#geometry-and-trigonometry
1 messages · Page 47 of 1
fair but you also didn't edit your msg as i asked
shapes, lengths, angles, and other measurements
very loosely
How to represent $y > x^2$ in graph?
I don't know where to begin!
KingDanger
graph x^2 as broken lines and then shade the upper part
Everything above the graph is greater and everything below is less than.
This
How to graph x² though?
Is it like graphing y = x²?
yes
The every area other than this is y < x²
less than or equal to
Ohh thank you @upper karma
np
@tiny finch
@zinc reef
Yes
why did yu ping me?
Did you need help with the Ramunjan Serie 1+2+3+4 etc = - 1 over Infinity
No, i need help with vectors
...
I can help you
I'm a thirteen year old prodigy in highschool, so I can help you
No, I didn't read the question
Its a geometry question so i think you can
Yeah, I study pre university math
Given a plane in space, the steepest directiom on the plane is the vector that makes the largest angle with x-y plane.
Okay
exoress this vector in terms of the normal n to the plane P
Okay
??
Pressure
Ok bye
What do you mean
i am going to find help on reddit
Ok
So, basically Write the vector, parametric, and symmetric equations of a line through a given point in a given direction, and a line through two given points.
Find the distance from a point to a given line.
Write the vector and scalar equations of a plane through a given point with a given normal.
Find the distance from a point to a given plane.
Find the angle between two planes.
Do you know about cross products?
the solution requires cross product
Okay, but I gave you a different solution
you didn't
if w points in the steepest direction, it perpendicular to some vector, hence you can find w using cross product.
Hi @dark sparrow
actually wait
can you restate the problem
cause i'm a little confused from this convo
Problem 3
For any plane P which is not parallel to the x-y plane, define the steepest direction
on P to be the direction of any vector which lies in P and which makes the largest
(acute) angle with the x-y plane.
a) Let P be the plane through the origin with normal vector n. Derive a formula, in
terms of n, for a vector w which points in the steepest direction on P
ok, and your progress if any?
if w points in the steepest direction, it perpendicular to some vector, w is also perpendicular to n because it is in the plane. hence you can find w using cross product. w = n × v
i cant find v
ok right so you have the right idea
the intermediate vector you called v needs to:
- lie in the xy plane
- lie in P
i.e. it needs to be perp to both n (the normal of P) and the normal of the xy plane
can you say what the normal of the xy plane is?
v does not need to lie in P(i think)
w is a special vector because it point in the steepest direction of the plane P. My intuition tells me that w is perpendicular to some vector. If i can find the vector i can express w as n × v
because w is also perpendicular to n
k = <0,0,1> is normal to the xy plane
yes
heres my idea
n is shown in red, and your target w is shown in blue
the intermediate vector in the calculation, which you called v (or so i thought), is green
blue and green are perpendicular to each other and lie in P
and green is also in the xy plane bc it is horizontal
well, as i said, it needs to satisfy two properties:
- v lies in P (grey plane) <=> v is perp to n (red)
- v lies in the xy plane <=> v is perp to k
here is the same setup from a different angle if you want, but i don't see much of a point.
this is enough(ish) to tell you what v is.
<@&286206848099549185> Does the line for the equation 3x-4y=12 correct?
Anyone ensure is this correct or not?(If not could you please explain?)
Yeah it’s correct
ABC is a triangle, |AE|=5 cm, |AC|=8 cm, m(CAD)=50° and m(BDE) given as 25°.
According to those info how tall is |BE| in cm?
assuming AD is the height of the triangle, it should make an angle of 90° on point D. you can find m(ADE) by subtracting both 25 and 90 from 180, and because it’s an isosceles triangle, it has the same value of m(AED)
No its all given
so then, we don't know that angle ADC is 90°?
Yeah we dont know if it is or not totaly
I m trying to search for its video resolving on the net
can anyone recommend any books on conics in euclidean geometry?
wait, how do we know AB = AC?
Hello! Given two points, let's say A(2,1,-1) and B(-3,0,2), how can i determine the equation of the bundle of planes passing through these two points?
how do i simplify this so that there is no sqrt symbol in the denominator
How would u normally try to simplify this if it was just like 1/rad1
yep lol
my brain was kinda fried at this point lol
🤣
I had already gotten the question wrong at that point so I was just trying to figure it out for the future
can i express tan²x as sin²x/cos²x
Yes
awesome thanks
Only when cos(x) isn’t equal to 0 though
that’s information i don’t rly have but im just gonna assume it isn’t and deal w the consequences later
if i have (sin²x-4)/cosx can i express it as two separate fractions?
sin²x/cosx - 4/cosx
then can i express that as
tanx • sinx - 4/cosx
?
Yeah you can
ok yeah im lost on this problem LMAO im gonna go open a help channel i think 💀
this is the first time i see them all occupied 😭😭😭😭
tan(x) isn't defined at those points anyway
Oh yeah
but that’s good thinking cause saying cot(x) = 1/tan(x) isn’t really true
like any trig identity, it's true as long as both sides are defined
it is seen more as a definition but it’s wrong
Triangle 
what makes it wrong?
KingDanger
<@&286206848099549185> anyone help me
Anyone help me 😭
@dark sparrow sorry for the ping!!!😔 could you please check my answer correct or not?
in your question you should replace the word "does" with "is"
otherwise seems approximately correct
idk why it passes through (-1, -2) though
,calc sqrt(2) * (-1) - sqrt(3) * (-2)
Result:
2.0498880527647
not far off i guess
this feels more #precalculus than geometry
This is exam practice for next Thursday
Ok ty
How would i check an inverse functions answer by composing the original model with the inverse i found
Hi there, I'm studying Birkhoff axioms for plane geometry and I can't really wrap my mind around the measurement axiom... In particular, I don't understand how should I interpret negative angles according to this description (or what it means for two angles to be the negative of one another)
hi everyone, can you help me? how use the rule "Horse" in trigonometry?
looks crude but probably ok? what do you need the drawing for?
Exercise for 1
ok then my earlier feedback stands
googling "horse rule geometry" doesn't give anything useful. you'll need to tell us where you heard that rule...
In Russia we use this rule in trigonometry, if its interesting for you, i try to explain it
да ну?
ты из россии?
я и по-русски такого правила не знаю...
да
ахахахах
так что это за правило
а тут спамить сообщениями можно?
в смысле спамить?
ну общаться
ну короче, я ща в колледже, скоро экзы и как-то надо готовиться постепенно, а я тригонометрию не знаю вообще, если кратко то там типа лошадь чет делает и кос на син меняется, что скажешь?
да я сам вахуе
ну и не еби себе мозг тогда непонятными мнемониками, а вникай в суть. на том же ютубе сто пудов есть видео, объясняющие, что такое тригонометрическая окружность, как ею пользоваться и т.д.
выучи разве что формулы $\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)$ и $\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)$
|Ann⟩
я тебе, конечно, прямо так вообще всю тригонометрию не объясню
могу тебе скинуть список тригонометрических формул собственного сочинения. я правда его сделала несколько лет назад и на английском. и еще я туда несколько совсем выпендрежных формул туда впихнула (параграф 6)
и там еще нет таблицы значений тригонометрических функций
если ты, как сказал ты сам, совсем не знаешь тригонометрию, то такая таблица мб хоть немного да поможет
Now this math is definitely gonna be horsing me around for an hour.
pay it no mind
it is a very stupid mnemonic
that i think would take more effort to remember than the thing it's supposed to help you with. lmao
Lol
Nah i guess this might be interesting though, right?
Like there is literally a horse.
And a graph is drawn on it
Imma look into this
The function doesn't change sides?
Interesting.
These many things to memorize?
Nvm.... This actually isn't interesting.
The rule of the horse in trigonometry says:
If you postpone the angle from the vertical axis, the horse says "yes" and nods, moving his head along the vertical axis. Then the value of the original function changes to a cofunction — a function opposite to this one. For example, for a sine, the cosine is a cofunction.
If you postpone the angle from the horizontal axis, the horse says "no" and shakes his head, leading along the horizontal axis. Then the function does not change.
Applying the horse rule, it is necessary to determine the sign of the new function from X. Its sign coincides with the sign of the original function.
спасибо тебе большое, я постараюсь вникнуть
а ты где вообще учишься?
I would never go and measure the vertical length of the face of a horse
Not very polite.
I didn't like this math
2/10.
its like rofl?
за рубежом
конкретнее не скажу
но заканчивала я матмех
а, ну понял, ты сейчас на высшей ступени образовании?
магистратура
капец, удачи тебе
спасибо
Is the arc length equal to the angle opposite of it?
Cuz then maybe 16) is 34, idk I need help with this
So would 15) be half of 190 and 16) be half of 34?
Or actually double 34 sorry for 16
whats this process called
wth
nvm
is it projection? im not sure
3d to 2d projection
Greetings
I am a sci-fi writer and I am trying to calculate/estimate the size of the Virgo Supercluster if it was a sphere that consisted of equidistant Hexagons of 400 thousand kilometers for each side and an area of 415.692 billion square kilometers.
The main problem I have got is that square roots do not convince me
Shouldn´t I use "a root of 6" to search for the area of a sphere that consists of hexagons?
I always thought I was kind of different
I will give enough data to have a start
The Virgo Supercluster is about 110 million light years in diameter, while its radius is of 55 million
And I could get with that information, the circumference of the Supercluster
The problem is that this is not a circle, but a sphere
And not any sphere; A hollowed sphere
I want to get the scale of the "surface" of this sphere
And I thought that by doing the square root of its circumference I managed to do it.
But I realized that I cannot do that because it consists of standarized Hexagons
Anyways, a light year consists of 9.461E+12 kilometers
I could divide that by 400000 and get the amount of planets that would fit in straight line and then calculate its square root
But oh
The same problem reappears
Square root has not the shape of a Hexagon
i think your problem is that you want a giant sphere filled up with hexagons right?
but the hexagons are equidistant (so the same size)?
Yeah pretty much
and you want the maximum amount of hexagons i'm assuming?
Yep
Yeah i got it
if you have the side length you could probably calculate the area of one
i kind of see an optimization problem coming
But that is not the problem, i do have the area
right
Or at least what the power of two gives me
my only problem is i don't know the behavior at the edges when it gets close to the edge of the circle
Oh that is not the problem
The problem is that the area is determined number by the SQUARE
And it´s a hexagon
I don´t know if I am breaking math, but I don´t think Squares is what I am searching for
What if I want to make a Hexagon out of Hexagons?
what "volume" formula did you use for the hexagon?
That´s another problem
There is no volume
i mean
area in three dimension is basically volume
it's not like you're looking for surface area
rihgt
OH
u mean theres no volume formula for a hexagon?
3d hexagon i mean
I AM looking for surface area
why?
oh
There are no inner hexagons inside it
you want the surface area size
They would be Truncated Icosahedrons
And they suck because they also add Pentagons
Precisely
because your 3d hexagon would have 6 sides, wouldn't it?
I think that is the perimeter
its the surface area
...
you add the area of the surfaces
To calculate the area i did the perimeter multiplied by the apotema
And then divided it by 2
Oh
Yeah, but i am just counting a single surface for the Hexagon
I would require that for the 3d Sphere
are you using a 3d hexagon to estimate the surface area of a sphere?
i'm confused lol
what's your big goal
No, a 2d Hexagon
To form a sphere with Hexagons
Like a cube with 6 squares except it is big enough to fill a universe
this is the area of a 2d hexagon...
kind of like reimann's sum
in 3d
lol
Yeah, that is the formula i used to search for the Area
and theres no problem there, right?
A single one
The side powered by the square
what do you mean
For the individual Hexagons that is ok, it just allows to give me the Area in square kilometers
But if i want to form a higher structure with those Hexagons
Shouldn´t I be doing it by the Hex?
Because there are 6 sides, not 4
And therefore the Area of a bigger structure wouldn´t have (?) by the square
That is the problem that is making my mind go insane
i'm not sure i understand what you mean lol
Squares are not Hexagons right?
right
this doesnt have anything to do with sqrt(6) or s^2, does it
from the pictur
Oh no not at all
ok
yes
That´s basically how the sphere consists of
at some point it ends though right
Multiple Hexagons together
and then there are small gaps
It does not
sorry for my poor drawing but uh
if the hexagons were inscribed in a circle
there would be gaps
It is a Sphere
Ah
there would be gaps
I never thought about it
because no matter how many hexagons you have you're never gonna get exactly a circle
oh lol
you can get smaller hexagons though to get even closer to the area of a circle
That´s really obvius now that i think about it
thats kinda what reimann sum about, except with rectangles
XD
i thought you were trying to find when the farthest out points hit the circle
sphere sorry
I thought i would avoid it because Hexagons are the best polygons to fill spaces
oh XD
They just have the best structure
Why do you think it is common in nature?
on the space that you're trying to fill
i would argue a lot of things are common in nature... 💀
nah but uhh
i guess yeah
Bees do their hives like that
Saturn´s storm is literally a Hexagon
that was the 1 thing i thought of XD
oh i didnt know that
It has something to do with Gravity
thats like how the fibbinacci sequences is randomly in pine cones or smthn
ah
Which is also the reason why the planets are exactly 400 thousand kilometers from each other
So that their forces do not make them crush
Equidistant, forming regular Hexagons
i would bet that circles would even be more common but they're also kind of irrational so its hard to use them sometimes 💀
Yeah
why hexagons though
They are futuristic and have a good structure as i said
Octogons leave more spaces
what do you mea
I could have used squares, but they are not as good
i mean if you inscribe an octogon in a circle it would leave less space than if you inscribe a hexagon into a circle
true
I meant if you match them together
But we´re leaving by the branches
wdym
We went far from the original topic
oh yeah 😭 🤣
About the Area of the Sphere
why don't you just calculate the area of a sphere without estimating?
like
the volume
or surface area, whichever you need
Because the problem is inside the formula of Area
huh?
Area basically consists of multiple little squares inside a bigger structure
And that´s not what i want
I want Hexagons
And there are even more problems because if I do it by 6 I am basically making a Sphere of 6 dimensions
do you want to fit in as many equal-lenghted hexagons into a sphere as you can, or do you want to fit as many as you can evenly?
i don't know how to explain what i mean
let me try to draw
I understood you
And yes, the first option
oh ok
But not in volume
it seems that you could use either volume or surface area to do that
why not?
it doesn't have to actually have mass for you to calculate it in that way
lol but like
the volume of an empty cube container can be calculated with the length of a side
it doesn't have to be full to be a volume
What matters is the "side" of the circle, the "Crust" of the sphere
and you're only using the volume to compare to the volume of the circle
surface area would be way more complicated than volume
I am basically searching for its perimeter
and i think you could just use volume to compare to the volume of the circle to find the maximum amount of hexagons
whether you have stuff in the hexagons or not
If that is how it should be called
i think its called surface area
basically perimeter but in 3 dimensions
Nice
like where you take the area of all the surfaces and add them
nah but like
how long was the side of each hexagon did you say?
ooh just realized i don't know how to find the volume
wait
wouldnt it be cubic kilometers 💀
No, because it is a Hexagon
wait my bad
Not a Truncated Icosahedron
you're doing surface area
a what now 💀
For the Sphere
The polyhedron equivalent of a Hexagon
looking at this, the bottom 3 sides will always be connected so they wont matter for the surface area right
wait
but in 3d
hold up
yeah
3d does not matter
you only have to worry about the outer half of the hexagon
right
They are 2d hexagons forming a 3d Sphere
so 3 2d hexagons per 3d hexagon
Is that for when i make a hexagon out of hexagons?
yeah, isn't that what you were doing?
Yeah
this is the surface of each 2d hexagon, there would be 3 2d hexagons for each 3d hexagon.
lets just say n is the number of hexagons
so n * 3 * this < surface area of circle
so you could divide the surface area of your circle by 3 * this
to get the number of hexagons
Oh yeah that is the Area i got
thats for each surface though
so you have to multiply by 3 for each hexagon
then by n (the number of hexagons)
WAIT
no
thats LESS than the surface area
and you're looking for the biggest integer n
right?
Right
so i would say
But why would you have to multiply it by 3 if you already found the Area?
because thats only one of the 3 outward-showing sides for each hexagon
Oh yeah
yeah
so
s = surface area of sphere(known)
n = number of hexagons (unknown)
x = surface area of hexagon with length 400,000
n < s/3x
you could solve the surface area of your circle
then the surface area of hexagon which is this
Oh that´s neat
Would s work for the surface area of a sphere?
i meant sphere btw
At this point that is what i thought so
yeah XD
Anyways, i think i have got everything needed to do the math
Thanks, i appreciate it
Indeed
Thank you
wait i forgot
?
yeah ok cool XD
gah i think we made a mistake
Hmm?
the surface area of the 3d hexagon being less than the surface area of the circle doesnt mean the hexagon is inscribed in the circle
what if the very tip is coming out of the circle, yet the surface area of the hexagon is still less than the circle
😭
like imagine the tips are a little bit out
LOL
i mean if complete accuracy isn't too important
than u can just do that method ^^^ XD
It´s just lore for a TTRPG
Also, i doubt that it might make a huge difference even if it was real
but if its just really deep lore then its fine XD
thats true, who knows tho XD
you might be off a few million miles...
🤷♂️
😂
Oh yeah, but it is not the same as building pillars in the poles of a planet
lool
Alright, there might have been a few mistakes
So, i basically putted S the value of 38 quadrillion and X the value of 1,24 trillion
It gave me only less than 31 thousand Hexagons
whats the radius of the sphere?/
um
does the units matter or can i just put in 55 million into the calculations
wait yeah they should matter
Oh crap i forgot
lol
Considering 9,461E+12 kilometers is equal to a single light year
The surface area of a hexagon is insignificant
wdym
Btw which app is this? it looks nice
google XD
Oh
a
hexagon
It does not appear with interactive stuff for me
yes, i believe it inputs the side length and outputs the 2d surface area
Wdym?
A = 3sqrt(3)a^2 / 2
is the formula
a is the input
and thats the side of the hexagon
Hm
is that not what u were asking
I honestly don´t know
lol
But i should try it
I mean
That was the result i got last time
And still it gave me only 31 thousand hexagons
did you mean 9461E+12 or 9.461E+12
Perhaps there is something wrong with the sphere surface area
9.461
ok
It is still a lot
that look about right?
Hold on, lemme do it
It looked real for a sec
Yeah
HOLY MOLY
what 🤣
That´s a lot
it is a lot
i know powers get weird and stuff
but isn't it like
theres 10^80 ish atoms in the universe
and yet we're dealing with kilometers XD
still 10^42 is so far off from 10^80 so yeah
Every planet is an atom in this big structure
thats the surface area of the sphere, s
Thanks for helping me once more
Don´t worry, i already have the hexagon area
I have to multiply it by 3 right?
yeah bro XD
anybody know where i can find a lot of hard trig questions? in looking primarily for identities
what difficulty??
im in 11th grade does that work or is it too easy?
like they take say 30 minutes each ig
im only 1 grade above, honestly it depends on the country
just send what you have ill check it
dm me like a pic
ohk
no one is going to be able to tell what difficulty that they should send so that it takes you 30 minutes, do you have any examples of such a problem? that would make it easier for us to send some problems like it
i wanted to include one but i dont think it let me send
Send it as a link then
you can type it out
You should be able to upload files though
this type of question
b4 i think it didnt let me send bc i was in main discussion
sry about that
@thorny juniper Something like this?
Catgod
Were you revoked of your rank?
?
I believe you used to have a role
Maybe not
I don’t think so, I just joined recently
Ahh maybe it was some other catgod then
How many catgods are there lol
Countless
i need to approximate the angles, but all the calculus i did, it was wrong
i did the a²=b²+c².2bc.cos
but it gave me a big number
i think i'm not using the right formula, or i'm miscalculating something
You dont need calculus for this
Do u know SOHCAHTOA?
Wait does he have to approximate or can he use a calculator?
Mb I didnt read this
So sounds like u need to approximate
Cosine rule honestly sounds like the only thing that's needed there
Yeah ur right
I dont think he used it correctly tho
Mhm
I need to brush up on geometry again sometime
I'm currently just working through calculus
no
i think ive mistaken this
help?
is this high school geometry? there does not seem to be questions this difficult for high school geometry on khan acedemy. anyone have any idea what i could search to find geometry lessons/problems that are as difficult as this?
it's a matter of writing down what you know, and discovering more things until you have found what they're asking for lol
what progress have you made
thank you for the completely noninformative answer
this is from aops introduction to geometry
whose problems tend to be more difficult than those in a standard HS geometry textbook
you mean me or them
probably not in your best interest to use an aops book or solve aops problems to study for a class
nah is it just me or is aops 10x harder than actual problems
theyre hard if you dont read their books
sure they have some typical and practical information but the books arent exactly the best to compare to a school curriculum
since its mainly meant for olympiads
that makes sense
naw like i remember i tried some problems recently from like idk maybe algebra or something?
a lot of information they supply you with wont really help you in regards to school
i have their intermediate algebra book
i realized the book wasnt for me cause it strayed far from our school curriculum and didnt really contain any useful information imo
yeah you have to start with like their prealgebra books
cause they use a lot of tricks from previous books
LOL
i don't remember exactly but i might have used aops prealgebra for a bit
it was kinda long time ago so i don't remember much of it besides the actual algebra that i use in actual math 💀
from my perspective aops is just working on an entirely different level 💀
lol
idrk any good geometry books
i think u can just learn it all online through youtube
and like khan academy or smthing
i mean i already passed them or "learned" the main material
but i may go through the aops like u said because
the aops problems are exponentially harder than the khan academy ones lol
geometry is substantially harder than algebra for me
maybe ill take a look at it idk
i just dont read the aops books because they take a long time to teach the most niche techniques for solving the most niche problems
and im not an olympiad
so
that makes sense
Hello guys
I noticed my photomath app always provides me with unified solutions of trig equation
Should i also try to unify my solution or it's not mistake if i dont do so
I mean instead of x=pi/2 +2pik and x=3pi/2 + 2pik, it shows just one solution x=pi/2 + pik
Where k is an integer
It unified those two solution into one
that specifically?
right
so that looks like it's specifically for the equations sin(x)=0 or cos(x)=0.
It was for cos
i mean the unification you're talking about generally.
no, it's not a mistake not to do it. but usually it is nicer to make your answer shorter.
Sometimes i just dont understand how to unify some solutions
Thats why i asked, but thanks
how to I sketch cos(x)?

I am still wondering
I think I might need an unit circle for this
cant believe they are asking for that without actually having taught us tho
they didnt teach you what a sine wave looks like?
it is a branch of geometry 
Yeah but it’s too advanced to talk about here
Go to #point-set-topology
could anyone help me with this?
a basic strategy to tackling this kind of stuff is to convert all trig functions to sin and cos
are you able to do that?
i got as far as 1/sin(t) * 1/cos(t) - 1/(sin(t)/cos(t) and then im confused
can i get rid of the ones by just flipping everything?
what do you mean flipping everything
instead of 1/sin(t) i make it sin(t)/1 which just means it's sin(t) and i do that for everythin?
you cant just arbitrarily flip fractions
what youre trying to do is similar to a student going:
ok i need to do 1/2 + 1/3
im just gonna flip the fractions
2/1 + 3/1
oh
ok the answer is 5
so a tip
what you need to think about is that the trig functions are still just functions
if you can interpret something like
1/x
then imagine 1/sin t the same way
sin t represents some value, like any other expression
you can imagine that value to be x if you want
so for now, if you want, do the following substitution:
y = sin t
x = cos t
and see if that helps
(note that i specifically set y to be sin t, i know the substitution is arbitrary but there is good reason to get in the habit of associating y with sin and x with cos, so i would use the substitution i have as written)
so if sin is y and cos is x. i end up getting 1/yx - 1/(y/x). the latter half can be simplified by multiplying the nominator by the reciprocal of the denominnator, so x/y meaning im now at 1/yx - x/y. then i tried to get a common denominator by multiplying x/y by x but i dont think im on the right track rn
you are
multiply the top and bottom of x/y by x, keeping the fraction value the same
now you have a common denominator
you will end up with:
(1-x^2)/yx
at this point it feels like youre stuck
the point of the substitution wasnt to get you all the way to the answer, it was to help you think about how these expressions can be manipulated
substitute your sin and cos stuff back
look for other properties you can apply
@amber hollow
oh yeah thats pretty much where i ended up and thought i was wrong lol
its right, sub the trig functions back
see if you can identify what to do with the numerator now
remember, this x y sub isnt something you have to do, and it will rarely get you all the way to the answer
it is only to help you organize your expressions so they are easier and more familiar to deal with
bruh
bc i noticed 1-cos^2(t) = sin^2
how do i memorize all this stuff. do i just try to remember the base stuff and how i get all the basic identities like that from them?
thx though
so all of the basic trig identities
so everything before law of sines/law of cosines/sum formulas
they are all either basic conceptual understanding of trig functions
or they are very simple, basically trivial algebraic manipulations of them
if you build a strong foundational proficiency with trig functions, their properties and behaviors, none of those trig identities ever need to be memorized
i have never memorized them and can pull them out any time i need to and can prove any of them easily
that might sound intimidating, but given the right instruction, its really not that bad at all
the question is simply do you have the resources and the time to work it out
i see. im only taking this class rn bc when i tried to do advanced calculus i noticed that they actually use a bunch of trigonoometry and while im good at calculus, i struggle alot with trig so im trying to build a foundation for it. its so weird and intimidating though, its like i dont get it at all
your last resort should be hard memorization
if you want
i can help give you that foundation here
it only takes about 30 min to go over some basics probably?
and the rest should follow if you just use my notes and practice
i do have a book thats supposed to be teaching it to me but i honestly feel like it doesnt cover everthing because i still dont understand hyp/opp/adj stuff except for like the very very bare minimum and every time i ask my professor for help its like im actually supposed to have a grasp of it already so
if you dont mind id like someone to go over basics with me and some notes to look at so i have a better understanding
also like how im supposed to know the unit circle without memorizing half of it x)
you don't. you memorize it lol
hm?
ok let's do it
let's start from square one
you said you don't quite get opposite/adjacent/hypotenuse
let's begin there
so start with a RIGHT triangle
now we are going to label one of the non-right angles our "reference angle"
i mark theta on the image, but to make it easier to type, i will type it as "t"
now ill label the sides
question: which side is opposite to t? adjacent to t? which one is the hypotenuse?
hypotenuse is the long one...opposite is, a because its opposite the angle? and b is adjacent cause its touching the angle but not the hypotenuse
correct
that's really it
so now we can define all of our trig functions:
we have
- sin t = opp/hyp
- cos t = adj/hyp
- tan t = opp/adj
- csc t = hyp/opp
- sec t = hyp/adj
- cot t = adj/opp
don't worry about the naming conventions of the last 3, we'll get there
as usual, hard memorize in the meantime, but once you understand, you will get to release more and more from memory as you'll just "understand" them
so what this is saying is that
no matter what kind of right triangle you draw
as long as the labeled reference angle is, for instance, 30 degrees
sin t, the ratio of opp/hyp, is always going to be EXACTLY the same value
in this case, 1/2
and that's why it's a function
as long you give it one particular value, and the output will never change
you can never have 1 input give 2 different outputs
following so far?
fixed spelling mistake
so by ratio u mean that you're always going to get 1/2 for the sin when you divide the lines for opp/hyp regardless of how long the lines are? as long as it's a right triangle.
!noans
The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.
okay i think i get it. at least on the theoretical level anyway
good, the reason why the value never changes is just a simple application of triangle similarity theorems
so if you have any doubts about that or want to be more rigorous in your understanding, literally just review similar triangles from geometry
ok moving on
so in summary trig functions you can think of like this:
This is for @cerulean turtle
for a given right triangle, and a marked angle t
input t into the trig function, output a side length ratio
trig functions: input angle measure -> output side length ratio
that's the key idea here
Ask help like how to do that @cerulean turtle
so now let's talk about to extend the domain of these trig functions
as you know, you can input things like 220 degrees into trig functions
in the context of the right triangle, this is nonsensical
but if we now reframe our definition, we can literally plug in any real number for our angle and have it make sense
so now I will present a new definition for specifically sin and cos
the previous definition you can think of as a motivation, why we might care about these functions
but this next definition i will provide is the one that will clarify literally almost everything
ready?
yes
here is the unit circle
start on the right side at 0 degrees
this is just definitional
now if we go counterclockwise, this will positively affect the angle
again, this is definitional
so now we can label things like:
but don't forget, we can label the other way going negative
so on this unit circle, we say that 90 deg = -270 deg
and there is no reason we need to stay within -360 and 360 either
we can say for instance that 0 deg = 360 deg = 720 deg = 360n deg for any integer n
again, this is just definitions, there is nothing to understand here except this is how we label a direction with an angle
this is really the only thing you will need to "hard memorize"
so far so good?
yeah im following
ok now the definition of sin and cos
pick an angle t
mark the POINT where the ray hits the unit circle
of course it has coordinates, as does every point on the plane
by definition, we say that:
x = cos t
y = sin t
wait so the points on the plane that is the circle are comprised of two ratios based on the right triangle formed by angle t?
i dont even understand that question
you could say that
so i think i can clear up what it is you're asking in that question and demonstrate it
i mentioned before that
being motivated from the original opp/adj/hyp definitions
we are now extending these functions so that their domains go beyond the 0-90 deg range
but in order to extend it, you first need to make sure you didn't mess up the definition that is already there
let's now show that our original first definition matches this new definition
otherwise what would be the whole point of this?
is that what you're basically asking?
i think so yeah
ok let's do that
you can see the right triangle in that diagram i drew
what's the side opposite of t? adjacent to t?
the side opposite to t is the purple part and the side adjacent is the red part
unit circle meaning that the radius is 1?
yes, that's what unit circle means
then yeah
perfect
so now we can see that sin t = purple/blue
and since blue = 1, sin t = purple
and isn't the purple line just the y coord?
can you see how it works for the adjacent side/cos too?
so the red line is the x coord. red/blue is red/1 so cos t is just red
bingo
would you now agree that this new definition lines up with the old definition?
yeah i understand now
great
now this is where it gets fun
armed with just this basic definition
you now no longer need to memorize ANY "basic" trig identity
like i mentioned previously
let's start with a simple example
sin^2 t + cos^2 t = 1
you know this one already, yeah?
yeah
do you know how it's derived?
the pythagorean theorum? a^2 + b^2 = c^2
ok let's go over this one then
can you see that the negative of an angle
is just the angle reflected over the x axis
does that make sense?
yes i understand that
