geometry-and-trigonometry

[wind heart]

It’s 16

[austere creek]

Send help

How tf do I do this

[thorn talon]

What about it?

[upper karma]

@austere creek Notice that it factors as $x^2 - 9 = (x-3)(x+3)$, $x \in \mathbb{R}$

[austere creek]

How did you get that man

[somber coyote]

all night:

@cat_the_dweeb Notice that it factors as $x^2 - 9 = (x-3)(x+3)$, $x \in \mathbb{R}$
```Compile error! Output:

! Missing $ inserted.
<inserted text>
$
l.54 @cat_
the_dweeb Notice that it factors as $x^2 - 9 = (x-3)(x+3)$, $x \in...
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.

LaTeX Font Info: Calculating math sizes for size <14> on input line 54.
LaTeX Font Info: Try loading font information for U+msa on input line 54.
(/usr/local/texlive/2018/texmf-dist/tex/latex/amsfonts/umsa.fd

[upper karma]

Where is the latex mistake..

[thorn talon]

@austere creek what factoring techniques do you know?

Do you know the difference of two squares?

[upper karma]

You don't need to know the difference of two squares

Just note that

[austere creek]

Ok but uhhhh I’m doing an error analysis and I’m struggling

[upper karma]

$(a-b)^n = (a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+...+b^{n-1}$

[somber coyote]

all night:

[thorn talon]

That seems a lot less useful than just applying the difference of squares

[upper karma]

The difference of squares comes from this, @thorn talon

[thorn talon]

????

You can easily prove the difference of squares without that

So no, it doesn't

[austere creek]

[upper karma]

Ok, whatever

[thorn talon]

It's just bad advice

Stop

[austere creek]

I just wanna understand this how did you get that answer

[upper karma]

@austere creek Multiply out $(a-b)(a+b)$

[somber coyote]

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

[upper karma]

Oh, it's your name that causes the latex error..

[austere creek]

but there is no b

[upper karma]

Yes there is

Why can't b = 3?

[austere creek]

How did you get three

Where did that come from

[upper karma]

$\sqrt{9} = 3$

[somber coyote]

all night:

[upper karma]

$3 \cdot 3 = 9$

[somber coyote]

all night:

[austere creek]

Oh right

[upper karma]

So if you multiply 3 by 3 in $(a-b)(a+b) = a^2-b^2$

[somber coyote]

all night:

[upper karma]

You will get $-b^2, or -3^2 = -9$

[somber coyote]

all night:

[austere creek]

Like this

[thorn talon]

Not really

[upper karma]

No..

[austere creek]

[upper karma]

$(a-b)(a+b)$

[somber coyote]

all night:

[upper karma]

Is equivalent to $a^2-b^2$

[somber coyote]

all night:

[upper karma]

Where $b=3$, in this case, because $\sqrt{9} = 3$

[somber coyote]

all night:

[austere creek]

I was out four days because I had surgery and I came back to work I’m struggling over

This is the original problem

But since this is a error analysis that answer is wrong

[thorn talon]

I mean

What do you actually have to do

?

Solve it

Or just spot the error

[austere creek]

Solve the original problem and described the error

[thorn talon]

I see

I mean

Factoring is one way to solve

Probably what I would have done

Or you could just like add 9 to both sides and square root

You probably should know both methods though

[austere creek]

Like this

[thorn talon]

Almost

Missing another solution

[austere creek]

The -3

[thorn talon]

Yes

[austere creek]

I already put that on

[thorn talon]

🤔

[austere creek]

After the picture

[thorn talon]

Ok

Good

The working out is good as well

[austere creek]

Ok I have another one I worked out but I got it wrong

[thorn talon]

Some people do x = sqrt(9) => x = +-3

Which is less correct

But yours is good

What is it?

[thorn talon]

Ok

I don't like that negative leading term

What can you do?

To make it nicer

[austere creek]

The wrong answers are -12 and 2

That’s the problem that’s the literal

Problem

[thorn talon]

I know

I'm asking you what steps you can take

[austere creek]

I can multiply everything else by the negative to get rid of it

[thorn talon]

Yep

Good

[austere creek]

Also whoops the 24 was post to be neg not pos sorry

[thorn talon]

That's alright

[austere creek]

Well it’s pos now since I multiplied it by beg

[thorn talon]

Ok good

x^2 - 10x + 24 = 0 right?

[austere creek]

Ye

[thorn talon]

Can you factor this?

[austere creek]

I got x=10 x=-24

Wait no

Whoops

I got -12 and 2 but it’s wrong

[thorn talon]

What two numbers add up to -10

And have a product of 24?

[austere creek]

Wait a second

-6 and -4

I feel stupid

Since there’s a plus at the end that means both symbols are the same

[thorn talon]

So those are what you can factor as

What are the actual solutions?

[austere creek]

_> x=4 x=6

[thorn talon]

Yep

[austere creek]

Did I do this one right

[thorn talon]

What's the full question even

[thorn talon]

Solve 2x + 3 = 0

[austere creek]

I would have to multiply the 3 by 2

[thorn talon]

Multiply?

Or divide?

[austere creek]

Wait what uhhh idk

[dusk wolf]

Anyone able to assist?

[dark sparrow]

nobody can assist you with anything until you post a question

[dusk wolf]

^ I like that

Tried to Google but no idea where I am going.

[dark sparrow]

which part is giving you trouble

[dusk wolf]

Honestly the whoel thing.

[dark sparrow]

also this isn't really geometry

so maybe we should move

[dusk wolf]

Ohh

yes

algebra?

[dark sparrow]

uhh yeah sure

[dusk wolf]

going to delete problem and repost in prealg-algebra

[upper karma]

,rccw

[upper karma]

how do I set this up in a proportion to solve for x? The main part I need help with is setting up the missing side.

so it would be 9/x = 6/10-x , and then I just solve it, is that right?

[sleek vine]

yes

[upper karma]

So once you cross multiply then it would become 6x = 90-9x right?

[eager kraken]

Can I write the 4 in place of the theta when calculating this or how would I find the cot from the tan theta = 4?

[thorn talon]

What's the relation between cot and tan?

[eager kraken]

tan theta = 1 / cot theta

[thorn talon]

Yes

So cot(t) = 1/tan(t) correct?

Where t is theta

[eager kraken]

correct

[thorn talon]

But you know what tan(t) is

[eager kraken]

so it is
cot(t) = 1 / tan(4)

[thorn talon]

No

[eager kraken]

cot(t) = 1/4

[thorn talon]

Yes

[eager kraken]

it just gets to confusing me when theta is there

cause like for this other problem I am not sure what to plug in where,
finding csc(t) given the cot(t) = -1/2

I believe i would use the identity
1+cot^2(t) = csc^2(t)

wait it would just be
1 + (-1/2)^2 = csc^2(t)

[thorn talon]

Possibly

Then you need to be careful with which signed version to take

[eager kraken]

what do you mean?

[thorn talon]

Keep solving for csc(t)

[eager kraken]

idk how to type root symbols

[thorn talon]

I think that's fine

Can you justify why taking the positive root is valid?

[eager kraken]

like choosing what quandrant it is in?

[thorn talon]

Yep

Ok, if you know what quadrant to pick

You're probably doing fine

[eager kraken]

well in this case I am not actually too sure

[thorn talon]

Would you agree that taking reciprocals don't affect the sign changes for the quadrants?

[eager kraken]

yea

[thorn talon]

So cot is negative where tan is negative right?

And similarly, csc is negative where sin is negative

[eager kraken]

ok

is that just a relation those pairs share or does it have anything to do with cot = 1/tan

[thorn talon]

The reciprocals

If x > 0, then 1/x > 0

And if x < 0, then 1/x < 0

[eager kraken]

oh yeah because sin and csc can both still be negative

if they start negative

[thorn talon]

You can apply this to your trig functions

[eager kraken]

i did have one other question that is a different area of trig here

[thorn talon]

Actually

🤔

[eager kraken]

This is the question, I got 4380, as my answer from multiplying 73 by 60

[thorn talon]

Did I do something stupid

[eager kraken]

oh?

[dark sparrow]

the question asks

how many rotations

[eager kraken]

so 4380/360

[thorn talon]

I think I said something stupid

[dark sparrow]

4280?

[dark sparrow]

but you said 4280 just now

[eager kraken]

oh yea

i meant 4380

[dark sparrow]

okay

[eager kraken]

@thorn talon the answer to that problem ended up being

just not positive

oh cause it said in the 4th quandrant

oof

[tardy junco]

Why are there two different formulas depending on the product of x and y?

Why are there two different formulas depending on the product of x and y?

And why that restriction on product of x and y in all the the formulae

[tardy junco]

Mention me @tardy junco if you reply thx

[austere knot]

https://prnt.sc/qrw4lo
For some reason it is not coming to me

[dark sparrow]

if the question asked for the length of the entire circle, would you be able to answer

[austere knot]

I’ve gotten this so far

[dark sparrow]

,rcw

[dark sparrow]

so... you have correctly found the answer as 7 * 2π/9 yards

now you just need to follow the problem's instructions on how to format it

and input accordingly

[austere knot]

So with 4 pi / 18 I just simplify it?

[dark sparrow]

no, you plug the whole thing into your calculator because the problem asks for the answer ROUNDED TO THREE DECIMAL PLACES

[austere knot]

I got it

Thx

[wind heart]

Would number 5 be correct

If not how do I do this

<@&286206848099549185>

[austere knot]

you would do a^2+b^2=c^2

[wind heart]

oh

Let’s see

Yeah I did that

But I got 6.5

Which is a decimal

I don’t know things with decimals are right or not

[austere knot]

i got a different number

it would be setup as 20^2 + 26.8^2 = c^2

[wind heart]

That’s weird

Because like

I thought you take half of the triangle and not the full one

[austere knot]

Well it gives you MQ = 20
then it gives you QO = 26.8
so you have to find the hypotensuse

So then you are looking for MO

[wind heart]

But like

They are asking for like two different triangles

[austere knot]

Number 5 says to find the measurement for MO

[wind heart]

Hmmm ok

I this a circumcenter triangle

[austere knot]

Yes it is

[wind heart]

So like

How would I add the congruent segments

Like how would I know

[austere knot]

that im not sure

[wind heart]

Ok

[quick perch]

How to find area of a triangle with vertices A(-1,2,3) B(3,0,2) C(1,4,5) ?

[thorn talon]

Hmm

You can take a vectors approach

Have you looked into those before?

[quick perch]

Nope

[thorn talon]

Never done vectors?

[quick perch]

I want just to see an example

[thorn talon]

What do you want to see?

[quick perch]

The method for solving applied

[wind heart]

I’m still very confused about this

[thick gulch]

Isnt it called the shoelace formula

Or is that different

Do you know determinants

[limpid snow]

Heron's formula

Find the distance between each vertex, plug em in to $Area=\sqrt{S(S-A)(S-B)(S-C)}$ Where $S$ is the Semi Perimeter: $S=\frac{A+B+C}{2}$.

[somber coyote]

Intel Core i7-8700:

[limpid snow]

Alternatively if you wanted to take the vector approach it's half the cross product of two vectors of that triangle.

[weary drift]

half the cross product norm

[upper karma]

so the answer is c

but I am confused on how angle Q is congruent to angle P

[thorn talon]

It's not

[upper karma]

yea iut is

[thorn talon]

🤔

How

[silent plank]

"angle P " is ambiguous

[upper karma]

idk what that means

but if you rotate triangle prs then p lands on q

[silent plank]

angle P could potentially refer to angles:
QPS, QPR or RPS

also, angleQ = angleRPS isn't being used in the proof

you can conclude the angles are equal after proving congruency

[unborn jacinth]

@silent plank why is the tan(x)'s period pi

isn't it 2pi

[thorn talon]

No

[unborn jacinth]

why not

[thorn talon]

Check out the graph

[unborn jacinth]

because for it to go around the circle, its 2pi

[thorn talon]

But it's the ratio of sine and cosine

[unborn jacinth]

wdym

im not sure I understand

[thorn talon]

tan(x) = sin(x)/cos(x) right?

[unborn jacinth]

yeah

[thorn talon]

Consider tan(x + pi)

What's sin(x + pi)

And what about cos(x + pi)?

[unborn jacinth]

like the graph?

[thorn talon]

What can you simplify them as

[unborn jacinth]

-sin(x)

tan(x)

-cos(x)

[thorn talon]

Yeah

And -sin(x)/-cos(x)

= tan(x) right?

[unborn jacinth]

yeah

[thorn talon]

That doesn't prove anything

But it should give some insight into how the ratio

Can result in a shorter period

[unborn jacinth]

ohk

thx

[regal void]

given a square abcd with a point p that is distance one away from vertex A, distance one away from vertex B and distance one away from the side CD. Calculate the area of the triangle APB

just a fun (albeit simple) geo problem I thought you guys might like

[native thicket]

hi guys why is Sin pie/6 = 1/2

[upper karma]

draw a circle

[native thicket]

ok

[upper karma]

make it 1 inch radius

[dark sparrow]

pi, not pie.

[native thicket]

Sorry

[upper karma]

then draw a line from the center to the end of the circle

rather, the edge

[native thicket]

Ok done

[upper karma]

ok, that line should be 1 inch long

[native thicket]

Radius

1 inch

[upper karma]

now go half way along that line, and draw a perpendicular line from it to the edge of the circle

does that make sense?

[native thicket]

ok

[dark sparrow]

maurice i think that'll make a π/3 angle, not π/6

[upper karma]

nope

now draw

a line parallel to that one

which passes thru the center of the circle

[native thicket]

ok got it

[upper karma]

ok

the last 2 lines u drew

there is a part of the edge of the circle

which reaches between those 2 lines

[native thicket]

Yes

[upper karma]

the length of that arc

is pi/6

inches

[native thicket]

wow

[upper karma]

god ordained things that way

it is as it is

[native thicket]

Never knew that

I feel happy now

😀😀😀

Thanks a lot!!

[dark sparrow]

uh

wow

[upper karma]

to be clear this is the part you should be looking at

[dark sparrow]

that last thing

why the fuck appeal to god

seriously that's just

no

big yikes

[native thicket]

I wish we had good teachers in our village. 😡

[upper karma]

nvm my phone battery is dying

i cant take a photo

i will paint it

[native thicket]

they tell us to by heart such things

[dark sparrow]

just draw an equilateral triangle with side length 2 and split it in half, you'll get two triangles with angles π/6, π/3 and π/2

whose sides will be 1, sqrt(3) and 2

[native thicket]

Ok

[dark sparrow]

okay let me draw that on paper to make it more clear

[native thicket]

thanks a lot bro

[upper karma]

the green portion

has a length of pi/6

if the circle has a radius of 1

nevermind my haphazard color scheme

[native thicket]

So when the angle rotates pi/6 the length of that arc is 1/2?

[upper karma]

nope, the length of the arc is always exactly the angle in radians times the radius of the circle

[dark sparrow]

here

[somber coyote]

Carter:

[native thicket]

Ok got it

[dark sparrow]

i think my picture may be more intuitive

as it doesn't rely on any "god ordained it to be that way" talk

[upper karma]

well then you've got to explain who ordained the sine function

[dark sparrow]

that π/3 angle is π/3, aka 60 degrees, because it's in an equilateral triangle

[native thicket]

Yeah

[dark sparrow]

what do you mean "ordained the sine function"

[upper karma]

my expression was only euphemism! the same sentiment applies here

[dark sparrow]

no???

[upper karma]

some degree of hand-waviness must be given unless you express precisely what is going on

[dark sparrow]

i can and will express precisely what is going on

[upper karma]

you've just relegated the waving from a circle's geometry to the sine function

which is all well and good

i encourage as much

[dark sparrow]

sin = opp/hyp lmao

[upper karma]

but it doesn't remove the sentiment

[dark sparrow]

you're trying to make me overthink it

[upper karma]

no

i'm trying to say

[dark sparrow]

there is no such sentiment

[upper karma]

"you must describe all parts of the system to demystify it"

[dark sparrow]

literally the definition of sin as far as geometry is concerned

[upper karma]

sure

[dark sparrow]

that ratio is what we call sine

[native thicket]

Please don't fight

[upper karma]

but it is not a workable definition

[dark sparrow]

that's all there is to it

[upper karma]

you have relied on the same mystery

[dark sparrow]

you're not suggesting that i whip out the calculus definition of it are you

that would be asinine

pun not intended

what mystery have i relied on

[upper karma]

no, you are!

the same one I did! the natural relation between the angle and the ratio of side lengths

[dark sparrow]

...

similarity???

[upper karma]

exact congruence, indeed

[dark sparrow]

so why all the stink

[upper karma]

well, that's what i'm wondering myself

i merely said we were saying the same thing

[dark sparrow]

over my rightful refusal to go all "gOd OrDaInEd iT tO bE tHiS wAy"

[upper karma]

you insisted not!

well, that's a common euphemism

[native thicket]

God created everything

[dark sparrow]

god doesn't exist, as far as i am concerned

[upper karma]

did god ordain that as well

[dark sparrow]

oh my god

[upper karma]

hehe i won't be so snarky

LOL

[dark sparrow]

can we end this discussion now

[upper karma]

im having a good chuckle

yes

lol

[dark sparrow]

this is going south really quickly

[upper karma]

im loving the puns

[native thicket]

But we don't know who created everthing yet as well

[upper karma]

shush

study your circles

[native thicket]

😃

I added you both as friends, please accept my request..

@dark sparrow I can't send you a friend request why?

It says error

[dark sparrow]

don't know

[native thicket]

Ok

[upper karma]

how would I set up this proportion

[quiet mason]

are EF and RQ parallel

[amber gull]

can someone

pls dm and explain how to do this

[dark sparrow]

that's a lot of problems

which ones do you need help with

...oh you said dm

[upper karma]

@quiet mason we dont know that they are parralel

@dark sparrow do you mind helping me with this if you have the time?

I do not know how to set up the proportion for this problem

[dark sparrow]

there is not enough information to solve this problem

E can be displaced as far as you want along PQ without altering any of the values explicitly fixed in the diagram, thereby making EP arbitrarily long

[upper karma]

well

its supposed to be 42/18 = 28/x

but idk how they know angle R is congruent to angle F

without imagining rotating the figure themselves

but the figures are so similar what if i make a mistake?

[dark sparrow]

ok wait

you yourself said EF and RQ were not known to be parallel

are you given any angle equalities to work with?

besides angle RPQ = angle EPF, which are equal by virtue of being vertical angles

[amber gull]

@dark sparrow all of them

[ivory vortex]

can someone help me with a BUNCH of trig

[knotty merlin]

@ivory vortex

ill do my best

[ivory vortex]

ok i got 50% on both of my trig tests despite me studying my hardest for them

and that was like 2 months ago

im only reviewing the tests because thats all the time i have before my exam in 4 hours

so i wont know the answers to some rly easy questions

ok first

$$sec(\frac{2\pi}{3}+x)=2$$

[somber coyote]

Voyager:

[ivory vortex]

how do i solve for x

i know sec is 1/cos but idk what to do with it

[thorn talon]

apply that

what if you take reciprocal of both sides?

[ivory vortex]

idk

does ls become cos

and rs become 0.5

[dark sparrow]

1/2

you get $\cos(\tfrac{2\pi}{3} + x) = \frac12$

[somber coyote]

Ann:

[knotty merlin]

ya

[dark sparrow]

are you able to solve this equation for x

[ivory vortex]

o i see

-60?

[dark sparrow]

uhhhh what no???

first off

radians, not degrees

,w cos(2π/3 - 60)

[somber coyote]

Display results online and refine query

Requested by Ann#0413

[dark sparrow]

there

definitely not 1/2 lmfao

60 radians is nowhere near a solution

[knotty merlin]

i think his mistake was

cos(pi/3)=1/2

therefore we now know

pi/3=2pi/3+x

this is of course only 1 solution

[ivory vortex]

well it becomes

2pi/3 + x = 60 right

[dark sparrow]

why 60

why are you so insistent on using degrees

[knotty merlin]

use radians please

wait

[dark sparrow]

cos(60 radians) isn't 1/2

cos(60 radians) isn't 1/2

cos(60 radians) isn't 1/2

cos(60 radians) isn't 1/2

cos(60 radians) isn't 1/2

cos(60 radians) isn't 1/2

[ivory vortex]

idk man its just how ive been doing things

[knotty merlin]

guys

[ivory vortex]

im not used to using radians

[knotty merlin]

whats the conversion rate from radians to gradians

[dark sparrow]

radians to gradians

who the FUCK uses gradians anymore

i mean... if you insist

how many grads are there in a full circle and how many radians

[knotty merlin]

i was joking ann

[dark sparrow]

you can get the conversion factor yourself from there

do not joke with me when i'm on a caffeine high

[knotty merlin]

,w cos(2pi/3 - pi)

[somber coyote]

Display results online and refine query

Requested by 7teen cant write punchlines#3535

[ivory vortex]

$$\cos(\tfrac{2\pi}{3} + x) = \frac12$$

[somber coyote]

Voyager:

[ivory vortex]

dont you take arccose of 0.5??

arccos*

[dark sparrow]

no you draw a unit circle

[knotty merlin]

,w x^2+y^2=1

[dark sparrow]

don't ever just take arccos blindly

[ivory vortex]

why

[dark sparrow]

because arccos isn't the true inverse of cos and arccos(cos(x)) is very often not equal to x

[knotty merlin]

doesnt it return the principle branch when given a non principle branch answer?

[ivory vortex]

how do i plot it on a unit circle,,,

[dark sparrow]

here lemme show

comme ça

$\cos\paren{\frac{2\pi}{3} + x} = \frac12 \ \frac{2\pi}{3} + x = \pm\frac{\pi}{3} + 2k\pi \ x = -\frac{2\pi}{3} \pm \frac{\pi}{3} + 2k\pi$ \ \ Solution: $\curly{2k\pi + \frac{t\pi}{3} \big| k \in \bZ; t = -3, -1}$

[somber coyote]

Ann:

[knotty merlin]

ann is that a screencap of geogebra?

[ivory vortex]

wtf

[dark sparrow]

desmos

[ivory vortex]

where did

pi/3 and 2kpi come from

what even is k

[knotty merlin]

because

[dark sparrow]

k is an integer parameter

[knotty merlin]

pi/3 = 60 degrees

[dark sparrow]

cos is periodic

with period 2pi

surely you're aware of that @ivory vortex

[ivory vortex]

yes of that last part

[dark sparrow]

so

[knotty merlin]

basically, cos(x) = cos(x+2pi)=cos(x+2pik)

right?

[ivory vortex]

ive literally never used this k thing

[dark sparrow]

what

[ivory vortex]

only parameter ive used is r i think

in xer and yer

[dark sparrow]

...

....

is "xer"\ meant to be the poor man's $x \in \bR$

[somber coyote]

Ann:

[dark sparrow]

bc that's not a parameter

jesus fuck

[knotty merlin]

oh my lord

[dark sparrow]

are you telling me you've never had exercises of the form "solve <trigonometric equation> over the entire number line"

[knotty merlin]

k refers to any integer

[dark sparrow]

did you only ever solve these over [0,2pi]

or what

[ivory vortex]

well ive never had it referred to as k

[dark sparrow]

did you use another letter

did you use n instead

the exact letter doesn't matter

[ivory vortex]

i havent used any of those letters in trig

[dark sparrow]

what matters, and what i admittedly didn't spell out, is that it's an integer-valued parameter

and

you

haven't answered my question

so please answer my question

are you telling me you've never had exercises of the form "solve <trigonometric equation> over the entire number line"

please answer my question

[ivory vortex]

idek

[knotty merlin]

...

is that a no?

[dark sparrow]

ok

[ivory vortex]

yep guess so

[dark sparrow]

let me put this more directly

have you ever gotten any exercises which said, "here's a trig equation. find all real solutions to it"

as opposed to "find all solutions to this equation that are in [0,2pi]" or something along those lines

yes or no.

[ivory vortex]

nope

think i had it once but it wasn't really taught in class

[dark sparrow]

... ok that explains everything

what about this question

[ivory vortex]

was like find the period of this function

or when it repeats

[dark sparrow]

can you post the question plus EXACTLY what you're told to do, including ALL INSTRUCTIONS

[ivory vortex]

my original question?

[dark sparrow]

yes

[ivory vortex]

that is it

[dark sparrow]

exactly as it was stated

please

take a screenshot

or a pciture

i want to see it

in its original form

[ivory vortex]

its literally

If equation, what does x equal

[dark sparrow]

please

is it

this hard

to take

a picture

or a screenshot

is that too much to ask

please

[ivory vortex]

my phone is dead and itll take 10 minutes to charge and take a picture

dude

im sorry but i just think i'll be better off studying on my own

thanks for the help but i guess i'm just too unable to even understand

[tardy junco]

Why are there two different formulas depending on the product of x and y?

And why are there restrictions on product of x and y?

Mention me @tardy junco if you reply. Thx

[dark sparrow]

it comes down to the fact that arctan's range is (-π/2, π/2) but the sum of two arctans can be outside that range in general

and you have arctan(x) + arctan(y) ∈ (-π/2, π/2) if and only if xy < 1

[tardy junco]

How did you deduce that XY < 1 btw
@dark sparrow

[dark sparrow]

well, arctan(x)+arctan(y) = ±π/2 iff xy = 1

[tardy junco]

If it's greater than 1 why do you add pi?

[dark sparrow]

tan is π-periodic, but arctan((x+y)/(1-xy)) ∈ (-π/2, π/2); for x, y > 0 and xy > 1, arctan(x) + arctan(y) > π/2

[tardy junco]

So arctan would be less than -pi/2.

That means arctan must have argument that is negative and greater than

Idk

[dark sparrow]

So arctan would be less than -pi/2.
no?

[tardy junco]

Otherwise it would be between pi/2 and -pi/2

Cus X and Y greater than 0

[dark sparrow]

x and y, not X and Y

[tardy junco]

Yeah sorry my phone corrected it

I meant that we were adding pi because arctan comes to be below -pi/2

But why is it less than -pi/2?

[dark sparrow]

why is what less than -π/2??

[tardy junco]

Yeah how to reason that out

[dark sparrow]

why is what less than -π/2??

[tardy junco]

Arctan when xy greater than 1

[dark sparrow]

arctan is NEVER less than -π/2

[tardy junco]

Then we are adding pi..

Why are we adding pi

[dark sparrow]

tan is π-periodic, but arctan((x+y)/(1-xy)) ∈ (-π/2, π/2); for x, y > 0 and xy > 1, arctan(x) + arctan(y) > π/2

[tardy junco]

I meant th sum of arctan is less than -pi/2

That's why we're adding pi right

So it's in the range again

[dark sparrow]

no, the sum of the arctans is GREATER THAN +π/2.

please READ what i'm saying.

[tardy junco]

K

[dark sparrow]

apologies for my messy handwriting

i hope it is legible

arctan((x+y)/(1-xy)) is too low

[tardy junco]

Yeah I got it. So subtracting pi to get it into principal branch

[sudden locust]

Today I posted a Geometry Problem that some of you guys might find interesting! https://youtu.be/f4LtTli6kI0 Can you solve it? How? 🤓

[near tusk]

@knotty merlin yes, cos(x) = cos(x+2pi)=cos(x+2pik) as long as k is an interger

[quiet mason]

lol some guy posted this somewhere and its actually nice

its from the sat

you guys might try

[rich wolf]

@quiet mason drawn to scale?

[quiet mason]

no

[rich wolf]

@quiet mason how solve?

I am stuck

[rare isle]

Kite is a rhombus

Ez

Lol

[knotty merlin]

..

[rare isle]

Also, fuck kites

[knotty merlin]

HOW ON EARTH IS IT A RHOMBUS

[rare isle]

Bruh

[knotty merlin]

its sides arent equal

[rare isle]

It's literally

In the figure

The lines in the figure

[knotty merlin]

ok am i reading it wrong

but

[rare isle]

Yeah lmao

Bruh

[knotty merlin]

but i believe ur wrong

[rare isle]

Wyf

Wtf

Ok so read what's a rhombus

[knotty merlin]

its not a rhombus

[rich wolf]

Rhombi have equal sides

[rare isle]

Oh shit

[rich wolf]

Lmao

[rare isle]

Maybe

[knotty merlin]

oh my god

[rare isle]

My math is fucking

[knotty merlin]

i thought i was going crazy

[rare isle]

Rusty

Maybe Ive been doing too much integration

Alright, I am gonna agree that it's a tough question

Also like I was thinking it wrong, all rhombuses are kites but all kites are not rhombuses

I feel like I never passed 8th grade being unable to solve this stupid question

[vague pagoda]

bruh stuff like that is taught to 16-17yos here

and its "advanced math"

so dw

[quiet mason]

if theta>=60,the third side is >=10

but 5+5<=third side

which disobeys triangle ineq

thus option 1

[tawdry pivot]

Seems impossible to determine $\theta$

[somber coyote]

EpicGuy4227:

[quiet mason]

its mcq

yes but you have to choose whats valid

[tawdry pivot]

$\varphi$ can't be less than $\theta$

[somber coyote]

EpicGuy4227:

[tawdry pivot]

I'm saying none of the answers are valid

Even if we ignored $\varphi<\theta$, "determine $\theta$" is ill-posed since $\theta$ cannot be determined

[somber coyote]

EpicGuy4227:

[tawdry pivot]

better would be to write "Which one of the following values can $\theta$ take?"

[somber coyote]

EpicGuy4227:

[dusty badge]

this isnt really a hard q, its just idk if this estimation is even accurate

https://www.ventusky.com/?p=16.1;145.6;4&l=rain-3h&t=20191007/2100
Weather Forecast Maps
The color and numbers on the map indicate the amount of rainfall in 3 hours (in inches). The map does not show longitude or latitude, but if you click on the map, the location coordinate pops up. The numbers are plotted approximately every 3 degrees in longitude and every 1.5 degrees in latitude.

By counting the 3-hour rainfall that is 0.05 inch or greater, roughly estimate the total rain that fell over this area in 3 hours in the unit of m3. (You can discount the rainfall greater than 0.05 inch but far away from the storm.) Hint: 1.5 degrees of latitude = 167 km. This is not an exact math — describe how you made this estimation.
so the way i did it is: The storm, if made into a square is roughly 5 numbers horizontally and 8 numbers vertically.5 horizontal * 1.5 = 7.5 latitude 8 vertical * 3 = 24 longitude7.5*24 = 180 units square180 167 = 30,060 km^2 is the rough affected area.30,060 km^2 = 30,060000000m^20.05inches = 0.00127m (0.05 inches is the 'average' rainfall level in the area).30,0600000000.00127 = 38,176,200m^3 of rainfall fell in those three hours.
does that seem roughly correct

again, its not rlly hard it sjust i want to make sure this estimation makes sense

[upper karma]

i think you're in the wrong channel

ask this question in #probability-statistics

[rich wolf]

Is the maximal area that can be enclosed by a given perimeter of rectangular fence always a square

Like for example if i have 200 meters of fence that means 2x+2y=200 so x+y=100 so you maximize x(100-x) and you take the vertex of that which is -b/(2a) and so you get 50(100-50)=2500 as the maximal area

Is that always true

[thorn talon]

yeah

i mean

you can kind of prove it

[rich wolf]

How

[thorn talon]

P = 2x + 2y

A = xy

do some work

and you can prove x = y in order to achieve maximum

[rich wolf]

So the same thing but with letters

[thorn talon]

yes

[rich wolf]

Lol

[dusty badge]

@upper karma could u help me out

[upper karma]

with what

[dusty badge]

the question i posted

[upper karma]

i don't follow the question or the answer

[grim cairn]

anybody know how to solve this?

nvm i found it

[sturdy ledge]

Can somebody make problems involving the properties of parallelograms, trapezoids, and kites please?I have a test on monday coming up and i wanna practice by having someone make problems for me to solve. Go easy on me tho im just a freshmen lol.

[upper karma]

Is khan wrong here?

How did they get (11/61)^2 = 3600/3721

[thorn talon]

they didn't

1 - (11/61)^2 = 3600/3721

[upper karma]

oh my god im dumb

thank you

[upper karma]

Hello I am just curious

How do you calculate the dihedral angle of this figure

Not flexing I just want to know how to do it

Why does the dihedral angle happen to be:
Pi-arctan(2) [degrees]

[umbral snow]

Between which two faces?

[upper karma]

White and red

They’re all the same

[umbral snow]

I see what you mean

[upper karma]

It is a dodecahedron

HOW

How do you type pi with the bot

[weak shoal]

$\pi$

[somber coyote]

Abhijeet Vats:

[grim cairn]

anybody want to help me with some trig homework?

if you want to just pm me

[silent plank]

[grim cairn]

alright ill post the thing then

its due at midnight so im kinda fucked

[silent plank]

which questions are you having difficulty with?

also question 1 is bad

[grim cairn]

well technically i dont really know anything but i really have difficulties with 1 2 8

[knotty merlin]

@grim cairn what have u tried for q1

[grim cairn]

nothing i dont even understand what that B symbol is

[silent plank]

$\beta$ is just a variable

[somber coyote]

ramonov:

[silent plank]

like x

used to represent an angle

[grim cairn]

so i could just do sin(4/9)

?

[knotty merlin]

...

[silent plank]

wdym by do?

[knotty merlin]

if sin(x)=a. what does cos(x) =

theres an identity which will help you

[silent plank]

no, you don't do sin(4/9)

[knotty merlin]

theres an identity

[silent plank]

4/9 is the ratio when you take the sin of B
i.e as the question stated
sin(B) = 4/9

[knotty merlin]

idk how to hint at it tbh

[weary drift]

it pains me that the hw lacks

[grim cairn]

yeah im fucked

thanks for the advice tho guys

[knotty merlin]

no ur not

[grim cairn]

nah im giving up

[knotty merlin]

whats does cos^2(x) +sin^2(x) =

@grim cairn

think of the unit circle

[grim cairn]

can I plug this in my calculator?

[silent plank]

hint: ||Q6||

you shouldn't use a calculator for any of these questions

[grim cairn]

how tho

[silent plank]

read the hint

[grim cairn]

question 6?

pathageorm theroam?

[silent plank]

technically yes

[grim cairn]

a2 + b2 = c2?

[silent plank]

but specifically that theorem using the trig functions
sin and cos

[knotty merlin]

note: Never say a2 instead of $a^{2}$

[somber coyote]

7teen20nine:

[silent plank]

as it is explicitly stated:
$$\sin^2(\theta) + \cos^2(\theta) = 1$$

[somber coyote]

ramonov:

[silent plank]

that is one of the most important/common identities you need to use in trig

[grim cairn]

oh it equals 1 lol

[silent plank]

i mentioned that the question was bad earlier because there's more than one possible value for
cos(B)

and it implied there was only 1

[grim cairn]

ok

[knotty merlin]

ya

unless its a trick question, which is worse

[silent plank]

as for questions 2 and 8,
do you know what radians are?

[knotty merlin]

ya

why are they using both radians and degrees

[grim cairn]

radians are the opposite of degrees

[silent plank]

no! not opposte
very very very ... very bad word

[grim cairn]

its a thing that is used instead of degrees on the unit circle

[knotty merlin]

theyre like

[weary drift]

so where do gradians fit in?

[grim cairn]

they fit on the unit circle

[knotty merlin]

LMAO

no

they fit in the dumpster

[grim cairn]

i think i fit perfectly in a dumpster

[silent plank]

google the definition of a radian

[grim cairn]

The radian is a unit of measure for angles used mainly in trigonometry. It is used instead of degrees. Whereas a full circle is 360 degrees, a full circle is just over 6 radians. A full circle has 2π radians (Roughly 6.28) As seen in the figure above, a radian is defined by an arc of a circle.

[silent plank]

do you understand all that?

that probably isn't the best explanation

[dark sparrow]

1 radian is the angle subtended by a circular arc whose length equals its radius.

that's the best definition imo

https://upload.wikimedia.org/wikipedia/commons/4/4e/Circle_radians.gif

[hot sable]

hey can someone solve me this problem sin(30+13,90) I've been trying to find an answer but i cant do it

Also could anyone explain me the way of doing it

I would be grateful

[silent plank]

are those in degrees?

use angle sum identities

[hot sable]

degrees yes

I don't have degrees on my keyboard sorry

<@&286206848099549185>

[silent plank]

look up and try to use angle sum identities

alt+0176

[hot sable]

I used angle sum identities but after that i totally blocked

I get into sin30° * cos13,90° + cos30° * sin13,90°

Then i transform it into => 1/2*cos13,90° + √3/2 *sin,13,90°

And I don't know what to do after that

[silent plank]

what does the question ask for?
decimal approximation?

[hot sable]

Ah it is from an old exam

[silent plank]

plug cos(13.9°) and sin(13.9°) into a calculator

[hot sable]

It requires an answer

[silent plank]

to get their values

[hot sable]

yeah but the answer requires

just a sec

https://gyazo.com/5f5180e7a5a8473b7e4f424383db2f68

[silent plank]

oh fk me

[dark sparrow]

...it's none of these

[silent plank]

13 * 90°

[dark sparrow]

wtf

[hot sable]

no, no dude its not *

its ,

[silent plank]

in this context,

for the question to make sense

its multiplication

[hot sable]

The answer is https://gyazo.com/abe28148ca5e7570fa1dc1ca8ac4740d

[silent plank]

dots can be used for multiplication, but the formatting is very bad

[hot sable]

Yes

😄

[silent plank]

$\sin(30^{\circ} + 13 \cdot 90^{\circ} ) $

[somber coyote]

ramonov:

[silent plank]

can you solve it if it's written properly like that?

[hot sable]

Me?

[silent plank]

yes

[hot sable]

I tried to solve it like that dude

i don't know how to transform sin,13,90° or cos13,90°

[silent plank]

not comma or decimal point

13*90

as i just stated

$\sin( x + k \cdot 360\deg) = \sin(x)$

[somber coyote]

ramonov:

[hot sable]

Oh I see now

[silent plank]

where k is an integer

[hot sable]

yes

[floral harbor]

This question is sort of based with the pythagorean theorem so, is there any possible value of x such that the square root of (2x^2) is in the set of natural numbers? If not why not?

Btw this is related to the isoceles right angled triangles.

[silent plank]

does x need to be a natural number?

[hot sable]

@silent plank Tried the way you told me still cannot do it

[silent plank]

what did you try after i gave that hint

[hot sable]

well i found how much is 13*90° which is =1170/360 so i find where i find that sin belongs in the 3rd graph

but i'm not even sure it's the right thing

[silent plank]

those aren't equal

13*90° = 1170°
NOT 1170/360

consider
13 = 12 + 1

[hot sable]

What do you mean

[silent plank]

which part?

[hot sable]

12*90° + 1 *90°?

[silent plank]

yes

$\sin(30\deg + 13\cdot 90\deg) = \sin(30\deg + 90\deg + 12\cdot90\deg)$

[somber coyote]

ramonov:

[hot sable]

so it would get to sin(120°) since 12*90 is equal to = 1080° and is not needed?

[silent plank]

"not needed" is a bad way to put it
sin(120° + 1080°) = sin(120°)

though you didn't really need to actually multiply the whole thing

12*90 = 3 * 360

[hot sable]

yeah

I just multiply it faster since it has 10 😄

Thank you very much !

[silent plank]

np

[blazing panther]

How to memeoewiae unit circle

[floral harbor]

@silent plank yes.

[rich wolf]

@blazing panther think about it in terms of turns
that is, fractions of 2 pi
like 270 degrees is 3/4 of a circle
so it is 3/4 * 2 pi
6pi/4
=3pi/2
also one of the reasons tau is superior to pi

[faint orchid]

we don't speak of tau here

[analog pumice]

It is actually illegal to use it in the US

[rich wolf]

$e^{i\tau}=1$

[somber coyote]

AMD:

[rich wolf]

look at it

beautiful

[weary drift]

lovely

[rich wolf]

If $e^{i\tau}=1$ and $e^{0}=1$ , then $e^{i\tau}=e^{0}$, implying $i\tau=0$

[somber coyote]

AMD:

[rich wolf]

checkmate euler

you were wrong

[eternal crag]

since i=sqrt-1 is imaginary it's clearly "nothing" in reality and hence it's 0, so math checks out still

[limber anchor]

So I actually have 3 questions but I think I should just start off with one, so I have all the values for the other angles but I’m unsure how that relates to B or A

[rich wolf]

@limber anchor

what shape is that

[limber anchor]

Ahh a pentagon and I just got it

[rich wolf]

use ur brain

[limber anchor]

i’m sorry

[rich wolf]

lol no don't be i'm just saying

[limber anchor]

You don’t think you could help me with the other problems in a moment could you?

[rich wolf]

sure

[limber anchor]

Alright thank you give me one second

[rich wolf]

been awhile since freshman geo lmao but i'll try

[limber anchor]

Alright so for this one I’m not really sure how I’m supposed to get the interior angles

I know the interior plus the exterior equals 180 but I don’t have a set number for either

[rich wolf]

what kind of shape is it

[limber anchor]

A trapezoid right?

[rich wolf]

which is

?

what type of shape is a trapezoid

[limber anchor]

A quadrilateral or a polygon

I believe

[rich wolf]

also it's not a trapezoid

[limber anchor]

ah

[rich wolf]

trapezoids have a pair of parallel sides

let me just say it for you

[limber anchor]

ahhh that’s right

[rich wolf]

the sum of the interior angles of a quadrilateral polygon = 360 degrees

[limber anchor]

so if i added them all up i would get 360, but how am i supposed to get their measures

[rich wolf]

180-externalangle1 + 180-externalangle2 + 180-externalangle3 + 180-externalangle4=360

do you see what i did there

and why it makes sense

you can express each interior angle as 180 minus the exterior angle

[limber anchor]

ohhh alright

[rich wolf]

and that should give you a linear equation in x

then you can find the value of x and plug it in

[limber anchor]

And I would separate them like (180 - External Angle) + (180 - External Angle) right?

[rich wolf]

addition is commutative

i personally would set it up like this

720 - (sum of external angles) = 360

because 720=180*4

[limber anchor]

And I would combine the like terms from the external angles and the 720?

[rich wolf]

combine the like terms of the sum of the external angles

[limber anchor]

ahh alright

[rich wolf]

this equation actually simplifies to: sum of external angles =360

which makes sense

[limber anchor]

So now I have 720 - 12x + 60 = 360

[rich wolf]

wrong

[limber anchor]

Why?

[rich wolf]

720-(12x+60)=360

[limber anchor]

Oh, so now what do I do?

[rich wolf]

actually my bad i'm stupid you were right

addition is commutative

so yeah it means the same thing

combine like terms

[limber anchor]

So 660 + 12x = 360?

or 780

[rich wolf]

you can also just say that the sum of external angles = 360

you got that 12x+60=360

solve that

[limber anchor]

Oh alright

So 25 is my answer

Alright now I just need one more problem but I believe I’m gonna try it on my own

[rich wolf]

👍

[limber anchor]

Thanks for the help

[eager kraken]

How do I find sec (t) =1.5625809?
sec(t) = 1/cos(t)
but when I type 1/cos^1 (1.5625809) into my calculator it gives an error

[rich wolf]

rad or deg