

r*sqrt(2)

@late frost

[upper karma]

help find != "give answer"

[rich wolf]

What is the factorial of "help find" and why is it equal to "give answer"

[weak shoal]

^I think he was joking lmao

[white cradle]

lmao

[weak shoal]

Anyways, there's a solution by trigonometry. You can find the length of half the blue line:

$L = R\sin(45) = \frac{R}{\sqrt{2}}$

You need twice the length of that, so we have:

$2L = \frac{2R}{\sqrt{2}} = \sqrt{2} R$

[somber coyote]

Abhijeet Vats:

[white cradle]

ew

why trig when pythagoras just fine

[weak shoal]

Because the angles have been given explicitly.

[white cradle]

use trig for unequal angles / non 90° sum

[weak shoal]

Don't tell me what to do, you're not my mom

❤️

[white cradle]

yes i am nd u better listen to me >:(

[rich wolf]

I just realized that trigonometry is trigon-metry which means the measurement of 3 sided polygons

I thought it was just one of those random words with no meaning

Like algebra

Idk that was kind of random but it literally just made sense to me

[upper karma]

actually I think the etymology is "corner angle measurement"

[rich wolf]

Trigon

Thats just fancy talk for triangle

[upper karma]

yknow what, actually, makes a lot of sense. I considered breaking it down into tri - -gon, .eg.: corner angle

wait

huh

wow, sin/cosin also have cool etymologies

very neat

[rich wolf]

@upper karma what is it

[upper karma]

sine alledgedly comes partially from the word "bowstring", tangent from "to touch", and secant from "to cut"

[rich wolf]

So cosine is cobowstring

[upper karma]

"complementi sinus", "the sine of the complement"

[rich wolf]

I dont get why sine is bowstring

Cuz sine wave looks like string ig?

[weak shoal]

No

Look very carefully at the circle

If you pay attention, you can see that the arc subtending the blue line, as well as the two radii, look like a bow. Then, the blue line looks like a bowstring

[civic nexus]

Can someone explain to me why the domain of arccos x^2 is -1<x<1?

[weak shoal]

Well, what do you think?

[civic nexus]

i know the domain of arccos is -1<= x <= 1

but i cannot square root -1

so somehow that changes <= to <

but i dont really know why

[weak shoal]

There are no square roots here, though

[civic nexus]

i tried working from -1 <= x^2 <= 1

so in order to get x i square root both sides?

[weak shoal]

[civic nexus]

gimme a sec ill write out my workings on paper

[weak shoal]

Well, okay, here's the thing:

-1 =< x^2 =< 1.

However, x^2 >= 0 for all real x. So, we have:

x^2 =< 1 => x^2-1 =< 0 => (x-1)(x+1) =< 0 => -1 =< x =< 1.

So, actually, the domain is just any value of x from -1 to 1.

[civic nexus]

woahhh whats that LOL

ill try to visualize that on paper

[weak shoal]

Yea

You can draw the graph out as well. If you don't want to do that, I guess you can use something called desmos

I've not used it but i've heard it's decent

[civic nexus]

oh never mind, the domain couldnt be 1 or -1 because the denominator was square root (1-x^4)

i was confused why it couldn't be 1 or -1

thanks!

[weak shoal]

What

[thorn talon]

🤔

[weak shoal]

Re-draw your diagram and make it bigger

Start labelling stuff

Okay

What have you tried?

Well, what did you try that was relevant to the problem?

Well, there are right triangles all over the place

What can you do with them?

[civic nexus]

Can someone help me with this

The answer is (1, pi/3 ) , (-1 , 2pi/3)

i dont know why is y = pi/3 instead of square root 3

and also why does the triangle start from x=0?

[distant sonnet]

Hm

Idek use calculus

[civic nexus]

wut

[distant sonnet]

Yes

What triangle?

[civic nexus]

the normal of the curve forms an angle with the x axis

so i formed a triangle with the axis and the curve

the angle is given in the question, which is pi/3

[distant sonnet]

Yeah and

It’s symmetrical so there is another point

Also you don’t know it’s square root of 3

Could be any multiple of it

And we don’t know that the triangle starts at x=0

Just solve square root of 4-x^2=cos pi/3

I mean tan

Then u get -1 and 1

Plug

Done

[civic nexus]

how did you get square root of 4-x^2=tan pi/3?

[distant sonnet]

Derivative of arccos(x/2)

I can’t see any way else

[civic nexus]

what does finding the derivative tell us tho

the gradient?

and why is the derivative = tan pi/3?

[distant sonnet]

Have you learnt calculus

[civic nexus]

yea

[distant sonnet]

Does this question require the use of it

[civic nexus]

i guess? not very sure

[distant sonnet]

I’m

Um

Yes

Tan(pi/3) is just the gradient of the normal

We know the gradient of the tangent

Then use the fact that their product is -1

[tardy junco]

Is this channel occupied?

[distant sonnet]

No

[tardy junco]

Okay then can it be proved that a = b? If so, then how to prove that?

[dark sparrow]

well, can it?

[tardy junco]

I don't know

😛

[dark sparrow]

what do you think

[tardy junco]

It just feels like a = b. But idk

[dark sparrow]

well then try to prove it

[tardy junco]

Can you give me a hint?

[distant sonnet]

Um

[dark sparrow]

what can you say about triangle OAB?

(that was the hint)

[tardy junco]

Two of its sides are equal

[dark sparrow]

pan don't spoil

[distant sonnet]

It’s quite obvious tbh

[dark sparrow]

yes. @tardy junco

and what do we call such triangles?

[tardy junco]

Isosceles

[dark sparrow]

indeed

and what can you tell me about isosceles triangles?

[tardy junco]

That two angles must be equal because the length of a triangle is proportional to its opposite angle?

The second statement is right?

That the length of a side is proportional to its opposite angle?

[distant sonnet]

No

[tardy junco]

Gosh I've forgotten everything in geometry. Then what is the relation between a side and its opposite angle?

[dark sparrow]

there's a thing called the law of sines but it's a bit of a nuke here

[tardy junco]

I've heard it before

[dark sparrow]

can you prove that in an isosceles triangle, the two angles at the base must be the same?

[tardy junco]

Hmmm

[dark sparrow]

without law of sines.

you can do it on your triangle, OAB. you know OA = OB.

[tardy junco]

Do I need to draw an extra line

And prove congruence

[dark sparrow]

you don't NEED to

you never NEED to do anything

but if you think it will help then go for it.

[tardy junco]

So there is more than one way to do it without law of sines?

[distant sonnet]

You can

Draw something

[tardy junco]

If you draw a line from vertex O that is perpendicular to the base. Is the base split exactly on half?

[dark sparrow]

i don't know, is it?

[tardy junco]

Hmm

[dark sparrow]

maybe it is, maybe it's not; maybe you can prove it either way

[tardy junco]

How to prove that it splits the base in half?

[dark sparrow]

or maybe it will be more straightforward to draw another line entirely! who knows

[tardy junco]

If it splits the base in half then we get two congruent triangles so then those angles would be equal.

But does it split the bade in half? Idk how to prove that or know when a line splits the base in half

It's called an altitude right?

[dark sparrow]

the line you suggested is an altitude.

but if you want the base split in half, then why not draw the line that does just that, i.e. the median?

[tardy junco]

Oh wait it doesn't need to be an altitude does it

It just needs to bisect the angle at O

I need to go offline now.

Thanks a lot Ann and Pan Pan for helping me. 😁

[late frost]

How to find length of blue line

PLEASE someone answer so I can go back to sleep

[white cradle]

@late frost can you label all the points

[late frost]

Ok

lol

[white cradle]

you missed a point

[late frost]

ik but it wont reupload

call it W

[white cradle]

kk

now

IG=EG-EI

and HF=KJ=IG (since all redangles are equal)

and you can find EG and EI by rsin(angle) in the respective triangles

[late frost]

oh shit

wait doesnt r sin(angle) give the opposite

[white cradle]

my bad

rcos

[late frost]

So I can get EG but how do I get EI? idk what EH is

[white cradle]

you know ED though

[late frost]

oh ok ok ok

[white cradle]

and that bigger angle

yeah

[late frost]

thx sir

[white cradle]

np

[late frost]

@white cradle So is it R cos (22.5 deg) - R cos (45 deg)?

[white cradle]

yes

[late frost]

yeet

Does that equal anything or do I need to write out each cos()

nVM I got oit

[white cradle]

aight

[late frost]

If I were to take the area of the rectangle that the blue line is in, would it be (R cos(22.5) - R cos(45)) * 2Rsin(22.5)?

rectangle HFJK

[white cradle]

@late frost yeah

[late frost]

My message was deleted, was it against the rules

[weak shoal]

Which message?

[late frost]

I asked something about the area of rectangle HFKJ above

[weak shoal]

It's still there

It certainly wasn't deleted

[late frost]

Oh, I can't see it

And I was pinged here

Weird

[weak shoal]

Strange, indeed

[civic nexus]

how do i find the range of y = x * arcsin x?

[dire rampart]

whatd u try

[civic nexus]

ive no idea, normally it is a number, but this time its x

logically speaking it is x* -pi/2 <= x <= x* pi/2?

[dire rampart]

sure

but theres only certain values that x can take

since ur domain is restricted by arcsin

[dark sparrow]

logically speaking it is x* -pi/2 <= x <= x* pi/2?
no

that makes no sense

[dire rampart]

ehh I guess it was their of trying to think of it

[white cradle]

@late frost

your question is still here though

[late frost]

dang

I couldn't see your answer either

[civic nexus]

so how do i solve it then? i'm not sure how to find range with x

[late frost]

I can see it now

[late frost]

same thing has happened

I cant see any of messages where we talked about me not seeing the message

¯_(ツ)_/¯

[late frost]

@white cradle But anyways, I was wondering if we infinitely sum up the areas of all of the rectangles in that direction, can we write that as the infinite series
sigma from n=1 to infinity of (2R * sin(90/2n)) * [R cos (90/2n) - R cos (90/n)]

Like if we continue these blue rectangles infinitely by halving the angle each time, is the infinite series for the total area I have above correct?

[white cradle]

@late frost yeah that seems right

but why?

what does that area give you

[late frost]

idk

let me find out

[white cradle]

aight

oh wait

I think it should be 2^n instead of 2n @late frost

[late frost]

o yea

[white cradle]

and 2^(n-1) instead of n

[late frost]

i only looked at first two terms

[white cradle]

oh lol

[late frost]

The rectangles get a little small for me to do it any fuurther

Sigma n=1 to inf of 2R sin(90/2^n) * (R cos(90/2^n) - R cos (2^(n-1))

Is that correct?

@white cradle

[white cradle]

yes

roughly converges to 2R^2 × 0.59567

[late frost]

Cool, I'm on my way to symbolab

Oh ok

[white cradle]

i ran it on desmos lol

but what now

[late frost]

OK

So now

I want to put squares in those curved cracks between the rectangles

[white cradle]

oh no lmao

[late frost]

Lol what

[white cradle]

yeah you're on your own now

[late frost]

I should probably study some stuff about arcs

Because I don't know anything about them

@white cradle This cant be that hard tho, right?