geometry-and-trigonometry

[umbral snow]

Alright, sure. I picked some random angle in this picture. Blue line is the value of cos(θ), green line is the value of sin(θ)

Here is the scenario flipped over the x-axis

[upper karma]

Ok

Ok

I think I have a better explanation

[desert pond]

wanted to ask if these are correct:

-cos(x) = cos(pi-x)

-sin(x) = sin(pi-x)

[upper karma]

The sin one is wrong

[desert pond]

why would the sin one be wrong ?

[silent plank]

try expanding using angle sum identites and see waht you get

[desert pond]

i see , thanks

can i ask what -sin(x) is equivalent to ?

[upper karma]

Deja vu lol

[silent plank]

-sin(x) = sin(pi**+**x)

[desert pond]

thank you

do you know if there exists a summary of all these identities somewhere ?

[silent plank]

google

has craptons

[desert pond]

ill have a look , thanks

if anybody would be able to simplify this expression:

$2cos(\frac{(k+1)\pi}{3})-2cos(\frac{k\pi}{3})\=2cos(\frac{(k+2)\pi}{3})$

[somber coyote]

txljk:

[desert pond]

i'm not sure if this is possible

i get stuck at $$2cos(\frac{2\pi}{3})\times cos(\frac{k\pi}{3})+sin(\frac{k\pi}{3})\times sin(\frac{4\pi}{3})$$

[somber coyote]

txljk:

[winged zinc]

@desert pond

I think you can

[winged zinc]

I managed to show that it is true for all k

[toxic cedar]

Anyone help with that?

[idle bloom]

@toxic cedar can you draw out a triangle for part A?

[mint bronze]

Can anyone help me with figuring out how to write an equation for word problems like this? I have no idea what I am doing.

[velvet flicker]

ok so first question. Do you know what all those terms mean?

@mint bronze

[mint bronze]

I know what they mean, I dont understand how to put it in a formula. my college is being generic.

[velvet flicker]

Right so do you know what function would be used to model a problem like this?

[rich wolf]

what trig function is 0 at 0

[mint bronze]

it would be sin or cos

[velvet flicker]

Right

They're basically the same function

[rich wolf]

use sine

[mint bronze]

okay

[velvet flicker]

yeah. you could use cosine, but that would be like saying 2/2 instead of 1

so what is the period of sin?

[mint bronze]

I dont understand periods like if its something I should figure out. My book says 2pi/b

[velvet flicker]

well the period of just sin(x) is 2pi

which means that sin "repeats itself after 2pi"

that is, sin(x) = sin(x + 2pi) for all x

[mint bronze]

okay that makes some sense with that explanation

[velvet flicker]

so how would you modify sin to change its period?

because we want a period of 7 not 2pi

[mint bronze]

We would have to find b to modify it which is 2pi/b=7 then make b=2pi/7

[velvet flicker]

right

so how does that change the function?

like how would we write "sin but with period 7"

[mint bronze]

sin2pi/7

or sin(2pi/7)

[velvet flicker]

right

so that's almost it

(for the problem that is. That is how you modify sin to give it a period of 7)

the issue is the amplitude

but that's easy to fix

[mint bronze]

amp is 2 which would make it 2*sin(2pi/7 t) or 2sin(2pi/7 t)

[velvet flicker]

yeah

and that's your answer

the only thing left to do would be if it didn't start at 0ft

[mint bronze]

So like why use sin and not cos? what about tan?

[velvet flicker]

cos works

sin and cos are basically the same function

because sin(x + pi/2) = cos(x)

[mint bronze]

But would be marked wrong in the software im using.

[velvet flicker]

convention I guess

it shouldn't be marked wrong

but it does make sense to use sin

just because 2sin(2pi/7 t) is simpler to write than 2cos(2pi/7 t + pi/2)

like I said, it's like writing 2/2 instead of 1

[mint bronze]

okay

[velvet flicker]

though in some problems the cos form might be better because it could allow you to use some trig identity

like if that function was from intermediate step and not the solution

[mint bronze]

Hey I really appreciate that walkthrough. really helped me out.

[velvet flicker]

np

and that same process works for any simple harmonic motion problem

[mint bronze]

you too @rich wolf

[velvet flicker]

the solutions are all a*sin(bx + c)

[mint bronze]

^ that would help out a lot.

[velvet flicker]

although in some cases you would use cos instead because it would be nicer

[mint bronze]

like if the point did not start where d = 0 and t = 0

[velvet flicker]

yeah let's try that

so suppose it started at d=0 t=0 but it was moving downwards

[mint bronze]

it would make the previous answer a sin^-1

right?

[velvet flicker]

no

[mint bronze]

oh

[velvet flicker]

that's where the +c comes in

the answer would be 2sin(2pi/7 t + pi)

[mint bronze]

if its moving downwards wouldnt it be - though?

[velvet flicker]

I guess it could also be -2sin(2pi/7 t)

or 2sin(-2pi/7 t)

they're all equivalent

there's a really nice way to visualize sin and cos that I'm not sure if you know

[mint bronze]

its actually making things a lot easier to visualize outside of a graph. Ive seen the -2sin or equivalent but like i said that it was generic, it didnt explain what was happening.

[velvet flicker]

so suppose you have a circle

centered at the origin

and you're moving around the circle at a constant speed

and you plot your x and y coordinates as a function of t

the x coordinate is rcos(kt) and the y coordinate is rsin(kt)

where r is the radius of the circle and k depends on the speed

assuming you start at the point (r,0)

[mint bronze]

honestly, when it comes to circles i hate it lol

[velvet flicker]

circles are weird

they're nice in a lot of ways

but also really annoying in many others

oh, you should check out betterexplained.com

they have a bunch of really good articles on trig

that helped me a ton when I was first learning it

[mint bronze]

I just pinned the website. Thank you. I'll look through it tonight.

[velvet flicker]

it's really really good

[mint bronze]

is KhanAcademy any good explaining these funtions?

[velvet flicker]

I'm not sure

I've heard good things about Khan Academy but I personally haven't used it much

I used Khan for calc practice problems and SAT prep and loved it for both

[mint bronze]

I used it a few years ago but thats when it was a free service, idk if its still free

[velvet flicker]

yeah I'm not sure

[upper karma]

Can someone explain phase shift to me?

<@&286206848099549185>

Anyone?

[upper karma]

@dark sparrow Can you help me

[dark sparrow]

could you NOT

[upper karma]

ok...

Sorry..

[velvet flicker]

why do people even @ individual helpers

[weary drift]

bc new members don't read rules

[eternal crag]

when are we ever gonna impose those react to access channels stuff

idk if that'll be any useful

[quiet mason]

ann is not even helper

[upper karma]

@weary driftI'm just useless.

[quiet mason]

.

whydont you search whats a phase shift and read it

and try to understand

[upper karma]

I did but I am braindead so I will never understand math

[quiet mason]

if you have doubts they can be cleared here

[dark sparrow]

I did but I am braindead so I will never understand math
bit of a defeatist attitude there eh

[quiet mason]

youll never understand shit f you say that

[upper karma]

Yeah but it's the truth too bad for me

[quiet mason]

ok nvm but read

and post what you domt understand

[upper karma]

I don't understand what it calculates. Like if you have 2sin(2x + pi/3)

You factor out the 2 to get 2sin(2(x+pi/6))

So apparently it's pi/6??? But why is the 2 forgotten about

I just don't know anything

Anyone know..

[weary drift]

the graph of f(x-c) is a horizontal shift of the graph of f(x) by c units

[upper karma]

Yeah

I know that

But that's not the phase shift

[weary drift]

consider $x\mapsto2\sin(2x)$ over real $x$

[somber coyote]

RokettoJanpu:

[weary drift]

if you want to shift the graph of this thing (phase shift) by c, we replace x with x-c

$x\mapsto2\sin(2(x-c))$

[somber coyote]

RokettoJanpu:

[weary drift]

plug in c=your desired phase shift and you're done

[upper karma]

I don't get it. What is phase shift then

The definition

[weary drift]

graphically it's a horizontal shift

[upper karma]

Not graphically

The formal definition

[weary drift]

displacement of a waveform from another waveform of similar frequency

[upper karma]

@weary drift Why waveform?

[weary drift]

don't dig the rabbit hole too deep

[upper karma]

@weary drift Ok but I mean I haven't learned about waveforms. Is there a way to generalize this to functions in math

[weary drift]

it's generally a healthy behavior but not what you need right now to answer this trig q

discussing phase shifts as horizontal shifts of graphs of periodic functions should be enough for now

[upper karma]

😦

If you have the function sin(3x/2) + cos 2x

How can you find the period of that

[weary drift]

least common integer multiple of the sinusoids' periods

[upper karma]

Well the period of the first term is 4pi/3 and the second one is pi

[weary drift]

least common integer multiple

[upper karma]

Yeah how

[weary drift]

how'd you find the lcm of any two numbers

[upper karma]

I mean 4/3 is not an integer

[weary drift]

ig the word integer threw you off

find the lcm of 4pi/3 & pi

[upper karma]

4/3pi???

?

[weary drift]

no

[upper karma]

Why

[weary drift]

first we're finding a number that's an integer multiple of both 4pi/3 & pi

[upper karma]

Is there an easy way to do this...

[weary drift]

you can brute force this by computing integer multiples of 4pi/3 & pi til you get smth

[upper karma]

Anything better..

[weary drift]

for this q it shouldn't take long

[upper karma]

I am just so confused.

[weary drift]

you may need to review what lcm's are and how to find them

[upper karma]

Yes I can with integers...

I don't know how to do it with not integers

[weary drift]

the process is very similar w/ non ints

[upper karma]

Why is it not 4/3pi?

[weary drift]

4pi/3 isn't an integer multiple of pi

[upper karma]

So can you help..

[weary drift]

you can brute force this by computing integer multiples of 4pi/3 & pi til you get smth
start here

[upper karma]

What do you mean multiples

Like 8pi/6, 2pi???

Or 8pi/3, 2pi??

[weary drift]

1*4pi/3, 2*4pi/3, etc

[upper karma]

Why don't you need to multiply the denominator?

Well it's 12pi

/3 so 4pi

[weary drift]

you don't know what integer multiple means

4pi

[tacit karma]

Anyone here

[eternal crag]

no

[tacit karma]

:(:(:(

Also on an unrelated note, I just came back from a test and I think I failed the last question. A road is built on a hill with a grade of 27%. However, it didn't fit with the standards of the state, so it had to be rebuilt with a grade of 14%. If the vertical height of the hill is 600 ft, what is the percent increase of the lengths of the old road to the new road. (Also, you don't have a calculator.)
I have literally no idea how to solve it without a calculator.

[umbral snow]

Let the length of the old road be x, and length of the new road be y. Then,

600/x = 0.27
600/y = 0.14

Divide one by the other:
y/x = 0.27/0.14

Now, getting that number exactly without a calc might be hard, so I'm wondering if you're expected to just leave it as is?

@tacit karma

Actually it's 2 - 1/14

[elder sedge]

https://youtu.be/eqTYHao5DK8

Watch this

And roast it my friend made ut

It

[tacit karma]

@umbral snow the x and y in your equation are representing the horizontal length, and the road is going at an angle

The length of the road is the length of the slope

The length of the road is not the distance from the road to the hill.

[eternal crag]

@elder sedge dude no one gives a shit stop spamming garbage

[elder sedge]

@eternal crag some people do give a shit so shut your garbage mouth

[eternal crag]

this is the fifth fucking time you posting it

if your friend wants to be educational i'm absolutely cool with that

but idk what your problem is

[elder sedge]

I posted it once

[eternal crag]

once you said?

quit your bullshit and stop spamming would you

[elder sedge]

After you

[eternal crag]

jesus you are unbearable

i hate kids like you

[elder sedge]

Thank you

Appreciate the love

[rich wolf]

Lmao

[dire rampart]

lol

[topaz hearth]

Can someone help me think through this? a regular Tetrahedron has a base area of 68.0m^2 and a height of 10.2m. I need to determine its volume but I have no idea how to find it?

[silent plank]

look up volumes of pyramids

[topaz hearth]

I have. its finding the unknown with a formula that im not confident with. whats it called when you use the formula to find an unknown?

[silent plank]

solving i guess

which formula were you able to get?

[topaz hearth]

sorry my internet went down.

[formal hamlet]

I dont understand this. how is A/B=A/B. Then 3/2=a/5 ? lolwat

[cobalt turtle]

so basically, 2/3 is a diffrent ratio of whatever tan setup you have

its like saying x=5 and 2x=10

both are correct, one is just diffrent numbers than the other

though if you need further explination i can work it, i sure as hell cant give a better explination cause im a bad teacher lmao

[formal hamlet]

okay whats confusing is that she has the corner points labeled as A, B, C

so Im looking at the corners like lolwat

[cobalt turtle]

oh lmao like comparing angle to angle i see you

[formal hamlet]

so how is the C side is 90 degrees, uhh I thought c was hypotenuse

OHHH, it is....

I see it now I think.

[cobalt turtle]

usually lower case letters are reserved for the legnth of a side that opposite to the angle of its capital letter

[formal hamlet]

THANK YOU.

[cobalt turtle]

c is hypotenuse, C is angle C

[formal hamlet]

FUCK that was hard to read/understand

[cobalt turtle]

b is leg, B is angle

[formal hamlet]

light goes on in head

[cobalt turtle]

yea its a small detail that is a little confusing, but in alttude and centriod fuckery it comes in use

[formal hamlet]

OHHHHHH shit. all makes sense now

Fuck I was like damn this shit is not making sense at all.

[cobalt turtle]

easier to say A and a and know the relation as it being opposite of this rather than A = BC like what

[formal hamlet]

Yeah. woooo. nice.

[cobalt turtle]

i hope it helps, i might be baby but at least i understand what ive leared so far

aight gl on your work / assignment / paper thing

[formal hamlet]

yeah it sure does bro thank you

Thanks dude. I was looking at it for hours last night trying to figure it out. said f this and started gaming lol

[cobalt turtle]

^ me with my life problems

[formal hamlet]

lol

[odd cairn]

I keep on getting these shitty partition type questions. I know how to partition two end points, but not how to find an endpoint given only one endpoint and a partition. Help?

[rich wolf]

@odd cairn ok so find the length of AB

[odd cairn]

i got square root of 61

[rich wolf]

Now length AC has to be 2/5 of that

[odd cairn]

i just found the length of ab tho???

oh

[rich wolf]

Do you know about similar triangles

[odd cairn]

what are similar triangles

[rich wolf]

Triangles which have the same angles

Their side lengths are proportional

[odd cairn]

like equillaterals?

[rich wolf]

No

Its hard to explain without pictures

[odd cairn]

oh like dilated triangles

[rich wolf]

But ill just walk you through the concept

Yeah

I think you know what im saying

[odd cairn]

ye ye

[rich wolf]

So if you want 2/5 of the hypotenuse you must take 2/5 of each side length as well

[odd cairn]

so you just increase the side lengths by 3/5

and use pythagorean?

[rich wolf]

No

Look at that original triangle

What are its side lengths

[odd cairn]

5 for the height

and 6 for the base

[rich wolf]

Now multiply both those by 2/5

[odd cairn]

2 for the height

2.4 for the base

[rich wolf]

Yeah

Now it only cares about the x value

What x value is point A at

[odd cairn]

other problems want both values, this is an easier one

-3

[rich wolf]

Now you start at -3 and go right 2.4

[odd cairn]

1.6

-1.6*

[rich wolf]

Check math

[odd cairn]

yes -1.6 is one of the answers

one of the choices

thats it?

[rich wolf]

-3+2.4

Use a calc lol

[odd cairn]

ohhhh

-0.6

thats the answer?

[rich wolf]

Sure

[formal hamlet]

A ladder leans against a building so that the angle with the ground and the ladder is 62 degrees. If the base of the ladder is about 10 feet from the base of the building, approximate the length of the lader to the nearest length of a foot.
Can someone help me with this one? I see that the lengths of Opposite and Hypotenuse are missing. Am I right on that?
I just don't understand why I have to select the CAH function, COS to complete this.

[umbral snow]

@formal hamlet
You know the adjacent, and you want the hypotenuse. The trig function that deals with this is cos

[formal hamlet]

Right, but how do I figure which function for other examples?? Im just trying to understand better

is there a table or something I can make flash cards out of

[umbral snow]

Know soh cah toa

[formal hamlet]

Yeah I know soh cah toa. but how do I apply that?

[umbral snow]

The one you want to use is the one that deals with the sides you have/want

[formal hamlet]

CAH=adjacent/hyp

I have the adj. Dont have hyp or op

[umbral snow]

You want the length of the ladder, so that's the hypotenuse. You have the adjacent. You don't care about the opposite, so don't use a trig function that deals with it

It's usually easier to eliminate the wrong choices, if that makes any sense

[formal hamlet]

...im getting a better understanding..

[umbral snow]

You have the adjacent, you want the hypotenuse. That's cos.

You don't care about the opposite. You don't have it, the question isn't asking for it. Don't use a trig function related to it

[formal hamlet]

ok

alright that makes sense 🙂 thank you

21'.

[dark sparrow]

might help to think about where the center can be to satisfy the constraints

[rich wolf]

@upper karma do you know the eq of a circle

$(x – h)^2 + (y – k)^2 = r^2$ where (h, k) is the center

[somber coyote]

AMD:

[rich wolf]

and r is the radius

[dark sparrow]

imo you shouldn't dive headfirst into a sea of algebra right away

there's a way to figure out a constraint on the radius geometrically

[rich wolf]

just by looking at it the midpoint seems like the most logical option

but that's just a visual guess and might be wrong

[dark sparrow]

it is wrong

i mean

passing through two points constrains the center of the circle to be on a line

and being contained in the first quadrant further constrains at least its center to be in the first quadrant

furthermore we know that the extremal circle (ie the one which maximizes the radius) will be tangent to one of the coordinate axes, bc any one that isn't can have its center nudged away from the midpoint of (1,2) and (4,5) and have its radius increase as a result

hm

well

yeah

well the perpendicular bisector of (1,2) and (4,5) has equation y = 6-x

that's something that can be figured out by graphing or algebraically

doesn't really matter how

so clearly we're gonna have to have our center be in the first quadrant

otherwise there's no hope of the entire circle being contained in it

[rich wolf]

is there such a thing as the furthest point equidistant from two points

[dark sparrow]

unconstrained? no

[rich wolf]

i think you might be on to something with the perp bisector then

making an isosceles triangle with one of the axes?

what is the point equidistant from three points called

like in the middle of a triangle

[dark sparrow]

circumcenter

[rich wolf]

tried it

didn't work

[dark sparrow]

can you show the first part you don't understand

bc it's easier to work with than the distance itself ig

[rich wolf]

it's part of the circle equation

$(x – h)^2 + (y – k)^2 = r^2$

[somber coyote]

AMD:

[dark sparrow]

i wouldn't call that the reason tbh

uh what

what formulas were you intending to memorize??

the only real "formula" at play here is the distance formula

if you're getting exhausted then maybe come back later when you have more energy

[languid ocean]

can anyone help me with geometry really quick

i have to deal with directrix and vertex and focus, etc.

"Focus and Directrix in Standard Form:
Recall that for a parabola in vertex form ( y - k = (+/-)1/4p(x+4)^2 ), the value of p gives the distance from the vertex to the focus.
Exercise: Circle the value of 1/4p in these vertex form parabolas. Then clear the parentheses and solve for y, which changes them into standard form. Finally, circle the value of 1/4p in standard form.

problems:

y + 3 = 2(x-5)^2

y - 4 = (-1/8)(x+4)^2"

[formal hamlet]

How do I find Sin A= ?

opp/hyp. but OPP is a missing value on the sheet.

[silent plank]

why type of triangle do you have?

perhaps a famous theorem could be applied here

[formal hamlet]

I have no idea if its a 45 45 90, or a 30 60 90

[silent plank]

neither

try finding the 3rd side without trig

[formal hamlet]

oh pyth theorem

[silent plank]

P___

[formal hamlet]

got it.

thank you

[formal hamlet]

Given Right Triangle ABC, if Tan A=3/2, and b=5, determine the lengths of sides a and c.

How can this question be worded better? It's honestly insane.

This is the answer key for that question

just a whole mountain of confusion honestly.

tan A=3/2 is obviously the ratio

2^2+3^2=c^2

c^2=square root 13

but the whole right side of the answer key is just excruciatingly confusing.

a/b=a/b lolwat

I just wish the question was asked in a better way. How would you rewrite that?

[silent plank]

they should've use a'/b'

or skipped that entirely

tan(A) = opp/adj = a/b = 3/2

[formal hamlet]

its the determine the lenghts of sides a and c that is confusing

[silent plank]

the tangent of A gives the ratio of a and b which is given to be 3/2

which gives the equation

$\frac ab = \frac32$

[somber coyote]

ramonov:

[formal hamlet]

right, but the next step of c/b?

there's like no clue to that

[silent plank]

pythag

[formal hamlet]

Okay I see it now I think

...no i dont

I just dont see how someone can deduce that you need to figure out a/b to get A

and to deduce that c/b to get C

because she gives the length of b and you have to work your way out?

😐

im thinking too much. time to sleep. thanks romanov.

[silk crown]

suppose in the plane 10 pair wise non parallel lines intersect one another. What is the maximum number of polygons (with a finite area) that can be formed?

hey can someone help me with this one

I'm not even sure how one must approach this

[blazing panther]

What does the zero mean above the 2pi

[silk crown]

@blazing panther bruh that is impossible to read in dark mode just post it with the background

it literally just means 0 radians = 0 degrees = the start = 2 pi = 360 degrees = the end

just imagine rotating something by 2pi(360 degrees)

it just ends up in the initial state i.e. = 0

[upper karma]

Please help me with both qs

[rich wolf]

it's talking about the line 6-x mentioned earlier

which is the perp bisector

[silk crown]

Say we have a triangle with two sides variable, if because the area is maximum at some specific {a, b} (the sides), does that mean the perimeter will also be maximum

The third side being fixed that is

ok let me just set one side as x and the other as 2x (for simplicity's sake)

the third side as some constant c

[rich wolf]

@silk crown nice teeth

[silk crown]

thanks

what about the triangle

[rich wolf]

those chord segments look to be equal

bad drawing but you should see the pattern here

[upper karma]

How does a calculator calculate something like arcsin(10)? Isn't this impossible, since it has to be 0 <= x <= 1?

[rich wolf]

@Plant#3864 no

@upper karma

Youre missing the point of my drawing

Its true all the way up

The bottom 2 are equal yes

Then the 2 above that are equal

The 2 above that are equal

And the 2 above that

Think isosceles triangles

[upper karma]

Can anyone help me?

[rich wolf]

K

[upper karma]

How

Anyone?

<@&286206848099549185> also reposting my question since it got buried: How does a calculator calculate arctan for example?

finds the inverse of tan thru algebra

How does that work

theres an algorithm for it

[river forge]

taylor series presumably

or that

[upper karma]

called CURCA or something

I'm confused.. 😦

idk what its called, but yeah successive approximations using partial taylor series are what is usually used

Ok, but for example

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

this is the algorithm i was thinking of

its really fast

I'm not really wondering about fast

I am just wondering how it's possible

it's what is actually used

well, you have a number of possibilities

do you know about taylor series?

No 😦

you will learn it in calc 2

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

But how does tan really work?

it's just sine over cosine

sine is

$\sin x = \frac{e^{ix} - e^{-ix}}{2i}$

cosine is

Okay, but for example tan(5). What happens?

[somber coyote]

Carter:

[upper karma]

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

[somber coyote]

Carter:

[upper karma]

$\tan x = \frac{i(e^{ix} - e^{-ix})}{e^{ix}+e^{-ix}}$

[somber coyote]

Carter:

[upper karma]

it's unclear what you're actually looking for here

you say "What happens when you run tan(5)"

well, the computer uses CORDIC or a taylor approximation at x=5

the algorithm tells you precisely what happens

there is no further simplification

you'll just have to read about how the algorithms work

they basically do a procedure with 5 as the input

and they do that procedure multiple times

I am just kind of confused, for example why is cot(arctan(a)) = 1/a

until they have enough precision

cot = 1/tan

Yes so

if you do arctan

of

sqrt(3)/2

you will get

,calc arctan(sqrt(3)/2)

[somber coyote]

The following error occured while calculating:
Error: (intermediate value)(intermediate value)(intermediate value) is not a function

[upper karma]

nice

Hmm

I'm just really confused..

just do more practice problems :)

Which

confusion is what leads to understanding

all of em

I don't know where to get good problems...

khan academy for a start

then ask around here for more

They don't cover this stuff

they don't cover algorithms for evaluating trig functions

but that isn't necessary

Ok but I am still confused tho..

your confusion will go away

Like my last question

before you understand how to actually implement trig functions

what was the last question you asked?

for example why is cot(arctan(a)) = 1/a

lets say

hmm

just draw an example point on a circle

and instead of working with cot and arctan

try working with

csc and arcsin

to simplify things

they have the same relationship

and just draw out each step visually

you'll understand then

I don't know how to draw csc and arcsin....

On a circle

lol

csc = 1/sin

arcsin takes a y value and spits out the angle

Yeah? How can I draw that on a circle

draw a point on the circle

draw a line from the center of the circle to the point

now draw a dotted line going vertically from the x axis to the point

that's your 'y value', aka the result of sin(angle)

if you do

arcsin(y value)

you get the angle

if you plug the angle into cosecant, you get 1/(y value)

by the definition

of cosecant

But that's not true though

The domain is restricted..

So it's not even that simple

:) it does indeed work

domain restrictions don't change anything here

Okay, for example

My triangle has horizontal base sqrt(73)

But the domain of arcsin/arccos is [-1,1]

ok

So it doesn't work?

it does, if you work within the proper modulus

What does that mean

you end up with modular congruences

What 😦

in radians, 0 = 2pi

= 4pi

= 6pi

they are all congruent mod 2pi

Yes but I am talking about horizontal length

So now what

you will have to actually show me what your moral dilemma is with it so i can address that

i can't think of any violations off the top of my head, except obvious ones

like cos(arcsin(x))

mod(sqrt(73);2pi), arccos(ans) = -1.45592595223391793376i

o_o

So it doesn't even work..

wat

of course it doesnt

I did what you just told me??

sqrt(73) is not an angle

no you didnt

ok

Say I have sqrt(73) as my horizontal base

u have to do

What now

• 1

mod 2

• 1

for modulus of lengths

on the domain [-1, 1]

really [-1, 1)

I am confused, I have sqrt(73)

yes

not [-1,1]

mod(sqrt(73) + 1, 2) - 1

is what you are looking for

to make it conform to the domain

What

And what does that do

"to make it conform to the domain"

Yes

But why mod 2???

-1

1 - (-1) = 2

[-1, 1)

it's the size of the domain

Yes... but why?

I don't get why you are doing mod 2

if you mod anything higher than 2 you will still possibly not conform to the domain

if you mod anything less than 2 you are not actually conforming to the domain, but rather invalidating your inputs

by making them not "wrap" correctly

so they are actually totally botched inputs at that point

it has to be exactly 2

I just don't get this at all...

i could show you graphically if i had video chat but im not at home rn

Will you be home later?

[upper karma]

i have to work unfortunately

i did not arrive at these ideas by instruction or reading, they are only intuitively obtained in my personal experience

[thick sonnet]

Assuming you aren't changing the slope angle, wouldn't you have 1/3 the height, 1/3 of the length, and 1/3 of the width, to maintain proper ratios to keep the slope the same? If I'm right, you would recalculate the SA formula using those new sizes and compare the ratio :)

[earnest acorn]

Posting again because it wasn't answered

(I'm in geometry)

[silk crown]

well

take a look at ΔKLH and see if you can notice something useful

do you know any trig

[surreal lark]

sin(4x)-sin(3x) = 0. aka sin(4x) = sin(3x). Currently i'm at ```
2sin(2x) x cos(2x) - sin(3x) = 0
4sin(x) x cos(X) x cos^2(x) - sin^2(x)- sin(3x) = 0
4sin(x) x cos^3(x) - sin^2(x) - sin(3x) = 0

[balmy lance]

Use the unit circle to see when sin(x) = sin(y)

[surreal lark]

Em

[balmy lance]

you havent had the unit circle?

[surreal lark]

I know what it is, I'm not sure what you meant.

sin-0.1 = sin 0.9 and sin0.9 = sin 0.9

This gives me nothing eh.

[balmy lance]

well yeah

you have two cases, of which the easier is $3x = 4x + 2k_1\pi$

[somber coyote]

Pappa:

[surreal lark]

You can just do that eh?

[balmy lance]

then the other one is $3x=\pi - 4x + 2k_2\pi$

[somber coyote]

Pappa:

[balmy lance]

well yeah

[surreal lark]

As long as they are the same type

[balmy lance]

type?

[surreal lark]

sin, cos

tg

ctg

[balmy lance]

well not quite

[surreal lark]

tg4x = tg3x should be 4x = 3x + pi * k

[balmy lance]

for example $cos(x) = cos(y) \iff x = \pm y + 2k\pi$

[somber coyote]

Pappa:

[surreal lark]

ah right...

all of them have negatives

[balmy lance]

drawing a unit circle and looking when the x- and y-projections are the same

is probably the best way to check these

[surreal lark]

tg4x = tg3x should be 4x = +/-3x + pi * k

And that'll be 2 answers.

Right?

Wait..

When did you use 4x = Pi - 3x + pi * k?

[silk crown]

$sin(x) = sin(α)$, then $x = 2n\pi + (-1)^n\alpha$, where $\alpha \in [-\frac{\pi}{2}, \frac{\pi}{2}], n \in \bN$

[somber coyote]

CoolShot:

[balmy lance]

if you know what the function tan(x) looks like, then you can see that for $\tan{x} = \tan{y} \iff x = y +\pi k$

[somber coyote]

Pappa:

[surreal lark]

tg4x = tg3x
4x = 3x + pi * k
4x = pi - 3x + pi * k
``` ?

[balmy lance]

the second one is wrong

its not the same for sin and for tan

[surreal lark]

For sin the pi - will work?

[balmy lance]

4x = pi - 3x + 2pi * k is the correct one

[surreal lark]

I just remember long ago something with that matter.

For sin

Okay, so just simply, with sin/cos You have equation and pi - equation. + 2pk for both.

And for tan, ctan it's equation and equation + pi. + pk for both.

?

[balmy lance]

its not the same for cos as it is for sin

[surreal lark]

Hmm...

[balmy lance]

do you know what cos(x) is on the unit circle?

[surreal lark]

equation + 2pk and -equation + 2pk.

for cos.

[balmy lance]

yes

great

[surreal lark]

Okay, kinda got it.

Thanks.

[upper karma]

If I have a hyperplane that divides a cube into two sides. Can I then describe the one side of the hyperplane as an inequality of the hyperplane's equation

[thorny cloud]

is there any website which contains geometry and trigonometry formulas? ❓ 😄 😅

[cinder portal]

there's a trig bible, google that

[upper karma]

When constructing an equilateral triangle, which step comes after constructing a circle?

[wind heart]

Hello

Would 9 be 26.8?

[blazing panther]

What does it actually mean

When you put sin or cosine of something in the calculator

[blazing panther]

Thankssss

[olive glacier]

How do you find coordinates for D?

[tacit karma]

I don't think you need point d

[rustic island]

^

[olive glacier]

How do you solve without using point D?

[tacit karma]

what were you gonna do with point d?

[olive glacier]

Find point E

Then find gradient of BE

[tacit karma]

lemme try to solve it

[olive glacier]

ok

[tacit karma]

i figured out if you draw a big square box around the square

you can find point d

[olive glacier]

a bigger box touching smaller box?

[tacit karma]

ye

[olive glacier]

elaborate please

[tacit karma]

i can't really take photos, but I will try my best

[olive glacier]

ok thanks

[tacit karma]

draw something like this

[olive glacier]

ahhh

actually no not ahhh

how does that help

[tacit karma]

those four triangles are congruent

right?

[olive glacier]

ye

[tacit karma]

and we know

that the triangle formed from the origin, b, and a

has legs of 5, and 2

right?

[olive glacier]

ok

ye soo

[tacit karma]

and so, since the square drawn outside the square creates congruent triangles

the other triangles have to have the same length

[olive glacier]

what did u get?

for point D

[earnest acorn]

@silk crown i see 13.4 but that's it. I don't know any trig

[tacit karma]

what?

[olive glacier]

what are your coordinates of D @tacit karma

ahh i see, thanks. (7,2) yeah?

[tacit karma]

yeah

7,2

[olive glacier]

brill

[tacit karma]

i was trying to draw something

on computer

[olive glacier]

using what software

[tacit karma]

microsoft photos edit tool

not that good

but still works

[olive glacier]

i use mc paint lol

[tacit karma]

you see 7,2 now?

[olive glacier]

ye, the use of congruent triangles. You can add the length of the sides of b and a onto the point A

[tacit karma]

k

[olive glacier]

what? u found it another way using congruent triangles?

if so, please tell me.

[tacit karma]

hmm

no

i was thinking of the exact same process you did

so @earnest acorn

just draw an altitude from K to HJ and label that point x

[earnest acorn]

what's an altitude

[tacit karma]

high of the triangle

height

it is the shortest distance from a triangles vertex to its side, so it forms a right angle

By linear pair theorem, we know that angle HLK is 90 degrees

and we know that an altitude creates right angles, so KXH is 90 degress also

By angle bisectors, we know that <LHK is congruent to <KHX

Therefore, by ASA, triangles HKX and HLK are congruent

[wind heart]

<@&286206848099549185>

Can one of you scroll up and check my answer plz thanks

[earnest acorn]

okay but what's the distance from point k to segment HJ

[tacit karma]

um

do you want an explanation or just the answer

(9.4)^2+x^2=(13.4)^2

[earnest acorn]

idk i'm just confused where this is going

ohh ptharhtyragreanmdmeigjwjreg theorm

[tacit karma]

sqrt(91.2)

thats ur answer

idk if you have to round

[earnest acorn]

i get it now

it makes sense

our teacher like didn't go over pythagrhberuwgyhrthqw4 theorem at all and assumed we remember from like algebruh which we didn't

[tacit karma]

lol

[upper karma]

Yo cb can u help me in #prealg-and-algebra

[tacit karma]

k

[upper karma]

Yo can anyone help me with a quadratic func

[tacit karma]

which quadratic lol

[earnest acorn]

thanks for the help

[upper karma]

Acc hang on

I might got this

Nope

I definitely don’t have this

[wind heart]

Uhhh

Is anybody there

[gritty sail]

What

Where

[fringe crater]

Hey @wind heart , if I read your problem sheet correctly, #9 is just asking for an angle measure, not a length, so it's not 26.8. Angle MQO is 90 degrees, as you have a 90 already drawn out on your picture for angle NQO

[rustic island]

$$tan(x) = sin(x)/cos(x)$$

[somber coyote]

kolt:

[weary drift]

$\tan(x)=\frac{\sin(x)}{\cos(x)}$

[somber coyote]

RokettoJanpu:

[rich wolf]

Tanx

[fringe crater]

=

[wind heart]

@fringe crater thanks

[fringe crater]

yw

[wind heart]

Do all circumcenters have perpendicular bisectors?

[fringe crater]

Yes

[mighty dune]

Hello! I have 2 questions about trigonometry (32 and 33) and I was wondering if someone could explain what my teacher put as the answer in the review guide.

[rich wolf]

@mighty dune

Zero product property

Do you know what that means

[mighty dune]

Never heard of it before

[rich wolf]

For number 32

If you have two arbitrary numbers A and B

Such that AB=0

That implies that either A or B must be equal to 0

So you analyze both cases

Set A equal to 0 and solve

And set B equal to 0 and solve

[mighty dune]

im not sure we are supposed to solve for this question, it looks like we just want to know how many sulutions are possible

[rich wolf]

You have to solve it to find how many solutions there are

Lmao

[mighty dune]

the way he did though looks simpler though

[rich wolf]

He did exactly what i am saying

[mighty dune]

oh wait i see what he did

for asin+a

he isolated sin

asin = -a

sin = -a/a

sin = -1

[rich wolf]

Yes

Now set the other expression equal to 0 then solve

[mighty dune]

i figured it out

its the same as the other side

bcos+c = 0

bcos = 0

cos = c/b

[gray ginkgo]

Hello! I'm confused about how they got cos(-330), did they use the unit circle? I'm lost

[upper karma]

2(-165) = -330

[gray ginkgo]

oooh duh

ok thank you so much

[upper karma]

my pleasure

[rich wolf]

Lmao

[eager kraken]

How do I find the angular speed of a small pulley that is connected to a large pulley given, the diameter of each pulley and the rotations per minute of the larger pulley?

PS Diameter is 3m
PL Diameter is 9m
PL rotates 47 times per minute

I dont understand how to find theta from the PL rpm to plug into the angular speed formula of w=theta/t

i can solve these problems when the given information is not the rotational rate, like if they give me the arc length or omega etc

i just dont understand how to work in the rate

[eager kraken]

(ping me)

[quiet mica]

Could somebody help me with a trig system

[dark sparrow]

[quiet mica]

Only the system dw about whats higher

[dark sparrow]

$\begin{cases} 2^{\cos(x)} + 2^{\frac{1}{\cos(y)}} = 5 \ 2^{\cos(x)} \cdot 2^{\frac{1}{\cos(y)}} = 4\end{cases}$

[somber coyote]

Ann:

[dark sparrow]

this?

[quiet mica]

Yes

[dark sparrow]

ok

so what is giving you trouble

[quiet mica]

The 5

[dark sparrow]

what about the 5

[quiet mica]

Actually im not sure where to begin

[dark sparrow]

well maybe you could, just for a moment, make a subsitution

u := 2^cos(x), v := 2^(1/cos(y))

the system then becomes the much simpler $\begin{cases} u+v = 5 \ uv = 4 \end{cases}$

[somber coyote]

Ann:

[quiet mica]

Im gonna try this and report back

Thank youu

Im left with 1/cos(y)=0,will this be equal to cos(y)=0 [the inverse]?

[cinder portal]

You should have 2 cases

What is ur u and v value

[quiet mica]

I do but i just need to know this one

Sry for bad quality i cant use my flash because of low battery

[cinder portal]

cos(y) is between -1 and 1, will that eqn rver give u 0?

[quiet mica]

Yes

[dark sparrow]

Im left with 1/cos(y)=0,will this be equal to cos(y)=0 [the inverse]?

can 1/(something) ever be 0

[quiet mica]

Nope

Im sometimes pretty dumb

This is the finished problem

I can continue the arccos but i dont want to bother right now XD

The 1/0 and 0/1 was me checking the y

Thank you all ill hit you up if i need more help

[quiet mica]

i came here in need of help once more

sin(x)cos(y)=1/4 and 3tan(x)=tan(y) [this is a system]

[dire rampart]

you're asked to solve it?

[gritty sail]

i think there are infinite solutiosn

yeah there are infinite solutions

first i would convert everythin in tan to sin and cos

then notice u can use sin(x+y)

in which u can write x in terms of y or y in terms of x (ur choice)

and then that solves the rest

[quiet mica]

Thank you

[night ridge]

Can someone please help me with this? “Calculate the length of x”

Been on this for an hour and I can’t sleep now cause I can’t solve it and it annoys me

Or just type the answer and how you came to that conclusion and then I can ask if I don’t understand something thanks 🙂

[idle bloom]

Similar triangles

start there

[night ridge]

But are they, one of the sides are just as long as the other how can they be similar?

I mean you’re probably right but I don’t get it

If they are similar shouldn’t all the sides of the other triangle be multiplied by for an example 2 but here only two sides can be multiplied by for an example 2

Maybe I’m stupid

[upper karma]

Can someone explain this solution for me please? it has been giving me trouble for the past 3 hours

the forth row should be equal to -(1/2)cos, right?

[idle bloom]

The hypotenuse of one triangle

is the leg of the big triangle

they're similar triangles

[upper karma]

ah, you're replying to @night ridge

[idle bloom]

@upper karma thye multiplied each side by 2

to get rid of fractions

[upper karma]

.

I spent 2 hours on it, just to know they multiplied both sides by two?

wow.....

[idle bloom]

brush up on your algebra then

[upper karma]

I really should

[idle bloom]

should have noticed that the 2 became a 3, sqrt[3]/2 became sqrt[3]

[upper karma]

imma have to improve upon it before I start any more maths

@idle bloom so if they all share a common factor, mutiply by the factor?

[idle bloom]

If you want lol

[upper karma]

like if it was sq.root(3)/5 + 1/5cos + 5Sin

[idle bloom]

to most people, getting rid of fractions makes it easier to work with

sq.root(3)/5 + 1/5cos + 5Sin = 0

you could multiply both sides by 5

sq.root(3) + cos + 25Sin = 0

[upper karma]

@idle bloom Thank you very much, You just motivated me to think effectively*

[night ridge]

Spamskin okay so you just flip em

But I still can’t believe they’re similar triangles

[idle bloom]

https://mathbitsnotebook.com/Geometry/RightTriangles/RTmeanRight.html

here

don't believe me, look at the prove yourself

much better way to learn

[night ridge]

Oh I see, thank you

So the answers is 6cm?

[eternal orchid]

@forest dove

[dim bronze]

@coral basin the point in each square at which the two lines intersect must move over 1/5 of the length of each square for each square

It's the same distance from the bottom as it is from the right

The distance from the bottom to where the line intersects each "wall" between two squares would be the same

Now you have a bunch of right triangles with the dimensions

Idk if there's an easier way, this seems a bit forced but it should work

[coral basin]

ok

thanks

also

buying winrar is an absolute alpha move

[wind heart]

I have a question

On 1-5, I’m kind of confused. I’m having a test tommorow, and I may mix up these definitions. Like how am I supposed to tell a circumcenter from an in center in definition format?

<@&286206848099549185>

Like, what does “vertices” even mean (for number 4) and how is number 2 incorrect?

[rich wolf]

@wind heart

Let's start with number 2

[wind heart]

Hi

[rich wolf]

circumcenter

circum means "circle"

think circle center

[wind heart]

ok

So like a incenter dosent have a circle?

[rich wolf]

that is a circumcenter

my bad i said the wrong thing earlier

[wind heart]

It’s aight

Ok so they all have the same vertex?

Equal vertex?

[rich wolf]

the vertices of the triangle are points on the circumcircle

[wind heart]

But like dosent incenter have equal vertex too

[rich wolf]

incenter is inside the triangle

"circum" is latin for round, or around something

circumcenter is around the triangle

[wind heart]

Hmmmmmm

Oh I see

But I still don’t get what number two is implying tbh

[rich wolf]

perpendicular bisectors of the triangle meet at the circumcenter

[wind heart]

But like aren’t incenter that too

[rich wolf]

no

[wind heart]

Because they have perpendicular bisectors I think

[rich wolf]

look at the picture. in the circumcenter, the vertices are points on the circle

but in the incenter, the MIDPOINTS are points on the circle

[wind heart]

Are the vertices the “vertex”?

[rich wolf]

yes

vertices is plural, vertex is singular

[wind heart]

What are the vertices though?

Like

https://media.discordapp.net/attachments/326138757474680852/676974053256331264/perpbis1ba.png

In this picture

What would the vertexes be?

Would the circumcenter be the vertex?

Or are M N and P the vertices which are an equal distance from the circumcenter

[rich wolf]

points A, B, and C are the vertices

because they are the points that define the triangle

[wind heart]

Also, for number 2, would the incenter not be a perpendicular bisectors because they don’t create congruent sides?

[rich wolf]

that's not the definition of the incenter

incenter is the point where ANGLE BISECTORS meet

[wind heart]

Then why do they have right angles

I thought 3 right angles = 3 perpendicular bisectors