geometry-and-trigonometry

[upper karma]

[dusty sleet]

yo

can someone explain me

when to only use degree /radian moded

cos i dont wanna fuck up lol

[silent plank]

if the question is in degrees use degrees
if the question is in radians use radians
formulae will state whether stuff is in degrees or radians

[dusty sleet]

okay thanks alot

[silent plank]

if they question asks for conversions, multiply by the appropriate value

[dusty sleet]

another question, my book says cosine rule= sqroot a^2+b^2+2abcos(c)

is this correctN

[silent plank]

no, wtf even is that

[dusty sleet]

wait

it says

C=sqroot A^2+b^2-2abcos(c)

does this make more sense?

[silent plank]

as written, that's wrong

math is case sensitive

and be clear what's supposed to be under the sqrt with parentheses

$c = \sqrt{a^2 + b^2 - 2ab\cos(C)}$

[somber coyote]

ramonov:

[dusty sleet]

yeah

thats it

[silent plank]

would be one of the forms of the cosine rule

[dusty sleet]

but its in big letters

[silent plank]

does what i typed match the book exactly?

[dusty sleet]

the formula, yes but all the letters except for cos c are big

[silent plank]

so its not the same as the book?

[dusty sleet]

look

[silent plank]

by exactly i mean exactly

[dusty sleet]

$C = \sqrt{A^2 + B^2 - 2AB\cos(c)}$

[somber coyote]

Pinacolada:

[silent plank]

ew

does the book have a diagram to go with it?

[dusty sleet]

no its written and sends me straight to do exercices

[silent plank]

by convention capitals are supposed to be used for vertices and angles
and the respective lowercase letters represent the respective sides opposite those angles

[dusty sleet]

yeah

[silent plank]

what book is it so I know what to not recommend to people

[dusty sleet]

russell c hibbeler, statics

it should be a reliable book thoug..

so the real cos rule is this i suppose, what i found on internet

$a^2 = b^2 + c^2 - 2bc\cos(A)}$

[somber coyote]

Pinacolada:
Compile Error! Click the reaction for details. (You may edit your message)

[silent plank]

i mean the letters are just variables, you just need to know what the variables represent

[dusty sleet]

with a^2 being the angle youre looking for ?

sides*

[idle swan]

you can look for any side as long as you know enough of the other values

[silent plank]

from the most common form:
$$c^2 = a^2 + b^2 - 2ab\cos(C)$$
$a$ and $b$ are two of the sides, $C$ is the angle between them, and $c$ is the side opposite $C$

[somber coyote]

ramonov:

[dusty sleet]

so the side youre looking for needs to have a opposite angle right?

ok nvm i get it

i just tried the formula i sent it worked though

[silent plank]

i mean the letters are just variables, you just need to know what the variables represent

[dusty sleet]

it worked with finding a side

it wouldnt work with angles i suppose?

[silent plank]

to get the angle, solve for C

just needs some algebraic manipulation

some texts may already do that for you

[dusty sleet]

what would be the correct formula for angles ?

[silent plank]

its effectively the same formula (in a different form)

[upper karma]

yes it is

[pallid shadow]

Hey, is there a way to solve this problem?

I have 2 objects, A and B.

I know A's actual diameter, I also know A's angular diameter (apparent diameter from my point of view).

I know my distance to A.

I know B's actual diameter

B's Angular diameter = A's angular diameter

What is my distance from B?

[mental vessel]

i dont get how to start this

[silent plank]

assign variables for your unknown lengths

(the height and the the distance from the base of the 51° angle to the tower)

apply tan in multiple ways and solve the system

[mental vessel]

i cant use sohcahtoa because its not a right angle for the 32 degrees and the 17.5

[dark sparrow]

there is a right angle in the picture.

[mental vessel]

yeah ik

[dark sparrow]

and there are two right triangles.

[mental vessel]

yeah i know but how can i do it if i only got the length of a tiny piece

if the 17.5 was under the 51 then i could use tan because its a right angle and i could do opposite over adj

but its not a right angle where the 17.5 is

[silent plank]

the 32° angle is also part of a right triangle

as mentioned above:

assign variables for your unknown lengths

[mental vessel]

yeah i know, but theres no side lengths besides the 17.5 so how can i solve it

[silent plank]

have you done any of the things i initially wrote?

[mental vessel]

yea

[silent plank]

what variables did you use?

[dark sparrow]

also may i suggest giving names to relevant points?

[silent plank]

^ that too

[silent plank]

tan(32°) = ?

[mental vessel]

z over adj

but theres no adj length

[silent plank]

segment addition postulate?

[mental vessel]

idk what that is

[dark sparrow]

what's the length of the lower side of the big triangle

[mental vessel]

dont know, i only know the length of apart of that bottom length

if i could find y i could add them

[silent plank]

a + b = a + b

[dark sparrow]

and what's stopping you from adding them without knowing y

would your fingers fall off if you wrote y + 17.5?

[mental vessel]

what would that acomplish

accomplish

[dark sparrow]

it'd let you write down an equation from using SOHCAHTOA on the big triangle

[silent plank]

and give you some information about your unknown lengths

$\tan(32\deg) = \frac{z}{17.5 + y}$

[somber coyote]

ramonov:

[mental vessel]

yeah

[silent plank]

similarly what can you get from tan(51°)

[mental vessel]

tan51 is z/17.5+y

right

[silent plank]

no

[upper karma]

[mental vessel]

bruh

[silent plank]

how are you getting that?

[mental vessel]

oh nvm

just z/y

[dark sparrow]

also note on fractions in plaintext

[silent plank]

$\tan(51\deg) = \frac{z}{y}$

[somber coyote]

ramonov:

[mental vessel]

eya

yea

but how can i get y

[silent plank]

ok since tan(32°) and tan(51°) are constants
you effectively have a system of 2 linear equations with 2 varaibles

[dark sparrow]

you have two equations and two unknowns

they aren't yet linear

[mental vessel]

im just starting this so idk

[silent plank]

have you solved any systems of equations before?

[mental vessel]

no

[upper karma]

ooof

[mental vessel]

weve just been doing sohcahtoa

[dark sparrow]

algebruh

[silent plank]

systems of equations is under general algebra

[mighty wharf]

[silent plank]

able to do stuff like substitutions?

manipulate equations?

[mental vessel]

ive done that 3 years ago so i forget all of it now

[dark sparrow]

bruh

[silent plank]

review it

[upper karma]

If JK = 7x + 7 and KL = 5x + 2 and JL =45
I found x = 3
But how do I find JL?

Would I set it up like 45 = 5x + 2 ?

Oh wait would I plug in three

And figure out what JK and KL is

Then I can figure out JL

[dark sparrow]

JK = 7x + 7

I found x = 3

[upper karma]

Yea that'd mean JK is 28?

[dark sparrow]

indeed.

[upper karma]

Haha lmao I think I know what to do know, I'll just apply it to KL now

And then plot it out

And subtract the total from JK

Thanks for the guidence

And help!

[slow void]

can some one please give me the answer to this? i've been on this question for 7 hours.

[stone salmon]

@slow void what does your teacher means by r.a. (deg)

[long wedge]

Can someone help me with this??

[patent plume]

matrices?

or what?

[slow void]

I need help with this last final one

[lusty gyro]

Do you know the amplitude and period?

[slow void]

nope

[lusty gyro]

The amplitude is the distance from the base line to the minimum or maximum, so in this case 3

[slow void]

okay

[lusty gyro]

Period is the distance between minimums or two maximums, so in this case 6

[slow void]

oh ok

[lusty gyro]

Since this graph has a maximum near x =0, and so does cosine, we will use a cosine function

However, this graph is a little tricky since the entire thing is offset horizontally

[slow void]

oh ok still confusing though

[lusty gyro]

We can see that the first positive x intercept is at x=1

But in cosine it is pi/2

[slow void]

ok

how would i form the answer

[lusty gyro]

So the function is pi/2-1 left from the cosine function. That is our horizontal displacement: -(pi/2-1)

And the formula is :acos(b(x-c))

[slow void]

ok

[lusty gyro]

a is amplitude

b is 2pi/period

c is the horizontal displacement

[slow void]

ok i remember i answered another question wrong and the format was wrong

thats how the format should have been

[lusty gyro]

Yes thats the form

But they used the sin function instead, which is not as good for that situation.

Do you understand how to do this question now?

@slow void

[slow void]

no i tried this way

[lusty gyro]

No, that is the answer from a different question.

[slow void]

3sin((pi/3)(x+2))

got it

[lusty gyro]

You got it correct?

[slow void]

yes 😄

[lusty gyro]

Good job!

[slow void]

@lusty gyro thanks

[lusty gyro]

You're welcome

I've got to say, I haven't done a question like that in a long time

[kindred fox]

Hello, is anyone available to help me with some trig problems tomorrow around 8pm EST?

[lusty gyro]

@kindred fox I can help you by discord chat tomorrow most likely.

[kindred fox]

OK sweet. Do you mean in this chat room right here or though DM?

@lusty gyro

[jagged relic]

@upper karma

[upper karma]

alright give me a bit to help the other guy

[jagged relic]

ok

thank you so much

[upper karma]

whats your question?

im not gonna do your homework for ya

[jagged relic]

yeah i know

i was just wondering like t general terms

the*

i received some notes but i looked at it and it didnt make sense at all

i was wonderig what are the 4 top terms

[upper karma]

do you have the notes you can post?

[jagged relic]

yeah i have this

its not letting me open the image

[upper karma]

@jagged relic do you still need help

[lunar mulch]

Where can I find the derivation of these equations? my textbook doesn't really have one

$\begin{array}{l}
2 \tan ^{-1}(x)=\sin ^{-1}\left[\frac{2 x}{1+x^{2}}\right] \
2 \tan ^{-1}(x)=\cos ^{-1}\left[\frac{1-x^{2}}{1+x^{2}}\right] \
2 \tan ^{-1}(x)=\tan ^{-1}\left[\frac{2 x}{1-x^{2}}\right]
\end{array}$

[somber coyote]

Exynouz:

[lunar mulch]

Hi

[acoustic jungle]

those equations aren't always true

but they can be derived I guess using a 1,x,sqrt(x^2+1) right triangle

and the double angle identity

@lunar mulch

[mossy narwhal]

,rotate

[mossy narwhal]

How would the graph B look like if the answer was C

[past geyser]

why is this in trigonometry

[silent plank]

there are shapes and an axis

[lunar mulch]

but they can be derived I guess using a 1,x,sqrt(x^2+1) right triangle
@acoustic jungle

Ohh hmm

those equations aren't always true
@acoustic jungle

How do you know the domain

The bounds I mean

@acoustic jungle

Ohh hmm
@lunar mulch

Got it!

[upper karma]

could i get some help please :)

<@&286206848099549185> (ouo)b

[upper karma]

no more help needed ty

[kindred fox]

How do i find the area of this? Lmao

[upper karma]

its 4 quarter circles

each with different radii

do you know how to find the area of a circle with radius r?

[lunar mulch]

Area of a quarter of a circle = (1/4)πr²

Now put the radius of each of the circles pieces and add them up

[dark sparrow]

yes, (x/y)^2 = x^2/y^2

[dark sparrow]

15 degrees you mean?

you don't need to calculate tan(15°) to calculate your thing

remember that $\tan(x)\tan(90\dg - x) = 1$

[somber coyote]

Ann:

[dark sparrow]

you don't need it

$\tan(15\dg) \tan(75\dg) = 1$

[somber coyote]

Ann:

[dark sparrow]

and $\tan(35\dg)\tan(55\dg) = 1$

[somber coyote]

Ann:

[dark sparrow]

it's this identity i gave you (which you ignored), with x=15 and then with x=35

R \ {(2k+1)pi/2} means the set of all real numbers which are not odd multiples of pi/2

,,,,,, no

if you mean degrees

then it'd be all real numbers that are not 90 times an odd integer

i.e.

everything except 90, 270, 450, 630, etc...

[dark sparrow]

i can't read this

[upper karma]

Sin^2 (x) + Cos^2 (x) = 1 for all x

It’s an identity

If I am correct

For all x means for any x out there

I mean

Just try it across the quadrants

Possibly?

Haven’t really tried this

But I’m pretty sure it works for all quadrants

[dark sparrow]

sin^2*x

[upper karma]

[dark sparrow]

@upper karma sin x does not mean the product of sin and x

the multiplication sign does not belong there

then why write it

yes!

ok so like... $\cos(x) \sqrt{1-\sin^2(x)} + \sin(x) \sqrt{1-\cos^2(x)} = 1$ is not always true

[somber coyote]

Ann:

[dark sparrow]

it's false for x = 5π/4 (radians) for example

[mighty wharf]

[dark sparrow]

idk. what are you asked to do?

oh

well ok

you might find it useful to remember that $\sqrt{a^2} = |a|$

[somber coyote]

Ann:

PRETIOSU:
Compile Error! Click the reaction for details. (You may edit your message)

[upper karma]

sinx |cosx| + cosx |sinx| needs a domain to open up those mods

Yes ?

yea it's just different ways of writing it

Um wdym ?

[somber coyote]

PRETIOSU:

[upper karma]

So ?

What do you want to do with it

It's already 2 multiplied terms

You want to expand it

?

No you can't do that

Np

[junior light]

Try writing the terms in terms of sines and cosines, then cross multiply.

Something about this question is fishy though.

[upper karma]

I think so but it's going to require so many calculations

[hallow crystal]

Is this right

[hallow crystal]

Omg thank youuu

I’ll re do it and show you

@dark sparrow M isn’t a point? But then whyd u say ok to the third to last question

Omg and for the coplanar point I meant to write n not m ahhh

[dark sparrow]

there are 3 different diagrams here, also m ≠ M

[hallow crystal]

Oh fuck my bad

[dark sparrow]

also in the first diagram there isn't even a point named N

[hallow crystal]

Ah it’s a line

Q is the right answer 😔

Ah for question b Q isn’t a coplanar point either ugh

And @dark sparrow what do u mean what’s up with the arrows?

[dark sparrow]

left of the bar

you're asked for the rays which start at J

[hallow crystal]

For all four rays or just JF?

[dark sparrow]

so the arrows should point away from J, but some of them do not

[hallow crystal]

Oh but it’s flipped around, J isnt the first point

[dark sparrow]

i'm talking about $\overleftarrow{JH}$

[somber coyote]

Ann:

[hallow crystal]

Oh so it would be HJ?

[dark sparrow]

HJ with the arrow to the left

or JH with the arrow to the right

[hallow crystal]

For the seventh question is my answer incorrect or did u want me to elaborate more

And for second to last question will there still be a bar over k? I just need to take the arrows out? But wouldn’t that make it a segment

I thought k was a line

And how is k the intersection between itself and plane A? @dark sparrow

[dark sparrow]

what do you think the intersection should be

the intersection of two geometric objects consists of the points which lie on both objects simultaneously

[hallow crystal]

Right, and M lies on k

@dark sparrow is it ok for me to @ u :

[dark sparrow]

and M lies on k
but k doesn't consist of just M

and yea ig it's ok to ping me...

[hallow crystal]

But M is still at least a part of k? It still doesn’t count?

[dark sparrow]

M isn't the only point that exists on k

if you are asked a question like "who among these 20 people smoke cigs" and you say "Jim" you are not only saying that Jim smokes but also that everyone else in the room doesn't

[hallow crystal]

For the seventh question is my answer incorrect or did u want me to elaborate more
@hallow crystal
@dark sparrow and what about this?

[dark sparrow]

the 7th? the one where you said the arrow pointed in the wrong direction?

[hallow crystal]

Yea, the open answer question

[dark sparrow]

i explained exactly how you were wrong

[hallow crystal]

:0

Was I full wrong or half wrong? Cuz ur explanation made me think I was half wrong

Cuz u were correcting me, but not answering the question directly I think

[dark sparrow]

full wrong

the arrow is not "in the wrong direction" because there is no ray you're trying to specify & could fail to specify properly

[hallow crystal]

So the direct answer to the question would be “no, the arrow is supposed to tell which the 2 points is the origin of the Ray”?

@dark sparrow

[dark sparrow]

no

[hallow crystal]

I mean the arrow’s doing her job but it’s pointing in the wrong direction right?

[dark sparrow]

"no, they are not the same ray. HJ starts from H while JH starts from J & they point in opposite directions

[hallow crystal]

And for second to last question will there still be a bar over k? I just need to take the arrows out? But wouldn’t that make it a segment
@hallow crystal > I thought k was a line
@dark sparrow
And this?

[dark sparrow]

single letter line names do not have anything above them

[hallow crystal]

:0 just line k

[dark sparrow]

just drop the arrows

yes

[hallow crystal]

I did nawt know dat

drop the bar too?

[dark sparrow]

yes

[hallow crystal]

@dark sparrow is this all right?

[dark sparrow]

point not coplanar with Q, S and T
Q

[hallow crystal]

WHA

SHDKFJSKFHSH

wait but there’s only one plane

And Q is the only point off the plane

n doesn’t count cause it’s a line

@dark sparrow

[dark sparrow]

you are asked for a point not on plane QST

the plane isn't drawn explicitly but it still exists

[hallow crystal]

:0

Changed it to V, anything else wrong? @dark sparrow

[hallow crystal]

@dark sparrow

[dark sparrow]

seems ok now

[eager ocean]

segment XY is congruent to YX, this is Reflexive Property of Segment Congruence, right?

[hallow crystal]

Ain’t that just another name for the line segment

[eager ocean]

right so it would be equal therefore its reflexive

[devout harbor]

cot(2x) = 1/(tan2x). Recall that tan(2x) = (2tanx)/(1-tan^2 (x))

[earnest echo]

Ctg(x)=1/tan(x)

[upper karma]

Hey

[lethal tulip]

hey

@upper karma @humble aspen

[humble aspen]

Hey

we basically did everything our teacher told us

[upper karma]

ok heres the problem

[humble aspen]

but like

[humble aspen]

the answer we have

isnt listed

[upper karma]

-2, -3.4

is what we got by using the equation

of

[humble aspen]

we did that formula

[upper karma]

(k (x2-x1)+x1, k(y2-y1)+y2)

[lethal tulip]

ok first what is the length of the line?

[upper karma]

i dont think you need the lenght

use the pythagorean theorem

[humble aspen]

mhm

[lethal tulip]

I would just find the length then the point that satisfies that

we can try it a different way if you want

[humble aspen]

i didnt try that

[lethal tulip]

so B is what? 7,2?

[humble aspen]

mhm

[lethal tulip]

A is -8,-9?

[upper karma]

and look

[humble aspen]

OML

LORENZO

i said that-

[lethal tulip]

your x point is wrong

[upper karma]

i looks exactly like 2/5ths

[lethal tulip]

a is -9

it looks like

no its not

i have old man eyes

[humble aspen]

wait so i counted wrong too

😊👍

[lethal tulip]

its -8, -7

[upper karma]

if you want distance

ill do it rq

[humble aspen]

i think i got it

the answer is b

(2,-1)

[upper karma]

wait how

im super confused

[humble aspen]

oops

i put a negative in there-

[lethal tulip]

you can check your work using the pythagorem theorem

find the length of each liens

[upper karma]

OH

[humble aspen]

yeah ill do that too

just to be sure

[upper karma]

THERE IS TWO RIGHT ANSWERS

[lethal tulip]

you might have found AP to BP

[upper karma]

omlll

[lethal tulip]

you might have found it backwards

[upper karma]

i did

[lethal tulip]

yea

[upper karma]

im dumb

[lethal tulip]

nah

just take your time

[humble aspen]

Lol

thanks

i spent so much time struggling-

[upper karma]

i dont think we needed distance tho

[north vortex]

Im studying for a test without a review and i know theres a question that talks about like ratios (3:1) the problems go like (x,y to a ratio of 4:1) im super loss

i really dont know what the problem is

@ me

[cloud spindle]

anybody here good at trig

[upper karma]

Just post

[upper karma]

Part?

[cloud spindle]

huh

part..

[upper karma]

Which part ??

[cloud spindle]

ALL.

[upper karma]

Do you know what's tan(x) in terms of sinx and cosx

[cloud spindle]

no.

[upper karma]

these are fundamental trigonometric functions

You shouldn't attempt before knowing what they are

tanx is sinx/cosx

Cotx is 1/tanx

Cosecx is 1/sinx

Secx is 1/cosx

[cloud spindle]

wh-

[upper karma]

Hm?

hey azazeel

kait has an exam soon

and she needs like

ALOT of help

how much for you to go through it with her

and help ;d

I didn't know her situation

she has an exam in like

a couple of days

and she needs LOADS of help

[cloud spindle]

OIWJDFHUGJ

[upper karma]

So ?

[cloud spindle]

REN.

oh man

I dont even know where to begin

[upper karma]

You want me to tutor ? I don't really do that

[cloud spindle]

pain.

[upper karma]

Read up basic things these things won't take much time

[upper karma]

Welp I'm confused on how to get angle m<KIH

[silent plank]

first set up an equation and solve for x

[upper karma]

Yea setting it up is what I'm struggling on

Solving shout be a problem

[silent plank]

you are given the measure of angle JIH as (4x+3)°
what are you told about KIJ?

[upper karma]

It's equal to/congruent to JIH

[silent plank]

so its measure would be: ?

[upper karma]

The measure of KIJ would be (4x+3)°

?

[silent plank]

yes

[upper karma]

Okie but now wut

[silent plank]

and then apply angle addition postulate

m<KIJ + m<JIH = m<KIH

[upper karma]

Ohhj

That makes sense so I add em together to get KIH

[silent plank]

and you are also given the measure of KIH in terms of x

[upper karma]

So 8x+6 = 12x?

Im a bit confused

[silent plank]

so far that equation is good

[upper karma]

Okie and now 6 = 4x

Then 6/4

X = 3/2

?

[silent plank]

lowercase x but yes

[upper karma]

Okie and I plug this back in

12(3/2)

18?

[silent plank]

yes

[upper karma]

Thanks for the help lmao

[tired knoll]

can someone hlep me with this

I think its easy but I cant find a way to get BK

<@&286206848099549185>

oh whoops just read the help info

[lusty gyro]

Its fine

[lusty gyro]

@tired knoll Do you still need help?

[tired knoll]

uhu

[lusty gyro]

Is that a yes?

[tired knoll]

yes

[lusty gyro]

Use similar triangles

[tired knoll]

but I dont know what ak is or ck is

[lusty gyro]

Do you know what AC is

[tired knoll]

yeah 20

[lusty gyro]

BC is to AC as BK is to AB

[tired knoll]

oh yeah

ohhhhh

yeah

ill try again

[lusty gyro]

You seem like you understand 🙂

[tired knoll]

thank youuu

[lusty gyro]

You're welcome

[grand belfry]

is this correct?

sorry for the resolution

[paper vale]

well kind of, the top one has correct answers but that is only because cos45=sin45, it should be the other way around

[upper karma]

6/5

no I'm just kidding

[earnest echo]

[upper karma]

can you write sin(2x) in terms of sin and cos

go ahead

uuuh?

sin(2x) = ?

yes

so you need to calculate cos(x)

do you know any identity involving cosine and sine?

yes

correct

BUT

remember that x is in the 2nd quadrant

aight

nice

np

[thorny bough]

how do i solve v * cos (vt) = -cos (t)
?

for t

[lusty gyro]

There are 2 variables

[drowsy rune]

Currently trying to find the trigonometric Function from point values on a graph, have these values but Webwork will not take any of the answers I've gotten for the formula
Amplitude = 5
Period = 20
midline = -1
This is a Cos Function, Shouldn't it be
y = 5cos(20x) -1
?

[thorny bough]

yes. i’d like t in terms of v

@drowsy rune period is inverse to frequency pretty sure

idk what midline is

[drowsy rune]

:0! 👀

[thorny bough]

so 5cos(x/20)-1

but im the idiot of the server so dont listen to me

[drowsy rune]

a new way to try is much better than the nothing I've got going! But webwork wouldn't take that either

[lusty gyro]

Midline is the line that the function oscilliates around for. most trig functions

[thorny bough]

well then ur equation seems right to me

[lusty gyro]

Without any offsett it is usually 0

[thorny bough]

what units did they give the period?

[high shell]

@drowsy rune for a period of T, you need the function to be cos(x*2pi/T), for example.

[thorny bough]

ye try 2pi

[high shell]

because then for x=T it's cos(2pi).

[thorny bough]

maybe they want in hz

[high shell]

maybe they want in hz
@thorny bough the period is a measure of time, not frequency

[drowsy rune]

i simply think... webwork is evil /lmao

[lusty gyro]

Sorry, I was wrong, nvm

[thorny bough]

@high shell in the function its frequency

[high shell]

@drowsy rune ah, that's because it's wrong

you're using cos, which is 1 at 0.

so at x=0 you get 5-1 = 4.

You need 5*sin-1 here

[drowsy rune]

Thank you, you are my hero

[pale cargo]

how do you find vertical asymptotes in tan functions

@high shell

[high shell]

find how?

[pale cargo]

$y=22\tan{(3x-8)}$ here is an example how would I find the Asymptotes

[somber coyote]

Albot1288:

[pale cargo]

for positive and negative closest to zero

[high shell]

I mean, the closest asymptotes of tan(x) is just at x = +-pi/2.

and then every pi.

so find the closest-to-zero x values for which 3x - 8 is equal to pi/2 + pi*k for some integer k.

[pale cargo]

I originally thought it was period/+-2

and is there a way without random integers? because it seems more like guess and check work

[high shell]

I didn't say "random integer". Each k gives an asymptote, but only one of them is going to be closest to 0 😛

$$ 3x - 8 = \pi/2 + \pi k $$
$$ 3x = \pi/2 + \pi k + 8 $$
$$ x = \pi/6 + (\pi/3) k + 8/3 $$

[somber coyote]

ConfusedReptile:

[high shell]

so now we just need to find the two k values for which x is closest to 0

[pale cargo]

can we set x to 0 to find it?

[high shell]

sure

you'll get some fraction for k, which you'll need to round up and down to get the two values

the rounding part may be annoying though

[pale cargo]

when I set x to zero I get $k=-1/2 - 8/\pi$

[somber coyote]

Albot1288:

[high shell]

that's right

[pale cargo]

so thoughs are our asymptotes?

[high shell]

so that's around -0.5 - 8/3 ~= -3.3 or something

so our k are -3 and -4 (for a more formal proof, prove that $-4 <= k=-1/2 - 8/\pi <= -3$)

[somber coyote]

ConfusedReptile:

[pale cargo]

-3.0465

[high shell]

one of them is going to be the negative asymptote and the other the positive one

[pale cargo]

but how do we find out?

[high shell]

we can just put one of them into the function to see which one it is.
Let's calculate them first:
$$x_1 = \pi/6 + (\pi/3) (-4) + 8/3 = 8/3 - (7 \pi)/6$$

[somber coyote]

ConfusedReptile:

[high shell]

$$x_2 = \pi/6 + (\pi/3) (-3) + 8/3 = 8/3 - (5 \pi)/6$$

[somber coyote]

ConfusedReptile:

[pale cargo]

where did the other fractions come from

[high shell]

I'm just substituting k = -4 and -3 into our formula for x from k.

[pale cargo]

nvm you simplified them

[high shell]

so now if we substitute $x = x_1$ into $3x-8$, the argument of the tangent, we get:
$$
8 - (7 \pi)/2 - 8 = - (7/2)\pi = -4\pi + 1/2\pi
$$

[pale cargo]

so then x1 = -0.9985 and x2 = 0.0487

[somber coyote]

ConfusedReptile:

[high shell]

indeed $3x_1 - 8$ is equal to $\pi/2$ modulo $\pi$ - it must be, since we found it as a solution to that equation. But it also is equal to $(1/2) \pi$ modulo $2 \pi$.

[somber coyote]

ConfusedReptile:

[high shell]

Our x is going to be above that - so we need to look at the graph of a tangent and see what asymptote is right after pi/2. It's a positive one, so the asymptote at x_1 is positive. The asymptote at x_2 must be negative, since it's the next one (they alternate).

So there we have them.

the plot proves us right:

(x_2 is around 0.04, so the right asymptote is just after zero)

[pale cargo]

oh ok Thanks for the help

[thorny bough]

can i solve v * cos (vt) = -cos (t)
for t in terms of v?

[lusty gyro]

@thorny bough Do you still need help?

[thorny bough]

@lusty gyro yes

please, thank u

[lusty gyro]

First of all, are you positive that that is the problem, because it is a little tricky

[high shell]

That looks like a nasty equation.

[lusty gyro]

I think that double angle formulas may be helpful, but I don't know

[pale cargo]

hey @high shell how do I know which numbers I need to round with when finding asymptotes

[thorny bough]

@lusty gyro yes

[lusty gyro]

Nvm, double angles won't be helpful

[high shell]

@thorny bough Is v real or maybe integer?

[thorny bough]

real

this should be an equivalent equation

[high shell]

Because I don't think it's solvable for real v

[thorny bough]

its the derivative of sin(t)+sin(vt)

[high shell]

@pale cargo Round? What do you mean?

[thorny bough]

@high shell uf that would make me very sad if true

[lusty gyro]

Sorry, got to go, maybe this is helpful? https://en.wikipedia.org/wiki/Chebyshev_polynomials

[pale cargo]

back when remember when we set our thing to 0 and got -3.3 something

then you choose -3 and -4 for us to find the asymptote

[thorny bough]

@high shell if i make v contant, say 1.93, can i solve and how?

[high shell]

@thorny bough Numerically or graphically, for example

[thorny bough]

numerically

[high shell]

back when remember when we set our thing to 0 and got -3.3 something
@pale cargo
We needed the two k-values for which x is closest to zero. So we found the (real) value for which is zero and took the two closest integers to that

[pale cargo]

but do they need to be whole numbers

[high shell]

Well, what are all values of x for which tan(x) goes to + or - infinity?

[pale cargo]

idk

and will it work for other functions like cos or csc as well?

[paper vale]

Think about the definition of tan

And note that as x->0 y/x -> ♾️

[earnest echo]

Write
ctg as tg and solve the Quadratic
Use the interval to determine the right value and then try to find sin(2x)

Yes

What?

Your interval is (0,π/2)

Can I see your work

👍

Yes

[quasi dove]

hmmm

isn't it faster to solve x for tg(x)=1

[earnest echo]

There is an identity involving sin(2x)

$\sin(2x)=\frac{2\tan(x)}{1+\tan^2(x)}$

[somber coyote]

The Godfather:

[quasi dove]

btw

$\tan(x)+\cot(x)=\frac{1}{\sin(x)\cos(x)}$

[somber coyote]

ThisIsMyName:

[bitter turret]

can someone show me how to draw this

[silent plank]

you don't even need to determine the value of tan(x)
let s=sin(x); c = cos(x)
s/c + c/s = 2
s^2 + c^2 = 2sc (by multiplying both sides by sc)

[quasi dove]

yeah, that's what i thought of

but that solution is also correct

[bitter turret]

anyone?

[silent plank]

i'm assuming this is related to construction with straight edge and compass?

[bitter turret]

yeah

its basically the lesson about planes lines segments etc

[silent plank]

well first draw a segment MN

and a point either P or Q depending on your preference

use the compass to copy the length of MN

and draw an arc using that from the point you drew

and connect your point to somewhere on that arc

[bitter turret]

how do i make them congruent

[silent plank]

use the compass to copy the length of MN

[bitter turret]

ik but where do i put them

[silent plank]

which steps in my outline don't you understand?

[bitter turret]

i dont get where i should place the P and Q

is it alright if u show me a visual example

[silent plank]

place one of the points anywhere you want

doesn't matter

you can even place it on a completely different sheet of paper

preferably keep it close by

by congruent they just want your lines to have equal length

[bitter turret]

so is it like parallel lines

[silent plank]

don't need to be parallel

[bitter turret]

where the lines are above each other with equal length

[silent plank]

just equal length

[somber coyote]

PRETIOSU:

PRETIOSU:

[summer spire]

what's tg?

[silent plank]

$\tan(2x) = \frac{2\tan(x)}{1 - \tan^2(x)}$

[somber coyote]

ramonov:

[silent plank]

tg(2x)=2tg(x)/[1-tg^2*(x)]
that ***** was extremely inappropriate

[dark sparrow]

no.

[high shell]

no.

only using the $\sin^2(x) + \cos^2(x) = 1$ identity.

[somber coyote]

ConfusedReptile:

[dark sparrow]

your question's kinda vague

i'm not sure what exactly you're looking for

is there a problem you're doing?

[high shell]

which gives two possible values:
$$
cos(x) = \pm \sqrt{1 - \sin^2(x)}
$$

[somber coyote]

ConfusedReptile:

[dark sparrow]

yeah the left hand side expands to sin^2(a) cos^2(b) - sin^2(b) cos^2(a)?

and then write cos^2(a) as (1 - sin^2(a)) and likewise for b

[slim saddle]

Hello

[upper karma]

Hello

What is your question?

[slim saddle]

I need clarification of what major arcs and minor arcs actually are

And how to get measure

[upper karma]

Well, if you select two points on a circle, it divides the circle into two arcs, the larger one is the 'major' arc and the smaller one 'minor'

[slim saddle]

Ok

Oh

[upper karma]

Major arc should be larger than the semi circle

[slim saddle]

Ok

Can u check if I’m right

For q3

[upper karma]

360-115=245 you are right

[paper mauve]

what are non congruent alternate exterior angle

[proven sentinel]

is it 45

helps m

e

ok yea it is

:smoothbrain:

i swear i am not usually this dumb

[stone valley]

how to solve this one?

[upper karma]

y = 180 - 2 * alpha

[stone valley]

thanks.

but then

how will i solve i

still problem

[upper karma]

tan^2 + 1 = sec^2

[stone valley]

sorry, but i literally started trig today.

i still didn't learn sec*2.

[hallow crystal]

Ight im confused 🤡 x can’t be negative right?

[silent plank]

600(km) isn't the hypotenuse of your right triangle

[hallow crystal]

And we’re not allowed to do this right? Cuz earth is a sphere

600(km) isn't the hypotenuse of your right triangle
@silent plank
That’s what I was thinking, that I flipped b and c

But how?

And we’re not allowed to do this right? Cuz earth is a sphere
@hallow crystal
Here I drew a line to make another triangle btw

[silent plank]

your hypotenuse would be (600+6370)km

[hallow crystal]

Nani?!

Shit then I’m guessing my B would be

Uhhh

your hypotenuse would be (600+6370)km
@silent plank
Woah woah woah wait, the hypotenuse for the big triangle right?

[silent plank]

yes

[hallow crystal]

But 6370 km is a horizontal side length on its own

And 600 km is part of the vertical side length

Wait, let’s rewind

[silent plank]

the sat is 600km above the Earth's surface

[hallow crystal]

Why is knowing the radius of earth important?

[silent plank]

the distance from the surface to the center is the radius

the distance of the satellite from the center would be r + 600km

[hallow crystal]

Yes, but we only need to find the distance from the sat to the surface

Wait wdym from the center, the sat isn’t in the center

[silent plank]

center of the earth

[hallow crystal]

Yes, the satellite isn’t in the center of the earth

It’s 600 km from the earth’s surface

[hallow crystal]

So wdym “the distance of the satellite from the center” :0

OHHHHHHHHHHH

ITS THE SAME LENGTH

SINCE ITS THE RADIUS

AHHHHHH

I SHOULDVE DRAWN IT BETTER

THANK U 😔😔😔

[silent plank]

the drawing was fine

well close enough

[hallow crystal]

Yea, but I feel like my brain cells would have clicked if it was more symmetrical

[silent plank]

wdym symmetrical

the only symmetrical thing should be the circle

[hallow crystal]

Like you know the radius is the same length on both sides

6370

On the triangle

In the earth’s center

The two lines, I wish I drew them more symmetrically

Especially the left one, above the right angle

It’s the bottom part of the red line u drew

[silent plank]

still not sure what you mean by symmetrical

[hallow crystal]

Oh my bad, it’s not above the right angle

Like I wish I drew it straighter, it was a little curved

[silent plank]

ah

[hallow crystal]

You see how it doesn’t fit perfectly with the red line u drew

Or maybe I’m just being petty lool

[silent plank]

could've just said used a ruler and/or line tool

[hallow crystal]

Ehh the 6370 km line was straight enough without a ruler

Wish I could’ve drawn the other side as straight as that uwu

Das all

[hallow crystal]

"no, they are not the same ray. HJ starts from H while JH starts from J & they point in opposite directions
@dark sparrow
My math teacher said Rays always have to point right

[silent plank]

the arrow indicating the direction can be in either direction

[hallow crystal]

I asked him “how come they’re opposite rays but they’re pointing in the same direction?”

And he replied with “because they always have to point right, look at this example. You wouldn’t wanna call that BA right?”

[silent plank]

although not as common, $\overleftarrow{\text{BA}}$ can be written to represent the same thing as $\overrightarrow{\text{AB}}$

[somber coyote]

ramonov:

[hallow crystal]

Yea that’s what I thought, I wonder why he said that then

[silent plank]

how come they’re opposite rays but they’re pointing in the same direction?
what did you mean by they

[hallow crystal]

JG, JH and JE, JF

Second pic I sent

[silent plank]

they're not pointing in the same direction

[hallow crystal]

Oh no

[silent plank]

the arrows above clearly indicate the direction

[hallow crystal]

Hhhhhhhhhh

I made myself look like a clown in front of the whole class then

Why the hell did he say “that’s a good question” -@:$/&$;/&$:

So it can’t be written as HJ arrow to left?

[silent plank]

it can, just less common

[hallow crystal]

Is there a reason why

Organization?

[silent plank]

people like reading stuff left to right

well in the west at least

[hallow crystal]

Like it’s rlly rare to see X before X^2

Ah

Yea, it’s like how ppl write expressions from greatest to least

[silent plank]

it can get really annoying to read if your text keeps switching between left and right arrows

[hallow crystal]

Yea!

I can see why

It’s good to keep organization

Especially with all those confusing ass arrows

[silent plank]

are you sure your phrased your query properly to them in class

[hallow crystal]

Yes, that’s exactly what I said in class 😭 I literally have it all recorded

Rip maybe he misunderstood

But could you look over my day 1 notes and tell me if they’re right 👉👈

[silent plank]

its one of those technically you can, but please don't things

is Q1 supposed to be 3 separate roots or their product?

[hallow crystal]

Nah, it just said “simplify the radicals” on top, idk why it got cut off

Basically 3 separate roots

[silent plank]

put a note about prime factorisation,
which will be a lot more useful if you're given larger numbers

[hallow crystal]

Ohh I was thinking about it and I thought it was useless-

Guess I was wrong

We haven’t done prime factorization yet but I’ll write that down

The method I used is the square root method right?

[silent plank]

yeh

for pythag, don't use Leg twice like that

if you insist on use the word leg, use subscripts to distinguish between the two

or just use the common representation a^2 + b^2 = c^2 (with an appropriate diagram)

[hallow crystal]

Ohh that’s what was written on the PDF, should I white it out

[silent plank]

yes

[hallow crystal]

What’s a subscript?

[silent plank]

$x_{\text{this}}$

[somber coyote]

ramonov:

[hallow crystal]

Isn’t the 2 above Leg the subscript used to differentiate between the two then?

[silent plank]

those are superscripts

indicating Leg squared

(but which Leg... do you use the same Leg for both...)

[hallow crystal]

Superscripts? :0

[silent plank]

superscript: above the baseline
subscript: below the baseline

look up their definitions

[hallow crystal]

OHHHHH I remember them from science class

Hm, aren’t they the same thing

[silent plank]

$x_{\text{subscript position}}^{\text{superscript position}}$

[somber coyote]

ramonov:

[silent plank]

subscripts are used for indexing

[hallow crystal]

Ohh I should put a superscript for one leg and put a subscript for the other leg?

[silent plank]

powers are placed in the superscript position

[hallow crystal]

What is indexing

[silent plank]

${{\text{Leg}}_1}^2 + {{\text{Leg}}_2}^2 = {\text{hypotenuse}}^2$

[somber coyote]

ramonov:

[hallow crystal]

ooooo

[silent plank]

square of first leg + square of second leg = hyp squared

[hallow crystal]

Leg 1 squared and Leg 2 squared, gotcha

[silent plank]

usually labels to distinguish between similar things

[hallow crystal]

I’m writing down the prime factorization note and, what exactly are the steps for prime factorization?

I’m rlly familiar with square root method, but we haven’t used PF yet bc were only working with small numbers for now

[silent plank]

identify prime factors starting from 2, reduce and keep going until you have a product of primes

[weary glade]

I ned help with my math

:c

plz help me with my 67 - 86

[upper karma]

67-86=-19

[silent plank]

you can use a combination of both if you can immediately identify larger square factors

[hallow crystal]

oo

[silent plank]

eg consider something like sqrt(3528)

[hallow crystal]

So in order to find the square root of 24 thru prime factorization, I would break it up into the two highest prime factors of 24?

[silent plank]

you'd use appropriate methods and eventually reach:
sqrt( 2^3 * 3)

[hallow crystal]

And what are these appropriate methods

[silent plank]

basic multiplication

in any order you want

[hallow crystal]

And the things I need to multiple thru basic multiplication are the highest prime factors of 24?

It can’t be 8 and 3?

They have to be prime?

Wait 24 doesn’t even have two prime factors,

[silent plank]

composite numbers can be expressed as a product of primes

[hallow crystal]

[silent plank]

you can first express it as 3 * 8,
and then keep reducing the 8

which is just 2^3

[hallow crystal]

Prime = Odd
Composite = Even?

[silent plank]

so 24 as a product of primes is 2^3 * 3

no, whether an integer is prime or composite depends on the number of factors

[hallow crystal]

Oh whoops

[silent plank]

positive integers have a unique prime factorisation

in this form: 2^3 * 3
you should see that the highest non-1 square factor will be 2^2

[hallow crystal]

I guess it isn’t making sense to me because it doesn’t work well with smaller numbers 😔

[silent plank]

try it with

sqrt(3528)

[hallow crystal]

That’s a pretty big number,

Which I’m guessing I probably don’t have the time to write down all the factors of?

Like if this was on a test

[silent plank]

there are several ways to do it, let me write it up

one of which is to use a tree

i'll use a combo of both

that utilises divisibility rules

,w 2^33^27^2

[somber coyote]

Results provided by WolframAlpha

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

Requested by ramonov#0921

[hallow crystal]

4?! :0

I was sure they’d both be double digits

[silent plank]

doesn't matter what route you take

easier to start relatively small

[hallow crystal]

I thought finding the highest one would be best

But I guess it’s different with bigger numbers

[silent plank]

regardless you'd reach 3528 = 2^3 * 3^2 * 7^2

[hallow crystal]

Hold on, 4 isn’t a prime number

[silent plank]

you'd end up reducing large numbers anyway so there's no point to find two large numbers

[hallow crystal]

I thought it was prime factorization?

[silent plank]

i used a combination of both

[hallow crystal]

Prime factorization and ?

[silent plank]

square factors, common divisibility rules

i knew it was divisible by 4, since the value of the last 2 digits was divisible by 4

and 4 could then be split into 2 and 2

alternatively you could divide by 2, and go from there

882 was divisible by 9 since the repeated sum of digits = 9

[hallow crystal]

How often is prime factorization used in geometry compared to square root method?

Repeated sum of digits? :0

[silent plank]

8+8+2 = 18
1+8=9

[mellow vector]

hey what are you talking about here

[silent plank]

prime factorisation

you won't really need it that much for geo

[hallow crystal]

[silent plank]

its more of an algebra thing

[hallow crystal]

oo

But at the same time there’s a lil algebra involved in geo

I’m not getting prime factorization rn, should I practice it rn so I can write down that note u told me?
Or should I not worry about it for now 😔

[silent plank]

its supposed to be a very basic concept

make a note of it and read up on it, maybe watch a video

if for some reason you were unable to spot that 4 was a factor of 12 when asked to simplify sqrt(12)
you can do:

$\sqrt{12} = \sqrt{2 \times 6} = \sqrt{2 \times 2 \times 3} = \sqrt{2^2 \cdot 3}$

[somber coyote]

ramonov:

[silent plank]

where it would be clear where the biggest square factor would be the 2^2

basic goal is to represent your number as a product of smaller numbers where its much easier to identify square factors

you don't necessarily need to reduce it to primes but its just usually what people say

[hallow crystal]

Doesn’t the sq rt cancel out the 2 superscript on 2 in the answer?

Making the answer just 2•3?

[silent plank]

the 3 would still be under the square root

[hallow crystal]

Whoops

[silent plank]

alternate route for: 3528
since its divisible by 4 and 9, its divisible by 36,
3528 = 3600 - 72 = 36(100 - 2) = 36 * 98 = 36 * 49 * 2
hence sqrt(3528) = sqrt( 36 * 49 * 2) = 6 * 7 * sqrt(2) = 42sqrt(2)

[hallow crystal]

Damn, these divisibility tests

I should get a hang of em

https://youtu.be/XGbOiYhHY2c
Ight so what I learned from this video is basically prime factorization is just breaking a big number up into multiplying smaller numbers 😄

[silent plank]

for 9, could've just said the immediate sum was divisible by 9,
repeated sum wasn't needed

yes

[hallow crystal]

What’s the difference between a repeated sum and an immediate sum

[silent plank]

immediate sum would just be 3+5+2+8

what i'm calling the repeated sum (probs not official) would be the continue the process until you reach 1 digit

[hallow crystal]

Repeat the prime factorization process?

[silent plank]

no, as in 3528 is divisible by 9 since 3+5+2+8 = 18 is divisible by 9

[hallow crystal]

Repeat the dividing process?

[silent plank]

as in you can then add the 1 and 8

to get 9, to confirm that it is indeed divisible by 9

[hallow crystal]

I still don’t get it 😔👉👈

[silent plank]

the "repeated sum" was to reference the divisibility rule for 9 only, nothing else

[hallow crystal]

Hmm I see

So should I use prime factorization only for bigger numbers

[silent plank]

since the sum is quite small, you don't really need it

factorise in general

if you can immediately identify certain factors, you don't necessarily need to use primes

[hallow crystal]

Alright! I think I’ll main Square Root Method and use Prime Factorization for emergencies

[silent plank]

if your divisibility rules are poor, you can systematically reduce a number to its prime factorisation by factoring out primes starting from 2,3,5,7 etc

but yeh, just factorise in general

[hallow crystal]

Alright so I won’t note prime factorization down on the first page bc square root method is better for smaller numbers like that

[silent plank]

now to read the rest of it

no parentheses around the (2x)

2sqrt(6) wasn't squared properly either