geometry-and-trigonometry

[vital frost]

Im having trouble with a, b, and c

I just really do not know what to do

[thorn talon]

well, for a: how can you show similarity?

[vital frost]

Idk AA~?

[thorn talon]

yeah

so how can you apply that?

[vital frost]

That's the problem

Idk

[thorn talon]

well triangles ABC and ACD both have a right angle right?

[waxen gorge]

AAA

angle chase

Set one to 90-x

One to 90

One to x

[vital frost]

What

ABC and ACD both have a right angle

[waxen gorge]

Ya

they're both 90

The other are x and 90-x

[vital frost]

So how do I make this as a proof

[jaunty plume]

Start with $\angle{ACB}=\angle{CDB}=\angle{ADC}=90^{\text{∘}}$

Idk how to get degree symbol

[somber coyote]

Gamedolf:

[jaunty plume]

Then let $\angle{ACD}=x$

[somber coyote]

Gamedolf:

[jaunty plume]

$\implies\angle{DCB}=90-x$

[somber coyote]

Gamedolf:

[jaunty plume]

$\implies\angle{DBC}=x$

[somber coyote]

Gamedolf:

[jaunty plume]

and so on

@vital frost

[vital frost]

What

What is this for @jaunty plume

[jaunty plume]

a

it's just showing that for each one of the triangles have the same angles

namely 90, x, and 90-x

[vital frost]

What is the question asking me to do

I have to evaluate the claims

How do I evaluate them.

What do I write

[jaunty plume]

the claim is that they are similar triangles

similar triangles share the same angles

so showing they have the same angles proves the claim

[vital frost]

Is that all I have to say

[jaunty plume]

yes

[vital frost]

But it already says that in a

Do o say they are similar triangles because of AA~

[jaunty plume]

the whole sentence after a is the claim

no it doesn't

"The altitude to the hypotenuse forms similar triangles"

[vital frost]

For B? And C?

[jaunty plume]

well b is false

you can show that by the same method as a

hmm

i'm not sure actually

idk what it means by pairs of proportionate sides but it asks if triangle ACE similar to triangle BCE

which is false because if you consider angle ACE = angle ECB = 45°

and let angle CEA=x

then CEB=180-x

But angle CAE=x-45

[glad ocean]

how can i memorize the pythagorean trig identities?

quickly

[umbral snow]

sin²(x) + cos²(x) = 1
Unless you're an absolute derp, that one should have no problem sticking

You can get the other two by dividing both sides by sin²(x) or cos²(x)

[glad ocean]

how did you get the exponential notation thing

by sin^2(x) you mean (sin(x))^2 squared right

sorry im new to trig and i just want to clarify because i dont have enough brain cells

anyways by pythagorean trigonometric identities i mean:
ratio of sides, calculated by trig ratios, in a right triangle

i should have clarified better i probably dont know the names very well

[thorn talon]

hmm

[glad ocean]

i mean like, for example, the ratio of sides in a 30 60 90 triangle and 45 45 90 triangle

[clear haven]

yeah sin^2(x) is (sinx)^2

notation tings

[thorn talon]

those special triangles hmm

[glad ocean]

yeah thats what i meant

[thorn talon]

you can always kind of derive them from scratch

[clear haven]

yeah

[thorn talon]

45 45 90 i believe

is a square cut along the diagonal

[clear haven]

mhm

then 30 60 90 is an isoceles cut in half down the centre bit

[glad ocean]

lets say the two legs have length x

[thorn talon]

hmm

equilateral specifically isn't it?

[glad ocean]

because opposite angles of an isosceles triangle's legs are congruent

[clear haven]

wait is it equilateral

yeah

[glad ocean]

no a 45-45-90 triangle

so according to the pythagorean theorem,

x^2 + x^2 (when i plugged x in to the leg values) = 2x^2

= 2*(x^2)

so square root that

[clear haven]

mhm

[glad ocean]

and you get sqrt(2)*x

[clear haven]

yeah

[glad ocean]

ok so x(sqrt(2)) is the hypotenuse's length where the legs of a 45-45-90 triangle have length x

[thorn talon]

yes

[glad ocean]

what about 30-60-90?

i know i can just google it but i like to interpret things before i jump straight into solving problems

[clear haven]

ok for that start with equilateral triangle

[glad ocean]

aight

[clear haven]

then slice in half along some altitude

[glad ocean]

ok lemme visualize this

[clear haven]

so you should have the new hypotenuse be x and the smaller leg be x/2

[glad ocean]

actually i'll just sketch it

[clear haven]

yeah gud

[glad ocean]

ok so the altitude's length is (x(sqrt(3)))/2

sorry it took me so long my pc froze lol

[clear haven]

na it ok

yeah i think that's right

wait

why /2

let me do a think

oh yeah cos we said it was x for the length not 2x

yeah yeah you're fine

this is good

ignore everythign i said i forgot what we said the hypotenuse was just 1x

[clear haven]

:(( i know sa dont like geometry

[waxen gorge]

Ur mom is geometry

[clear haven]

my mom is actually kowalski analysis

smh

[waxen gorge]

Oh

[upper karma]

damn

what a burn

[acoustic fiber]

Hi, can two line segments intersect at more than one point?.. what if we put one on top of the other? (just a beginner)

[hot scaffold]

At least not on a flat plane, I think.

[acoustic fiber]

ah okay, one more question.. rays coming a torch light is parallel or not? i mean if we extend it backwards, they will intersect, right?

[plucky marlin]

ummm

parallel lines dont intersect

thats why they are parallel

https://prnt.sc/m1lpem

that's a torch btw

[worthy igloo]

that's assuming it has a parabolic mirror

usually they do not

[plucky marlin]

[acoustic fiber]

thanks @plucky marlin

[hot scaffold]

I think you can use some sines to determine that CAB=CFG

And also CGF=CBA

@vital frost

[vital frost]

We haven't learned Sim yet

Sines

[hot scaffold]

...Ah.

[vital frost]

So does triangle cfg ~ cab

[hot scaffold]

It does, but I think you're meant to find how.

[vital frost]

I have to evaluate the claim

Aa~? Sas~?

[hot scaffold]

Ok I havent learnt that

[terse holly]

I’m pretty sure if you just prove all the angles are congruent you can prove the triangles are similar

[vital frost]

how

[terse holly]

Don’t quote me in this, but assuming FG and AB are parallel, you could use those parallel line rules. And all right angles are congruent

[hot scaffold]

I think you cant use that yet, actually.

[terse holly]

Can someone help me out with 215? I just need an accurate drawing

[hot scaffold]

Have at thee.

[glad ocean]

so much stuff to DOOOOO\

[hexed lintel]

Hey I am having trouble at this equation
cos(4x) = sin(2x + π/4)

Could someone help me out?

I think I need to do -cos(4x) so that there's a 0 at the end but I don't know where to go next

[waxen gorge]

@hexed lintel well the first step

Is getting them to the same trig function

So u want to use

sinx = cos(pi/2 - x)

so cos4x = sin(pi/2 - 4x)

Then u want to set

pi/2 - 4x = 2x + pi/4

so 6x = pi/4

or x= pi/24

Since sin's period is 2pi

pi/24 + 2(pi)n

and then check if it works

[hexed lintel]

Thank you very much!

[hexed lintel]

Sorry again for bothering but I have an inequation to solve.
It goes like this: cos x < rootof3/2

I did something like this: cos x < cos pi/6
x < pi/6

But i am not sure if i am correct

[serene field]

This?

[somber coyote]

Yue:

[hexed lintel]

Yes

[serene field]

Think about your unit circle

[hexed lintel]

Well if x is pi/6 then cos x is equal to rootof3/2

x must be equal to everything between pi/4 and pi/2 right?

Also I forgot to mention that x must be between 0 and pi/2

[umbral snow]

Where do you get π/4 from?

[waxen gorge]

ur statement means cosx's range is [-1, sqrt(3)/2)

look at the unit circle

And u'll see all the possible points

[split laurel]

something something graph it

[left folio]

Show that arccot(x) + arctan(x) = π/2?

[gritty siren]

are you familiar with cos(π/2-x)=sin(x) and sin(π/2-x)=cos(x) ?

[left folio]

Not really but keep going please

[gritty siren]

From that you can get tan(π/2-x)=cot(x) and cot(π/2-x)=tan(x)

[left folio]

I see

Thanks

[gritty siren]

You're welcome

[vital frost]

What..

[waxen gorge]

Similarity

[upper karma]

What does that mean

[vital frost]

@waxen gorge uhh help on my question

[wild hamlet]

2y-27=y+15

3x+4=5x-10

use a boat

or...

have the left dock be dockA

and the right one be dock B

extend the length of dock B upwards until it reachs a point where you can draw a straight line from it to dock A

make sure this line is perpendicular to the line that would connect docks A and B

[vital frost]

Huh

[wraith pumice]

@vital frost

[vital frost]

Yo

[wraith pumice]

pick a number(don't tell me): 1, 2, 3, or 4.

k?

[wild hamlet]

Sorry, Your mario is in another castle

[vital frost]

Ok

[wraith pumice]

3

[vital frost]

No

[wraith pumice]

2

[vital frost]

Yes

[wraith pumice]

thanks

[vital frost]

Yw

[wild hamlet]

,rotate

[empty moss]

the ratio when she was 6 months old is 9:42 (Factor)->3:14

I did it by (Head Diameter in inch)/(height in inches)

Use the ratio to find the remaining answers in parts a

Part B you require a different ratio, the one from when she is 20, which is found the same way

Part C you just make them fractions and find the difference and do part D on your own

@vital frost

You get it?

[vital frost]

Yes I do that k you

[empty moss]

👍

[upper karma]

<@&286206848099549185> I’ve waited more than 15 minutes so I’m reposting with a tag

[keen tangle]

do the green arcs mean that the angles are equal?

[limpid basin]

yeah

[wild hamlet]

Yo, I already gave you the answer

So scroll up

[waxen gorge]

@wild hamlet no u

[wild hamlet]

!15m

Nonouu

[glad ocean]

@stable horizon

to rotate a triangle:

first make sure you have all the points

https://schoolwires.henry.k12.ga.us/cms/lib08/GA01000549/Centricity/domain/2896/7th and 8th grade math/8th grade flexbook/unit 1 sections 1-23/1.7. Rules for Rotations.pdf

skip to page 2/13

btw

the rotations are counterclockwise by default\

so 90° clockwise = 270° counterclockwise

[stable horizon]

I can’t click that link, it says runtime error 😔✊🏽

[glad ocean]

mmmmm

it should be on google anyways dw

actually rotating something is easy

but combined with all the other transformations

[stable horizon]

I’m stupid when it comes to geometry ngl

[glad ocean]

its annoying af to memorize all of the functions

i have a 100 avg but most high school mah is just concrete thinking

math* not mah

[stable horizon]

I’m great at algebra

I just cannot understand the not algebra parts of geometry for the life of me

or apparently rotations

how do I know what 270 degrees counterclockwise looks like?

[umbral snow]

90 degrees → one quarter turn
180 degrees → two quarter turns
270 degrees → three quarter turns
360 degrees → full rotation

[stable horizon]

oh okay thanks

wait

so if I rotate it around a point

I’m so confused

will that not not necessarily be a point in the figure after I rotate it?

[empty moss]

The original point of rotation?

[stable horizon]

ik it might be a dumb asf question

[empty moss]

when you are rotating points, you have original figure

Say it is a triangle

Call it A

when you rotate the triangle, you changed it so they are not the exact same. This triangle that was the result of the rotation will be called A' (A prime)

Did that help or confuse more

[stable horizon]

I understand that part

kinda

[empty moss]

Ok so are you confused about the points defining the figure, or the point around which the figure is rotated?

[stable horizon]

don't mind all the tabs open but I'm trying to do this and I kind of understand more with all the help but I still don't understand what to do exactly

[empty moss]

Ok most of the time it just says "rotate around the origin"

or something right?

or just "rotate (blank figure) 90 degrees counter clockwise"

[stable horizon]

I’ve never done this before in my 15 years of living before rn so I don’t know

thank the American education system and my state for that

[empty moss]

You haven't rotated any figures yet

ok

Mind if I link you a page?

[stable horizon]

yeah go ahead

[elder oak]

Hi for some reason my brain is telling me this is wrong but it feels right; if you draw a line from a vertex of a scalene triangle to bisect the opposite side of that vertex, and multiplied the length of that line times half the length of the side bisected, that would give you the area of the triangle right?

[glad ocean]

draw it out

with the sides 3, 4, and 5

[chrome mica]

How can I calculate the blue zone area? It's the first time we do this kind of geometry problem, so I don't know.

[thorn talon]

Hmm

There will be a couple methods

I don't know the best

But I would fill in from the line CA

To get a full triangle

Find the area of that using standard ab/2

And then subtract the part we added in

[crisp hornet]

area of triangle using coordinates

[upper karma]

[warm pine]

I'm assuming it's a Rearranging version of the shoelace formula

[narrow sleet]

Yez

[rose monolith]

If i wnat to ask questions somethjing relating about speed, travel and distance and also bearing

do i ask here?

[narrow sleet]

Don't think so

[rose monolith]

where should i go then?

pls tag when you can answer

[jaunty plume]

just ask in one of the questions channels

[vital frost]

Didn't get any help with this one really

[clear haven]

let me do a thinking bc this one is actually kind of weird

since you're going to only be able to work on land here is what i suggest

2 points of both triangles must be on the docks

and then (kind of like drawing an ellipse with string and pencil) you can let the third points of each triangle vary freely

however

you want to set up a proportion with these similar triangles in order to do this

i suggest making one triangle have the length from dock to dock across the water (the unknown value) be the longer leg (or even "hypotenuse" if you can do that, it's kind of a cheap way out but i dont think you're allowed to tell if there are any right angles anyway so dont)

hmmmmm

mate if they allowed a protractor this would be easy

but only with similar triangles ??

rip

[jaunty plume]

just walk north from both docks and then measure horizontal distance

ez

[clear haven]

🤦 i didnt realize that there was more land haha

[rose monolith]

Cant

[clear haven]

wOT

[vital frost]

wtf

[rose monolith]

Confused I’ll just watch, i need help afterall

[jaunty plume]

yeah idk how it expects to use similar triangles

was thinking maybe like this

[clear haven]

yeah cos like you need one of the legs to be the unknown one ???? somehow and you cant really do that unless you have a protractor to get AA similarity

smh

mega confuse

[rose monolith]

Lol ok

[clear haven]

so i suppose in this case you know that they're right triangles w/o needing a protractor

the only ones you'd be able to knwo are the vertical lines tho bc you're only allowed to measure on land

hmmm

unless!!

[jaunty plume]

wait

it actually says sketch

[clear haven]

the lake is conveniently shaped

[jaunty plume]

yeah lol

just draw lines with a ruler

[clear haven]

and you can get the hypotenuse to pass strictly through land lol

in which case you technically wouldnt need another triangle bc you could just use pythagoras but lol whatever

[vital frost]

I hope my teacher says we don't have to do this part

I go back tmrw

And its due monday

The 7th

[jaunty plume]

i think I got it

sec

[clear haven]

oH

[jaunty plume]

where the distance from dock to dock is the base of the purple triangle

but hypotenuse of the red+purple

but like

if they have a measure that long they can literally just walk around the lake to the other side

and hold the measure over the lake

lol

[clear haven]

i see

[vital frost]

So how do u answer a, b, c, d, e

[vital frost]

Ok f that question

@jaunty plume this one?

[jaunty plume]

ratio is just divide one by the other

at 6 months old she was 21 in tall and 4.5 in diameter head

[upper karma]

Hi! Can anyone help me with this one?

If sin78 = x, what is the equal of sin66 as in x?

[jaunty plume]

that's a ratio of 1:4.5/21

[upper karma]

If sin78 is x, then so is cos12

[jaunty plume]

do you mean write sin66 in terms of sin78

[upper karma]

Yes, the question is asking me to find sin66 as in x

and x is equal to sin78

So I tried opening them as in:
sin(78-12) (since it is equal to sin66) and got the following:
sin78.cos12 - sin12.cos78

2
First part is x but I couldn't find the other half

I cant write 2 over x

but you get it

[jaunty plume]

hmm

[vital frost]

What @jaunty plume

What numbers do I plug in

[jaunty plume]

?

it's 21:4.5

the ratio

[vital frost]

Ok

[jaunty plume]

which is the same as 1:4.5/21

for every 1 in height her head diameter increases by 4.5/21 in

[upper karma]

There, this should make the job easier

[jaunty plume]

you can draw a triangle

with angles 12 78 and 90

and use pythagoras and trig

[upper karma]

But we don't know the side length of the triangle

[jaunty plume]

you do

[upper karma]

Oh, the x.

[jaunty plume]

yes

[upper karma]

Yeah let me try

[jaunty plume]

and 1

[upper karma]

But this doesn't work, sin12 is x and so is cos 78

so it becomes x square once again

[jaunty plume]

no

sin12 is not x

[upper karma]

What is it?

[jaunty plume]

that is the information you have

right

[upper karma]

Ah yes

[jaunty plume]

then you can work out the third leg length

to get cos78

or sin12

[upper karma]

I can find sin12 but how can I find cos78 without knowing 3rd leg's length?

Wait I cant find that either

[jaunty plume]

sin12 and cos78 are the same

you can tell by the triangle

[upper karma]

I know but

[jaunty plume]

what is the third leg

in terms of x

using pythagoras

[upper karma]

Thats what confused me, if we call it A

A square + x square = 1

[jaunty plume]

yes

so A=?

[upper karma]

1 - x square?

[jaunty plume]

no

square root

sqrt(1^2-x^2)

[upper karma]

Thanks a lot for the help, I will try to understand this.

[jaunty plume]

so you have sin66 = sin(78-12) = sin78cos12 - cos78sin12 = x^2 - sqrt(1^2-x^2)^2

[upper karma]

I don't know why, I am struggling to understand it even though it is simple.

[jaunty plume]

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

= 2x^2 - 1

[upper karma]

yeah thats the correct answer

Thanks a lot friend, this means a lot!

[jaunty plume]

np

@vital frost still need help?

[vital frost]

I'm fine thanks

[uneven ruin]

Hi, how do I prove that angle EAC equals angle BAD ?
All that’s known is AE = 2R
and AD is height in ABC.

[glad ocean]

ok so if AD is height in ABC, AD must be perpendicular to BC i think

[clear haven]

^

ive been staring at this all day today and i cant

im bad tho so someone might be better

no IS

[wild hamlet]

AE = 2r, then AE is diameter

[clear haven]

yeah...

sad problem :((

[hearty sinew]

Why was trig so accurate with giving answers even when they were very large ?

[elder oak]

?

[hearty sinew]

I am wondering how was ancient civilization able to answer questions with accuracy using trig ?

[clear haven]

hmmmm

[hearty sinew]

given they had no calc

or tables?

[elder oak]

They had methods for calculating roots

and most sins and cos' and roots (afaik)

like they could spend a day doing arithmetic and calculate root2 to 30 decimal places

or whatever

[hearty sinew]

👀 with trig?

[clear haven]

im sure you can break it down to a bunch of elementary sums??

[elder oak]

I mean

[clear haven]

but like

a bu n c h

mega tedious

[elder oak]

sin45 =root2/2, like they could spend a long time getting root2 to a couple decimal places then divide by 2 and have their measurement

[hearty sinew]

Trig just seem to have so much real life applications

[elder oak]

it does

[hearty sinew]

its kinda crazy how long ago it was discovered and yet how widely used it still is

XD

[spice raft]

just like numbers themselves

[glad ocean]

is khan academy a good source to learn algebra ii/trigonometry

and precalculus

to prep for ap calculus bc

its a long story i'll explain upon request

[clear haven]

yes i think it's fine

although there will be a few missing parts

i recommend learning from several sources if you're going to self teach

[glad ocean]

ok

i spent a couple hours learning law of sines and law of cosines, applying them to different triangles and word problems, also proving both the law of sines and cosines

proved it by drawing altitudes

but thanks; what free and preferrably online sources do you recommend

if you are wondering, i live in new york state in the united states

we take final exams called "regents exams"; i'll send you a pdf of a recent one just to demonstrate the kind of questions they aks

ask*

http://www.nysedregents.org/algebratwo/818/algtwo82018-exam.pdf

man trig has legit brought back my addition to math

its like the perfect combination of algebra and geometry in a way

[wild hamlet]

addiction or addition?

[glad ocean]

addiction*

ye

[wild hamlet]

math often loops itself and it's nice when it does that

[glad ocean]

ik

so i've been doing khan academy

and you know how he does a problem or two to demonstrate how to apply what you've learned

[wild hamlet]

ye?

[glad ocean]

he sometimes says "i encourage you to pause this video and see if you can figure it out on your own"

i always do it and its so satisfying

imo its a good way of learning

if you had some background in the area

cause it also tests your ability to apply what youve learned instead of simply plugging in the formulae that you've explicitly learned

[wild hamlet]

I find that I learn best by deriving the formulae that are used

[glad ocean]

what do you mean by derive?

like differentiate?

holy i've been spending the past two hours binging trig

is that even a thing

[wild hamlet]

Derive like as in make the formulae by intuition and logic

not as in differentiation

[glad ocean]

ah

so like analyze the formula

like rigorously try to go through it, genuinely dissecting the meaning, and make it better kinda thing?

[timber hinge]

Start at one thing and end at another is to derive it

basically a show that question tbh

[wild hamlet]

ye

e^*(ix) and heron's formula were fun to derive

that is a simple challenge problem on math 2 btw

[timber hinge]

ugh Herons formulas derivation is a mess

[wild hamlet]

?

[timber hinge]

Didn't you use the polar coords method?

[wild hamlet]

I did

I thought it was fun

I knew about trig subsitution so it could've simplified things

[timber hinge]

Herons formula is messy, probs why its not taught tbh.

Plus other like

"better" methods I guess

[wild hamlet]

What math do you know?

[timber hinge]

well thats not at all vague

Idk I don't keep a list lmao

[wild hamlet]

Like, the order isuaully alg1,geo,alg2,trig,calc,

[timber hinge]

I'm from the UK we don't sort things into the US categories like that

I'm an undergrad

so have all the normie knowledge

[fluid hawk]

herons formula is not messy

[wild hamlet]

Can you solve this: $\frac{dy}{dx}$ at $x=3$ for$ y = xln(x) $ for instance?

[fluid hawk]

take that back

[somber coyote]

Yeat:

[timber hinge]

Differential equation?

not done them in a while but don't see why not

[fluid hawk]

that's not a differential eqn

[timber hinge]

no its not

its just calc

kek

its 5am I saw the dy/dx and just said differential lmao

[fluid hawk]

ans is ln(3) +1

[wild hamlet]

I would think that's a yes. What about this? $\int_{-\pi}^{2\pi}{sin(x)cos(x)dx} $

[somber coyote]

Yeat:

[timber hinge]

its just double angle m9

come on

[wild hamlet]

yep

I'll give you somethin less easy then

[timber hinge]

fite me

[wild hamlet]

$\int_{0}^{1}{\frac{-ln(1-x)}{x}dx}$

[somber coyote]

Yeat:

[wild hamlet]

hehehe

[timber hinge]

thats one of those special functions

[wild hamlet]

dammit u know....

[timber hinge]

Ive seen it but never played with the Li stuff

[wild hamlet]

$\int_{0}^{\frac{\pi}{2}}{ln(2cos(\frac{x}{2}))dx}$

[somber coyote]

Yeat:

[timber hinge]

hmm

[wild hamlet]

Good luck with that one

It's just another form of Li

Both those are unfair though, $\int_{0}^{\frac{\pi}{5}}{cos(x)^{3}dx} $

[somber coyote]

Yeat:

[timber hinge]

pi/5 is not nice

but thats trivial

1-sin(x)^2

then inspection p. much

[wild hamlet]

'inspection'?

[timber hinge]

yea thats what we call it here

cos(x)sin(x)^2

can be integrated via inspection

since its just like

sin(x)^3 differentiated

but with some other junk

or just

u-sub

[wild hamlet]

$\frac{1}{\pi}\int_{-\pi}^{\pi}{e^{\frac{x}{p}}cos(xn)dx} $

Do that then

[timber hinge]

complex form of cos can take care of that no?

[wild hamlet]

Can't remember

[timber hinge]

you can do by parts but its messy

[wild hamlet]

o wait

[timber hinge]

$\frac{1}{2}\left(e^{ix}+e^{-ix}\right)=\cos\left(x\right)$

[somber coyote]

Colen:

[wild hamlet]

one other thing i forgot in there

[somber coyote]

Yeat:

[wild hamlet]

heheh

[timber hinge]

doesn't rly change the problem

Just makes it more messy

$\frac{1}{\pi}\int_{-\pi}^{\pi}e^{\left(\frac{x}{p}\right)}\cos\left(xn\right)dx=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{\left(\frac{x}{p}\right)}\left(e^{ixn}+e^{-ixn}\right)dx$

[somber coyote]

Colen:

[wild hamlet]

You haven't done fourier series yet

[timber hinge]

no

[wild hamlet]

You would recognize this if you did

[timber hinge]

I've heard of them but not studied them

[wild hamlet]

You should 😉

[timber hinge]

shhh i'll get around to it

more fun things to do

[wild hamlet]

I wonder what it is...

[timber hinge]

What what is?

[empty shoal]

@wild hamlet so, sorry if this is a bit off topic, but why is it that there are integrals of functions that cant be represented by elementary equations?

like Li for example (I know there is a proof in Galois theory somewhere, but what is the intuition for this?)

[timber hinge]

Galois theory afaik doesn't deal with proving there are some integrals with no elementary solutions (Its polynomials of degree 5 and up not having solutions or something I believe). Imagine you have the integral, you'd need to differentiate it to get it back into the form of the integrand. You just can't do that with everything 🤷

would be nice if you could

in the real world they just use numerical analysis for a lot of integrals

[wild hamlet]

Sometimes the elementary equations can be defined as integrals or it can lead to new insights to define them as such rather.

In chill I put a proof of the basel problem using the elementary function : x/2 and using integrals with it

[timber hinge]

Isn't that the sum to pi^2/6 thingy?

[wild hamlet]

yep

[timber hinge]

Hmm I don't remember the proof for that tbh

just realised we're not even in chill lmao

[wild hamlet]

I used fourier series in my proof, it would cut down on half the proof if I didn't care to rigorously find the fourier series without using the integrals for it

[timber hinge]

I think I do Fourier series next year as part of my course but i'll likely study them on my own before then

[wild hamlet]

Fourier series ❤

[timber hinge]

A lot of uni maths can be done by HS students tbh, its just they don't need it so its not taught

Maths only looks hard if you don't have the pre reqs and tell yourself its hard

~~with exceptions because some maths is just 🔫 ~~

[wild hamlet]

Matrices... are kinda forgettable for me

It's not fun to calclulate by hand and using a computer to do it kinda lets you lose the actual feel and understanding of doing the math

[timber hinge]

Computers are used in a lot of research now, kinda unavoidable

[wild hamlet]

I shall revert everyone to the paper age!

[timber hinge]

The whole you'll never have a calc thing kinda fell apart

[wild hamlet]

📰

[timber hinge]

print wikipedia tbh

[wild hamlet]

I think... I should get some sleep. I'm currently interested in smooth surfaces and using that for calculus for differentiability and smoothness. Analytic continuation too

Cya, ttyl8r

[timber hinge]

bai

do complex analysis tho

[lapis wolf]

$\begin{bmatrix}
x' \ y'
\end{bmatrix} =
\begin{bmatrix}
\cos{\theta} & -\sin{\theta}\
\sin{\theta} & \cos{\theta}
\end{bmatrix}
\begin{bmatrix}
x \ y
\end{bmatrix}$

[charred spear]

Command disabled

The sever owner has disabled that command in this location.

[lapis wolf]

can someone explain where does this formula for 2d rotation about the origin come from

x,y are coordinates of the given point and x',y' is the required point

theta is the angle

[somber coyote]

stormblessed:

[umbral snow]

Describe a point by
x = rcos(α)
y = rsin(α)

Then you can describe a rotation by θ using:
x' = rcos(α + θ)
y' = rsin(α + θ)

x' = rcos(α)cos(θ) - rsin(α)sin(θ)
y' = rsin(α)cos(θ) + rcos(α)sin(θ)

x' = xcos(θ) - ysin(θ)
y' = ycos(θ) + xsin(θ)

Then rearrange to matrix form.

@lapis wolf

[lapis wolf]

oh that's it?

I thought it required complex

[umbral snow]

That's it lel

[lapis wolf]

thank you

[umbral snow]

It requires the sum/difference identity, which can be proven easily using complex

But no need

[lapis wolf]

yeah I get it now

i could have found out the value using it if I tried

the matrix confused me

using the identity i mean

[umbral snow]

Remember that a matrix is legit just a system of equations, if the notation is confusing you, then un-matrix it

Matrices have a ton of perks, so recognizing this CAN be written as a matrix is huge

[lapis wolf]

It doesn't always its just this was stated as a formula and I couldn't trace it back

[indigo tartan]

how do I dot unit vectors from two different systems like z_hat and r_hat (spherical)

example: $\hat{z} \cdot \hat{r}$

[somber coyote]

noonoo:

[lament bay]

#2

[lament bay]

<@&286206848099549185>

[waxen gorge]

Hi

[lament bay]

Do I just find the slope of BC=AD?

[waxen gorge]

,rotate 90

[lament bay]

Hehe

[little osprey]

Oof

[waxen gorge]

,rotate 180

[little osprey]

You have to find the gradient

gradients

[waxen gorge]

All u have to show is

The sides are equal and the angles formed on the bottom are too(by using slopes)

[lament bay]

The hell is gradient

[waxen gorge]

slope

UK term

[lament bay]

Oh lol

Smart people in UK😔

[little osprey]

Um

No

You have to show that the sides have equal slops

Parallel lines always have the same slope

[waxen gorge]

Which sides....

The top and bottom yes

[lament bay]

I found the slope for AB=DC but teacher says I have to also use BC=AD

[waxen gorge]

The right and left no

[lament bay]

“Not just for specific numbers, but for any numbers”

[little osprey]

Yeah cuz you have to prove it's isoseles

[lament bay]

How to I prove it using variables then?

[little osprey]

Wait

Let me attempt first

Ok tell me what you've done so far

[lament bay]

found the slope for AB and DC

both were 1=1

used distance formula

[little osprey]

m1xm2=-1/2 right

[lament bay]

AD=6 BC=6

[little osprey]

Oh shit

Yeah

Sorry

That's other stuff

Tbh I don't see where you went wrong

[lament bay]

idk

[little osprey]

What did you teacher tell you to do again?

[lament bay]

"Also need to prove BC=AD not just for specific numbers, but any variable"

Thats what he put on my paper

[little osprey]

Did you sub in values?

Because if you didn't then I have no clue as to what he mean't by Specific Numbers.

[lament bay]

well since i found out the slope was 1

i plugged a=1

[little osprey]

Uh yeah, I think that's what he wanted to talk about

He wanted you to solve using the distance formula algebraically

Im guessing

[lament bay]

How would I do that?

[little osprey]

Use the distance formula on all sides

[lament bay]

(2a-(-1a)^2?

[little osprey]

Yeah

It should work algebraically, even if you dont have values to sub it in with

[lament bay]

its still gonna get the same answer lol

the a's subtracted each other

thats wack

[little osprey]

Yeah

But Im guessing that's what the teacher wanted you to practice, just looking at his 'feedback'

[lament bay]

So i should also find the slope for BC and AD?

[little osprey]

Yeah

[lament bay]

do I still gotta do distance formula? for the sides?

[little osprey]

Then you can say that it's an isosceles trapezium cuz of the special property stuff

:/ Like I said.

From the wording of your teacher it seems that's what he wants.

[lament bay]

first i do slope AB and Dc

then do BC and AD

I would only have to include the distance formula of AD and BC because that's the sides which decides if its a isosceles trapezoid

[little osprey]

Yeah

That's what I would do tbh

I would conclude it's a trapezium first though

And then say since...

[lament bay]

ok

ty

[little osprey]

np

[glad ocean]

gg bois

[rare scarab]

How do u draw a tangent for 2 equal circles , again ?

[pulsar dirge]

How do I find the trig ratios for sin 60

[umbral snow]

Make an equilateral triangle, split it in two, and use pythag

Gives 60 and 30

[lament bay]

#4

[lament bay]

<@&286206848099549185>

Any ideas?

[upper karma]

@lament bay

like this?

[lament bay]

@upper karma

[upper karma]

i can't read your teachers writing for shit

[lament bay]

uh I think it ays

says

"these are only true if DC=DB

@upper karma

[upper karma]

@lament bay ?

[lament bay]

I think my drawing is right

but my proof is wrong

I dont get what I didnt wrong in my proof

[upper karma]

https://www.algebra.com/algebra/homework/word/geometry/Median-in-a-right-triangle-drawn-to-its-hypotenuse.lesson

@lament bay

[lament bay]

oof thats a lot

[glad ocean]

is it a problem that trigonometry was the subject that brought back desire for math

[lament bay]

idk

[glad ocean]

its so fun

[lament bay]

https://cdn.discordapp.com/attachments/326138757474680852/530563217722966016/image0.jpg

ideas on #4?

https://cdn.discordapp.com/attachments/326138757474680852/530574307861069844/image0.jpg

my original work btw

[glad ocean]

like primary trig ratios (SOHCAHTOA), reciprocal trig ratios, inverse trig ratios

and the use in both right and general triangles

[lament bay]

<@&286206848099549185>

[lament bay]

pls

[spice raft]

Ill try to help with the first one

[lament bay]

yea I just need help with #4

[spice raft]

lmao idk

[dawn stone]

@lament bay here's how you solve #4

You've drawn DE||AC, and CD is the median which implies AD=DB
Hence BE=EC by intercept theorem

And then prove DEC and DEB are congruent triangles using SAS

And from that, you get CD=DB(CPCT)

So CD=AB/2

[pallid comet]

Hey, can anyone help explain to me the hyperbolic trig functions? I'm trying to derive them from the ground up and everything I've found online explains things circularly. Its pretty frustrating tbh

I understand this diagram. But I'm not really sure how you define cosh and sinh using it exactly.

[woeful flame]

hi

cos (2x+3)=sin (3x−5)

how can i do it

?

[pallid comet]

@woeful flame When doing these problems, you have to use an identity to rewrite both sides in terms of the same trig function. Kind of like what you have to when solving log equations, if you've done those before. The identity you wanna use is cos(x) = sin(pi/2 - f(x)), so you can rewrite cos(2x+3) as sin(pi/2 - (2x+3)) -- noticed I substituted the inside of the f(x) for what you had inside the cosx before. Now you have both sides in terms of sine, and you can solve it like you would a normal equation, by look at the inside of both sines and setting them equal to each other.

@woeful flame once you solve it, remember that the sinx and cosx functions are periodic, so you have to append a 2pi * n to your answer

[clear haven]

damn what a chad he didnt even get his original question answered and still helped someone out 🖤

[woeful flame]

hello sir

thank you but I already know that

but still not access

[clear haven]

???

[woeful flame]

im trying these 3 question

im trying to solve these 3 question for 18 hours already

@pallid comet in question 2 i did as you say

sin(π/2-(2x+3)) = sin(3x-5)

so

π/2-2x-3 = 3x - 5 + 2π*k

[clear haven]

[woeful flame]

so

-5x = -2-π/2+2π*k

so

5x = 2+π/2-2π*k

so

x = 2+π/2-2π*K / 5

[pallid comet]

that should be right

[woeful flame]

it say that there is another solution

i tried

[pallid comet]

The 2nd solution is due to another symmetry of sine. sin(x) = sin(pi - x).

[woeful flame]

ohhh

I found

[pallid comet]

yup

It can be pretty tricky

[woeful flame]

but I also tried the first question and cant do that

I made

2x+3 = t

cos(t) = 0.7

so

t = +0.795 or t = -0.795

[pallid comet]

are you allowed a calculator?

[woeful flame]

Yes

sure

[pallid comet]

cause here you're gonna have to use arccos

[woeful flame]

what is arccos?

[pallid comet]

the arccos is the inverse of cos, and so it cancels it out the same way, say, 2^x would cancel log2(x)

arccos(cos(x)) == x

[woeful flame]

i dont know

[pallid comet]

and cos(arccos(x)) == x

[woeful flame]

dont get it

like when I click on shift->cos on the calculator?

[pallid comet]

yes! its just in a different notation

[woeful flame]

yes im using it

[pallid comet]

arccos(x) and cos^-1(x)

yeah

[woeful flame]

sure

that how i found t

cos(t) = 0.7

so i clicked on shift->cos 0.7

and got the result

t = +0.795 or t = -0.795

then i did

[pallid comet]

but wait, did you actually solve it? the inside of the cosx is still there. so you have 2x + 3 = arccos(0.7)

[woeful flame]

2x + 3 = 0.795

hmm

[pallid comet]

oh and make sure you calc is in radians

[woeful flame]

i dont get you sir

[pallid comet]

look let me show you what happens

[woeful flame]

sure im doing in rads

now im doing 2x + 3 = 0.795

[narrow sleet]

S/he is in radian mode

[pallid comet]

yes that is correct

now just solve for x

[woeful flame]

im getting wrong answer

are you sure its the correct way?

[narrow sleet]

But that's not the arccos (0.7) we want, cuz if u use 0.795, the ans is out of range

[woeful flame]

maybe its a mistake in the book

[pallid comet]

x = (0.795 - 3) / 2

[woeful flame]

what wait what?? o_O

wich step is wrong?

[pallid comet]

what he's saying is you're right, but since your calculating arccos(0.7) with your calculator at the beginning and rounding/cutting off the decimal, you're getting an error

[narrow sleet]

Not that

[pallid comet]

the best thing to do imo is keep the arccos(0.7) when your solving your equation and then plug the x in the end into your calc

[woeful flame]

i really doont get it

[narrow sleet]

I'm saying that he should use 2pi - 0.79/ instead of 0.795

[woeful flame]

my english is bad

that is so hardddd.....

😭 😭 😭

[pallid comet]

hold on

Are you sure about that? i think what he's doing is fine. Just solve for x like you would normally

[narrow sleet]

But if u use 0.795, x = - 1.102 which is out of the range 2<x<4

[pallid comet]

Yeah, just noticed that. true

[woeful flame]

so I can add 2pi after getting - 1.102

not?

[narrow sleet]

No

U say u have two answer of t = 0.795 and -0.795

[woeful flame]

yes

[narrow sleet]

U can make negative rad into positive rad by adding 2 pi

So -0.795 = -0.795 + 2pi = 5.488

[woeful flame]

its out of range

maybe it should be +1pi

cuase its 2x+3

[pallid comet]

What's probably confusing you is the periodicy of the functions, which is something that always got me when I first learned this. See, you actually have multiple answers, your calculator is just giving you one. Because the trig functions are "periodic" they repeat and have multiple solutions so - like Bad Wolf is saying - you can use a symmetry in the trig function to find another answer.

[narrow sleet]

^

[pallid comet]

I have to check, but that may be what you need to do

I think sin(x) = sin(pi + x), don't remember @narrow sleet

[woeful flame]

but If I had not range limit

wait guys please

believe me i understand

but

its not the thing

cause if I had no range limit

i should get the result first

and only then i should add the 2pi

to make it in the range

like if i had no range limit

i would get - 1.102

[pallid comet]

indeed. but 2pi isn't the only periodic symmetry you can exploit. You tried doing that earlier I think and it didn't work

[narrow sleet]

Etc etc etc

[woeful flame]

and then if someone will say it must be 2<x<4

so I could add 2pi to the result i got

[pallid comet]

Yes

[woeful flame]

you dont get me guys

please read what i say

I know that cos(5) = cos(5+2pi) = cos(5+4pi) and etc etc etc

but you add the 2pi AFTER you get the result

like, if I had no limitation of range

so I would get - 1.102

and after that, i can search again in the range 2 < x < 4

but I already has the result - 1.102 so I only need to add +2pi or -2pi

but im doing it after getting the result - 1.102

[pallid comet]

did adding 2pi get you in the range you needed?

[narrow sleet]

That means u don't my diagram

[woeful flame]

what?

[pallid comet]

lol what

[woeful flame]

I understand all the cos in your picture are equals that is abvious

you really dont get my point

[narrow sleet]

Basically I'm saying arccos(0.7) = 0.795 = 2pi - 0.795 = 2pi + 0.795 = 4pi - 0.795.....etc. So u try 0.795 and 2pi - 0.795 and thry didn't work. So u should try 2pi + 0.795 next

U need to kept doing the same pattern until ur x value is in the range

[woeful flame]

😭

can you tell me the answer?

[narrow sleet]

No

[woeful flame]

i will check if youre right

[pallid comet]

I think what you're saying is that you're adding the 2pi too late and thats why its wrong?

[narrow sleet]

Yes

[woeful flame]

I need to find X and not T

[narrow sleet]

That's one

[woeful flame]

look at the question again

[narrow sleet]

U need t to find x

[woeful flame]

T = 0.795

[narrow sleet]

If u use the wrong value t, u find wrong value of x

[woeful flame]

forget the range limitation

just leave the limitation

for one second

t = 0.795 right

?

[narrow sleet]

That's one of the answer for t

[woeful flame]

yes, and the second one is -0.795

right?

[narrow sleet]

Or 2pi -0.795 if u want to make it positive

[woeful flame]

ok

so 2x+3 = 0.795 +2pi*k

right?

or

2x+3 = -0.795 +2pi*k

thats right?

[narrow sleet]

Yes

Where k = 1, 2, 3, 4...

[woeful flame]

yes

k = integer

so

x = -1.1025 +pi*k

or

x = -3.795 + pi*k

right?

[narrow sleet]

-3.795

[woeful flame]

fixed

so,

NOW i want to search for the range limitation with the result that I got

can i do that?

[narrow sleet]

Yes

By just substituting k with a integer until u find one that fall within the range

[pallid comet]

basically

[woeful flame]

why im still geting wrong?

[narrow sleet]

Wait

Let me check

[woeful flame]

thank you so much sir

im trying for 18 hours

i so need help

[pallid comet]

it's actually pretty easy once you have a grasp of it

that's normal, don't feel bad about it.

[woeful flame]

how long it takes?

[pallid comet]

yeah. i'm in calculus and i used to practice whole nights doing derivatives, etc.

[woeful flame]

how old are you?

[narrow sleet]

x = -3.795 + pi*k is wrong. U forget to divide -3.795 by 2

@woeful flame

[woeful flame]

oh no

x = -1.897 + pi*k

so there is nothing in the range

[narrow sleet]

Yeah

[woeful flame]

only this -1.1025+\pi

[narrow sleet]

There is only 1 ans

Yeap

[woeful flame]

but still wrong

[narrow sleet]

Did they say write in how many decimal places or significant figure?

[woeful flame]

no

[narrow sleet]

I swear that's ths only answer

Oh

Rounding effect

For trig, it's safer to use the whole number on the screen instead of taking part of it

So it will be better if arccos (0.7) = 0.79539883 instead of 0.795

[pallid comet]

I think you guys may have your order wrong, like you're adding 2pi*k too late in the process and its throwing off your result? You got x = (arccos(0.7) -3)/2 and then you're adding 2pi *k to it -- I don't think that's correct.

[narrow sleet]

Nah we added the 2pi*k when we evaluate arccos (0.7)

[pallid comet]

Ah good good because that's what I'm reading here

[narrow sleet]

Guess I press in my calc it should -1.1023 instead of -1.1025

@woeful flame

[woeful flame]

Still.....

[pallid comet]

@narrow sleet by the way, I'm also seeing this solution set in my online solver. Are you sure he's getting this too? Sorry i'm in another thread checking on this one and I wanna make sure @woeful flame understands