geometry-and-trigonometry

[rapid valley]

I did it wrong tho

I tried to shortcut

You need to mutiply 68 by 2

6.7

[kind bronze]

3.37 is correct but it is only the value for the left half of the isosceles triangle

You need to multiply it by 2 if you want to get x

[mellow merlin]

finding tan in quadrant 3 is confusing, i tried finding tan(2/5) but i was actually supposed to find tan(5/2)

like that what ppl probably intuitively do since tan(2/5) is next to the origin

well me atleast

maybe im supposed to find the purple triangle instead of the yellow one?

purple one makes sense

thats where i probably went wrong

[haughty dragon]

i mean i always think of tan y/x

since in q3 both are negative

becomes positive

if you think of it the angle of q3 is the same as the tan + pi

same as q1 angle

rotated 180 degrees

[mellow merlin]

tru

[carmine swift]

Anyone here to help on proofs 😭

[tulip pulsar]

What kind of proofs

[carmine swift]

this

[median plaza]

someone help me please

its special right triangles

[neat wigeon]

the amount of lines means that the angle or side is congruent to another angle or side that has the same amount of lines

so angles D and N are congruent cuz they both have 2 lines
angles C and O are congruent cuz they both have 1 line
line segment BD and PN are congruent cuz they both have 1 line (and obviously a line segment can't be equal to an angle so line segment BD isn't congruent to angle O, etc)

so we have 2 congruent angles and 1 congruent side so the triangles are congruent, proven by angle-angle-side (AAS)

[neat wigeon]

it says the triangle is equilateral, which means all it's angle is 60°,
but the line segment with 11 splits the triangle in half, so now there's 2 triangles with 30-60-90 degree angles

the side opposite to the 30° angle, in this case x, is half of the hypotenuse (but we don't know the hypotenuse so it doesn't matter)
while the side opposite to 60° angle is x multiplied by sqrt(3)
so 11 = x•sqrt(3)
so in order to find x we just divide sqrt(3) on both sides
so x = 11/sqrt(3)
but it asks us to put the denominator as rational
so we just multiply sqrt(3) on the denomination and numerator
(11•sqrt(3))/sqrt(3)^2 = (11 • sqrt(3))/3
so (11 • sqrt(3))/3 is the final answer

[median plaza]

this?

[neat wigeon]

yes

[median plaza]

yessssssssssss

[carmine swift]

@neat wigeon ur such a genius omg

ok i need 1 more and im finish

[neat wigeon]

AC ≈ OM
CB ≈ MN
BA ≈ NO
side-side-side (SSS) congruence theorem

[carmine swift]

omg ur such a life saver 😭

Now these two ixls i need to do

And got to get them at a 80% :/

and they are hard too

Well 1 of them is cause i wasnt here for it

The other 1 I forgot how to do

[neat wigeon]

I'm about to sleep tho

use chatgpt ig

[carmine swift]

yes

[royal charm]

Hi! I need help with trigonometry

[spiral lodge]

!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

[mellow merlin]

;cos(45)

[tame barn]

...

[obsidian harness]

what's the third angle? (angle C)

great, now use the sine rule with sin(C)/15 = sin( ) / ( ) ....

the sine rule can find both side lengths

[obsidian harness]

how is that relevant lol

[obsidian harness]

the key is you need an angle and the side opposite to it
there are 3 such angle-side pairs in a triangle

[vivid silo]

The third angle should be 65°

We can find the measure of side BC by using the value of cos60

[obsidian harness]

you mean AC

also it's the sine rule........

you don't have all three sides, or two sides and the angle in between

so it's not the cosine rule

[vivid silo]

Yes

[obsidian harness]

I just had to correct all of that, nothing personal against you btw

[vivid silo]

Ayoo no worries

[woeful moon]

use sine rule

use cosine for 60

[obsidian harness]

ah I mean I see now

instead of doing sine rule twice you could do sine then cosine

I feel applying sine rule twice is easier though

[vivid silo]

Thanks for making everyone understand

[woeful moon]

ur welcome

[woeful moon]

yeah it can be

[vivid silo]

Which grade btw?

[split river]

How does one find x

[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

[split river]

Why hello bot, I don't know where to begin

[exotic yarrow]

What are some properties about rectangles that you know of?

[split river]

Hmm....

Well all four angles are 90 degrees and they have parallel sides

[exotic yarrow]

Anything else?

[split river]

Nope

[exotic yarrow]

Alright then you may just not have the content required

You need to know that opposite sides are equal

the diagonals are equal to each other

And that the diagonals bisect each other

See where you can get with that

As a check, the sum of the digits of x^2 is ||18||

[split river]

Hmm

[neat wigeon]

every rectangle has congruent diagonals
if u don't know what a diagonal is, it's the sides inside the triangle that go through the opposite angles, line CA and BD
the 2 diagonals meet each other in the middle, which makes 4 sides inside the triangle, each of all these sides are also congruent

all u need to do now is to pay attention to the angles
look at the 60° angle at the top of the center, with this one angle u can figure out all the angle measures
the angle at the bottom of the center is also 60° cuz it's a vertical angle
which means the angles to the right and left are 120°

then all u have to do is pay attention to special triangle rules (30-60-90 and 60-60-60/equilateral)

[fickle thorn]

ABCD is a rectangle => A = 90 degress, B = 90 degrees, C = 90 degrees, D = 90 degrees
Name the center O
BOA has all of its side equal, meaning all of its sides are 15 (because all of the corners of a equal-side triangle are 60 degrees)
COD has all of its sides equal to 15 for the same reason
CO + OA = CA
15 + 15 = CA
CA = 30

And then use pythagoreum theorum on the CAD right triangle to find x

also if u wanna prove that BOA has all its sides equal

The diagonals of a rectangle are cut equally in half at the center(BO = OD) (CO = OA) but only in a rectangle or a square the diagonals are the same (CA = BD) => (BO = OD = CO =OA)

[fickle thorn]

since its a rectangle that is proven by the "theoreums" of parallelogramms

basically they state that the parallel sides are equal

and the diagonals are cut in half in the center (bo = od) (co = oa)

and since its a rectangle the diagonals are equal

sorry for my poor english...

[vivid silo]

Sorry for poor handwriting but I hope this helps

[neat wigeon]

angle abd is 60° cuz triangle BAO (O is center) is an equilateral
also diagonals BD and CA are 30 cm, not x

[vivid silo]

But we can use the value of sine to find out 'x'

[neat wigeon]

not necessarily, we can just use 30-60-90 triangle rules(which i guess is just very simple trigonometry)
the answer for x should be || 15 • sqrt(3) either way ||

[vivid silo]

Just an alternative method

[dense carbon]

That’s an equilateral triangle so you if I label the point in the center as y, then AY, BY, CY, and DY are all 15. Multiply 60 by two and subtract from 360 means mDYA and mCYB is 120. Bisect that to turn it into 4 right triangles on both sides.
The right triangle has the internal degrees of 30 60 90, which makes a special ratio to not need trigonometry to solve.
AY is equal to 15, so the hypotenuse of all the triangles is 15. Divide by 2 to get the base of 7.5, then multiply by the square root of 3 to get the adjacent side to be 14 times the square root of 3 over 2. Multiply by two to get the whole side length as 14 times the square root of 3

Sorry I did a typo

It’s 15 not 14

Fun little problem

I miss when math was simpler

[gritty topaz]

well

know this

and this

[gritty topaz]

by this, we can know this is true for every rectangle

[dense carbon]

Kewl

[sharp marsh]

Okay so i got a question, im just curious about. I was doing chemistry and saw this shape and I was wondering if there was an alternative shape where all 5 "vectors" would be an equal angle between them all

bc like the tetrahedral has 120degrees between all of them and the octahedral has 90degrees

my hypothetical OCD doesnt like that some are 90 and some are 120 lol

I think the problem would be the dimesnsions... like in 2d, itd be possible, and in 5d, i think itd be possible, maybe... am i trippin?

[boreal tundra]

honestly a great question

[sharp marsh]

but is there a great answer 🥲

[boreal tundra]

https://en.wikipedia.org/wiki/Triangular_bipyramid

not sure

[sharp marsh]

huh

[boreal tundra]

I'm trying to visualize this

I dont have blender on my laptop

[sharp marsh]

thats better than im doin, i was using a white board

[boreal tundra]

So just trying to think about it, this shape would need to be two "trigonal planar" shapes orthogonal to each other. So all angles are 120 degrees. Since the "trigonal planar" shape is symmetrical we just pick any one hydrogen then add another trigonal planar essentially on top of it, but orthogonal to the existing one.
Or another way, we just take the existing trigonal bipyramidal then try to expand the 90 degrees to 120. As we increase the angle of the top hydrogen with regard to the left hydrogen, this must decrease the angle with regard to the front hydrogen, as the front hydrogen is not orthogonal to the top-left hydrogens. Due to symmetry this also applies if we do it to any other combination of hydrogens. Since any increase in the angles in one place decreases the angles elsewhere, we cannot change it into a shape with all equal angles.

Not quite a full proof

[sharp marsh]

I seee..... kinda 🥸

what if you made like... 5 equal pyramids connected to a center point?

i think thats how itd work

[boreal tundra]

I mean isn't that just exactly what this is already

[sharp marsh]

uhmmm wait maybe i think so

so then.. could you use that for a proof? or smth

[boreal tundra]

Another thing I'm thinking is that you can essentially draw tetrahedrons around the center atom and 3 vertices. Then it's a matter of showing you cannot create 6 equal tetrahedrons around this central atom.

[sharp marsh]

my intial thought, was taking a more of a "physics" approach, and solving for theta

just have to have the magnitudes of the vectors equal

[boreal tundra]

oh yeah that could work

[sharp marsh]

but like.. thats math... and this is a side task lol

im lazy lol

[boreal tundra]

suppose 5 vectors where the dot product is equal between all pairs of vectors

[sharp marsh]

hmmm wouldnt you be solving for a WHOLE lot of variables?

bc its the i j and k of each

so 15 unknowns

well i guess you would be solving for their relation to each other, rather than their value

[boreal tundra]

I just asked ChatGPT and it said it's impossible

[sharp marsh]

🧍‍♂️

its just not goated like us

does chatgbt have a proof?

[boreal tundra]

it involves linear algebra which I need to take a moment to understand

[sharp marsh]

mannnn im a freshman in college

we havnt done that yet

ive taken phys I and im taking statics and MATLAB

and thats all my relevent knowledge

[boreal tundra]

it's just talking about the gram matrix and saying that you cannot have more than 3 linearly independent vectors in R3

but this cannot be right since in an octahedron this works fine

hmmm, but this formulation that I tried doesn't work though

because it's not quite that every pair of dot products has to be the same

opposite atoms don't need to have a 90 degree angle, because there's an atom in the middle between them that's at 90 degrees

[trail tendon]

stemming from a single point? thats four...

right?

does that count tho

it doesn't work for an octahedron because the vectors "opposite" each other make a greater angle than the "adjacent" ones

[sharp marsh]

another way of interpreting it is to have 6 dots. 1 dot in the center and the other 5 have a radius away from the center and nearby dots. so all dots have to be a r radius away

i think thatd also work... maybe

i just cant do the math for any of this 😔

[trail tendon]

those dots could be anywhere on the sphere, that doesn't tell you anything about the angle

unless those dots represent vectors at the origin that all have dot product 0 with each other

or at least the same dot product

they dont hvae to be perpendicular

[sharp marsh]

have you used a compass before? if you make two points, and draw two circles, there is only two points where a third dot can go (so that they are evenly spaced)

so i think itd be the same concept but with spheres

and where they intersect, you put dots

maybe?

when you draw two spheres, their intersection is a circle

maybe lol dunno

the sphere thing makes the trigonal bipyramid

[vivid silo]

This is all very intresting but a bit confusing can you guys please tell me about this in an easy way I'm a 10th grader though

[sharp marsh]

whats your highest level of math education?

[sharp marsh]

i think the general gist is like... dividing a circle into equal slices

but with a sphere

[vivid silo]

I'm a 10th grader still completing my school

[sharp marsh]

i think bro wants to talk to everyon

[neat wigeon]

huh

[cunning lion]

<@&268886789983436800> spam

[silent plank]

my 👀

[woeful moon]

grade 9

[split river]

I understand the angle part, but what are the special triangle rules? 30-60-90 and 60-60-60? So does that mean the triangle at the bottom is an equilateral?

AFAIK that basically means all sides are also equal then

[obsidian harness]

Yes you can break any equilateral triangle into two congruent right triangles

60-60-60 into two 30-60-90s

[obsidian harness]

That's correct, you find the base angle is 90 - 30 = 60 and hence all sides are equal

[split river]

South

What if instead of 60º, it's 50º?

[obsidian harness]

Would be harder

It's still right angled trig if you break any isosceles triangle into two tho

[deep wadi]

tengo una duda

podeis xplicarme las ecuaciones trighonometricas

[obsidian harness]

si tienes un ejercicio puedes subirlo aquí

si no, hay tantos videos que las explican en YouTube

[deep wadi]

no es un ejercicio es si alguien puede darme ejercicos y yo los hago pero lo que no entiendo es cuando hay dobles soluciones o cuadruples como las pongo

[obsidian harness]

sin x = 1 por 0 <= x <= 360, por ejemplo

la única solución será x = 90

[deep wadi]

vale y estos ejercicoos sen(2x+45)= 0,5 por ejemplo

[obsidian harness]

ah, un momento

[deep wadi]

que haya una ecuacion en el seno o coseno

[obsidian harness]

básicamente debes recordar que sin(30) = 0,5, además sin(150) = 0,5

[deep wadi]

si eso si pero es hacerla

[obsidian harness]

los dos están relacionados por 30 + 150 = 180

2x + 45 = 30, 2x + 45 = 150

la próxima sería 2x + 45 = 30 + 360

[deep wadi]

vale osea tengo que poner la sol q sea igual en dos cuadrantes

claro 360k por el n vueltas

[obsidian harness]

ah sí, la solución general es 2x + 45 = 30 + 360k, o 2x + 45 = 150 + 360k

[deep wadi]

voy a hacerlas mejor ç

[obsidian harness]

sen(x) = sen(180 - x)

la identidad correspondiente por cos es cos(x) = cos(-x) = cos(-x + 360)

y por tan, solo tienes una solución cada 180 grados
las equaciones con tan son más sencillas

[obsidian harness]

espero que mejores, qué tengas un buen día

[split river]

Hey guys

[obsidian harness]

hey

[split river]

I sketched the position of the terminal side in a unit circle when theta = 50º and phi =130º

[obsidian harness]

btw my native language is English!
I really got a good work out trying to talk to this guy in Spanish

[split river]

And they seem to have the same sine value

[obsidian harness]

omg we were literally talking about that

[split river]

And the cosine is the same but negative

[split river]

Dang

[obsidian harness]

have you seen the unit circle before?

[split river]

Yes

[obsidian harness]

great, so you should be familiar with the concept

that cos theta = x-coord
sin theta = y-coord

[split river]

But my teacher did an ass job at explaining it

[split river]

Well I know that, but I had to figure out why myself

[obsidian harness]

it's a definition; don't overthink it basically

it's how we extend cos and sin to all 4 quadrants

[split river]

Yes

But why sin 130 = sin 50

[obsidian harness]

then I think this diagram should explain itself

if you think about it for long enough

[split river]

Ok let me se

[graceful talon]

they have the same y

same level

[obsidian harness]

the important thing is that left of x = 0 means negative

yes the purple lines have the same length, but one is negative and the other is positive

for the cos identity

[split river]

So, simply put, they have the same y, but since the other triangle is in quadrant 2 and the cosine is negative?

[graceful talon]

exactly

[split river]

Interesting...

[graceful talon]

they mirror along the y-axis

[split river]

But how do I put into words that sin(130)=sin(50)

[graceful talon]

wdym

hmm

[obsidian harness]

they have the same height, so they must have the same y-coordinate
by the unit circle, the sines of both angles (alpha, pi - alpha) must be the same

[graceful talon]

does he know radians?

[split river]

No, no I do not

My teacher did not teach that

Nor did she give any resource to learn that

[graceful talon]

it's just another units for angle nothing complicated

π radians is 180° in degree units

[graceful talon]

there's a reference in the pins, 2nd pin

[split river]

This one?

[graceful talon]

yeah

[split river]

Well it doesn't matter anymore

[obsidian harness]

please don't memorise all of that

[split river]

I have only 2 hours left of studying time and tomorrow is summative

Good luck to me ig

[obsidian harness]

just keep drawing triangles on the unit circle so you can relate any angle back to the 1st quadrant

[obsidian harness]

if you flip these 2 triangles down

you get triangles in all 4 quadrants

[obsidian harness]

good luck!

[split river]

Man

Now I have to explain why sine theta is = to cos(90-theta)

[obsidian harness]

another identity, another picture

[split river]

I'm officially cooked

[graceful talon]

I'd draw it as the rectangle

,rotate

[split river]

😭

Thanks

[obsidian harness]

yeah the point is the sides reflected across the line y = x are equal

y = x is the line of symmetry through the 1st quadrant

[graceful talon]

i made a mistake: right side should be sin theta = cos a, not sin a

but it should be the same as the left side

[split river]

I'm failing regardless

Thanks for the help anyways

[graceful talon]

i think it's pretty intuitive

[split river]

Yh but I'm not a math guy

[graceful talon]

like i think it's all visual

[graceful talon]

maybe u could rephrase this as: why the y-side of the triangle of a given angle is equal to the x-side of the triangle of its complementary angle

it's because the angle's x and y sides are switched for the complementary angle

[graceful talon]

you can pass it

goodluck

[split river]

Does me changing the equal sign in the bracket change anything

Typo'd it

[graceful talon]

i think we knew it was a typo

[split river]

...

[graceful talon]

so it was assumed to be a minus sign

[split river]

I'm gonna have to justify everything based off of intuition

And speculation

[graceful talon]

bc it just doesn't make sense to have an equal sign there lol

[split river]

Every single online source uses radians and surds

[graceful talon]

if u see surds, just remember pythagorean theorem

for this topic

[graceful talon]

just reimagine π as 180° and it won't be hard

ok

[split river]

Good night bro

I'm not dealing with this anymore

[graceful talon]

most of the time the triangles used in problems are 30-60-90 and 45-45-90 triangles anyway

[split river]

I hope I dream of some magical theorem or justification

Thanks

[graceful talon]

yeah. i mean it's not the end. after it, you can see where you missed

[obsidian harness]

count right triangles in your sleep

zzz

[mellow merlin]

why is when x = 0, y = -1/2?

wait nvm i think i know

cos0 = 1

i thought it equals 0

[undone vigil]

Hi guys, does anybody have resources for basic trigonometry please?

[silent plank]

khan academy

[vivid silo]

You can also go for pw's lecture on trigonometry by ritik sir

[upper karma]

That would only be useful if she's from India since the lectures are in Hindi.

[timber trout]

2023: SOH CAH TOA
2024: SOH CAH TUAH

[solar python]

that's sin

For any integer n:

cos(2n*pi) = 1
cos((2n+1)*pi) = -1
cos((n+1/2)*pi) = 0

sin(n*pi) = 0
sin((2n+1/2)*pi) = -1
sin((2n+3/2)*pi = 1

And for any x in R: sin(x) = cos(x+1/2*pi)

[upper karma]

what is the formula to find the measure of each interior angle of a regular polygon

[left plover]

Is khan academies geometry and trig course good?

[vivid silo]

Ah yeah I forgot UwU

[upper karma]

Yes.

[left plover]

Thanks I’m trying to refresh my knowledge before I take ap precalc

[upper karma]

Nice! Gotta keep those basics up.

[viscid lance]

(Context: D&D and other such tabletop tile-maps consisting of squares that are 5 feet in length)
How would I determine the total size of a square that consists of 3000 5-foot squares?

Supposedly, the new D&D 5e Tarrasque can be killed by 3 thousand commoners in a single round of combat. So I wanna know how far away the furthest commoner is from the tarrasque in the center (to see if it's even possible for 3000 people to make an attack every round with regular weapon ranges)

It's been a long time since I did basic geometry so the solution seems like it's fairly simple, but I can't easily think of one.

Margin of error here is pretty generous, this is napkin math.

Not when the difference between a longbow's short range is 150 feet and its long range is 300 feet.

I think maybe the solution looks something like this?

• 3000 people, 5 ft length per person, so side-length is 15,000 ft
• sqrt(15000) = 122.47 ft
  So it doesn't fill a perfect square, there'd be empty spaces, which is fine.

[dark sparrow]

uhh no

if you wanna arrange 3k people into a square and look at how big it is you would be interested in AREA not length

so 25 ft^2 per person, 75000 ft^2 for the whole horde if you wanna think of it that way

but also you can just work directly in tiles and find the approximate number of tiles on each side of the square as sqrt(3000)

,calc sqrt(3000)

[somber coyote]

Result:

54.772255750517

[dark sparrow]

so they would all fit into a square 55 tiles wide

surrounding the bad guy in the center means the people at the edges are 28 tiles out, which seems well in range for your ranged weapons

[viscid lance]

I understand that I was wrong, but I'll at least explain how I came to the wrong answer: I was thinking that if it were just a 5-foot-wide line like for a tile map, it'd be the area of 1x(3000x5ft). Which is 15000. So what's the square whose area is 15000.

[viscid lance]

If I had started out properly with base formulae, I could have worked it out better instead of the number of shortcuts I made which introduced error.

[mellow merlin]

sin(2(x-1.5π)) is kinda weird, i need an explaination on why it works

context:

[dark sparrow]

wait

so im confused

[mellow merlin]

also if i have something like sin(3x + 15π) do i do sin(3(x+5π))? does that mean it shifts it -5 units? why tho and why does it work

[dark sparrow]

do you want your 3k people to go single file or do you want them in a square shaped crowd

[viscid lance]

A square shaped crowd. I'm just explaining that when I was thinking of the area of the line, I wasn't converting it into a proper dimension of 5-by-5 squares. Basically only one term of the LxW area (1x3000) line was multiplied by 5.

[dark sparrow]

oh you were explaining where you went wrong w your thought process ok

[viscid lance]

Yeah yeah

[graceful talon]

you could do it to any kind of graph

1. addition to x: translation to left
2. subtraction to x: translation to right

and same with y. addition moves it downward and subtraction moves it upward

then for scaling, same thing. Same analogy

multiplying contracts it or zooms out, dividing zooms in

[graceful talon]

-5π units

[sturdy sand]

how would i go about proving that BC^2 = CE *CA?

[ruby lichen]

Tangent secant theorem.

[ruby lichen]

https://proofwiki.org/wiki/Tangent_Secant_Theorem

[ruby lichen]

Read this for more information.

[sturdy sand]

you are a life saver thank you

[ruby lichen]

Welcome.

[graceful talon]

BC²=CA²-AB²

BC²=CA(CA-AB²/CA)

then CE must be equal to CA-AB²/CA or (CA²-AB²)/CA

[obsidian harness]

triangles ABE and BCE are similar

[sturdy sand]

can you explain?

[late shoal]

How do I solve this?

[graceful talon]

side view, front view, top view, …

[trail tendon]

ugly

[graceful talon]

needs more shading (?)

[sturdy sand]

nvm i got a vision that proved it

[worn stag]

Heron's hyperpigmentation formula?

[obsidian harness]

what do you know about similarity?

ah okay I see

[azure helm]

help please

[harsh ginkgo]

does anyone know the point of Simplified Radical form

the book i'm using is asking me to put it into simplified radical form, and i've done it before but i don't really understand it fully

a problem for example would to use the Pythagorean Theorem to find the hypotenuse (ok, not hard at all) but then it says "Leave answers in simplified radical form"

[obsidian harness]

ah, it's based on prime factorisation

[harsh ginkgo]

why not just leave it in decimal form!?

[obsidian harness]

so if you know that 52 = 4 * 13

$\sqrt{52} = \sqrt{4} \sqrt{13} = 2 \sqrt{13}$

[obsidian harness]

exact form is nicer and easier to recognise

[somber coyote]

south

[obsidian harness]

I mean the advantage of doing this is questionable

[harsh ginkgo]

but i would think that the average person wouldn't know what that means just by looking at it

[obsidian harness]

a simplest form is something that is inherently subjective

[obsidian harness]

what are you referring to?

[harsh ginkgo]

i'd think that a number rounded to the nearest hundreth would be fine

[obsidian harness]

well given we have calculators everywhere it doesn't matter too much I guess

what the actual value in decimal is

[harsh ginkgo]

i'd think that an average person wouldn't know what number radical number means

[obsidian harness]

that's just a "you haven't seen the half of maths" issue

[harsh ginkgo]

neither has the average person, right?

[obsidian harness]

what do you gain by comparing yourself to the average person

[harsh ginkgo]

i don't know

[obsidian harness]

you're in school to learn and of course you'll forget most of it when you leave school

[harsh ginkgo]

i feel like maybe i'm treating this like a document and not like a math problem

[obsidian harness]

hmm

[harsh ginkgo]

like a paper you would do in english class or whatever

that people would be reading

[obsidian harness]

do you know how to find the inradius?

[harsh ginkgo]

what did i do to deserve the 😭

[obsidian harness]

if so then simply use the fact that tangents to the circle must have equal length
you can find AF this way by adding all 3 equations, say (a + b = 5, b + c = 6, c + a = 7, and find a), then it's easy cause you have two congruent right-angled triangles

[obsidian harness]

I mean it's just a convention that everyone who uses mathematics uses

I don't think there's much point overthinking why we write answers in this form

[harsh ginkgo]

it confuses me

[obsidian harness]

okay, what about it confuses you?

[harsh ginkgo]

i have trouble doing it because i'm used to doing something like 13.53

if 13.53 was the answer, i'd be ok with that

[obsidian harness]

might I suggest a video on this topic then

[harsh ginkgo]

finding simplified radical form instead feels like too much of a hassle

[harsh ginkgo]

get this

i do have one

but i can't find the one where the guy explains why simplified radical form exists

[obsidian harness]

I guess it's useful when dealing with similar triangles

it's easier to see that 2 sqrt(7) and 6 sqrt(7) are related

than

,calc 2 sqrt(7)

[somber coyote]

Result:

5.2915026221292

[obsidian harness]

and

,calc 6 sqrt(7)

[somber coyote]

Result:

15.874507866388

[harsh ginkgo]

ohh so like if the decimal is different after the hundreth but the same before?

[obsidian harness]

I mean it's subjective, but in the heat of the moment you probably wouldn't think they are multiples of each other

[obsidian harness]

that's something unrelated

[azure helm]

how

I think there is a formula

[obsidian harness]

indeed

if you can find the area of triangle ABC first

[azure helm]

maybe herons

[obsidian harness]

then consider the three triangles ABI, BIC, CIA

[obsidian harness]

yep!

[obsidian harness]

and the heights of all these triangles is the inradius

[azure helm]

can you help

[obsidian harness]

how so

[azure helm]

Iam stuck

[obsidian harness]

on Heron's?

[azure helm]

ye

and finding inradius

[obsidian harness]

the semiperimeter is just half the perimeter

[azure helm]

True

[harsh ginkgo]

how do i find the simplified radical form

am i supposed to use the largest number that can divide into the radical or does it not matter or what

[obsidian harness]

it doesn't matter if you keep simplifying the radical

[harsh ginkgo]

so then how do i do it?!

[silent plank]

you want to try identifying larger perfect square factor(s)

[lyric halo]

Use the pythagorean theorem

Get the third side

Compute sin cos tan

[hasty latch]

Okay

[lyric halo]

Estimate the angle

[graceful talon]

don't get cooked, chicken

[loud rock]

how do I find trig ratios using a coordinate?

[tired quartz]

yo wtf is this

[opaque meteor]

math

[tired quartz]

help me

[opaque meteor]

bro i need help too 🤣

Funnily enough its with a similar equation

plz halp

[stray gulch]

Fish.

[proven lantern]

It's not that hard

Just google

@tired pine i needs ze help

Can u help me figure out why the values of the trig ratios are what they are.
Like the reasoning behind forumulas

[dark sparrow]

you are kind of setting yourself up to get hit with the very same "it's not that hard, Google it" line that you just said yourself :P

which angles are you interested in tho?

[vivid silo]

We can find the value of AB by using the sine ratio and then by using the pythagoras theorem we can find the length of BC

We can also find BC by using the cosine ratio

[proven lantern]

Not really

He's missing a formula I'm missing an explanation

And I specifically asked a person as I trust their judgement

[proven lantern]

Like the reasoning for values of the ratios

[dark sparrow]

ok but which angles?

[proven lantern]

Any

[dark sparrow]

even ones like cos(69°) which have very ugly exact forms?

[proven lantern]

Like why is sin 30° 1/2

[proven lantern]

No

Standard ones

[dark sparrow]

so like what. 30, 45, 60 degrees?

[proven lantern]

Yeah

[dark sparrow]

that's why i asked. maybe you wanted something more exotic.

[proven lantern]

Extending to 2pi

Oh

No

[dark sparrow]

you can get 45° by looking at a square with a diagonal drawn inside it, and 30° and 60° by looking at an equilateral triangle with one altitude drawn

[proven lantern]

Yeah but

Why is sin 30° = 1/2

[dark sparrow]

sin(45°) is not 1/2

[proven lantern]

Typo

[proven lantern]

I should watch a vid on this

Thanks for illustrating this

[dark sparrow]

if something is not clear in the diagram you can ask me

[proven lantern]

Yeah, can you explain the 1st diagram

[dark sparrow]

i am looking at a square with side length 1

which of the measurements are not clear to you?

[proven lantern]

Ok i get it, you have to know the exact value beforehand to draw this out right?

[dark sparrow]

wdym

[proven lantern]

Could you maybe show me how you calculate the exact values

[dark sparrow]

im just taking the side length of the square as 1

[proven lantern]

Like how are the values calculated

[dark sparrow]

are you asking why the diagonal is sqrt(2)?

if so, you can get it via Pythagoras

[proven lantern]

Sure

But just values in general

[dark sparrow]

...

[graceful talon]

formula?

[dark sparrow]

from my perspective it appears as if you are flip-flopping between asking me about generality vs specifics

i cannot possibly tell you in full generality how to find any length or angle on any diagram

[graceful talon]

zooming in, zooming out

[proven lantern]

No I'm asking how you get the formula

[dark sparrow]

what formula

[proven lantern]

Well the values of the trig ratios

[dark sparrow]

ok right...

do you know the SOH CAH TOA mnemonic

[proven lantern]

Yes

[dark sparrow]

for the definitions of trig functions as ratios of sides

yeah so

you use that

with the angle you want and the relevant sides

which you read from my diagrams

[proven lantern]

How

To get the exact values

[dark sparrow]

sorry i am once again confused as to what you are asking

[proven lantern]

Like can you solve it

[graceful talon]

does he mean deriving the trig formulas from Pythagorean ?

[dark sparrow]

are you asking

"So how did that sqrt(2) happen?"
or
"So how do we get sin(45°)=1/sqrt(2) from this?"

[dark sparrow]

she*

[proven lantern]

"you can get it via pyth"

[graceful talon]

sorry, the pfp

[dark sparrow]

ok so, do you know Pythagoras' theorem

[graceful talon]

sorry again. didn't see it well. it's a woman

[proven lantern]

Both, but you can explain any you like.
I asked for the latter before

[proven lantern]

Yes

[proven lantern]

Yes

[dark sparrow]

ok right so

if you look at the right triangle that makes up the bottom right half of the square

its legs are 1 and 1

hyp^2 = 1^2+1^2

hyp^2 = 2

[graceful talon]

maybe you don't

[dark sparrow]

hyp = sqrt(2)

does that make sense to you?

[proven lantern]

Yes but, why are the sides 1 & 1

[dark sparrow]

because i started with a square of side length 1

those are its sides

[graceful talon]

in many examples tho, they choose the hypotenuse to have the length 1. By doing so, the trig ratios become more simplified

[proven lantern]

So are the sides & adjacents always assumed to be 1

[dark sparrow]

no not always

i do it that way because it makes things simpler

[graceful talon]

maybe just use variables lol

[dark sparrow]

it is a matter of convenience

[proven lantern]

Could you explain the 2nd diagram

[dark sparrow]

in the second diagram i start with an equilateral triangle whose sides are all equal to 2

[graceful talon]

well there are like 2 commonly used right triangles: the 30-60-90 and the 45-45-90 (those are angles)

[dark sparrow]

(i chose 2 again for convenience)

in an equilateral triangle all interior angles are 60° as marked in the bottom left

do you follow so far?

i am not done with the explanation but i want to make sure we are on the same page

[proven lantern]

Yeah, I'm curious is this the actual way the values are derived ?
If you chose a random number say, 57, would you still get similar results?

[dark sparrow]

you can start with 57 if you want yes

[proven lantern]

Wait what

[dark sparrow]

you would eventually end up seeing those 57s cancel out

[proven lantern]

Wait can u solve that

[dark sparrow]

it would create like twenty times more work than necessary

[proven lantern]

Umm, maybe a smaller number

[dark sparrow]

you want me to write down the same things for a triangle of side 57?

[proven lantern]

I'm not sure, i just want to see a random prime number be used in place of 2

[dark sparrow]

i am not sure what you mean by "the actual way" also. do you think that every single math result has One And Only One Correct Derivation™️ and that all others are Just Wrong Before The Math Gods Say So™️?

[dark sparrow]

57 isn't prime

[proven lantern]

Shiet

[dark sparrow]

do you specifically want a prime number just for the sake of signing me up for tons of extra work

[proven lantern]

I don't know I'm just confused and want to know how the actual values came about

[dark sparrow]

yeah so i am telling you...

no matter how you go about it there's gonna be one point where you have to pin down one specific length in a diagram and you can do it arbitrarily

i am presenting to you the choice that i think is most convenient for me personally

[proven lantern]

No LOL, it just seems weird for a prime number to give similar results in place of an even number.

[dark sparrow]

you are now trying to nitpick your way about it

[graceful talon]

we be always understand examples more

[proven lantern]

You can pick 7 idk

[dark sparrow]

look, swapping out the 2 for 7 will only have the effect of scaling the diagram up by some factor

the angles will not change

and the ratios won't change either

[proven lantern]

Yeah but
I just want to see how it's done

[dark sparrow]

ugh ok fine one moment

[proven lantern]

Thankssss

[dark sparrow]

is this what you wanted to see?

[proven lantern]

I think so. What's the result change to now?

[dark sparrow]

what result?

[proven lantern]

Like, the change of the values by replacing 2 for 7

[dark sparrow]

...

sorry, i really do not understand what you are asking

WHICH values?

i have put all lengths on the diagram.

you have to be more specific in what exactly you're asking me for

[light fable]

in a 45-45-90 triangle, the legs are always equal because the two 45 degree angles are equal. It's a basic fact of isosceles triangles

[proven lantern]

Like the sin/cos/tan values, when you took it as 2 the values were identical to what's used but 7 alters that

[spiral lodge]

It doesn't alter the ratios

[light fable]

and in a 30-60-90 triangle, the hypotenuse is always twice as long as the short leg

[proven lantern]

I mean not technically but the result isn't the same now is ir

[spiral lodge]

It is the same...

Calculate it and tell us of it is or not...

[dark sparrow]

no. sin(30°) = (7/2)/7 which still simplifies to 1/2.

[light fable]

7 ÷ 7/2 = 1/2

[proven lantern]

Yeah that's what i was asking

[dark sparrow]

no, the result still is the same. we just get there by a different, more circuitous route.

[dark sparrow]

ok so even just saying "the trig values" would have clued me in.

[dark sparrow]

(7/2)÷7

[proven lantern]

I mean it's obviously a different number before simplification

[light fable]

oops right, of course

[dark sparrow]

it looks different but it is the same number

[proven lantern]

Yeah i know

[spiral lodge]

Then why did you ask? 😅

[light fable]

I wonder if they were asking about like sin(57 deg)

[dark sparrow]

if you want to say that numbers which differ in form are different themselves, then you have to say that English "five", Spanish "cinco", Russian "pjat' ", Japanese "go" etc. all refer to different numbers

[dark sparrow]

no

[light fable]

yeah you're right

[proven lantern]

? What i asked was different, I obviously know it simplifies to the same thing but wanted to know what the non-simplified result was

Oops

@spiral lodge

Anyway, thanks y'all

[spiral lodge]

👍

[smoky halo]

I'm wondering how is sin of 120 degrees equal to √3/2?

[wise pawn]

I'd draw a picture

[mystic crest]

Same
It’s according to the root3,1 and 2 right angle triangle, with 30 and 60 degrees

Sin60 = sin120 = root3/2

Cba to draw a diagram rn

[maiden brook]

Something I thought of:
If A, B and C are three angles in a triangle, what is the maximum value of sin^2(A) + sin^2(B) + sin^2(C)?

[light fable]

I imagine it's when A = B = C = pi/3

[maiden brook]

Yeah but why

[light fable]

feels like a good guess. It's probably provable with something like Lagrange multipliers. I don't expect there's a very elementary proof

it feels like a good guess because we want to make all three angles as large as possible at the same time

and the form of the function is symmetric

so the triangle that does it is probably symmetric too

[maiden brook]

rlly?

for the last part

[light fable]

yeah, you're dealing with a smooth function of three variables, looking for a max

with a constraint of A + B + C - pi = 0

A,B,C all in (0,pi)

this is the kind of setup Lagrange multipliers are good for

uh did one of my sentences not go through somehow?

I was saying you're looking for a max of a smooth three variable function

[opaque meteor]

Thabks

[sacred nimbus]

I need help with number 30 please

[daring bronze]

can someone help me out with this one please

last question i need to be done with the assignment

[upper karma]

It's fairly simple... Since you know de and cd ...using tan alpha' ... You can figure out alpha ... And then alpha = alpha'... So using tan alpha ' and AC is given ...so you can find the height of tree

[mystic crest]

Use sin rule?

[obsidian harness]

that is possible but then you know, sin(90 deg) = 1

[mystic crest]

That’s very true, but also sin 50 from 180-(40+90)?

[obsidian harness]

yeah so you only needed to use right-angled trig

the other angle is 50 correct

[mystic crest]

🙏

[outer hazel]

equation to find radius of red circle? (centre is intersection)

[outer hazel]

only points are given no extra knowledge

[pastel berry]

how do you find the equation for a circle given 3 points on the circle?

[outer hazel]

(x2 + y2)A + xB + yC + D = 0

[fading dove]

The given coordinates are the center of the 3 circles

Let the blue circle be s1, black one be s2, and the green one be s3

Circle red should be assumed s4

S2 might be useless in the question

S1 and s3 are having two common tangents

[fading dove]

Isn't the equation of a general circle x^2 +y^2 + 2gx + 2fy + c = 0

[outer hazel]

Nice!

[outer hazel]

Tbh I just copied off google idk much yet

[fading dove]

Ohh

[outer hazel]

I’m in the middle of the circle topic rn

[fading dove]

Since it's two common tangents c1c3 = |r1-r3|

[fading dove]

Oh i see

[outer hazel]

Do you mind drawing this? I think I get it, cause the two tangents will meet at centre s4?

[fading dove]

Nope, the two tangents will be at the top, the midpoint of c1c3 will be the centre

Neglect the tangents and s2 as of now

[outer hazel]

Oh ok I think I might need to see it drawn 😆

[fading dove]

Draw it but without s2 aka the black circle you have drawn, i bet you added that to traumatize people

[outer hazel]

Ha, I honestly drew it to show that the two points of intersection of each circle, intersect at the centre of the red circle

Aka the unique circle that connects the three points

[fading dove]

Ohh i see

Since the two centres are extremities and forming the largest chord aka the diameter

Their midpoint will be centre of the 4th circle

Once you get the mid point aka the centre

For radius we use the formula root(g^2 + f^2 - c)

Since it's general circle

[outer hazel]

Good! Also fun little fact, if we use this radius for the other 3 circles, all 3 circles will meet at the centre of Red

[fading dove]

Yes

Asper c1c3 = |r1-r2|

[outer hazel]

How did you find the value in terms of x and y of the other point on the diameter?

[fading dove]

There are certain formulas

In circles chapter

With that we manage to do it

They must have told you about the equation of standard and general circles

[outer hazel]

Great! I’ll try and find the algebraic equation for the entire process

[fading dove]

These formulas were derived from their equations

[fading dove]

Cool

[outer hazel]

Yeah (x+x1)^2 + (y+y1)^2 = r^2

[fading dove]

And there are more such forms

For standard centre is (0,0)

So x^2 + y^2 = r^2

I have mentioned for general circle above

[outer hazel]

How do you find where two circles intersect?

[fading dove]

Intersecting circles means they aren't just touching so we know that 2 poi is there

[outer hazel]

Do you use simultaneous equations to find their co-ordinates or something like that?

[fading dove]

You can do that but it will be extremely lengthy if you directly substitute the circle equation

So there will be 2 common tangents

Assume those tangents as y=mx+c

You can find the equation of the tangents by substitution

Then you have to find the perpendicular lines that intersect them both

This distance of perpendicular lines might vary depending on the scenario of the circle

The line will intersect both of the poi

From that you can find the coordinates by using simultaneous equations

[meager glacier]

Should i be concerned if i think my trig class is too easy

[cloud yarrow]

no

[meager glacier]

Theres a new lesson every class but then an entire week for just reviewing and each lesson has like only 5 homework questions

I gotta take all the calculus i just hope im gonna be prepared next few semesters

[cold sun]

is there a reason why 23 degrees is explicitly crossed?

this is my teacher's work, i wasn't at school so i missed his explanation

[trail tendon]

can you see that the angle theta is obtuse?

[dense carbon]

Someone please check my work my 8th grader friends don’t understand any part of it

[cold sun]

omfg u right

[dense carbon]

Oh this makes sense
It imagines a bigger triangle where the one shown is one part of it
You find the side opposite of 15 degrees by doing sin(15) then multiply by 30 to find the length
You divide that length by 20 because you can imagine a smaller right triangle there, then do arc sin to it to find the supplementary angle to theta. However since it is not equal to theta, you subtract it from 180 to get theta

I’m going to sleep now

[cold sun]

Alright

[left cosmos]

can someone help solve for x and y here

[lunar jackal]

Are you aware of the geometric mean

[sudden harbor]

u could use pythagorus theorem and make diff equations

[worn scroll]

or euclid

x/4 = y/(20-4)

[manic belfry]

hi

please if anyone has a cheat sheet about trigonometry

[meager glacier]

Yes write down SOH CAH TOA

[remote dune]

can i get help for my geometry work?

so confused

[light fable]

what do those little curvy things on angles X and Y indicate?

[terse stirrup]

It’s supposed to represent an angle

[terse stirrup]

What have you done so far?

(Wanna try #❓how-to-get-help )

[light fable]

yes but that's not all they indicate when they're both a single arc like that

as opposed to one being a single arc and another having two arcs

[upper karma]

why u write in green

[dense carbon]

Idk

[left cosmos]

Pretty sure c = 16 because the angles show it's an isosceles triangle.

[forest hemlock]

anyone knows how this works

[upper karma]

Im confused with the 90°

[silent plank]

90° appears to be the measure of arcUV

consider

[upper karma]

Do you guys know how to prove cos(a + b) = cos(a)cos(b) - sin(a)sin(b)

[mystic crest]

Could try by induction maybe

[solid needle]

if you knew complex numbers, you could prove this plus sin(a+b) simultaneously

do you accept that when complex numbers are written in polar form, their arguments are added?

this is imo the most important, most intuitive derivation for these formulas but does require a bit of prerequisite here

[solid needle]

if you accept this, or are happy to give this approach a try (watch https://www.youtube.com/watch?v=-dhHrg-KbJ0 starting 9:20 ish to understand why this is true) ping me and ill fill in the rest of the details

[upper karma]

Thanks

[lost lotus]

dude

wher arctrigonometry

[maiden brook]

first prove cos(a - b) = cos(a) * cos(b) + sin(a) * sin(b)

[lost lotus]

....

[maiden brook]

Hint: Draw angles a and b on the unit circle and think of another, more convenient way of expressing the angles

[lost lotus]

wait bye i gotta sleep its 11pm in my time

im in asia

[maiden brook]

gn

[maiden brook]

what country just curious

[obsidian harness]

picture proof

ah you can get the identity for cos(a + b) from cos(a - b)

just replace $b \mapsto -b$ in the formula

[somber coyote]

south

[maiden brook]

but that doesn’t work for non-acute angles so u have to do some more work

[upper karma]

I think I need an algebraic proof rather than a visual one

But sure, Ill try your method

[upper karma]

Also, I understand now

[solid needle]

to be clear, are you saying that you know the full derivation now or do you mean you're ready for me to explain the next steps?

[upper karma]

I didnt understand the derivation but I understood the concept

[solid needle]

ok, would you like me to finish the derivation?

[upper karma]

Sure why not

Also thanks

[solid needle]

fyi i have an appointment in 5 min

so if i leave ill be back later

so we know complex numbers can be written either in rectangular form (a+bi) or polar form (there are standard ways of writing this but i will write it as r cis t, where r is the magnitude and t is the argument)

you may also write it as $r \angle \theta$ in your notes to make it more intuitive

[somber coyote]

Cozmogrgdfschkipkhrshtensi

[solid needle]

so for now lets set r to 1, we dont need it

do you at least understand from just definitions of sin and cos

that:

[upper karma]

Yes

[solid needle]

cis t = cos t + i sin t

[upper karma]

Wait what is cis

Is that a typo

[solid needle]

i defined it as this

polar form of a complex number

where t is the argument, or angle

the direction of the vector when you plot the complex number

[upper karma]

Ah sorry I wasnt reading

Ok cool cool

[solid needle]

ok so

we also just learned that

cis t * cis s = cis (t+s)

because when you multiply complex numbers, the angles are added

that was the key insight from the video i linked

so far so good?

[upper karma]

Im tryimg to catch up

Hol on

Im kinda slow

[solid needle]

np

[upper karma]

Sorry about that

[solid needle]

take your time

[solid needle]

one additional comment before you read past this

this is also why we call this function cis

[upper karma]

I really froze around what cis is representing

[solid needle]

cis x = cos x + i sin x

[maiden brook]

It’s not a visual proof

[solid needle]

Cos I Sin -> CIS

[upper karma]

Ah

Damn

Is it really just cos times i(sin)

[solid needle]

brb half hour

[upper karma]

Cool

[upper karma]

More like a proof where I dont need to draw anything

Just jumbling around variables

[solid needle]

yeah

[summer ferry]

Is this how De moivres theorem was proved, Or was it the other way round?

[solid needle]

imagine r was the hypotenuse of the triangle

can you see that x is cos and y is sin

[solid needle]

basically yes

[upper karma]

Im really lost

But Im still gonna try to understand it

[solid needle]

type out any questions, ill respond when i can

we take it one step at a time

[upper karma]

I got this one

[upper karma]

This one, eh idk

Wait actually yeah I understand now

Im sort of piecing out the puzzle pieces

Im gonna try to solo understand this if I can

Nah I couldn't really do it

Ill try youtube vids

Sorry ill be sleeping now

If you ever want to explain it to me, just continuously chat it and ill read tomorrow

Again, thank you very much for discussing this to me

[solid needle]

np, this kind of stuff always takes time

ill try to explain in more detail, a bit slower this time

so complex numbers can be represented as a+bi, where a and b are real numbers

they can also be plotted on the complex plane as (a,b)

so for instance, the number 2+3i could be represented as the point (2,3) on the complex plane

dont worry too much for now about why this plotting is useful or makes sense, itll click later

the question now is, is there a different way to represent a complex value?

well if we can say a+bi can be represented as a point (a,b), then the opposite is also true

any point (x,y) in the plane can then be x+yi

so if we have another way to represent the point (x,y), then that would be another way to represent the complex value x+yi

instead of giving x and y

we can say, starting from the origin, move a distance of r in the direction t

this also gives us the exact position of the point

for example, move sqrt2 units in the 45 deg direction (up-right)

you land on the point (1,1)

so this distance r and direction t can therefore also be used to represent a complex value

we call this polar form

lets write this "distance r direction t" as r cis t

[solid needle]

and thats what this is

where i use t instead of theta

now the question is just

how do you convert from a polar form to a rectangular form?

this is easy if you know the definition of sin and cos

look at that same picture

use t as the reference angle

sin = opp/hyp = y/r

so if sin t = y/r, then r sin t = y

the same steps can get you that r cos t = x

so we can now substitute x and y in x+yi

x+yi = r cos t + i r sin t

factor out the r, and this gives us that

r cis t = r (cos t + i sin t)

where, once again, r is the distance to the point representing the complex number

t is the direction

(note: this next comment about the orientation of direction will only be relevant once you start plugging in values for sin and cos, you can ignore for now)
this direction is oriented such that positive x direction (right) is 0 degrees, clockwise is negative, counterclockwise is positive

so straight up is 90 deg counterclockwise from 0, so it is 90 deg

straight down would be -90 deg, or 270 deg

once you understand polar form, we can proceed with the next steps

take your time, and have a ping for bookmark reference @upper karma

[upper karma]

No, I reallt understood what you meant

[upper karma]

I understood the polar form from here

Im sorry if I made you say a lot

I mean, I could go for a faster one

[sturdy gyro]

Hey, so I took Geometry around 10th grade but I forgot pretty much everything lmao, does anyone have any textbooks or YT courses or something like that they could recommend? Or just specific topics I should focus on?

[solid needle]

good, no worries

do you get why cis a * cis b = cis (a+b)

[torn ridge]

EGMO OF EVAN CHEN

[glass bronze]

guys ive done trig at highchool level and pre calc level but not further anyway you got any help regrading my concept of trig?? its a bit rusty

[upper karma]

Not really. Actually that was I was trying to prove

Also, I understand why e^iπ = cosπ + isinπ

e^iπ is -1

cosπ is just -1

isinπ is 0

I still dont know the algebraic proof though

How its derived

[upper karma]

Clarification

(cos a + isin a)*(cos b + isin b) = cos(a + b) + isin(a + b)

Is this the correct expanded form?

[obsidian harness]

yes, and if you accept this fact, here are the next steps:

you're using cos a cos b - sin a sin b = cos(a + b), because i^2 = -1

and sin a cos b + cos a sin b = sin(a + b) for the imaginary parts

[obsidian harness]

the idea behind this is that angles add and magnitudes multiply

[upper karma]

Cool

[obsidian harness]

yeah compare real and imaginary parts

[upper karma]

Is it actually possible to prove cos(a + b) = cos(a)cos(b) - sin(a)sin(b) without any triangle or drawing anything?

Purely by using trigo definitions

[obsidian harness]

you could use the exponential forms for cos and sin and do it algebraically

$\cos x = \frac{e^{ix} + e^{-ix}}{2}$ for example

[somber coyote]

south

[upper karma]

Woah

[upper karma]

e^(ix) = cos(x) + isin(x)

e^(-ix) = cos(-x) + isin(-x)
e^(-ix) = cos(x) - isin(x)

(e^ix) + (e^-ix) = cos(x) + isin(x) + cos(x) - isin(x)

(e^ix) + (e^-ix) = 2cos(x)
[(e^ix) + (e^-ix)]/2 = cos(x)

I'll try it on cos(a + b) identity now

e^(ix) = cos(x) + isin(x)
e^(ix) - {[(e^ix) + (e^-ix)]/2} = isin(x)
[2e^(ix) - e^ix - e^-ix]/2 = isinx
(e^ix - e^-ix)/2 = isinx
(e^ix - e^-ix)/2i = sinx
-i(e^ix - e^-ix)/2 = sinx
???

Is this right

Also, if its like, lacking or still needs more simplification, how?

[maiden brook]

alr. this is probably my favorite proof

(of only the two ones i have seen..)

[upper karma]

Amazing

[solid needle]

so this is what i was saying in the beginning

[solid needle]

oops i meant this

[solid needle]

this video and the timestamp I gave

gives a visual intuition of why when you multiply complex numbers, you add the arguments

if you can accept this very important core idea of complex numbers, you get

cis (a+b) = cis a * cis b

for free

this prerequisite might not be entirely algebraic, but like south said, you can derive the e^it form using calculus and stuff

that is a little bit more advanced, the geometric way is, at least imo, by far the easiest way to intuit it

the rest can be totally algebraic if you want

an extra comment that is totally not necessary but for fun:

so we have

$\cos x = \frac{e^{ix} + e^{-ix}}{2}$ as south mentioned

[somber coyote]

Cozmogrgdfschkipkhrshtensi

[solid needle]

there are these uncommonly used variants of the trig functions, called the hyperbolic trig functions

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

[somber coyote]

Cozmogrgdfschkipkhrshtensi

[solid needle]

similarly, any hyperbolic trig function can be defined in a similar way:

sinh x = -i sin ix (thanks ann)

so if you ever see those in the wild, even if you dont yet understand anything about them, at least you rigorously know what they are

[dark sparrow]

sinh(x) = -i sin(ix)*

[lucid thicket]

how to prove this?

[solid needle]

oops hehe

how did i mess that up

i think i need to review

[dark sparrow]

do you know the definition of "odd function"?

[lucid thicket]

f(-x)=-f(x) right?

[dark sparrow]

yes

[solid needle]

easiest thing to do is ||plug in both sides and show both sides are equal||

alternatively, ||show that the base function is odd and the transformations applied turn an odd function into an odd function||

oh i see what i got wrong

cosh x + sinh x = e^x

no i in front of the sin term here

[upper karma]

@solid needle I figured it out now

prove cos(a-b) = cos(a)cos(b) + sin(a)sin(b)

cos(a-b):

cos(a-b) = e^[i(a-b)] + e^[-i(a-b)]/2

[e^(ia-ib) + e^(ib-ia)]/2

cos(a)cos(b):

[e^(ia) + e^(-ia)][e^(ib) + e^(-ib)]/4
[e^(ia + ib) + e^(ia - ib) + e^(-ia + ib) + e^(-ia - ib)]/4
{e^[i(a+b)] + e^[i(a-b)] + e^[-i(a - b)] + e^[-i(a+b)]}/4

e^(ix) = cos(x) + isin(x)
e^(ix) - cos(x) = isin(x)
e^(ix) - {[e^(ix) + e^(-ix)]/2} = isinx
[2e^(ix) - e^(ix) - e^(-ix)]/2 = isinx
e^(ix) - e^(-ix)/2 = isinx
-i[e^(ix) - e^(-ix)]/2 = sinx

sin(a)sin(b):

{-i[e^(ia) - e^(-ia)]/2}{-i[e^(ib) - e^(-ib)]/2}
[e^(ia) - e^(-ia)][e^(ib) - e^(-ib)]/4
[e^(ia + ib) - e^(ia - ib) - e^(-ia + ib) + e^(-ia - ib)]/4

cos(a)cos(b) + sin(a)sin(b):

e^[i(a+b)] + e^[i(a-b)] + e^[-i(a - b)] + e^[-i(a+b)] + e^(ia + ib) - e^(ia - ib) - e^(-ia + ib) + e^(-ia - ib)/4

e^[i(a+b)] + e^[i(a-b)] + e^[-i(a - b)] + e^[-i(a+b)] + e^[i(a + b)] - e^[i(a - b)] - e^[-i(a - b)] + e^[-i(a + b)]/4

2{e^[i(a+b)] + e^[-i(a+b)]}/4

{e^[i(a+b)] + e^[-i(a-b)]}/2

{e^[i(a+b)] + e^[-i(a-b)]}/2 = cos(a + b)

cos(a + b) = cos[-(a - b)] = cos(a - b)

This took so much time just to prove

[solid needle]

uhhhhhhh

cos(a+b) cannot equal cos(a-b)

unfortunately you made an error somewhere

think about it

jf they were equal

[upper karma]

Aww man

[solid needle]

then i could set a to be 45 deg

b to 1 deg

[upper karma]

Lmfao

I did it wrong

I rushed

[solid needle]

and you would be telling me that cos 46 = cos 44

yeah so

heres the thing

pause

the e^it stuff is good

this is important eventually

but for now you are just making this way harder than it needs to be

lemme walk you through where you need help

first of all, did you read my response

do you get why:

[upper karma]

I havent, let me one sec

[solid needle]

cis (a+b) = cis a * cis b

that is step one

[upper karma]

Rewatched the video

Yeah

I get it now

[solid needle]

ok good

now

we know two things now

cis (a+b) = cis a * cis b

cis x = cos x + i sin x

for convenience let's set some variables:

ax = cos a
ay = sin a
bx = cos b
by = sin b

so far so good?

@upper karma

cis a = ax + i ay
cis b = bx + i by

[frosty kraken]

Helpp

[upper karma]

Mhm

[solid needle]

ok now sub everything into this

and expand

let me know what you get

[frosty kraken]

I got a≈35

[upper karma]

(ax + iay)*(bx + iby)