geometry-and-trigonometry

[real sparrow]

Yes.

[exotic meteor]

hmm

[gritty siren]

so you can't $\iff$ here

[somber coyote]

Tuong:

[gritty siren]

$(x-1/2+iy)^4=64$ translates into $x-1/2+iy\in{2\sqrt2,i2\sqrt2,-2\sqrt2,-i2\sqrt2}$

[somber coyote]

Tuong:

[gritty siren]

And then you can get 4 distinct solutions I guess

[exotic meteor]

$1^{1/3}$

[somber coyote]

Autistic Hoodie:

[exotic meteor]

As a complex number

$r = 1$ and angle is 0

[somber coyote]

Autistic Hoodie:

[exotic meteor]

$(1(cos0 + isin(0)))^{1/3}$

[somber coyote]

Autistic Hoodie:

[exotic meteor]

That would be like

$cos\frac{2k\pi}{3} + isin\frac{2k\pi}{3}, k = 0, 1, 2$

[somber coyote]

Autistic Hoodie:

[exotic meteor]

That's good

What is the trick with

$\sqrt{-3(cos\frac{\pi}{4}+isin\frac{\pi}{4}})$

[somber coyote]

Autistic Hoodie:

[exotic meteor]

Is it not

$\sqrt{-3}(cos\frac{\frac{\pi}{4}+2k\pi}{2} +isin\frac{\frac{\pi}{4}+2k\pi}{2})$

[somber coyote]

Autistic Hoodie:

[gritty siren]

i think you're confusing $x^{1/n}$ and the $n$th roots of $x$

[somber coyote]

Tuong:

[exotic meteor]

What??

It says using the trigonometric form of complex number solve

[gritty siren]

I mean, the notations

[exotic meteor]

How?

[gritty siren]

$x^{1/n}$ is the image of $x$ given by the function $t\longmapsto t^{1/n}$, it's ONE $n$th root of $x$, but there are other

[somber coyote]

Tuong:

[gritty siren]

when you're asked to find the $n$th roots of $x$, the question isn't about finding $x^{1/n}$

[somber coyote]

Tuong:

[exotic meteor]

I am using this formula...

[gritty siren]

the notation in this is bad

[exotic meteor]

What do I do now..

[gritty siren]

use the formula, but keep in mind you're allowed to judge the material you're given

$\sqrt[n]r\left(\cos\frac{\varphi+2k\pi}n+i\sin\frac{\varphi+2k\pi}n\right)$ for $k$ in ${0, ..., n-1}$ where $r=|x|$ and $\varphi$ is an argument of $x$ are indeed $n$ distinct $n$th roots of $x$

[somber coyote]

Tuong:

[exotic meteor]

I have found a trick

I can convert the complex number into trigonometric form and back to normal form

and it will looks like

$3(cos\frac{3\pi}{4} +isin\frac{3\pi}{4})$

[somber coyote]

Autistic Hoodie:

[exotic meteor]

And I can then find the root of it

[zealous bough]

Can someone help with these please

[frigid creek]

use cyclic properties

[meager acorn]

is this correct?

hello?

[void osprey]

find x

i cant do this for the life of me

even though i've done problems similar and been fine

15/9 = 1.667

so

1.667(2x - 1) = 3x?

that gives me x = 5

but how can the hypotenuse = one of the bases

does that not defy the pythagorean theorem

still stumped

[void osprey]

OK so it's an isosceles triangle

thats gay

[wind gale]

@void osprey have you tried setting both sides to equal eachother?

[void osprey]

i already figured it out @wind gale

wasn't confused at how to do the problem

i was confused since i assumed it was two right triangles

[wind gale]

Ah I see

[upper karma]

hello i need help

[empty moss]

just ask away bro

just post the problemo and somebody will solvo the problemo

[empty moss]

find the distance between A and B

[upper karma]

ok then

[empty moss]

then you gotta think about the ration

[upper karma]

rations?

sorry im really stupid

[empty moss]

ratio

sorry ratio

[upper karma]

yo how bout this one

[old seal]

How to evaluate??

[hard gale]

well you are summing up 2n arcsines, which all sum up to n*pi

and arcsin always outputs a value between -pi/2 and pi/2

so the only way of getting that is : all the arcsines output pi/2

ie all the x_r's equal 1

@old seal

[hallow smelt]

If I have f(x)=2tan(x) since tan doesn’t have a amplitude how do you label the 2?

[dry glen]

can anyone help me with a question?

<@&286206848099549185>

[copper valve]

read #info @dry glen

[dry glen]

ok sorry about that

[upper karma]

Y u no post question

@dry glen

Post the questioonnnnnnnnn

[dry glen]

sorry

[upper karma]

Y u ping b4

You know sine and cosines?

Draw a line from B to D

Fill in unknowns

[dry glen]

yes

I was trying to do that

but the answer was incorrect

sorry if I am not as smart as you

[upper karma]

[dry glen]

cute

[upper karma]

Well

[dry glen]

welll

[hard gale]

[dry glen]

this is what I wrote
I called the line that DB as y
and wrote that
y^2= AB^2+16^2 - 2 * 16 * ABcos120
y^2= (AB+5)^2+25^2-225(AB+5)cos60
then I solved it but I could not get the answer

help me you geniuses

[hard gale]

well your shit should work

[hallow smelt]

In the function f(x)=3tan(x) what do you label the 3?

[drowsy spoke]

that isn't even geometry m8

post it in questions

or precalculus

[hallow smelt]

It’s trig

[drowsy spoke]

woops, sorry

what do you mean by label?

[hallow smelt]

It’s not amplitude it’s something else

Did that clarify?

[drowsy spoke]

yeah

well, you're right, it's not the amplitude

it's just a scaling factor

or the coefficient of tanx

[glass relic]

when do you use inverse sohcahtoa?

[dry glen]

damn it

when you know the side lengths

but not know the angle

[hallow smelt]

Thanks @drowsy spoke

[glass relic]

oh ok thanks sorry lol

[dry glen]

onion art

can you solve it for me/

[raven relic]

does anyone know why there are so many variatiosn of Menlaus thereom...

is there a general case or do I have to learn all the individual ones

[drowsy spoke]

@dry glen where'd you get stuck? upload your working

[dry glen]

y^2= AB^2+16^2 - 2 * 16 * ABcos120
y^2= (AB+5)^2+25^2-225(AB+5)cos60

I can't solve this

for some reason

@drowsy spoke

[hard gale]

y^2= AB^2+16^2 - 2 * 16 * ABcos120
y^2= (AB+5)^2+25^2-225(AB+5)cos60

the two bold things are equal to y², ie the bold things are equal to each other

AB^2+16^2 - 2 * 16 * ABcos120 = (AB+5)^2+25^2-2 * 25(AB+5) * cos60

[dry glen]

then?

I did that

and then I can't solve it

cos 120 = -1/2

i substituted that value and I can't find it

is there a way to not put in -1/2

and solve it?

@hard gale

[hard gale]

[dry glen]

I'm sorry

[hard gale]

why would you not ?

[dry glen]

cause I can't find the answer when I did that

i'm sorry

you guys seem angry

can you please solve it for me onece

@hard gale

[hard gale]

~~AB²+~~16² +16AB = ~~AB²+~~10AB+25+25² - 25(AB+5)

(expanding everything + replacing the cos)

31 AB = 25+25²-125-16²

AB = 269/31

ugliest shit ever

[dry glen]

aright thank you

HOW

OMG HOW DID YOU GET IT RIGHT

LIKE WTF

[hard gale]

is it really the ans?

WTF

[dry glen]

YES

WOWOWOWOOWO

[hard gale]

i was like 269/31 : something ain't right boi

aaaaaaaaaa

[dry glen]

yea... IB math is rarted

[exotic meteor]

What's the easiest way to find z^4

with z i mean (x+iy)^4

where I do not know x and y

[upper karma]

can anyone help with this

[exotic meteor]

With what @upper karma ?

[exotic meteor]

It's easy

Do you know the formula for the distance between two points

That's the formula @upper karma

You following @upper karma ?

[upper karma]

yes sorry i was eating

[exotic meteor]

ALright

Do you know the coordinates of the ship?

You can get them form the graph

[upper karma]

i cant do it dude ima drop out

[exotic meteor]

It's super easy -_-

Look

It's (-3, 8) where x = -3 and y = 8

Do you understand how I got the coordinates?

[winged saddle]

bruh

[patent echo]

https://media.discordapp.net/attachments/532776486969540609/545458620314288138/image0.jpg?width=788&height=591

how do u do this

@tropic shard

[old seal]

@hard gale thanks

[lucid sentinel]

may someone please help me understand this

[dire rampart]

u know the double angle formula for tan?

[gleaming egret]

How do you solve 1.25sin(x) = x , for other than x=0 ?

[dire rampart]

analytically ? dont think its possible

you can use numerical methods to get an approximation tho

,w solve 1.25sinx-x=0

[somber coyote]

Display results online and refine query

Requested by lemon catto#7133

[dire rampart]

[gleaming egret]

Thanks

[red leaf]

i need some help

if you know cos of theta is 9/41 how can you find the inverse sin and tan

[upper karma]

pythagoras' theorem

[red leaf]

@upper karma im just tripping on how to start it

[upper karma]

the adjacent is 9 and the hyp is 41

so use pythag for the other side

== sqrt(41^2 - 81)

oh lol mathbot is gone

,calc sqrt(41^2 - 81)

[somber coyote]

Result:

40

[red leaf]

@upper karma tyty

have a math test i need to clutch today

2 sqroot of 13 i

is it the same as 2i sqroot of 13

[dire rampart]

wot

[red leaf]

https://youtu.be/s03qez-6JMA?t=239

[gritty siren]

you want the square roots of 13i?

[red leaf]

at the end of the very vid he said thats the most simplified you can get to

i wrote 2i sqroot 13

@gritty siren ooh i meant 13

oof

@gritty siren basically im asking are these the same

2sqrt(13)i = 2i sqrt(13),

[gritty siren]

yea they're the same

[peak sluice]

Hi does anyone know how to do this problem

[sacred cradle]

On a coordinate plane A(0,0) and B(20,20) what is the distance of AB note each unit is 10 feet

Also round to nearest feet

I got 283 ft can someone check that for me?

[timid saffron]

Does anyone have a trick or mnemonic for remembering that

$\int \sin(x) = -\cos(x)$

[somber coyote]

cornelius:

[past siren]

Is there a specific way to calculate this graph on the calculator? I could get rid of everything except B and D

[dire rampart]

what's the value of the function at x=0

see which one fits

[hallow bramble]

what he said

[steady sleet]

@timid saffron just know your trig derivatives tbh

[void osprey]

is this right?

[void osprey]

it was

[upper karma]

can someone help me solve this

[dull egret]

@upper karma how many dimensions? and how many solutions can there be? cause even in two dimensions you ll have infinitely many if i m not mistaken cause rotating u and v simultaneously does not affect the three conditions

[upper karma]

no idea

i just tried solving it

[dull egret]

@upper karma the three vectors from a triangle with those side lengths. the angles they form can be calculated and that between u and v must be kept constant. since you know their lengths you can then get expressions for u and v dependent on some angle lets call it phi. just add them then and you ll have the geenral expression interms of phi

[glad falcon]

So i did all my calculations accurately

but when i put it in desmos i get this

i have been look for mistakes the past 30 mins. I am brain ded. Can someone tell me what i am doing wrong.

( question is calculate the circumcircle and the equation of the circle )

tag me when someone replies

[tropic shard]

@glad falcon That moment when you say 13/14 instead of 3/14.

[glad falcon]

@tropic shard 🤦 Thanks my man ! 😃

[sacred cradle]

is there a website or youtube channel where I can learn some geometry?

[glad falcon]

@sacred cradle proffesor leonard

[tropic shard]

https://www.youtube.com/playlist?list=PL0o_zxa4K1BVkRxCZubMPcCJ5Q5QwZdEM

[glad falcon]

Or Organic Chemistry

hes pretty fast paced

[tropic shard]

Professor Leonard does Geometry?

[glad falcon]

umm maybe. i just said his name cuz i have seen him cover nearly everything. 😅

@tropic shard btw u know when it says what is the closest horizontal distance between point a and b

i thought u would have to calculate the distance so that the slope is none

[tropic shard]

Hm.

[glad falcon]

but when i read the answers it said u would have to take the perpendicular line to that

and then see the distance

i know i didnt explain properly but heres the question

Question D

[tropic shard]

I'll graph it and see how that works, lol.

[glad falcon]

aight

I thought it would be something like this

but apparently not

[tropic shard]

I think they meant this

[glad falcon]

Yes. i think that is what the mean too

but i need an explanation on why that is the closest distance

[tropic shard]

Hm.

I'm not sure how to explain it besides imagining moving the point along the function and measuring the distance xD

There's a video of it using calculus, but you'll need to know derivitives for that video.

Since we have a line, you wouldn't need to use calculus, so lemme think of a way to translate from calculus to algebra.

nvm, you know what, just imagine a circle, I'll graph one.

[glad falcon]

Ok..

[tropic shard]

Did you post something?

[glad falcon]

oh no nvm that.

Ohh I think i understand.

[tropic shard]

Oh ok lol, I saw it for like a split second

[glad falcon]

Since that line is a tangent to the circle

[tropic shard]

Yup

[glad falcon]

That is the only point that would touch it . ( at one point )

so that is the place where the radius meets the line

[tropic shard]

You got it

[glad falcon]

so if it was some other point on the line. It wouldn't be the shortest !

Yea ! 😃 Thanks Man That is way more clear now

[tropic shard]

And all the other points in the line are a larger radius, exactly

np :)

[glad falcon]

Oh yea and the rule.

when the tangent meets the rad. It is 90 degrees therefore it is perpendicular !

[glad falcon]

Aight So i have another question

So in this problem

Ab and OC clash

if i changed the route so that it went from A to C to B to O

would the distance remain the same

[astral hornet]

you can use point slope form to get the equations of the lines then find their intersecting coord, and use dist formula to compare dists

[zinc grail]

does any one have time to help a struggling boi

to solve this I would need to prove thatthe smaller triangle is similar

what would that be?

[sudden lance]

I would say 5/3. You see, if AD is half of DC, then AC=3*AD. So you use Thales theorem and you have 1/3=DE/5.

[zinc grail]

this is SAT stuff

I dont know thales theorem

is there a simpler method to find that solution?

[sudden lance]

But the thing is... that's the simplest method, according to me. Maybe ask the helpers ?

By the way, what is SAT ?

[zinc grail]

its an American standardized test

for college

[sudden lance]

I understand

[wanton eagle]

The Angle-Angle postulate will help you prove the two are similar. They share an angle, and each has a right angle. Therefore, they have two (ultimately meaning they have three) congruent angles and, thus, are similar.

[zinc grail]

what would be the shared angle?

[wanton eagle]

Angle A

[zinc grail]

oh and the 90?

[sudden lance]

And the right angles.

[wanton eagle]

Angle EAD equals Angle BAC

And they each have a right angle, so those are congruent as well

[zinc grail]

and so the big triangle has legs of 5 and 3x

how would we use that to solve the smaller one

with the ratios

[sudden lance]

By using the ratios, you would get 5/3. However, what is the length of AC in fact ?

[zinc grail]

3x

oh so the smaller triangle will have the same ratio as that

[sudden lance]

So that was an x.

[wanton eagle]

5/3x = ED/x

[zinc grail]

so 5/3

[wanton eagle]

X cancels in each side, leaving ED = 5/3

[sudden lance]

Yes. I know that in my country, we're used to learn that theorem which tells that in fact.

[zinc grail]

that makes sense

thanks

[sacred cradle]

can you prove a quadrilateral is a parallelogram but not a rectangle given two adjacent right angles?

[quasi steppe]

Hello?

Someone mind helping me with this problem I’m struggling to solve it

<@&286206848099549185>

[upper karma]

what shape is that Lol

[quasi steppe]

serious question?

[drowsy spoke]

Is T on the circle of radius R + Rh?

[quasi steppe]

looks like i made a mistake with the drawing

T is suppose to be alittle bit above the top of the circle

and R is the distance from the center of the circle to the top of the shape(not circle) and Rh is the radius

[livid light]

any tip how to learn all the values?

[gritty siren]

for sin and cos, there's a

$\frac{\sqrt 0}2,\ \frac{\sqrt 1}2,\ \frac{\sqrt 2}2,\ \frac{\sqrt 3}2,\ \frac{\sqrt 4}2$

[somber coyote]

Tuong:

[gritty siren]

pattern

and if you have sin and cos, you can have tan

[upper karma]

the volume describes this what is enclosed by the edges
and area describes what the figures that are formed by the edges cover
if so why the volume of a right triangular prism is smaller than it's total area

i understand why is it so but i can't possibly imagine it

[jovial sierra]

consider a cube of side length 1 meter

its volume is 1 m³, but it has 6 sides which are each 1 m², so its surface area is 6 m²

[sudden lance]

Well, you can't really compare cubic metres and square metres...

[marsh spruce]

Could anyone help me memorize the trig identities like cos(a+b) and cos^2 something is one? Help me in like an hour please

[sudden lance]

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

[somber coyote]

TSM64CM:

[sudden lance]

I would rather say that cos changes the plus to a minus... and that it gathers cos and sin on the same side...

And about the other one... Which one ?

Because there are several formula for cos²(x)...

[sweet hazel]

There's a good bit of formulas involving the cosine squared

$\cos^2(x) + \sin^2(x) = 1$

[somber coyote]

BuilderBudCarl:

[sudden lance]

I was thinking about that one too.

[sweet hazel]

that's the Pythagorean identity

[sudden lance]

This one is just Pythagoras theorem on the trigonometrical circle.

[sweet hazel]

there's another one that you can get from the identity TSM posted above - the case where a = b

$\cos(2x) = \cos^2(x) - \sin^2(x)$

[somber coyote]

BuilderBudCarl:

[sweet hazel]

and if you use Pythagorean on that and substitute the sine squared, you get this

$\cos(2x) = 2\cos^2(x) - 1$

[somber coyote]

BuilderBudCarl:

[sudden lance]

No need to go on complex numbers, or should I still do that ?

[sweet hazel]

wait, did he also want Euler's formula?

[sudden lance]

I don't know.

I'm asking if it was worth putting that...

[sweet hazel]

maybe we could tag him

@marsh spruce awaken from thy slumber and tell us what identities thou desire

[sudden lance]

Wherefore art thou, Rising_Lamp ?

[proper echo]

Hello everyone, I have a question about maths

geometry*

This is an assigment that I have to do

And Im having problems with the object Nr. IV

I know that the side next to the right trapezoid is equal to 4

[fleet skiff]

Anyone know the formula of slope?

[glad ocean]

yeah rise and run

(change in y)/(change in x)

[marsh spruce]

Sorry @sudden lance @sweet hazel I was busy but I'm here now

[sudden lance]

Fine. So what was the formula you were talking about ?

[marsh spruce]

The phythegorean identity, even-odd identities, the simplicity identity (the ones that factors stuff like (cos^2x + sins)(cos^2x + sins), the formulas to prove identities, the sum and difference identity, and the double angle formula

[zinc grail]

I hate unit circle

:/

[glad snow]

Hi what is a nice phrase to describe this

Equidistant? Contour lines? Loci? Polygons?

The idea is the linear interpolation of points that maintains constant ratio of distance AX/XP where A is an arbitrary point in the square, P is every point on the perimeter of the square and X is the locus/trajectory to describe

X happens to create those other
Squares

[astral hornet]

right nested?

[glad falcon]

If i want to find the distance of the altitude
how would i do that
I know how to find Point A
A = intersection point of B and y intercept & c and y intercept

[tropic shard]

wdym altitude? As in the yellow line? Like you said, it's when the line associated with point B (f₁)
and the line associated with point C (f₂) intersect (i.e. 0=f₁-f₂).

And to find what the f (function) of each line is, you'd need to use y=mx+b, where m is the rise over run.

f₁-f₂=0 will let you solve for x, but you want y, so you'd plug it (the x) back into either f₁ or f₂ (it'll give the same y value) and voila, you'll get the altitude.

[bronze harbor]

can someone explain real quick whats arcsine arccos and arctan are

[tropic shard]

$sin(θ)=\frac{Opposite}{Hypotenuse}$, $arcsin(\frac{Opposite}{Hypotenuse})=θ$

[somber coyote]

MathLyfe:

[proper echo]

So can nyone help me with my assigment?

[dusty coral]

Anyone help me with the following problem?

*equidistant

[valid pike]

Can someone explain how I find x here

For 1

[dusty coral]

@valid pike

[valid pike]

Yes

[dusty coral]

First you have to solve for BC.

Then you have to solve for CD.

After which you do:

the law of sines

[valid pike]

Ohhh

[dusty coral]

Set DBE = 90 - x

here let me type it out

sin (90 - x + 20) / (16 + CD) = sin (x) / (BC)

I did this after noting that BCE is a triangle

[valid pike]

Alright I see

Thanks

[dusty coral]

np

[valid pike]

Have I done this right

[acoustic peak]

I found this problem online where they ask you to find the volume of the glass using the measurements provided in cm. Just thought I'd leave this here in case someone would like to give it a try.

[valid pike]

No habla

[acoustic peak]

@valid pike I explained what the problem says. Find the volume using the numbers on the drawing.

[faint orchid]

@acoustic peak is this question possible without some form of calculus? geometry seems to be pushing it

though i', sure it can be done

using geometry alone

[acoustic peak]

@faint orchid No idea, I don't know how to solve this and I don't know calculus, just thought some people here might enjoy giving it a try.

[faint orchid]

@acoustic peak is this a problem you need to solve for homework or just for fun?

I think I can solve it if given enough time

[acoustic peak]

Just for fun, I found it online.

[faint orchid]

Rip. not really worth my time then

[acoustic peak]

https://tenor.com/view/gary-tay-baby-laugh-gif-7517608

[limpid basin]

Without further instructions the problem is incredibly ill defined.

[faint orchid]

solving for the first bottom half of the glass seems really simple. You have a cone with a radius of 2.5cm and height of 2.5cm as well, using the formula for the volume of a cone leads to 16.36cm cubed

[timber hinge]

okay

simple Q guys

wat is the volume of a hypericosahedron in 29 dimensions?

[umbral snow]

Ugh I'm getting tired of all these trivial questions @timber hinge

[timber hinge]

sry

[faint orchid]

Yeah pls go back to middle school

[timber hinge]

I just get stuck

have mercy mighty Kaynex

[umbral snow]

Like, start with the volume of the 32 dimensional genus cuboid and apply the isomorphism Louiville theorem to reduce the determinant of the regularized angle. In radians, of course

Don't forget to settle with an orthogonal ideal ofc

[dire rampart]

assuming convergence on a tridemnsional radius btw

[umbral snow]

Oh derp

Wow, excuse me I must be tired

[upper karma]

hi

anyone have a second for a person who is very mathematically challenged?

this is literally very basic i just can't seem to learn it. it's like geometric trigonometry if you can call it that

i have to find missing angle measurements and i'm like at a loss of what to do

[faint orchid]

Just post the problem my man

[upper karma]

this is gonna sound strange but i just figured it out like 5 seconds after u posted this

[faint orchid]

Welp

Ok then

[upper karma]

i think it was ur words of wisdom

thank you ❤

[dire rampart]

yw

[upper karma]

Ah here we go

How would I solve this? Literally looked everywhere and am at a brick wall

I solved thankfully the past 28 but this one just doesnt relent

I need to find X

Using tan, sin, cos

[tropic shard]

@upper karma You needed help with 29?

[upper karma]

21.3 is given as the answer

i just had no idea how to get there

[tropic shard]

Most calculators are set on Radians, but to get 21.3 as the answer you'll need to set it to degrees.

[upper karma]

and indeed filling out ur form worked

thank you 😃

[tropic shard]

Oh nice :)

[upper karma]

but my question is how do i decide if it's tan/sin/cos in these situations

[tropic shard]

You know SOHCAHTOA?

[upper karma]

i mean i get theres an opposite and an adjacent

yeee yeah it's just

how did you decide this

it's based on the angle?

[tropic shard]

Yup.

[upper karma]

ahhhh so why does the 'y' side have to be involved in finding the 2nd part

[tropic shard]

?

[upper karma]

like the left side

why's it need to be involved

wouldn't it be easier w/o

[tropic shard]

Since we don't know what opp is.

So you have 2 variables, which is not possible to solve by itself.

[upper karma]

okay, that makes sense

can i post a picture of the next and for me final problem, then solve it and check it with you real quick?

[tropic shard]

Sure

[upper karma]

thank you

okay so 32

this appears to be an opposite and hypot

so it's sin?

[tropic shard]

Nope,

[upper karma]

rip lmao

[tropic shard]

Opposite is the side that the angle doesn't touch

[upper karma]

oh adjacent and hypot

[tropic shard]

Yup

[upper karma]

okay thus cos

okay let's see here

[plucky marlin]

just use sine rule tbh

[upper karma]

well shit

i just realised..

[plucky marlin]

wat ?

[upper karma]

there adjacent and hypot

*the

is the same wait

okay now i'm tripping myself up

oh wait

[plucky marlin]

??

[upper karma]

nope im lost

[plucky marlin]

[tropic shard]

The dashed line is the same side for both triangles, which side is it (opposite, adj, hypontenuse?)

[upper karma]

Oh its adj

Thus cos(42) = 30/19 ?

No

I mean

Lmao, I said i was bad at this

Didnt think this bad

Welp, lost again

[plucky marlin]

https://prnt.sc/mmgmxa

[upper karma]

oooh okay

that actually makes sense tho

just a hidden y

thank you guys, i'll use such methods to work out some more and check them now i have a basic grasp

i just used that logic and got the right answer

wow thanks so much ❤

[plucky marlin]

np

[upper karma]

@plucky marlin uhhhh

okay so turns out one is still plaguing me

https://cdn.discordapp.com/attachments/326138757474680852/546867996489023488/20190217_193754.jpg

30.)

i'm doing O/H

but the answer isnt coming to be what is expetced

expected

[plucky marlin]

its the same process

[upper karma]

ikr

i tried and it's like

nope, sorry sacro, you're dumb

i did

sin(67) = y/44

which gave me 40.502

so i thus applied the same over there

sin(46) = x/40.502

and i'm pretty sure this is where i went wrong

[plucky marlin]

https://prnt.sc/mmgt4i

[upper karma]

oh it can be seperate processes

i'm stupid af

[untold whale]

can somone help me with trig

[glad falcon]

Need help with all

[glad ocean]

9. altitudes are perpendicular and medians bisect, so yes

10. uh what

11. isosceles triangle I THINK?

[subtle spear]

10 may be that the intersections form a line known as Eulers line

That is, the circumcenter, orthocenter, and the centroid

Those three points are all colinear, and it is known as eulers line.

I would say 11 is equilateral

@glad falcon

[hard gale]

10 would be more like the fact that median, perp bisector and altitude of the base are actually the same line

We're not even talking about intersections of all of those lines, since we're focusing only on those of the base

and iirc Euler's line isn't even specific of an isosceles triangle

[subtle spear]

I didn’t even read the isosceles bit lol, my bad

[merry knoll]

Idk where to put this

a) 29 right?

for b you go to 20 and across then down?

so 91?

and c) 20?

I think this website is broken

[umbral snow]

@mighty sand
This is trigonometry. Have you not been taught any? And you're still expected to know it?

[mighty sand]

im in alg 2

idek

my pp not that bigg

[plucky marlin]

[mighty sand]

when u take online school and ur teacher keeps trying to call u about plagarism

yeet

[upper karma]

pro tip

csc = 1/sin = h/o

[plucky marlin]

sssssssss

[upper karma]

wait im a day late

[plucky marlin]

[upper karma]

In a rectangle ABCD points E and F are on the sides BC and CD respectively in such a way that the triangle AEF is equilateral. The point M is the middle of AF. Prove that the triangle BCM is equilateral.

[raven sequoia]

how would you go about doing number 59?

[serene field]

Try getting the entire equation in terms of just sine or cosine.

[amber raft]

anyone up for this one?

idk if this is the right channel, nobody could figure it out in the other one

[buoyant oasis]

The line $y = 2x+7$ is dilated through the center $(-2,-1)$ with a factor scale of $b$ giving $y=2x-13$. Find $b$!

[somber coyote]

Rumoden:

[buoyant oasis]

Can anybody give me any clues regarding this?

[hallow smelt]

Does the graph of cotangent have a hole in the x intercept?

Everywhere I want to plot the x intercept of cotangent there is a hole. How do you properly graph cotangent on paper?

<@&286206848099549185>

[upper karma]

Lmao, is that b factorial or is your teacher enthusiastic

Try to imagine unwrapping the unit circle onto the x-axis

[hallow smelt]

Me?

[upper karma]

Yeah

I'm not too experienced in graphs of trig functions but I think that that would be valid

It's defined within integral multiples of pi<x<2pi if I understand correctly

on those two bounds it goes undefined

[hallow smelt]

0<x<Pi

[upper karma]

Ah

Okay

[hallow smelt]

Same as you but starting at origin

[upper karma]

Ah yesyes

So try to trace out the asymptotes, and then fill in the rest of the function between them

[hallow smelt]

But when I look for the x intercept I got a hole in the calculator

There is no asymptote of pi/2 in the functions f(x)=cot(x)

But there is a hole there

[upper karma]

Tangent within 0<x<pi looks the same within -pi<x<0

It shoots up at the end and starts the interval shooting up from the bottom

I thiiiiink

Lemme check and make sure I'm not feeding you false information

Oh shit nvm ignore everything I said lmao I'm retarded

On the graph there appears to be a point along the y-axis

so Idk why you keep getting holes

sorry m8 ):

Oh the x-intercept

[hallow smelt]

Yes

Pi/2 there is a hole it’s in the middle of the function

I should say period*

[upper karma]

Hmm

I don't even know how to help you even though I'm looking directly at the graph

Do you know what the general image of f(x)=tan(x) looks like?

[hallow smelt]

Ya but I need 1/tan(x)

[upper karma]

OOOOH

[hallow smelt]

Tan x does not have holes in it

[upper karma]

It is asymptotic though

So you can consider those holes

Because it's undefined at 2pi, it should be undefined within 2kpi where k is an integer

[hallow smelt]

Ok well I’m trying to graph this accurately

You don’t include a point for a limit do you

?

[upper karma]

Never. Only the points infinitely close to it.

[hallow smelt]

Ok here’s the scratch of my graph.. you are saying just make a random mark and say that’s where it crosses?

[upper karma]

Uhhh

Not exactlyyyy'

[hallow smelt]

Where do I make the points

[upper karma]

Uhhh... honestly man

I'm not too sure, so if nobody else answers you, I'd find a video

Sorry my dood

[hallow smelt]

I just drew it like that hopefully it’s right hahaa

[upper karma]

Looks pretty accurate to me

You can check with Desmos graphing calculator if you like

and if it's wrong, you can still ask why it's wrong

[hallow smelt]

Ok @upper karma haha asked my prof

There is no hole. Calculator does not have cotangent only 1/tangent

1/undefined = undefined, but if you set cos/sine = 0

You get 0/1 so there is no hole/asymptote

[exotic meteor]

A plane can be defined by three points, a point and a line and two lines

I understand how to find the plane formula with three points

But how do I find it with two lines?

[umbral snow]

They'll need to intersect somewhere. Let that intersection be the point

Now it's a point and a line again.

[drowsy spoke]

I was just thinking about that, if they don't intersect it doesn't define a plane, does it? That statement is not entirely correct then?

That statement - a plane can be....

[hard gale]

If you take 3 collinear points you're pretty screwed that's for sure

[drowsy spoke]

Oh that too

[exotic meteor]

Found it

The normal of the plane is the cross product of the two lines

Then I just pick a point of the lines and that's it

I believe this works each time?

[drowsy spoke]

I think you mean vectors and not lines

[exotic meteor]

Well yeah, the vectors of the lines

[drowsy spoke]

Yeah that works

[exotic meteor]

How about if a line and a point is defined

I guess I could make a new line with the first line and the point and use the same method afterwards

since the first line has a point starting point

[drowsy spoke]

Yeah

[exotic meteor]

Awesome, I like how one method mixes with the other one ^^

[drowsy spoke]

Well, that's chapter 1 of vector calculus

[exotic meteor]

Yeah, I am repeating everything for the semester, linear algebra, vectors, analytical geometry, derivatives, sequences and series

I have my last test in Monday for this semester, I failed the last one :/

[zinc grail]

hello

so this concept is really messing me up

[round veldt]

Well

Congruence just means something is equal, so just look and see what's equal.

[astral hornet]

Based on parallelogram props

[upper karma]

can anyone show methis?

[split raven]

Does anyone know how to prove that alpha + beta + gamma = 90 degrees?

[weak zenith]

maybe use arcsin and smth like arcsin(AE/AF)+arcsin(AE/AG)+arcsin(AE/AH) = 90

[limpid basin]

sin B = 1/sqrt5
sin Y = 1/sqrt10
sin(B+Y) = (1/sqrt5)(3/sqrt10) + (1/sqrt 10)(2/sqrt5) = 3/sqrt50 + 2/sqrt50 = 5/sqrt50 = 5/(5sqrt2) = 1/sqrt 2. Since B+Y < 90, B + Y = 45 and a + B + Y = 90

(Y = gamma :p)

[split raven]

Thank you

[fleet wolf]

argh

so much algebra

numberphile has a nice video on this exact problem

[upper karma]

how great is numberphile tho

[swift thorn]

numberphile is very good at explaining things three to six times as slowly as they could be explained

[spiral bear]

@swift thorn exactly like most professors in my univ

[kind breach]

edit: i'll not be lazy and will return when i need help for things i cant do

[upper karma]

Can you help me with Geometry?

[serene field]

[upper karma]

ok

[upper karma]

solve for the letters

[fleet wolf]

@upper karma construct a cyclic quadrilateral

then use inscribed angle theorem

for the second ine

[upper karma]

yeah i got it

thanks

[dense tundra]

would this be (1 - cos^1(x)(1 - sin^1(x)) ?

[tranquil wigeon]

don't think so

Did you use any trig identity to solve It? @dense tundra

[desert heart]

nope it wouldn't be that js is that not a math sin?

because that expanded would not be the original expression

need help solving this identity

[hallow smelt]

f(x)=3tan(x). the 3 there is not an amplitude is it? Im asking for the label of the 3 not what it's doing to the graph of the function. Just what we call the 3. Is there a proper label?

[tranquil wigeon]

@hallow smelt I don't think it has a label, not sure tho

[hallow smelt]

@tranquil wigeon its not an amplitude though?

[tranquil wigeon]

@hallow smelt not an amplitude, it doesn't have any

[hallow smelt]

@tranquil wigeon would you call that 3 the amplitude if we swap tan for sec and csc?

i know thats why were we start the graphs for both of those, but unsure what to call the |A| of those functions either

thats where we start the graphs**

@tranquil wigeon the guy on khan academy called it a scalar

[tranquil wigeon]

@hallow smelt for csc and sec are considered not having amplitudes either. A scalar works, but they really don't say alot

[dire rampart]

@dense tundra use the angle sum formula for cos

[twin prawn]

,rotate

[sick veldt]

Is this correct?

I think its either C or B

[swift thorn]

well, look at where x+2 and 3x-14 are in the diagram

they're part of the same diagonal

if x+2 equals 3x-14, then they're both the same size

they're both half of the diagonal

[sick veldt]

So C

[swift thorn]

yep

because bisect means cut in half

[sick veldt]

yeah

[swift thorn]

👍

[sick veldt]

could you check 2 more?

if its no bother

its another 50/50

[swift thorn]

[sick veldt]

I don't think a and b are reasonable

answer C I did consider

but D sounds better to the ear

[swift thorn]

what is D in the diagram? a point?

which point

[sick veldt]

no answer option

blue bubble is my answer

[swift thorn]

no i mean the diagram has a label "D" in it

what is that labelling

[sick veldt]

line d?

with an arc

[swift thorn]

D is a line? option a and c make D sound like a point

[sick veldt]

hmm

so it couldn't be those two

cause its a line not a point

[swift thorn]

no, i think this is just a very awful diagram

[sick veldt]

schools haha

I'll just leave that one as a guess

let me grab the next

[swift thorn]

i would not be able to solve this question!

[swift thorn]

well, what does the diagram tell you

what kind of quadrilateral is that

[sick veldt]

trapezoid

and we know

that point x

si the same as w

[swift thorn]

and it's not just any trapezoid, it's a symmetric one

[sick veldt]

oops

x*

[swift thorn]

yeah, because it's a symmetric trapezoid, angle W is the same as angle X

and likewise angles Y and Z are the same

[sick veldt]

yeah

[swift thorn]

i think we both know you know how to get the answer

[sick veldt]

trapezoid formula?

[swift thorn]

you have four angles in a quadrilateral

112, 112, Z, Z

[sick veldt]

so 224 = 2z?

divide by both sides = 112

[swift thorn]

?

where'd that come from

[sick veldt]

I'm not sure how to solve this

[swift thorn]

you have four internal angles in a quadrilateral

don't they add up to something?

[sick veldt]

180?

no

[swift thorn]

triangles are 180

[sick veldt]

360

[swift thorn]

that's the one

112 + 112 + z + z = 360

[sick veldt]

= 68

[hallow smelt]

So cos=1/sec and sec=1/cos right?

And sin=1/csc and ccs =1/sin?

[foggy zealot]

yes

[hallow smelt]

@foggy zealot can you also say since Sec = 1/Cos and Cos = 1/Sec

that Cos(x) * Sec(x) = 1?

[foggy zealot]

yes

[hallow smelt]

thats cool we went over all the identities and i don't think our prof mentioned that cos(x)*sec(x)=1

must also mean sin(x)*csc(x)=1

wait

so do cos(x)*sec(x)=Sin(x)*csc(x) @foggy zealot

that means sin(x) = cos(x)*sec(x)/csc(x)? hahaaa

[plucky marlin]

sin(x) = x tbh

[hallow smelt]

no it doesnt lol

sin(pi/2) = 1

and pi/2 =/= 1

[plucky marlin]

,w pi/2

[somber coyote]

Display results online and refine query

Requested by soap#7503

[plucky marlin]

almost 1

i'll take it

[hallow smelt]

do not do that

[plucky marlin]

[hallow smelt]

lmao thats not how it works at all

[plucky marlin]

wdym

that IS how it works man

im a physicist

thats how we do it in my town

also cos(x) = 1

[hallow smelt]

stop that

[plucky marlin]

$\sin(x) = x - \frac{x^3}{3} + \frac{x^5}{5} + \dots$

[somber coyote]

soap:

[plucky marlin]

wat u gonna say now

😏

[hallow smelt]

i dont know what that is

[plucky marlin]

okay lemme write that in sum notation

[hallow smelt]

restrict sin to 0-2pi

[plucky marlin]

$\sin(x) = \sum_{n = 0}^{\infty} \frac{(-1)^nx^{2n + 1}}{(2n + 1)!}$

[somber coyote]

soap:

[plucky marlin]

wait

wdym restrict sin to 0 - 2pi ?

[hallow smelt]

because my brain is small

im in trig

[plucky marlin]

oh right

[hallow smelt]

infinity is not something we deal with regularly

[plucky marlin]

my bad

anyways

[hallow smelt]

i dont even know how to read what you wrong as a whole

[plucky marlin]

sin was domain of all reals tho

[hallow smelt]

i can kinda real each symbol piece by piece

yes im aware, but you don't really look at the entire domain just a period

[plucky marlin]

[hallow smelt]

when you are trying to figure out what the graph is doing i mean

[plucky marlin]

hmm

i have lots of other definitions of sine

wanna see ?

[hallow smelt]

why would you need to look at sin

how you posted it

[plucky marlin]

idk

im just showing it

what was even the question ?

[tranquil wigeon]

Well it's 1 only close to zero

[plucky marlin]

mhmm

u see he gets it

[tranquil wigeon]

We just over Taylor series

Went

Pretty cool I should say

[hallow smelt]

if sin(x)*csc(x) = 1 and sec(x)*cos(x) =1 then sin(x)*csc(x)=sec(x)*cos(x)?

that was my question

[plucky marlin]

yeah

[hallow smelt]

and then sin(x)*csc(x)/sec(x)=Cos(x)?

[plucky marlin]

,w sin(x)csc(x)=sec(x)cos(x) true or false ?

[somber coyote]

Display results online and refine query

Requested by soap#7503

[hallow smelt]

thats pretty cool

you can algebra them all to infinitiy then right?

[plucky marlin]

[hallow smelt]

keep replacing them with eachother

[plucky marlin]

sure

[hallow smelt]

thats really cool hahaa

[tranquil wigeon]

@plucky marlin yo can you explain big o notation, I really don't understand the meaning of in, in Taylor series.

[plucky marlin]

i dont either

[tranquil wigeon]

Uhg

[plucky marlin]

lel

[hallow smelt]

is that trig?

[tranquil wigeon]

Whoops

[hallow smelt]

big o notation? that sounds made up

[tranquil wigeon]

No

[plucky marlin]

it's literally called big O notation

[hallow smelt]

you know im a tutor at my college

for math

hahahaa

[plucky marlin]

damn

really ?

[hallow smelt]

ya lol

[swift thorn]

i'm the president of mit

[hallow smelt]

i mainly answer the stats questions, but im trying to really understand trig

[plucky marlin]

[swift thorn]

i rejected your app personally b/c i'm evil

[drowsy spoke]

president of mit? o_O

[tranquil wigeon]

How did that mistake happen

I'm joking

[hallow smelt]

is there a name for the fact you can algebra the trig functions into infinity

them hiring me? lol

[drowsy spoke]

what do you mean by 'algebra them to infinity'?

[tranquil wigeon]

Not you, tubular

Maybe limits?

[plucky marlin]

"algebra them to infinity"

t!yt bruh sound effects

[loud cedar]

📽 | ** https://www.youtube.com/watch?v=9MJ-RuNYILo **

[hallow smelt]

if sin(x)csc(x) = 1 and sec(x)cos(x) =1 then sin(x)csc(x)=sec(x)cos(x) and then sin(x)*csc(x)/sec(x)=Cos(x) i can keep doing that

[tranquil wigeon]

LMAO

sry caps

[hallow smelt]

im assuming i can do that to infinity

[plucky marlin]

wat do u think infinity means ?

[tranquil wigeon]

I love how mysterious you are @plucky marlin

[plucky marlin]

wat ?

[tranquil wigeon]

Just shooting these questions

[hallow smelt]

replace the cos in sin(x)*csc(x)=sec(x)*cos(x) with sin(x)*csc(x)/sec(x) and then solve for sin(x)

[tranquil wigeon]

One after another

[drowsy spoke]

m8

[hallow smelt]

plug in the solution then refill the cos you get from that

[drowsy spoke]

you're just squaring everything

[hallow smelt]

and so on replacing strings with other strins

strings*

[drowsy spoke]

the third step is the equation in the first step squared

[hallow smelt]

[sin(x)*csc(x)]^2=[sec(x)cos(x)]^2 = sin(x)*csc(x)=sec(x)*sin(x)*csc(x)/sec(x)

is that what you are saying?

[drowsy spoke]

sin(x)csc(x) = 1
sec(x)cos(x) =1

sin(x)csc(x)=sec(x)cos(x)
sin^2(x)csc^2(x) = 1

[hallow smelt]

i dont understand how you get sin^2(x)*csc^2(x)=1

[drowsy spoke]

sin(x)csc(x)=sec(x)cos(x)
multiplied both sides by sin(x)csc(x), or if it makes sense I took the denominator of the fraction on the RHS to the left

[hallow smelt]

you can do that because we established it equals 1 and multiplying by 1 is ok

right?

[drowsy spoke]

,$ \sin x \csc x = \sec x \cos x\
\sin x \csc x = \frac{1}{\sin x \csc x}\
\sin x \csc x (\sin x \csc x )= \sin x \csc x (\frac{1}{\sin x \csc x})\
(\sin x \csc x )^2 = 1\
\sin^2 x \csc^2 x = 1\

[somber coyote]

Kelsier:

[hallow smelt]

oh wait

[drowsy spoke]

i think i made a mistake

[hallow smelt]

i actually started writing it out on paper

turns out im just dividing by one over and over hahaa

[drowsy spoke]

-_-

nvm

i made a mistake

seem to be making a lot of those recently

[hallow smelt]

do you see what i meant though?

[glad falcon]

Ok so i have practised alot for my Maths Coordinate Geometry Internal
and have nearly finished all the resources i had
i found out that the hardest question on the test is a word problem related to perpendicular stuff.
what should i do

[glad falcon]

Help

[twin prawn]

@glad falcon what do you need help with

Funniest thing I've written all day

[glad falcon]

S

So I calculate d all the distances

But it says each square is 10k so do I multiple by 10k? @twin prawn

[twin prawn]

No, each unit square area is 10k, so each unit length is only the square root of that, so 100

You need to multiply by 100

[glad falcon]

Ohhhh yes .

I need to keep that in mind during the exams thanks ma dude

@twin prawn

[twin prawn]

Np 🍮

[slate phoenix]

Problem:
A and B are antipodal points on a circle.
C and D are two other points on the circle.
Prove that the projections of AC and BD on CD are equal.

[young echo]

Is it possible to find/estimate the angles of each vectors, where two rotating vectors will intersect each other, given the vectors and rotation axis?

I'm currently using the rotation formula, https://en.wikipedia.org/wiki/Rodrigues'_rotation_formula

[twin prawn]

@slate phoenix What have you tried?

[zinc grail]

how would we prove the triangles are similar