geometry-and-trigonometry

[thorn narwhal]

wait for the trig values of special angles

do we just gotta memorize them

[left mauve]

What angles do you mean

[thorn narwhal]

like pi/6

[left mauve]

Yes

[thorn narwhal]

okay

[left mauve]

0,30,45,60,90

[thorn narwhal]

yeah ik those

[civic rock]

37

[left mauve]

I would also say 37 and 53 except people keep cancelling me

Smh

[maiden brook]

Nah u don’t need that

[left mauve]

Yet again

[native pier]

you really don't, except when you're going to take the jee

which is the same as saying we don't, ever

[loud echo]

why would u

[hard knot]

someone help me with graphing this;

graph y = sec x, -2pi greater than equal to x greater than eqal to 2pi

i dont understand how to like start the points 😭

[dark sparrow]

-2pi greater than equal to x greater than eqal to 2pi
if you can't type the symbols ≤ and ≥ you should replace them with <= and >=

and in this particular case you could've also said "x goes from -2pi to 2pi"...

[dark sparrow]

do you still need help w this

[hard knot]

nah im good thank you though.

[civic rock]

<@&268886789983436800>

good riddance

[worthy eagle]

without context, both angles look like π/2 to me

[mighty pulsar]

hi

[zealous pike]

Hi

[civic rock]

hiii seremnya

[zealous pike]

Why side BQ is equal to BA?

Triangle BPQ is neither congruent nor similar to triangle BPA, as some angles are different

So I'm confused

[civic rock]

∆BQA is isoceles triangle since angle BQA = BAQ = 50°

[zealous pike]

Oh

Mb

[civic rock]

oh god not these types of problems 😭

[zealous pike]

😭

[civic rock]

with weird angle transformations

my 8th grade nightmares

[zealous pike]

I still don't understand much of the constructions to help solve for x in these problems. Id like to learn more

[civic rock]

basically just try and try again till you hit an isoceles or equilateral triangle

in this case I also notice that triangle BPQ is isoceles

[zealous pike]

I thought that AC was parallel to BQ

[civic rock]

?

there's no evidence to show that

[zealous pike]

Indeed

Is it possible to solve this without trigonometry?

[civic rock]

many of these problems are designed to be solvable without trig

only congruent triangles, isoceles and equilateral triangles

<@&268886789983436800>

[left mauve]

<@&268886789983436800>

[zealous pike]

I did nothing

[civic rock]

nah some mf just sent a not-so-appropriate link here

[zealous pike]

Why am I unable to solve this one 😭

[civic rock]

it involves many extra constructions and weird angle transformations just to find a key to solve the problem

[upper echo]

I found a construction that solves this problem, but it is different from your 1st method. Actually it goes this way: ||reflect P across AB to P'. Then continue P'A and BC till they intersect at D. Then take a point E on AD such that PE=PB.||

[leaden sorrel]

Shapes

[untold coyote]

Hi can someone help me with this/explain it to me? I don’t really understand it.

[ashen crystal]

nice

[neat bridge]

Do you know what sin and cos represent on the circle ?

[untold coyote]

Yes

[neat bridge]

Which one are you struggling with ?

(Also, can you assume that for example DB is perpendicular to AD ?)

[untold coyote]

Most of them, but mainly the negative ones like question f and d

[untold coyote]

Yeah

[neat bridge]

Do you have any idea of how to find it ?

For $sin(-\theta)$ you need to find the « height » at which the line drawn at an angle $-\theta$ intersects the circle

[somber coyote]

StaYin

[neat bridge]

I’m going to sleep so I’m putting the answers behind spoilers, ping me if you want more explanation ||sin(-theta) is DC, and tan(-theta) is (I think) AF but I’m not sure||

[untold coyote]

Alr

tysm

[neat bridge]

Is that fine ? Are you sure ?

[untold coyote]

Wait I kinda get it now

[neat bridge]

I messed up while typing :(

[neat bridge]

Please recheck for the correct solution

[neat bridge]

Awesome 🤗

[zealous pike]

How do I learn this better?

The geometry book I'm reading is not helping me with this ;-;

[upper echo]

Practice — solve as many problems as you can. If they feel way too hard, switch to slightly easier ones, so that you can fully solve about half of them on your own. If a problem seems interesting to you, spend a long time thinking about it. In general, all of this takes time, which will hopefully turn into expertise.

[zealous pike]

Is there any tips to deal with constructions? It all ends up in extending / reflecting sides and points to create isosceles and equilateral triangles?

[upper echo]

I try to seek necessary constructions by connecting parts of the problem. Suppose, you have a condition in a statement that some segments are equal, then your construction should use this fact, it should somehow connect those segments into a triangle or some chain of triangles or circles or something that makes these segments closer.

[zealous pike]

understood

Is any triangle problem solvable without the use of trigonometry? If no, is it easy to identify if I can not use trigonometry?

[upper echo]

Don't avoid trigonometry 🙂 Some construction which leads to a short and beautiful solution is much easier to find after you solve the problem with some bash method. Because when you have one solution, though complex and technical, it shows what really works for the problem. Btw in your above problem I found that construction exactly this way. Solving with the law of sines it clearly shows that the fact 50*2=100=180-30-50 gives the solution. So, it is probable that good construction should double the angle of 50 somewhere.

[zealous pike]

I just want to improve my geometry intuition, and I still don't have a strong basis in trigonometry ;-

[upper echo]

Ok, learn trigonometry.

[zealous pike]

I'm doing

I already know stuff like sin = sqrt(1 - cos²), 1 + cos² = csc², sqrt(tan² + 1) = sec

[upper echo]

Probably so, because all properties of trig functions can be transferred to geometry statements. But this fact doesn't make the life easier. Having a trig. solution does not mean that you can come up with some reasonable geometric one.

[zealous pike]

I still can't demonstrante the sin and cosine laws, so I still struggle to remember them

[upper echo]

I've got a problem of my own for which I failed to find a geometric solution. Tho I have a trigonometric one

[zealous pike]

I struggle a lot to remember formulas and such, but if I figure out why, such as in demonstrations or proofs, it becomes easier for me to understand and use them

[upper echo]

My opinion is that trigonometry is an easy part of math. Just a set of rules to memorize. Usually everything is straightforward. Unlike geometry.

[zealous pike]

I find trigonometry hard because, like, there is no easy way to calculate sin, cosine and tangent, and the other functions. While it is kind easy to demonstrate their values for angles like pi/3, pi/4, pi/6, it isn't as easy for other angles.

I understand their definition as the ratio of the sides of a triangle given an angle, but still 😭

[upper echo]

it's ok, just some practice is needed. And time.

[zealous pike]

Coming to think about it, trigonometry is sort of related with similar triangles isn't it?

[upper echo]

yes, cos, sin, tho they can be defined as functions of the segments ratios, are still functions of angles. This is because similar triangles have equal angles.

[soft quiver]

wait I think I got it

cos theta is ob's horizeontal leg

sin theta is ob's vertical leg

tan theta is probably the segment from a to intersection of ray OB with vertical tangent

cos (negative theta) same as cos (theta)

sin(negative theta)

wait so isnt it just tan(-theta)

which is f

[zealous pike]

just remember that in the unit circle, coordinates (x, y) = (cos x, sin x)

[soft quiver]

ok

[maiden brook]

which “laws”

[maiden brook]

No, u don’t needa memorize very much

[zealous pike]

Law of sines and law of cosines?

[maiden brook]

oh. Those are a bit hard yeah to derive

[waxen river]

hello

[zealous pike]

Here it is showing the proof for the addition formula of cosine, and the proof of the other formulas are left as an exercise. The addition and subtraction formula for sine and tangent comes from the same equation shown?

[obsidian harness]

what's the diagram?

[obsidian harness]

ah it should work for sine but not tangent

[zealous pike]

how is the proof for the tangent ?

[obsidian harness]

the tangent one is derived as follows:

[zealous pike]

clever

[zealous pike]

For sine, do I substitue cos(s+t) by its addition identity and solve for sine?

[obsidian harness]

actually I think there's a more direct way using the fact that cos(pi/2 - x) = sin(x)

[zealous pike]

[obsidian harness]

so you can write cos((pi/2 - a) - b) = cos(pi/2 - (a + b)) = sin(a + b)

then use the cos addition formula with pi/2 - a and -b

[zealous pike]

understood

thx

[obsidian harness]

np!

[copper halo]

Guys help

[upper echo]

you can substitute tan(theta)=3tan(x/2) then it gives theta=0 and theta=+-pi/4

[zealous pike]

Hi

[dapper goblet]

Hai

[zealous pike]

Given an equilateral triangle, the line that bisects one of the angles always goes through the barycenter. The length of the line to the barycenter is half the length to the midpoint of the opposite side?

Suppose this is an equilateral triangle and P is the barycenter. d(AP) = d(AB)/2?

[summer cradle]

no

d(AP) = ⅔d(AB) actually

[dire lotus]

how do i do triangle proofs

[mystic umbra]

start from the basics like proving that an isoceles triangle has equal base angles

then just

ramp up the difficulty

[zealous pike]

2/3d(AB)?

[summer cradle]

oh yes I mean AB

[zealous pike]

Understood. Still in this, the triangle is AMN, the angle formed by ABM is a right triangle right?

[civic rock]

it is only true if the big triangle is isoceles with apex A

[zealous pike]

anm is an equilateral triangle here

[civic rock]

oh then that's true

if you want to verify it yourself you shouldnt rough-draw it

[zealous pike]

how do I prove/demonstrate the length of AP here?

[civic rock]

since the big triangle is equilateral, P is the intersection of 3 median lines of the triangle

[eternal wagon]

Anyone?

[snow karma]

say u have a triangle where you know two of its side lengthes a and b

then the third one lies between a - b and a + b

[eternal wagon]

I’m honestly confused abt the whole question

[left mauve]

Welcome to the server kys!

[eternal wagon]

Ty

[snow karma]

ur given two lenthes of two sides of the triangle

what are the possible values

of the third length

thahts what it asks

[eternal wagon]

Ok I understand that what would I do to start the question?

Do I divide them ?

[snow karma]

use this rule

that rule is general for any case

of triangles u have

[eternal wagon]

4<x<28

Would that be the correct answer ?

[snow karma]

yes

[eternal wagon]

Ok thank you

[eternal wagon]

Sorry to bother but is this correct?

[obsidian harness]

ah no

XY = ZY is the given one

and then angle XYW = angle ZYW goes below it

[eternal wagon]

Ohh I didn’t know you could switch that

[obsidian harness]

the rest is correct

[eternal wagon]

Thank you

[obsidian harness]

no worries!

[zealous pike]

hi

if this chat was a classroom I'd be that one classmate who practically lives here

[obsidian harness]

hi

[plucky radish]

I did ext angle property on the figure in the middle and got y as 46 and x as 44. Am I correct?

[obsidian harness]

wait

[obsidian harness]

q20 part a right? that's not an exterior angle

this situation is not what you have

the right way to approach this question is: what must angle ACE be?

then you'll know

[plucky radish]

Aw man. Thanks for telling me

Google said it's correct 😂

[zealous pike]

U wrong

Angle ECB is the one that is 136°. If you find angle DCB then u can find out the rest

Or subtracting angle ACE from 136°, u find x then

[obsidian harness]

lmao it seems they have a ton of progress to make on AI

it's not just Google actually: all of them can be this shit given the right conditions

[zealous pike]

Gemini, ChatGPT, etc, can't reason guys, remember that

[obsidian harness]

yeah

[zealous pike]

Their answers are just a mix of other people's answers for questions that'd look like yours plus a small mix of "humanity", which is also derived from other people texts

[sly horizon]

Hi. I have interesting geometry problem in my whatsapp group. Who can solve it?😁

Can l send pic of problem here

[tawny rapids]

yes

[tawny rapids]

me I can 😁👍

[sly horizon]

That figure is regular pentagon.

[upper karma]

can someone help me in dms with my ixl.. everytime I get even one wrong, it takes me back so much 😭

[tawny rapids]

dyou still need help?

[upper karma]

yea lol

[tawny rapids]

you can dm me the problems and I'll help you with them

[upper karma]

tyy !

[zealous pike]

Hey

[zealous pike]

What those two ">" on the line means? they are parallel?

[left mauve]

yes

[drifting ermine]

Are inverse trigonometric substitutions and derivatives of Inverse trigonomtric functions related?

[silent plank]

yes and if you're talking about derivatives,
that is in calculus territory
not basic geo/trig

[drifting ermine]

Trigonometric substitutions are part of trigonometry not calculus

[dark sparrow]

???

[drifting ermine]

Yes

I am talking about questions like Sin 90 - x/Sin 30+x

[still geyser]

Is it 108 degree

[obsidian harness]

what do you have to do in these questions?

[still geyser]

Construction

[obsidian harness]

not replying to you

[still geyser]

Sorry

[drifting ermine]

Find the easiest route to solve by substituting values of x. Introduce a constant a if there is any constant in the question

[obsidian harness]

solve what?

there is no equation in what you wrote; 'solve' is used for equations

[drifting ermine]

Yes, when you have equations then you do what I said

[civic rock]

what you said doesnt make any sense

[obsidian harness]

what do you substitute then to solve sin(90 - x)/sin(30 + x) = 2?

[drifting ermine]

Questions like these

,rotate

[drifting ermine]

15 and above

Knowing substitutions makes differentiation easy

Again, I have two topics at the same time

[obsidian harness]

then don't ask differentiation and integration here

[drifting ermine]

So the channel is still valid

[upper karma]

hi i am good at math but i had a bad teacher the last year and at my exam of june i had 7/20 .because i don't understand geometry and functions is to go from something visual to equations and how that relates to the concrete. chapters that i really don't understand are plane analytic geometry and functions of the second degree. anyone know how i could find out about these subjects?

[zealous pike]

Stewart precalculus book is a go ig

[upper karma]

Thank you

[sage bison]

15 is just (x + y)/(1 - xy)

you can set $x = \tan{\frac{\theta_1}{2}}$

[somber coyote]

impract1cal

[sage bison]

and $y = \tan \frac{\theta_2}{2}$

[somber coyote]

impract1cal

[sage bison]

and it becomes clear that $\sin^{-1} \frac{2x}{1 + x^2} = \sin^{-1} \left(\sin \theta_1 \right)$

[somber coyote]

impract1cal

[sage bison]

etc

for number 16, try to tan both sides and use the fact that $\tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}$

[remote shale]

[somber coyote]

impract1cal

[drifting ermine]

OK thank you

We equate the three functions to y one by one and find principle values, then solve (with identities if needed)

In Q15

[sly horizon]

Hi. How are you guys? I have an interesting geometry question in my whatsapp channel. Who can solve it?😁

[grave pond]

You can ask it here, but be sure to be clear whether you already have a solution yourself, or you want help with finding a solution.

Wait, is is the same you already posted yesterday?

[sly horizon]

Yeah

That is it

[zealous pike]

Why alpha + beta = 180°?

[grave pond]

PCB and CPA are a pair of SSA triangles; if they're not congruent then the indicated angles are supplementary.

(In other words, if you rotate triangle CPA to make C'P'A' where A'=B and P'=C, then BP is the same infinite line as A'C', and CPC' is isosceles with two x sides.)

[mystic umbra]

or just sinB = sin180°-B = sinA ===> A+B=180°

maybe?

[grave pond]

It's not clear to me whether sin(alpha)=sin(beta) is a given or a goal here. :-)

[zealous pike]

It's because a ≠ b, if alpha = beta then a = b which is a contradiction

I mean, the exercise says that a ≠ b

[drifting ermine]

What are the quadrants of a circle used in trigonometry for anyways? Like I get that sin tan and cos have their own place in the quadrants where they are positive but what do we use this concept for?

OK, for checking whether cos sin or tan are negative

At different degrees

[grave pond]

Mostly as a conventionally understood shorter way to say things like "to the left of the y-axis and above the x-axis", I'd say.

[drifting ermine]

Pythagorean theorem

[civic rock]

shocking

[grave pond]

What?

[pearl hinge]

This is used when u supposedly express an angle which is in quadrant format,to a normal angle

For instance
If we go for sin(pie+theta),pie as in 180

Then it would mean that the angle is in 3rd quadrant and since the value of sin(theta) is negative there ,
We can write
Sin(pie+theta)=-sin(theta)

[drifting ermine]

Sin 1441° = Sin 361° = +ve

"pie as in choco pie"

[pearl hinge]

Choco pie? 180 is represented as pie

How did u go 361 fro 1441

Okay i got it and yeah that's how it works

[viscid nimbus]

it's just a circle in normal euclidean plane

though the radius is 1

so you start at (0, 1)

if you go counter-clockwise

it's (1, 0)

(-1, 0)

(0, -1)

[zealous pike]

Hi

[heavy oriole]

hi pi

[tribal canyon]

hi house

[tight dock]

hi Josh

[dark sparrow]

!redir, y'all.

[lime crown]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[zealous pike]

I understand why the first rectangle base is in the interval [1, 5/4], but I don't understand why it's height is 4/5 if the base length is 1/4

Can someone give me a light?

[cunning lion]

the height of each rectangle is the value of the function at the right endpoint of its interval

[zealous pike]

oh

I can see it now

thx

[proud jacinth]

gente, esto es simple y basico

Expresión	Significado	Resultado
$a^{\frac{1}{2}}$	raíz cuadrada de $a$	$\sqrt{a}$
$a^{\frac{1}{3}}$	raíz cúbica de $a$	$\sqrt[3]{a}$
$a^{\frac{m}{n}}$	raíz n de $a^m$	$(\sqrt[n]{a})^m$
$a^{-\frac{m}{n}}$	lo mismo, pero invertido	$\frac{1}{a^{\frac{m}{n}}}$

[somber coyote]

Migi

[median locust]

Confused on how to solve this

What would be the formula for N?

[obsidian harness]

Consider only the x-coordinates

[maiden brook]

dont need a formula, do u remember for instead how to find the midpoint given two points?

[obsidian harness]

Yeah then change in LN = 2 units
change in ML = 4 units = 7.7 - (-2)

So it's the unitary method

[summer cradle]

which is M and which is L here?

[median locust]

Just solved this with a method someone from a different channel told me

But still confused on how it was solved in the explanation?

[median locust]

I think second blue point and fifth blue point from the top

[summer cradle]

if looking from their method M seems like to be the topmost blue point

actually you don't need a formula here

you can use similar triangles

[median locust]

I haven't learned that yet :/

[summer cradle]

similar triangles?

[median locust]

Yeah

[summer cradle]

oh

[median locust]

Do yk if you can explain how the problem was solved in the explanation?

[maiden brook]

what grade r u in?

[summer cradle]

they used formula

and just plug values in

[median locust]

Freshman

[median locust]

How was the ratio figured out?

[summer cradle]

by counting manually from the diagram

[median locust]

So would point M be the top most point?

[summer cradle]

probably

it's not really clearly labeled in the given diagram

[sleek stratus]

are special angles only used so that you don’t need to memorize values for trig ratios?

[silent plank]

wdym

[knotty quiver]

you mean the angles 30°, 45°, 60°, 90°, etc?

[sleek stratus]

yeah

[silent plank]

wdym by

used so that you don’t need to memorize values for trig ratios?

[sleek stratus]

oops wrong wording I found the answer tho I meant to ask why are they called special angles

do they help find exact values?

[knotty quiver]

what's the answer

[silent plank]

they have simple constructions to get concise exact value representations for their ratios

[obsidian harness]

the trig ratios of other angles get progressively nastier

[inland yarrow]

0 15 30 45 60 75 90 are still okay

[obsidian harness]

the sin, cos, or tan of the vast majority of angles cannot be expressed using the four operations or radicals: https://en.wikipedia.org/wiki/Trigonometric_constants_expressed_in_real_radicals

you can use the half angle formulae on the special angles and you'll get surd values

[inland yarrow]

or you can construct, 75/15 is particularly nice

[lean blaze]

How is sin (210) = -sin(30) im confused

[round beacon]

Radians or degrees?

[lean blaze]

degrees

[maiden brook]

sin(210°) = sin(-30°)

[round beacon]

$\sin\qty({210}^{\circ})=\sin\qty({-30}^{\circ})\Rightarrow-\frac12=-\frac12$ The statement is true.

[somber coyote]

NerdInGlasses

[round beacon]

I think it has to do with the placement of the degrees on the unit circle

[sleek stratus]

can anyone properly explain the unit circle? Is it just that the x axis is cos theta and y axis is sin theta? What is its usecase and does it only apply if the triangle has a hypotenuse (radius) of exactly 1?

[viscid nimbus]

the unit circle is a circle on the xy plane many people are familiar with

yes it has a radius of 1

it's centered on 0, 0

and yes cosx is x, sinx is y

[sleek stratus]

so it only works if the triangle is exactly at 1?

[viscid nimbus]

the thing is

you don't need the triangle's hypotenuse to be at exactly 1

angles are preserved

you can kind of see it as a fraction

let's say in p/q, p and q are even

then they would be equal even in the lowest term because of the proportions

for example 1/2 = 2/4

because 1/2 = 0.5

2/4 = 0.5

what this is asking is

"what is half of 1"

which is 0.5

while the other is asking

what is "half of 2"

which is also 0.5

[sleek stratus]

Oh

that makes sense

tysm

[proud jacinth]

people

if the half o 2 is 1 an 1 is 0.5 and 0.5 is 0.25

whats the half of 0.25?

[jovial maple]

Anyone tutor or help with math by chance

[visual flume]

help pls

aaaaaahhhhhhhhhhhhhhhhhhhh

i got

lim(x->0-) 3sqrt41-3+x

qwq

or like

3√41-3-(1/∞)

sybausybausybua now i got 15>x>15

arghhhhhhh

[jovial maple]

Anyone know how to do this

[visual flume]

use pythagorean theorem

just kidding

try it out

RAHHHHHH

yeah use pytha theorem

[jovial maple]

Okok ill see what I can do

[visual flume]

(side 1)^2 + (side 2)^2 = (longest side)^2

[jovial maple]

Im guess approximately 17.34 what i got

[visual flume]

√301

[native pier]

!nosols

[lime crown]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[native pier]

What is your progress?

[zenith vector]

Hello,this is my first time in this server and I saw this section for geometry so I wanted to ask whats the best way to self teach yourself geometry? I just finished alegbra 1 and I have Geometry next year and plan to study in the summer so I just wanted to ask for any resources or tools to help me learn(I have IXL for practice btw)If you see this later plz@ me since I might not see a response and its very important to me

[mystic umbra]

there's resources

like erm

khan academy?

that's the only one i know

i'd suggest focusing on studying the fundamentals and practising lots

it gets pretty easy after that

:3

[zenith vector]

Alright, will khan academy teach me most of what I need to know because my biggest problem is algebra was sometimes I would have a gap somewhere and I would fully have to relearn the topic😭

[mystic umbra]

algebra is

important

for a lot of things you will be learning

in the near future

better practise more

[zenith vector]

Ok

[mystic umbra]

but

just practise more

[zenith vector]

Ok

[mystic umbra]

you'll get the hang of it

[zenith vector]

Ok thx

[native pier]

I think IXL is really good too when it comes to practice. Teaching, however, it's not so good at.

[zenith vector]

Yeah I've noticed that lol

[native pier]

Welcome, by the way!

[zenith vector]

Thanks

[full plaza]

well

have you guys heard of......

AOPS ONLINE !

https://tenor.com/view/bifrost-bifrost-wallet-bifrost-meme-doge-doge-coin-gif-15065845624597504181

[acoustic hollow]

Can someone help with this?

It's an SAT pratice question and using desmos I got it down to 2 possible answers

How do I know which one it is?

[knotty quiver]

both are correct

[acoustic hollow]

But if it’s open ended which should I pick

[knotty quiver]

anything i guess

[acoustic hollow]

Ic

[knotty quiver]

lemme try

[candid bobcat]

Anyone here into Eulers formula and complex trig?

[knotty quiver]

what did you get

[knotty quiver]

I'm forgetting it

i only got one answer and it's (3,-6)

[knotty quiver]

,,e^{\pi i}+1=0

[somber coyote]

glass

[candid bobcat]

YESSSS

[knotty quiver]

I'm forgetting the formula for cosh and sinh

[candid bobcat]

Except

[knotty quiver]

i remember now

[candid bobcat]

cosh = e^x+e^-x
________
2

[knotty quiver]

[\cosh x = \frac{e^x+e^{-x}}2]

[knotty quiver]

ok i was gonna ask if it's correct

[candid bobcat]

I dont know LaTeX

im sorry

[somber coyote]

glass

[candid bobcat]

How do I use TeXit

[knotty quiver]

i needed hyperbolic trig to remember the formula for representation of trig functions in e

[candid bobcat]

no you don't

Thats eulers formula

[knotty quiver]

try the help command

imean

[candid bobcat]

@somber coyote e

.

Nvm

[knotty quiver]

lemme derive it first

[candid bobcat]

You dont have to

e^iø = cos(ø) + i sin(ø)

[knotty quiver]

oh yeah

[candid bobcat]

I can prove it if you need a proof

Calculus based

[knotty quiver]

i don't

[candid bobcat]

Kk

[knotty quiver]

I'm just trying to recall the formulas

ok

i have to guess first then proved it

guessed formulas of cos and sin bc i know it looks similar to cosh and sinh

so indeed,
[\cos x=\frac{e^{ix}+e^{-ix}}2\quad\text{and}\quad\sin x=\frac{e^{ix}-e^{-ix}}{2i}]

[somber coyote]

glass

[candid bobcat]

Yes

[acoustic hollow]

-5

I got -5 😭

[knotty quiver]

probably the second one?

i probably did mine wrong

graph shows -5

sorry

wait

thi is the correct one. it also matches the upper intersection's graph

[knotty quiver]

did you have a solution or just by graph

[knotty quiver]

(-6,3)

wait wtf my original answer was right. i just swapped the x and y

[acoustic hollow]

Graph

Well ummm

Best I can hope rlly is that the SAT screwed up or multiple answers anew allowed 🤷

[lean blaze]

can anyone explain easily what the ambiguous case of sine is

[obsidian harness]

given two sides and an acute angle not in between, two triangles are possible

[lean blaze]

does it only apply if h < a < b?

[obsidian harness]

indeed!

you must also have angle A being acute

[lean blaze]

so theres 3 diff posibilites right? one for NO triangle one for ONE triangle and one for two triangles? but i dont get why can’t we just place side a and make the triangle? Why do we say it can swing and make two?

[obsidian harness]

cause it depends on:

1. the angle A
2. side a, opposite angle A
3. another side, side b

[lean blaze]

so there can be two possible sides opposite of a thats why?

wait that makes sense

ty

[obsidian harness]

no worries, also you should learn the congruence rules for triangles

and in fact there are only four of them that make a triangle unique

1. side, angle, side
2. side, side, side
3. angle, side, angle
4. angle, angle, side

[lean blaze]

oh yeah

it only applies to SSA?

where A is not an included angle

[obsidian harness]

yeah, cause SSA is not one of the 4 congruency criteria

the ambiguous case shows you why

[obsidian harness]

oh since you know more than I thought, I'll tell you that cases 1) and 2) are the cases for the cosine rule

3. and 4) are for the sine rule

[chrome token]

I was just trying to prove pythagoras rigorously, when i realised i didnt rven know how to define it, i thought maybe we could say pythagoras like a/b + b/a = cc/ab, but how do we even define ratio of two line segments when it isnt an natural ratio, so what are the minimal set of axioms/defns to prove pythagoras?

[mystic umbra]

the pythagorean theorem is an iff statement that says that a^2+b^2=c^2 iff the triangle is right angled

[chrome token]

Ik that

[mystic umbra]

then you know how to define it

[chrome token]

Not rigorously

What does it mean to square a line segment

[mystic umbra]

sorry what?

you mean like squaring the length of a segment?

[chrome token]

How do you define "a^2"

[mystic umbra]

not the same thing

[chrome token]

Yep, how do we define it

[mystic umbra]

well that just gives you

the area of the square

[civic rock]

area of a square side length a?

[mystic umbra]

duh, it's in the name

[chrome token]

Idk how to define addition and congruence of area

[civic rock]

wdym by natural ratio?

[chrome token]

b/a = 1 or 2 or 3, ...

[mystic umbra]

no idea what you're talking about that's just called divisibility

if you meant natural numbers

as the quotient of division

[chrome token]

Well 0 is divisible by 0 but i cant really say ik the ratio

0 as in line segment between two same points

[mystic umbra]

anyways

if you're asking about

minimal set of axioms

erm

euclid's axioms

that's it

[chrome token]

Thats definitely not enough, i need a definition of area

[civic rock]

Didnt Euclid's Elements already include that

[obsidian harness]

it's not defined

[civic rock]

oh

[obsidian harness]

Euclid talks a lot about how certain areas are the same as each other

so equivalence relations

https://hsm.stackexchange.com/questions/17456/ancient-greek-definition-of-area

[chrome token]

Does he define an equivalence relation explicitly? If so i dont remember it

[obsidian harness]

no Euclid is definitely on the wrong track

https://en.wikipedia.org/wiki/Area#Formal_definition

to define 'area' you definitely need some kind of measure theory

[chrome token]

Oh, what about ratio of length of line segments?

Given two arbitrary line segment how could i assign a real number to there ratio?

[obsidian harness]

you need to define a notion of distance then

such as the Euclidean metric on $\mathbb R^n$

[somber coyote]

south

[mystic umbra]

all roads lead to measure theory 😔

[chrome token]

That would make pythagoras trivial?

Can we say parralel postulate + existence of a length function of line segment is enough?

[obsidian harness]

ah yes it would

(I forgot I was specifically talking about the Euclidean metric)

[mystic umbra]

so much to study🥀

[obsidian harness]

you seem to be talking about the metric-space axioms

[chrome token]

Yes

Is that enough for pythagoras? Alongside the ueclidean axioms?

Also, is it possible to have a finite non singleton/empty space which follows euclid stuff?

[lean blaze]

what are reciprocal trig functions used for are they just to look cleaner?

[mild cargo]

I guess so, we arent even specifically taught about them in my education system theyre just in the log books

[knotty quiver]

also more compact to write

[knotty quiver]

in calculus you'll use them a lot i guess

i guess it's just like tangent

instead of writing sin(x)/cos(x) everytime, you just write tan(x)

[maiden brook]

they’re used to give calculus students headaches with derivatives and integrals

[knotty quiver]

there are also ones that don't give you headaches and make you happy instead

like $\int !\sec x,dx$

[somber coyote]

glass

[maiden brook]

yeah

Why does sec make you happy though

[knotty quiver]

idk maybe because it looks complicated at first until you learn the technique to integrate it

and everything starts yo make sense and

you see a logarithm and trig function in the integral

[maiden brook]

Nah there are pretty straightforward ways

[knotty quiver]

i didn't know it. i think i learned it pretty recent

after i learned it, i learned to multiply (a+b)/(a+b) whenever i see an inverse trig function to integrate

[maiden brook]

?

multiply by (a + b)/(a + b)?

[knotty quiver]

where a and b are trig functions

[knotty quiver]

yea

how would u integrate sec x

[maiden brook]

just use partial fractions

that’s the easiest way

[sinful marten]

how do I memorize all the trig identities for calculus?? 🙏

[knotty quiver]

for me, i multiply (secx+tanx/(secx+tanx) then proceed to u sub

[maiden brook]

that’s the “black magic” way

[maiden brook]

don’t need to

[knotty quiver]

learn the few ones then practice deriving others

[sinful marten]

like the sum identities?

[knotty quiver]

i might not know how partial fractions is done to secx

[maiden brook]

try to think about it then

[knotty quiver]

ya, u can learn that. many other identities derive from that

[sinful marten]

Alr thank you so much

[maiden brook]

Ok yeah I would remember the two angle additions ones. But then u don’t needa both with double angle or some others

[sinful marten]

Kk

[knotty quiver]

also deriving/proving the sum identity might help you remember it

[maiden brook]

There’s not rlly a fast way to prove it though right?

But actually the best way might be Euler’s identity

[sinful marten]

I have the sum identities ingrained in my head rn

[knotty quiver]

idk. i tried it once by drawing triangles

[knotty quiver]

that's good

[maiden brook]

Well that’s not rlly a full proof

[sinful marten]

I havent used/learned about euler's identity yet though

[knotty quiver]

sincos cossin
coscos sinsin

[knotty quiver]

i see. then just deriving then

[maiden brook]

u haven’t taken precalc?

[maiden brook]

cause that’s only the acute angle case, at best

[knotty quiver]

true

[sinful marten]

I have, they didnt talk about it though

Im getting ready to take calc bc in a month after school starts

[knotty quiver]

oh but i did it in a unit circle

so the hypotenuses are always 1 unit. makes the calculation simple

[maiden brook]

Did u use the distance formula?

[maiden brook]

Oh

[maiden brook]

Ok

[knotty quiver]

it's been a while I don't remember. i also get errors often, it always take me a few tries before i derive the correct identity

but the method i always happen to come up is something like, u form a smaller similar triangle to the other triangle, and yea computations happen

(two triangles produce a child triangle)

i might draw it

i start with something like this

circle's radius is 1

but yeah, always have to figure out things on the spot. it's not ingrained in my head

maybe that inner small triangle should actually be outside

oh that dotted line should continue further down and that would be your sin(a+b)

[knotty quiver]

i tried solving it and it took me look because of a mistake

i kept writing cos b as 1-cosb for some weird reason

i think I'm comfortable/used to it now at this point

I've forgotten most of highschool geometry terms so I can't mention the stuff there properly

like how those smaller inner triangles are similar

and congruent to the first lower triangle, one with angle alpha

the substitution might be confusing bc it's not top to bottom. I did it on paper that has top to bottom substitution. I did it after many attempts with mistakes

so yea that's how i make sense of the sum identity

because the first time i saw it, it bothered me. idk how it makes sense

felt like i can't go through calc without understanding it

probably bc we had one prof said that calc is hard if u don't know trig

[novel wing]

Here’s an awesome problem for geo lovers

It’s an Olympiad style problem

What you’ll need is just some basic angle chasing and itll do the job

It’s why I like this problem, it tests creatovity more than theory

[inland yarrow]

mods ahh daily problem

[civic rock]

mf left the server funnily enough

also the problem is very easy lol

[pale sentinel]

joins server
drops a geometry problem
doesn't elaborate
well actually they do elaborate it a little bit, saying it involves some "creatovity"
leaves

[civic rock]

"creativity"

difficulty is nowhere near olympiad difficulty

[pale sentinel]

I recognise the word lol, I'm joking abuot the misspelling

[civic rock]

nah the problem doesnt require drawing any extra stuff

just basic properties of cyclic quads

[pale sentinel]

A UK Junior Maths Olympiad problem? sure

[pale sentinel]

[but for context that would be aimed at 14-year-olds at the oldest]

[upper karma]

can someone check my answers in dms 😓

[rocky thicket]

isn't it better to send them here tho

[candid pivot]

Hello

[remote shale]

hello

[undone vortex]

Im on my second attempt of unit test

help

[golden sigil]

Guys I got 146.52 for this question but I want to be sure it's right

Can someone plz check

[knotty quiver]

correct

[golden sigil]

Thank you

[knotty quiver]

,texsp||$x=h(\tan\alpha-\tan\beta)$||

[somber coyote]

glass

[ripe trench]

thun

[frigid dome]

Please help me solve B and explain how you did it. I'm stuck. I know that a triangle equals 180, so that means A in Triangle ADC is 30. I know D in triangle ABD is 110 as BDC is a straight line (180-70). I can't figure out what B or A is though the triangle ABD. Please help. I don't just want the answer based on the angles in the imagine, I want toe maths answer. Thanks

[dark sparrow]

this looks like not enough info

[civic rock]

not enough info

[dark sparrow]

also

a triangle equals 180
bad

[indigo nexus]

im close if i havent make any mistake

its 40

[civic rock]

wrong

[indigo nexus]

what how

[civic rock]

someone already explained it in #help-5

[dark sparrow]

how do you know it's 40° and not 39° or 41° or 1°

[civic rock]

i want toe maths answer
lmao

[civic rock]

I've found the original source of the problem

There's an extra given which is AC + CD = AB

[left mauve]

[civic rock]

||solution: extend DC and let E lie on the ray so that CE = CA
you can prove that DE = AE = AB and from there find the angle ABD||

[maiden brook]

why are DE and AE equal to AB

This shouldn’t make sense cause there shouldn’t be a unique answer

[civic rock]

^

original source is from some Japanese tiktok video

[maiden brook]

Oh

Oops

Yeah then u got it

[ivory ivy]

k

[scarlet rune]

so I was bored and I made a few problems and I would like to 1. make sure they are possible 2. the logic makes sense and 3. someone can solve them

[snow karma]

well everythings good with it

[scarlet rune]

i'll send the rest ig

imma be honest I just started geometry and haven't officially learnt some of these concepts so yeah

[zealous pike]

:3

[snow karma]

the one that asks for the circumference of the circle is good

[civic rock]

this seems like a bunch of stuff thrown together

[snow karma]

im not too sure if its possible to solve the two other

[civic rock]

@scarlet rune error in 4th problem

angle CTR = 138° but you said TV bisects that and CTV = 56° which is wrong

[scarlet rune]

ah

[civic rock]

I haven't solved problem 2 yet

seems fairly ugly

but I like problem 1

[hasty rune]

how do I find the measure of the far right angle

[civic rock]

!1c

[lime crown]

Please stick to your channel.

[hasty rune]

ok

[lime crown]

Please stick to your channel.

[civic rock]

@scarlet rune how is problem 2 even possible

I can modify y to be anything

there's 0 connection here

problem 3 is fine (tho a bit disappointing)

but for a person who hasnt learnt geometry you didnt do too bad of a job

[zealous pike]

Book says that the sum of the angles A and B equals angle 1. Is it because of the exterior angle postulate?

[maiden brook]

Yep

[grave pond]

The supplementary of angle 1 forms a triangle together with A and B -- so 180°-A-B = 180°-"1".

[pale sentinel]

I would argue that you shouldn't use numbers as names for your angles

[grave pond]

Just stick a slightly larger theta to the northwest of the number, and you're good,

[zealous pike]

The question is to find angle(A+B+C+D+E+F+G)

[grave pond]

Then I'd say the sum of the exterior angles is clearly 2×360° because the polygon winds two times around the center.
The sum of the interior angles is then 7×180° - 2×360°.

[zealous pike]

Wind?

[grave pond]

"goes two times around" would be another equivalent wording.

(Consider walking along the polygon until you're back to your starting point. The direction your nose points in will sweep around the horizon two times).

[zealous pike]

Oh

[zealous pike]

Hi

GOT IT GOT IT

OMG

[rich wagon]

is this the right place to ask for help on how to use Desmos? for trig

[civic rock]

sure go ahead

[rich wagon]

how do I find what the x coordinate of this line is?

[civic rock]

x = cos x has a solution called Dottie number

[rich wagon]

ok

never heard of "Dottie number" before

what is it?

[civic rock]

https://en.m.wikipedia.org/wiki/Dottie_number

[civic rock]

well I think desmos doesnt allow you to see the x intercept of that

to find it you should draw the graph y = x and y = cos x

and hover over their intersection

like this

[rich wagon]

how come mine is different?

ohhhh

wait lol

now I just need to add the wave again

wait how doI do that?

it won't show anymore

[dark sparrow]

you're zoomed way in

[rich wagon]

yeah

I just realized lol

thanks I got the problem right

[sharp parcel]

bre jst zoom in, and click on the intersection of it and the x-axis

oh nvmd

i jst realized it dont work

[civic rock]

<@&268886789983436800> scam + spam

[spice nebula]

why is this wikipedia so modern

[civic rock]

?

[spice nebula]

it doesnt have the sidebar

on the link

idk how to explain it

[sand steeple]

cant you just split the shape up

[hasty rune]

that is what i did

[sand steeple]

Since blue is a rectangle you’ll notice that x° is just 90 + something

[hasty rune]

yeah i understand it know

[sand steeple]

And the something is the angle on the green triangle

[sand steeple]

Alr

no worries

[hasty rune]

i need help with this

[sand steeple]

these are where your points are

[hasty rune]

pk

ok

[sand steeple]

now you need to find the other point so that it is isosceles

[hasty rune]

so we need to find the distance

between the bottom pints

[sand steeple]

an isoceles trapezium just means the two slanted lines are equal in distance

[sand steeple]

no no

the last corner (the one we're solving) will be up

[hasty rune]

yeah

so how do we do that

[sand steeple]

well

the light blue point is 1 big square down from the black point

(it is 2 down)

so we need to keep the light blue's x-coordinate, but the final point will be 2 up from the red point

[hasty rune]

id there a way to do this algabraicly

[sand steeple]

not really

This is the shape btw

[sand steeple]

the question is just finding the final corner of a shape, you cant really use algebra to find it

you just need to visualise or draw it out

[hasty rune]

ok

thx

[civic rock]

hmm there are actually 3 solutions to this

[sand steeple]

is there?

[polar kayak]

Am I in the right place chat

[grave pond]

Depends on what you're going to say ...

[zealous pike]

The problem is to find the sum of all marked angles (A+B+C+D+E+F+G). I'm still reasoning through it

A + B + C = 180° + 2 and E + F + G = 180° + 1, right

If so, I think I found the answer:

A + B + C = 180° + 2
E + F + G = 180° + 1

Substituting, we have:

180° + 180° + (1 + 2 + D = 180°)° = 540°

[grave pond]

Just like the other 7-gon that turned around two times in total.

[zealous pike]

What about this one

[grave pond]

Also 540° (but it's a bit harder to see that this one "winds around two times".

[zealous pike]

I think that these problems would be better if they didn't bring up the extended angles, so we can improve our view

[zealous pike]

wait a minute, you got the answer pretty fast. Is that what experience makes with us?

[obsidian harness]

1 + 2 + 3 = 540 - 180 = 360

So a + b + g + 180 - (1) = 360
a + b + g = 180 + (1)
e + f = (2)
c + d = (3)

Hence 180 + 360 = 540

But yeah imagine yourself on point A and then follow the path

That's what Tropo means

If you go from A to G, between D and C you will have rotated in the same direction as from A to G

[twin hedge]

Hey y'all, I am starting geometry soon, any tips?

[midnight sun]

When you see proofs for the first time just know they’ll come back for you

[twin depot]

You’ll be fine. Just try your best remembering theorems. For me I got them down within a day but some people process differently which it’s okay.

[twin depot]

^^^Yes this proofs are indeed a pain but don’t try to be intimidated

[fierce mango]

does anyone have good resources for this topic? like good yt videos

[dark sparrow]

what topic

[left mauve]

Geometry and trigonometry as a whole I suppose

[fierce mango]

geometry

why else would i be in this channel

[left mauve]

[dark sparrow]

you want YouTube videos on all topics in the entire subject of geometry?

ig Khan Academy and The Organic Chemistry Tutor exist -- but you gotta be a good deal more specific about what you want

[fierce mango]

the basics really

[silent plank]

the resources mentioned above are great for the basics

[wispy whale]

yeah unless u wanna go balls deep into geometry

[wary thicket]

Guys I need help

If a= sin10°. What value of sin10°sin80° is

[civic rock]

sin 80° is just cos 10° right?

[wary thicket]

Yes but

The answer is like this

[exotic yarrow]

This is just dumb

Note that cos 10 sin 10 = 1/2 sin 20 = 1/2 sin(30-10)

[upper karma]

can someone help me in dms 😓

[left mauve]

You can open a help channel if you want a channel just for yourself #❓how-to-get-help

[wary thicket]

Hmm so this problem is strange no?

[exotic yarrow]

I just don’t see the point of it

[civic rock]

hmmm

[exotic yarrow]

You’re uglifying it for no reason

[civic rock]

one thing I see is you can divide by 2 in the root

[exotic yarrow]

[civic rock]

so you get the √3/2= sin 60° and all that stuff

[grave pond]

And by this we'd get minus option A, which doesn't even look like it matches any of the options.

[wary thicket]

Alright guys thanks for the explaination

[exotic yarrow]

,w x-(1/4)(x*sqrt(3)-sqrt(1-x^2)), x=sin(10 deg)

[somber coyote]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[exotic yarrow]

,w 1/4(cos(pi/18)-sqrt(3) sin(pi/18))+sin(pi/18)

[grave pond]

Very helpful, Wolfram.

[somber coyote]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[grave pond]

$ calc
C-style arbitrary precision calculator (version 2.12.7.2)
Calc is open software. For license details type:  help copyright
[Type "exit" to exit, or "help" for help.]

; pi=atan(1)*4
; ten=pi/18
; c=cos(ten)
; s=sin(ten)
; (c-sqrt(3)*s)/4
    ~0.17101007166283436652
; s*c
    ~0.17101007166283436652
; 

[fierce monolith]

why is there trigonometry in limits

i hate trigonometry

i hate identities

u either a genius and can derive shi easily or u memorize

[dark sparrow]

trigonometric functions are functions like any other and their behavior deserves to be studied, regardless of your individual feelings on the matter.

are you solution-oriented about this, or do you just want to vent for a bit?

[fierce monolith]

just vent

[dark sparrow]

aight

#chill then

[fierce monolith]

alr

[upper karma]

i have my adv geometry final exam this friday
are there any topics in particular they focus on the most so i can prioritize what to study

[gilded crown]

Try writting sin80 as sin (90-10)

And expand

That doesn't give the answer

Your question is wrong

I think

[grave pond]

Yes, that's what we concluded above too.

[hasty rune]

how do you do this

[civic rock]

use sine theorem to find angle ACB

[hasty rune]

how do you do thiz

[civic rock]

SAS

[hasty rune]

but the angles are not given

oh

angle X

ok

[mystic umbra]

are XYZ and XWV collinear though?

i assume they are because they're depicted that way

[brave tinsel]

ratio of lengths ig?

[final aspen]

Hello is there a software or web app that i can display a 3d surface with limitations, and it calculates de volume of it?

[upper karma]

wolfram alpha? eh

[velvet carbon]

• GeoGebra 3D calculator
• Calc Plot 3D
• WolframAlpha

[hasty rune]

does anyone know the value of z

[zealous pike]

I do

[silent vector]

Get DC from pythagorus then ecludian

Hold on something is missing

[native pier]

no, I think there's enough information

[native pier]

!status

[lime crown]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[silent vector]

Yes, looks like it can be solved with similarity

[native pier]

and if you need help on specific problems, the best way is to go to #❓how-to-get-help

welcome to the mathcord btw @hasty rune and @silent vector

[silent vector]

👋

Always get 45.5625, I don’t know if it is right or wrong

[zealous pike]

The question is basically "there are points $Q_n$ where n = [1, 2, 3, 4, 5, 6]. Find the point $Q_n$ whose angle $AQ_{n}B$ equals angle $AQ_{1}B$

How do I start this?

[somber coyote]

Pi, a future fluent jp speaker

[zealous pike]

[silent vector]

That would take a while, sigh

[zealous pike]

really?

the question is designed for an exam where each question has to be solved under 3 minutes

._.)

[silent vector]

I actually don’t know where to start, maybe something I didn’t study although I have finished high school

[zealous pike]

ah

oh

not really

yea

point $Q_3$

[somber coyote]

Pi, a future fluent jp speaker

[zealous pike]

angles subtended by the same arc are equal

[silent vector]

Oh I get it now

It is really simple but needs to look at the correct circle containing this point

[zealous pike]

now I wonder

If the point $Q_3$ didn't exist or, in other words, there wasn't any point on the same arc as $Q_1$, would it be possible to guess if there is a point with the same angle ? What about guessing that every other point has a different angle?

[somber coyote]

Pi, a future fluent jp speaker

[silent vector]

My eyes hurt

[grave pond]

That arc contains all the points from which you see A to the right of B, separated by the angle at Q1.

[zealous pike]

Wdym?

[grave pond]

Every part not on the arc A-Q1-B has a different angle between its direction to A and B.

[silent vector]

I think that would depend if a circle somehow managed to have the same arc and alpha, but I can’t see which one …. Maybe I am thinking wrong

Or my brain is on power saving

[zealous pike]

So it doesn't matter where in the circles the other points are, their angle to points A and B will be different from point Q1 and Q3? I think I can see why: As the points gets closer and closer to AB, the angle gets wider. As the points gets far away, the angle becomes tighter?

[grave pond]

Suppose you have a point Qn that is not on that circle. Then (unless Qn,A,B are collinear) there's some circle containing A,B,Qn. Its center Cn must be on the perpendicular bisector of AB, but it cannot be the same as the centerC1 of the A,B,Q1 circle, because we assumed that Qn is not on that circle. But that means that angle ACnB is different from angle AC1B, and angle AQnB is half of that, just as angle AQ1B is half of angle AC1B. So angles AQ1B and AQnB are different.

[zealous pike]

interesting

I wish someday I'll develop that intuition

[grave pond]

Writing it down makes it look much more technical than it really is.

The intuition is just:

1. Every point is on some circular arc through A and B.
2. Different arcs have different common angles towards A,B.

[zealous pike]

[pallid osprey]

Guys I’m having trouble understanding the concept in chapter 14 D in methods 1/2 Cambridge textbook, like can someone explain the concept ? How is it different then evaluating sin(theta) etc? And what’s its purpose thanks. The image is basically 14D

[restive solstice]

this is the Unit circle

Its radius is the unit length (the unit could be m, km, or anything). It's used to provide a simple proof of the values of the trigonometric ratios.

[pallid osprey]

Yeah ik but I don’t understand all of this symmetry stuff what’s it’s used for and when do we use it like the textbook gave examples like sin x = 0.6 find sin (180 - x) etc

[fierce monolith]

all students take calculus

A ll
1st quad: all positive
S tudents
2nd quad: only sin positive
T ake
3rd quad: only tan positive
C alculus
4th quad: only cos negative

and for pi - theta or pi + theta n shi like that, search up reference angles in a unit circle on youtube

[restive solstice]

The ASTC rule

[fierce monolith]

yo im lowkey smart at formatic n shi

[restive solstice]

1. All
2. Sin
3. Cos
4. Tan

[pallid osprey]

Dyk when we use these formulas and why?

[left mauve]

switched

[fierce monolith]

which formulas

lets say we wanna find the sin and cos of 135 degrees

its in the second quadrant

so u use the formula for finding ref angle in the 2nd quadrant

which is pi - theta, or 180 - theta is working with degrees

180 - 135 =45

and sin of 45 is sqrt2/2

and cosine is sqrt2/2

but now we apply the astc rule

and since we are in the 2nd quadrant, only sine will be positive

meaning the cosine will be -sqrt2/2

so the sine of 135 is sqrt2/2 and the cosine will be sqrt2/2

but i advise u to watch a youtube vid so u understand and visualise it better

[pallid osprey]

So we use this when it’s not so simple as sin(9pi/2) where u can just simplify using 2pi?

[fierce monolith]

thats coterminal angles

[fierce monolith]

9pi/2 is the same on a unit circle as pi

[obsidian harness]

No, pi/2

[fierce monolith]

oh ye

sorry

[pallid osprey]

This is intro to circular functions so we just evaluating simple sine and cosine like that to find (0,1) etc

[obsidian harness]

Yes so by this logic, 9pi/2 - 2pi - 2pi = pi/2
And so sin(9pi/2) = sin(pi/2)

||= 1|| (don't click the spoiler if you want to quiz yourself)

[fierce monolith]

ye so 9pi/2 is coterminal angle of pi/2

[true musk]

Are you asking what's the difference in doing sin(π-θ) instead of just doing sinθ?
Or are you having trouble understanding why?

[pallid osprey]

Yeah mostly this and why? And what does the answer mean - does it tell us the coordinates in the unit circle?

[obsidian harness]

The unit circle definition tells you that cos θ is the x coordinate and that sin θ is the y coordinate for a point on the unit circle

If you look at the positions of the angles θ and 180 - θ
Imagine θ to be in the first quadrant for example, so 180 - θ is in the 2nd quadrant

The diagram in 14D tells you that these 2 angles have the same height and thus the same y coordinate

Thus sin θ = sin(180 - θ)

[grave pond]

14D? Gosh, that sounds hard to visualize.

[obsidian harness]

12 more hidden dimensions, yes

Jk

[pallid osprey]

But why are we doing all that sin(pi - theta) why not sin(theta)

[obsidian harness]

If you start with sin(135 deg)

Using that identity you get that equal to sin(45 deg)

So it becomes easier

[pallid osprey]

Ohh ok so it’s just faster?

[pallid osprey]

You can also use it to solve for x in similar situations? Like my textbook gave if sin x = 0.8 find sin (180 -x)

Sorry if I’m asking too many questions

[idle goblet]

look at this cool function i found

[wintry ibex]

i made a new constant

[silent plank]

don't troll

[dark sparrow]

go to #chill with this

[wintry ibex]

sorry guys 😦

[fierce monolith]

if it ask u for the reference angle

[obsidian harness]

Immediately sin(180 - x) = 0.8
That's it