geometry-and-trigonometry

[hoary prism]

it follows from how we define distance in R^n

idk if you need all of this even but like if you want some technical way to view it

[faint pasture]

I read the math stack exchange page they didnt list this as an axiom? (Their proof was also the same as yours)

[hoary prism]

which page?

i can take a look

[faint pasture]

I definitely dont lmao but curiosity

[faint pasture]

Lemme find it

[hoary prism]

by need i meant like, to convince yourself that this is logically consistent

curiosity is always good

[faint pasture]

https://math.stackexchange.com/questions/350383/formal-proof-that-area-of-a-rectangle-is-ab

[hoary prism]

first answer?

[faint pasture]

No shit intuitively its obviously true and since ive been using it for like 9 years already itll be hard to believe its false but i still kinda wanted a proof

[worn stag]

I smell measure theory

[hoary prism]

i mean here what im assuming is like, we are taking the properties of real numbers for granted

cause that's what it boils down to

[faint pasture]

They ended up having to assume integer too

[hoary prism]

going further than this as catgod said would probably have us talking about measure theory

[hoary prism]

yea i dont think it has anything to do with integers honestly

like

if we accept that we can get from one real number to another by some scaling factor

then i dont see why the distinction of integer/non integer is relevant

[faint pasture]

💀 uh i definitely cant do that rn maybe ill revisit this when i can

[hoary prism]

LMAO yeah i mean its cool stuff

i just meant like

its hard to reduce it past what i said without it idk

[faint pasture]

I mean that kinda assumes scaling factor doesnt break shit

Which we dont exactly know rn

But ye

I get what you mean

Thanks

[hoary prism]

np

[dawn flame]

Is anyone around to help with some math understanding?

[obsidian harness]

just ask!

[dawn flame]

My friend is trying to understand inverse trig functions

the problem he is working through is sin(arccos(1-x)

He needs to know how to do inverse trig functions when there is a function involved (such as 1-x). He needs to understand how to solve them when its not just something like sin(arccos(1))

Can anyone explain how to do problems like that?

[obsidian harness]

well if your friend can find sin(arccos(u))

just sub in u = 1 - x

[dawn flame]

Is it correct that the Pythagorean identity needs to be used?

[obsidian harness]

yeah

[graceful talon]

the fn and its inverse must match

[trail tendon]

fn?

[graceful talon]

function

[trail tendon]

oh

wdym it has to match?

[graceful talon]

the inner function must be the inverse of the outer function

[trail tendon]

in order for you to get the original value yea

but you can simplify without them matching

with trig functions

[graceful talon]

i see

[flint ingot]

Skibidi

[modest hull]

skibidi pop

[graceful talon]

skibidi is just scoobydoo in disguise. there's nothing new under the sun

[golden crag]

toilet

[finite wolf]

is a right triangle with a hypotenuse of 3, and adjacent leg of 1, and an opposite leg of 2.8284271247461903 possible?

[golden crag]

,w 2sqrt(2)

[somber coyote]

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[golden crag]

yes

,w 1^2 + (2.8284271247461903)^2

[somber coyote]

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[golden crag]

if you're not being highly concerned about sigfigs then it is possible

,w arcsin(1/3) in deg

[somber coyote]

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[proper ivy]

Hi i need help with number 1

[fathom ocean]

!status

[lime crown]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[upbeat jungle]

My test is next period

[obsidian harness]

the shape is not convex: you can draw a line that passes through two points in the polygon which also passes outside

[proper ivy]

Its asking for a reason why its not a polygon , not whether it is concave

[obsidian harness]

a polygon has to be convex by definition

I mean a simple polygon

[hard smelt]

Can someone explain why this is false? Wouldn't the left side of the equation basically be (-1) * (2) * (sin(x)) * (-1) * (3) * (cos(x)) and then I can move those values around however I need to simplify?

[silent plank]

its false because you are introducing multiplication when that isn't present

-a - b isn't the product of -a and -b

@hard smelt ^

-1 - 2 isn't 2

[rare trout]

hi

[hard smelt]

Are you saying that cos(x)sin(x) is different than cos(x) * sin(x)

[silent plank]

no

i'm saying that
$$-2\sin(4\deg) - 3\cos(4\deg)$$
is different to
$$-2\sin(4\deg) \red{\times} - 3\cos(4\deg)$$

[somber coyote]

ℝαμOmeganato5

[hard smelt]

oh duh. idk how that didnt register in my brain

[flint ingot]

How to remember trig formulas

[fathom ocean]

by solving as many questions as you can

[flint marlin]

Can anyone get me what do we get if we use sterwarts theorem to find angle bisector length?

I cant simply it

Please Help

[upper karma]

Be born smart.

That's the best advice anyone can give.

[graceful talon]

nah

[graceful talon]

look at the pinned chart

and u make sense of it

there's a reason for their grouping

u could start with the Pythagorean identity

as an example

[flint ingot]

So I should crawl back in the womb

[flint ingot]

Ok thanks

[upper karma]

It's better not remember them and get better at deriving them instantly

In your head

A more valuable skill

[graceful talon]

do u know how the other Pythagorean identities were derived?

[flint ingot]

What are the other ones

[upper karma]

No you are already contaminated by this world.

[graceful talon]

,,1+\tan^2\theta=\sec^2\theta\
1+\cot^2\theta=\csc^2\theta

[flint ingot]

Ok

[somber coyote]

0_א

[flint ingot]

No idk those

[graceful talon]

but do you know
$$\sin^2\theta+\cos^2\theta=1$$
?

[somber coyote]

0_א

[flint ingot]

Yeah

That’s just Pythagoras theorem basically

[graceful talon]

you just divide it by $\sin^2\theta$ or $\cos^2\theta$

[somber coyote]

0_א

[flint ingot]

Oh ok

Oh ok

[graceful talon]

if u divide it by sin² u get the first one. if the other one, the 2nd

[flint ingot]

Yeah

Thx

[graceful talon]

yea, as zaid said, it's better if u know how to derive instead

it also shows how much u understand about them

which is good if u do

[flint ingot]

Ok

[graceful talon]

@flint ingot an advice, u should know the sum of product identity

bc most other identities derive from it

[flint ingot]

I just want to know them in case one question need them

[flint ingot]

What’s this

[graceful talon]

,,\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta

[somber coyote]

0_א

[graceful talon]

and

[flint ingot]

Oh I know that one

And for cos a+b

[graceful talon]

yeyea

[flint ingot]

It’s sinasinb + sinbsina

Ok

[graceful talon]

no,

coscos minusplus sinsin

[flint ingot]

Yeah

[upper karma]

Check if what you say makes sense.

[flint ingot]

It does

[upper karma]

By giving values to stuff

[flint ingot]

??

[graceful talon]

,,\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta

[somber coyote]

0_א

[flint ingot]

Ok

[graceful talon]

my eyes are blurry I can't see the ± well

i see it as double +

[flint ingot]

[graceful talon]

what's with those emojis

[upper karma]

is a genius emoji

[upper karma]

Bro is high on trig

[flint ingot]

Think thonk thinkies thinkeusrein

Thinkestein

[graceful talon]

@flint ingot do u know the double angle formula?

[flint ingot]

[flint ingot]

What

Like the circle theoermen

[graceful talon]

sin2x=2sinxcosx?

[flint ingot]

No

But yes now

Thx

[graceful talon]

it's derived from sincos+cossin

[flint ingot]

Ok

[graceful talon]

u can solve it urself to see how it derived

try it. it's fun

[flint ingot]

Ok

[graceful talon]

sin(a+a)

sin(a+a)=sin2a

[flint ingot]

Yeah

[graceful talon]

yea

[flint ingot]

[graceful talon]

if u practice deriving them, soon u will memorize them

[flint ingot]

[golden crag]

sus

[graceful talon]

meanwhile the—i forgot what's it called but

[somber coyote]

@golden crag

[graceful talon]

,,\sin^2x=\frac{1-\cos2x}2

[upper karma]

Wrong

[somber coyote]

0_א

[golden crag]

that's just cos double angle rearranged right

[upper karma]

Yep

[graceful talon]

correct now?

ok

[graceful talon]

yup. nice

[upper karma]

Quick bruder

[graceful talon]

i think it's a better definition. bc i always use Pythagorean and double angle for it

when i derive it

[upper karma]

Trig is so boring tbh

[graceful talon]

idk what im saying at this point

[golden crag]

cos2x has these identities

[graceful talon]

nani

oh wai

[golden crag]

lmfao

[graceful talon]

i thought the tan one is the sum of products for tan

idk the tan one

i don't have to know them all

[golden crag]

good

[graceful talon]

last one yea

[upper karma]

Pretty easy to derive

[graceful talon]

yup

[upper karma]

In yer head

[golden crag]

these were messy to remembr

[graceful talon]

,,e^{ix}=\cos x +i\sin x

[upper karma]

The last one

I hate it

[golden crag]

yeah

exceptions in math 🤡

[somber coyote]

0_א

[graceful talon]

@upper karma is this also boring?

[hard smelt]

trig identities are killing me. I'm trying to memorize them

[upper karma]

Not really but it's not pure trig

[graceful talon]

not familiar but looks derivable

[golden crag]

sucks i know

[graceful talon]

fair

[graceful talon]

learn to derive instead. start with Pythagorean and sum of of prods

[upper karma]

Do the same question again and again you'll get used to it.

[upper karma]

This was the only thing I liked in trig

[golden crag]

derive the sum identities too

trol

[graceful talon]

,,a^2=b^2+c^2-2bc\cos\alpha

[somber coyote]

0_א

[golden crag]

very deriveable

[graceful talon]

where is 1

[golden crag]

1?

[graceful talon]

let x = 1

[golden crag]

ig

[upper karma]

$r = \frac{{1}{∆}}$ oh fuck

[graceful talon]

,,\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dotsb

[golden crag]

taylor swif

[somber coyote]

0_א

[golden crag]

series

[graceful talon]

mclaurin

right

[somber coyote]

گل کاک
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[graceful talon]

isn't mclaurin taylor with x=0

[upper karma]

Ye

[graceful talon]

idk much abt series

[golden crag]

macdeez

[graceful talon]

i learned it very briefly

I'll get back on it later, after a while

[upper karma]

In uni?

[graceful talon]

no

[upper karma]

Oh I thought you were in uni.

[graceful talon]

first time i heard it the word mclaurin is from trisectrices of mclaurin

[graceful talon]

I was

[upper karma]

They really cool

Cool names too

,w trisectrix

[graceful talon]

,w trisectrices of mclaurin

[somber coyote]

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[upper karma]

Quadratix is pure genius

[graceful talon]

,w maclaurin trisectrix

[somber coyote]

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[upper karma]

,w quadratirx

[somber coyote]

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[graceful talon]

ee

[upper karma]

You an old single father?

[graceful talon]

oof i forgot im gonna watch something

[graceful talon]

i wish

but not single

2 is better than one

2 > 1

[upper karma]

4 is better than 2

[graceful talon]

i think im being misunderstood at this point but i gotta go

[upper karma]

See ya

[fickle citrus]

can any1 help me w my proofs for hon. geometry?

😓😓🙏

[keen robin]

JM=KL, as a parallelogram always has a pair of two equal and parallel sides.
JK=ML "__"

and MK is a common side

so they are both congruent by the SSS condition

[zealous pike]

Each point Z is a complex number and a vertex of the hexagon on a complex plane, which each intersects a point on the circle of radius r > 0

What is the distance of z_6 to the origin?

Yeah it is r

But

Why is it that

$r = \frac{z^2_{6}}{z_5}$

[somber coyote]

Pi, a future fluent jp speaker

[keen robin]

if it's along a complex plane, wouldn't the axis be real against imaginary numbers

[keen robin]

also, the angles for each vertex would be z1=re^0i, z2=re^i.(𝜋/3) etcetera. So the angle for vertex z6 would be z6=re^i.(5𝜋/3). Using euler's formula, e^i.(5𝜋/3) = 1, so z6=r . 1 which is equal to the radius of the circle (i think)

[graceful talon]

Given 3 points, there is a plane containing them.

[graceful talon]

,w euler's formula

[somber coyote]

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[graceful talon]

are u sure $e^{\frac{5\pi}3i}=1$?

[somber coyote]

0_א

[graceful talon]

squaring z_6 makes the angle the same as z_5

at the same time r is squared. so when u divide them, r remains while the angle cancels to 1

in complex numbers, raising it to a number multiplies the angle of the complex number to that number

[somber coyote]

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[graceful talon]

assuming you know $$a+bi=r(\cos\theta+i\sin\theta)=re^{i\theta}$$

[somber coyote]

0_א

[winged plinth]

Is it possible to solve 5b without implicit and parametric differentiation? Because the book is saying I'll learn them later.

:_( how to solve. I tried solving but go it wrong

[winged plinth]

<@&286206848099549185>

[cyan steppe]

I have tried it but you gotta do the calculations without the parametric form.

It is given that The normal is parallel to PQ: 3x - 2y + 12 = 0,

Thus it's slope will be the same as that of PQ which is 3/2.

Then the slope of tangent at that point A will be -2/3
(as it is perpendicular to the normal so the product of slopes must be -1).

From there you can write the equation of tangent at the point A as 2x + 3y + λ = 0
(where lamda is constant).

Then you gotta substitute either x or y from this equation in hyperbola xy = 48, to obtain a quadratic equation. And since it's the tangent, you put the discriminant of the equation to 0, to obtain 2 values of λ, with which you'll get 2 tangents, one in 1st quadrant and other in 3rd.

Then you gotta find the point of intersection of the two curves.

Dang that's long ass work.

I got the point in 3rd quadrant though, it is (-6√2, -4√2).

[winged plinth]

🥺thank you so much for explaining. It makes sense 😭

[cyan steppe]

The other one will be in the 1st quadrant (6√2, 4√2).

[winged plinth]

YES

Thx Ayush

[cyan steppe]

Welcome

[winged plinth]

Real I must have been in drugs when I tried solving it

💀I got 4, 18

[cyan steppe]

Dang they don't even satisfy the equation homie

Ayoo no problem it happens too often

[winged plinth]

Yeah : (

[winged plinth]

Is the question easier with implicit differentiation?

[cyan steppe]

Can you tell me what were u trying to do? Like using differentiation in what way?

Oh I see you were trying to equate the slopes and obtain the value of t?

[winged plinth]

I differentiated normally then found the gradient of the normal subbed points and tried solving it simultaneously 💀

( I'm on drugs)

[cyan steppe]

Wait lemme try using differentiation and parametric form. I guess it's easier than this long ass calculation

[winged plinth]

👀

[cyan steppe]

Bruh it's so short and easy with parametric form

[flint marlin]

Anyone got some good questions?

[obsidian harness]

differentiate $y = 48/x$ to get $dy/dx = -48/x^2$

so $-48/x^2 = -2/3$ or $x^2 = 72$ which means $(x, y) = (6 \sqrt{2}, 4 \sqrt{2}), (-6 \sqrt{2}, -4 \sqrt{2})$

[somber coyote]

south's secret twin brother

[obsidian harness]

it's actually not bad at all

[zealous pike]

Where can I see a demonstration for this identity?

[graceful talon]

they're just special cases of sum of product identities

,,\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta

[somber coyote]

0_א

[graceful talon]

^ first one is like that but operates on the same value

\begin{align*}
\sin(\theta+\theta)&=\sin\theta\cos\theta+\cos\theta\sin\theta\\
\sin(2\theta)&=2\sin\theta\cos\theta
\end{align*}

[somber coyote]

0_א

[graceful talon]

I leave to you the derivation of cos2x as an exercise

[zealous pike]

I still dont understand

From where comes that cos

[graceful talon]

@zealous pike u mean this?

im eepy

[graceful talon]

or the cos here

[zealous pike]

In both

[graceful talon]

ok start with this
$$\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$$

[somber coyote]

0_א

[graceful talon]

||the cos2x derives from there||

im too eepy to continue

[magic verge]

||\begin{align*}
\sin(\theta+\theta)&=\sin\theta\cos\theta+\cos\theta\sin\theta\
\sin(2\theta)&=2\sin\theta\cos\theta
\end{align*}||

[somber coyote]

MRrineew

[deft cedar]

hello can sb help mme with this task?

I've already made the sketch to this task

[upper karma]

Guys for these formulas, i just plug them in when it says "cos" "sin" or "tan" right?

[graceful talon]

do you mean plug the sin and cos to tan since tan is sin/cos ?

yea, you can

[graceful talon]

are AX and BY line segments?

[upper karma]

Thank you

[bleak holly]

Help please

[rich scroll]

In number 8, i think it can be verified by counting the triangles formed, but in number 11 it cannot be counted since the diagonals must come from one vertex therefore it cannot be counted. Am i correct?

[flint marlin]

Drop a perpendicular to BC from E and use Pythagoreans theorem and base area ratio theorem

[flint marlin]

What exactly do you mean by counting the no. Of triangles ?

For eleven sum of angles of n triangles would be 180n but we do not want to count the angle around the centre so we will subtract 360 hence its 180(n-2)

This is the proof for 12

11"

[upper karma]

Could sm1 help me with my geometry quiz

[limpid nest]

He just said "Which can be rewritten as..."

[upper karma]

The term "Origin power of 2" doesn't make any sense

[coarse patrol]

i did it it was 10

[maiden brook]

i know this is a bit basic but i cant exactly figure it out: suppose CB is a chord of circle A, and that chord EF is perpendicular to BC at D. how do you show precisely that ED is maximized when EF is a diameter?

[rich scroll]

Why do we have to subtract 360

[flint marlin]

We are counting the centre angle

Of all The triangles

Which we dont need

[mental yew]

Bros how do I prove this formula works?

[obsidian harness]

subtract 3 right triangles from the big rectangle

[real vine]

Hi how can you prove sin(x-y) and cos(x-y)?

[faint pasture]

just use complex numbers

[upper karma]

Sing "Brave sir Robin ran away..."

It will come to you automatically

i have this

How did we know that angle POQ is n delta ?

[obsidian harness]

the exterior angles are delta, 2 delta - delta, 3 delta - 2 delta.... = always delta

since the tangent is perpendicular to the radius

the angles of the triangles at O will also be delta

[obsidian harness]

technically it's like (omega t + delta) - omega and so on, but omega t always cancels

[upper karma]

can you mark the exterior angls

angles

[upper karma]

yes

those would be delta

[obsidian harness]

now make a right triangle with the black side (hypotenuse) and the points

there would be 90 - (90 - delta) = delta

cause tangent and radius are perpendicular

[upper karma]

oh but how would the black side be the tangent ?

[obsidian harness]

yes it's tangent to the circle

[upper karma]

this side would be inside the circle circumscribing it

so its elongation would also pass through the circle twice

[obsidian harness]

ah no, there, the tangent line would be the line S

[upper karma]

okay,
im not able to imagine it properly so i will just remember its true

yeah

[obsidian harness]

I should just say the line that is the continuation of the previous side

[upper karma]

i get it

so

n sides means n delta

nice

thanks

[upper karma]

can you also explain how we got angle OPS And OPQ

OPQ would be (180- ndelta )/2

OPS ?

[obsidian harness]

wait I think I explained it wrong
so the interior angle would be 180 - delta for all of them
by isosceles triangles, the base angles would be (180 - delta)/2

so 180 - (180 - delta)/2 * 2 = delta

there we go

[upper karma]

i thought that is what you said lol

[upper karma]

angle OPS?

[obsidian harness]

ahhh so by this logic, OPS would be (180 - delta)/2

and OPQ would be (180 - n delta)/2

[obsidian harness]

cause the interior angles are all equal

[upper karma]

OH!

i get it

90- delta / 2

nice

thanks!

[obsidian harness]

np!

[upper karma]

💀

using complex numbers was way easier

[mild patio]

how should i go on about proving that
${cos(\frac{x}{2})=\sqrt{\frac{1+cos(x)}{2}}}$

[somber coyote]

kazooha

[obsidian harness]

try rearranging for cos u in cos 2u = 2 cos^2 u - 1

after that sub in u = x/2

[obsidian harness]

also there should be a plus minus sign on the RHS

[mild patio]

okay thanks for the help

[obsidian harness]

(sqrt is positive by definition, that's why)

[mild patio]

yeah so i guess it depends on the interval in which x varies

[obsidian harness]

exactly basically whenever cos(x/2) is positive or 0

[heavy bobcat]

O is the centre of the circle. If AC = 28 cm, BC = 21 cm, angle BOD = 90 and angle BOC = 30, then find the area of the shaded region

[obsidian harness]

since AB is the diameter

what circle theorem can you use to find angle ACB?

[heavy bobcat]

oh! yeh forgot the theorems completely now I got will take it from here!, ACB = 90 (2 chords from the diameter to the circumference make 90) thank you :D

[obsidian harness]

np!

[jagged venture]

The frustum of the regular quadrilateral pyramid ABCDA'B'C'D' has AB = 7sqrt2 cm, A'B' = 4√2 cm, AA'=3sqrt10 cm. We denote 0 and O' the centers of the bases. Compute OM, where M is on OO', so that triangle C 'MC is isosceles from the point M

[weary plume]

Can someone walk me through the geometry here?

[dusk dock]

is this the right formula to get the area of a square using its diagonal sides?

(d^2)/2

i kind of forgor 👉👈

[dusk venture]

yes

[dusk dock]

alright thanks

[undone quest]

Just rember like this

Area of rhombus is d1d2/2

So can we say every square is a rhombus because it has same properties as rhombus it is just a special
Yes we can say that

So area of square is also d1d2/2

Here d1 or d2 are equal

So area goes like

d1²/2
Or d2²/2

[weary plume]

Can someone still prove it?

[undone quest]

Send me the figure dude

😦

[weary plume]

s

[ocean coral]

Could someone help me understand this problem? The solution is C but I can't figure it out with transformations and why the graph would end up like that.

[pseudo nebula]

I knew it @rancid veldt

[pseudo nebula]

any guesses?

[rancid veldt]

Ill come up with worked sol

[ocean coral]

ty

[rancid veldt]

@ocean coral done and dusted

[rancid veldt]

Finding the y intercept most of the time is a dead give away

[ocean coral]

tysm that helped a lot

[rancid veldt]

Glad I could help

[rancid veldt]

Are you by any chance an aussie? maybe year 11?

[rich scroll]

Help numbers 12,13,14 i have no progress

[upper karma]

someone answer with clear explanation

<@&286206848099549185>

do you know what an altitude is?

Yes

define

Altitude is the perpendicular drawn from a vertex to the opposite side of any triangle

Good

now grab a pen and a paper

Ok

draw a rough triangle

K

draw altitudes from two of the vertex

[obsidian harness]

hint: area = 1/2 * base * height

if the two heights are equal, what must also be equal

[upper karma]

.

in such a way that the altitudes are equal in distance

[upper karma]

Wow

Thanks ❤️ bruh

Thanks Tyler u too

Welcome

whats the answer?

Isosceles

Good

What stopping it to be an equilateral triangle 🤔

think

[obsidian harness]

you've shown that 2 of the sides must be equal

but then the 3rd side could be anything

[upper karma]

yah

[obsidian harness]

like it can be equilateral ofc

it just doesn't have to be

[upper karma]

dude let him think, dont spoon feed him

Yeah but if isosceles wasn't mentioned in options i would click on equilateral

yup

that would be the correct answer

if isoceles is not in the options

[jagged venture]

One angle

[zealous pike]

If I'm given that cos x = 4/5 and sin y = 5/13, can I use the identity $1 + \tan{x}^2 = \sec{x}^2$ to find sin x?

The question is to actually find tan(x + y), but I wonder about this

[somber coyote]

Pi, a future fluent jp speaker

[obsidian harness]

you don't need to find sin x

just find tan x and tan y

and actually you can draw a triangle with adj = 4 and hyp = 5 to find tan x

and similarly for tan y

[terse parcel]

any1 know the answer?

[zealous pike]

Just say me if it works or not

😭

But does it works

[obsidian harness]

actually yeah I see, you're finding sin(x + y) and cos(x + y) first

it works

so you just need to find cos y and sin x

you already have sin y and cos x from the question

[zealous pike]

For me it is easier to first find sin x and cos y to find tan(x) and tan(y)

[obsidian harness]

okok

[placid pilot]

Can anyone explain me this and how can i get the answer

[obsidian harness]

what circle theorem does number 1 look like?

you can search up pictures of circle theorems online

https://images-ext-1.discordapp.net/external/TUqi26LoFqbg1awE2l4LEuWIoDYKa5e-SPbWhJNdpic/https/thirdspacelearning.com/wp-content/uploads/2022/05/circle-theorems-poster-722x1024.png?format=webp&quality=lossless&width=351&height=498

[placid pilot]

Ohh tyy

[obsidian harness]

yeah

no worries

[upper karma]

Really needed something like this thanks Brody @obsidian harness

[upper karma]

This i mean

[obsidian harness]

no problemo!

[regal hawk]

7/24

[somber coyote]

greenballticker

[regal hawk]

or simply 22.5

no 3 is obviously 45

i mean 90

because it intercept semi circle arc

no 2 is 27

same arc inscribed angle have equal measures

ez

[weak plaza]

what did I do wrong here?

[trail tendon]

cos^2(4x) is a trigonometric function with a power greater than 1

[weak plaza]

so what would be the answer

[trail tendon]

i think the last term should be positive and you need to get rid of the square but otherwise it looks correct

3/4 - cos(2x) + 1/4 cos(4x) I believe

[weak plaza]

https://tenor.com/view/ace-ventura-pet-detective-jim-carrey-comedy-duh-gif-11333313905875935459

I was almost there

I got my steps mixed up

ill go try a similar question

https://tenor.com/view/exterminador-do-futuro-gif-20118743

[trail tendon]

😂

[weak plaza]

Alright I figured it out

https://tenor.com/view/tom-and-jerry-jerry-the-mouse-jerry-shake-hands-handshake-gif-17827738

[trail tendon]

😎

[runic pasture]

cnidarian emoji

[trail tendon]

bro people claiming emojis now? :l

[runic pasture]

I know yours

[trail tendon]

💀

[runic pasture]

☠️

is yours

[trail tendon]

☠️

[runic pasture]

Fair

[trail tendon]

💀

[paper token]

hey anyone need some help for geometry

i got oyu

[hoary prism]

yeah i got a few questions about etale cohomology

[junior mesa]

Hey is anyone on right now?

I need help with trig compositions

[hoary prism]

!da2a

[lime crown]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[junior mesa]

very lost

[junior mesa]

whats that

[hoary prism]

like if you have a question just ask the question

[junior mesa]

ohh

[hoary prism]

rather than asking if you can ask

yea

[junior mesa]

I'm lost on this problem

How do I solve for i t

theres more of them but I just don't understand how to solve for it

[slender flint]

Is 5π/3 in sin's principal branch?

[junior mesa]

uhhh

I actually

don't think we learned about

the principal branches yet

[trail tendon]

arcsin(sin(x)) = x
sin(arcsin(x)) = x
by definition of inverse functions

[junior mesa]

hmm ok

[hoary prism]

well

nah not here there's a domain restriction

the domain of arcsin is [-1,1] so youll have to shift the 5pi/3 back by a period to get the answer

[junior mesa]

hmm ok

wait should I use

a graphic calculator for this question

[hoary prism]

no need

[junior mesa]

graphing*

oh

[hoary prism]

what's the period of sine

[junior mesa]

like

the full Circle

[hoary prism]

which is

[junior mesa]

2 pi

[hoary prism]

yeah

so

sin(5pi/3 + 2pi*k) are all gonna be the same for any integer k

[trail tendon]

oh yeah mb 😂

[hoary prism]

you need to find the shift that lands you in the domain of arcsin which is [-1,1]

[junior mesa]

hmm ok

[hoary prism]

so with that in mind what do you think

[junior mesa]

lets see

let me put this into my calculator

[hoary prism]

put what into ur calculator

[junior mesa]

the uhh

the formula

[hoary prism]

nah just like

we know 5pi/3 > 1, how many factors of 2pi do we need to shift it back by

to fall within [-1,1]

[junior mesa]

hmm

+2?

[hoary prism]

no thats going in the wrong direction

[junior mesa]

oh

mb

minus 2

[hoary prism]

minus 2pi

[junior mesa]

so we got

5pi/3 - 2pi

[hoary prism]

yeah

arcsin(sin(5pi/3)) = arcsin(sin(5pi/3 - 2pi)) = 5pi/3 - 2pi

[junior mesa]

so

would it be

5pi / 3 - 2pi / 1

then Keep change flip

cross out the pi's

[hoary prism]

what

[junior mesa]

I'm so lost

sorry

[hoary prism]

like to simplify it?

[junior mesa]

yea

[hoary prism]

5pi/3 - 2pi = 5pi/3 - 6pi/3 = -pi/3

[junior mesa]

but like

[hoary prism]

thats the simplest form

[junior mesa]

actually

that makes a lot more sense

I have a question

[hoary prism]

yeah i initially thought so too when i saw the question then i realized

lmfao

[junior mesa]

is it like

[hoary prism]

whats up

[junior mesa]

How can I visually tell if theres a domain restriction

[hoary prism]

well

[junior mesa]

like how do I know

[hoary prism]

yk what the graph of sin looks like right

[junior mesa]

yea

Unit circle right?

[hoary prism]

no i mean like a wave

sin wave

[junior mesa]

oh

that

yea

[hoary prism]

yeah so

the reason there's a domain restriction on the inverse sine is because if we try to send the outputs of that wave back to the inputs we don't have a well defined function because each sine value has infinitely many inputs that map to it

[junior mesa]

so is it like

everytime theres a inverse sign on it

or on the front

[hoary prism]

no not always for example like the inverse of sin and cos will look like this due to the periodic behavior but the inverse of a straight line or something will just be the line reflected across y=x so we can say the inverses cancel each other out and youre just left with the original value for something like that

for all inputs

another example of a domain restriction ur probably familiar with is the square root, so when we say f(x) = x^2 this sends x and -x to x^2 but if we try to make an inverse for it we need to choose a unique value to send x^2 to, so that's why we make it the positive square root

therefore if f(x) = x^2, then f^-1(f(x)) = |x| rather than just x

[junior mesa]

ah ok

[hoary prism]

so like if we had f^-1(f(-2)) or something this would be 2

yeah

so its the same type of idea for this as well

idk if that answers your question but

[junior mesa]

a little bit

[hoary prism]

thats the conceptual idea at least

[junior mesa]

so basically

one of my biggest problems

in math is just

decyphering

what the question is asking

[hoary prism]

yeah tbh my biggest tip for that is just practicing

which like

is maybe not the most helpful advice cause it sounds obvious but there isn't really a shortcut

most people would tell you the same

[junior mesa]

ah ok

just to be sure

this is asking for

this value right

like the value

inside of both sin's

the X

[hoary prism]

well depending on the question it'll either be equal to that inside value or shifted by some period like we did for the problem you showed me

[junior mesa]

ah ok

[hoary prism]

like if we had started with sin^-1(sin(-pi/3)) then the answer would just be the inside value yeah but in our case it was different

yea

[junior mesa]

could u help me with B

for this problem

[hoary prism]

well try and figure out what it would be from what we did, like you need to see what the domain restriction you need to apply to tan is for arctan to be a perfect inverse of it and then adjust the inside value if need be because tan is periodic

look at the graphs for each, try to trace through what we talked about above and see if you can figure that out

[junior mesa]

alright yeah

ima try that

I ended up getting

the same answer

for some reason I feel like its wrong

[hoary prism]

No ur right lmao

This one is also -pi/3

[junior mesa]

Thank god!

I'm getting better at it

[hoary prism]

Nice

[junior mesa]

How do I do

the next one

It has no pi in it

or fraction

[hoary prism]

Well

What we did didn’t depend on any of that

[junior mesa]

your right

[hoary prism]

Yeah it still will be the same for plus or minus any multiple of 2pi

Regardless

[junior mesa]

wait

what do you mean by that

[hoary prism]

Like the period of cos is 2pi

[junior mesa]

oh

[hoary prism]

So cos(a+ 2pi*k) = cos(a) for any integer k

It doesn’t matter what a is

Like

It doesn’t have to have pi in it itself

[junior mesa]

oh

so this one I technically just have

[hoary prism]

Granted that if we just have some rando integer as the input it would come out to be some weird value that nobody really uses but it’s still defined

[junior mesa]

oh

this one I got

archcos(cos(4))

should I turn the 4 into

a 4/1

and then

hmm

[hoary prism]

Like

[junior mesa]

the part I'm lost on is like

what can I even subtract or add to this one

the shift

[weak plaza]

I keep getting the answers wrong
Can someone tell me this one is so I can back track what im doing wrong

[hoary prism]

Actually you can still find it but it’s gonna need a little bit of trickery cause for the domain restriction here we need the input to land in [0,pi], and there isn’t any integer multiple of 2pi such that 4 + 2pi*k is within that (cause 4>pi and the smallest shift backwards we can do is 4-2pi which is negative), but we can to make use of the fact that cos(x-pi) = -cos(x) cause going half of a period back just lands you on the correct absolute value but with the opposite sign

[junior mesa]

so would we just

go back 1 instead of 2?

[hoary prism]

No that won’t be the exact answer but that’s like the insight here you can use to get to it

Youd have to do it like

[junior mesa]

ah okay

[hoary prism]

pi-(4-pi))

So 2pi-4

Like cos(pi-x) and cos(x-pi) are both -cos(x) so to get back to something you can input into cos(x) you need to do that process of finding the right answer with the opposite sign by shifting with pi twice

And making it fall into the domain of arccos

Idk it’s hard to explain in words but if you look at the graphs it’ll make sense I think

[junior mesa]

hopefully

let me check the graph

to be honest

I'm still lost

[swift vale]

im gonna be honest im so lost can someone please help me.

[swift vale]

<@&286206848099549185>

[honest topaz]

tbh I have forgotten these

there used to be congruence and similarity criterions

https://www.geeksforgeeks.org/congruence-of-triangles/

https://www.cuemath.com/geometry/similar-triangles/

I'd wager you would need to exploit one of these criterions

[upper karma]

you aren't supposed to ping them here

[rugged flint]

How to calculate the length of bc using the cosine formula? H is 20 cm.

[ancient kindle]

you know what the formula is, right?

[rugged flint]

The adjacent leg/hypotenuse, right?

[ancient kindle]

okay wait sorry what cosine formula do you mean

like... the cos(x) = adjacent/hypotenuse?

if it is that one, give me a moment

wait pardon my stupidity how is h 20?

[rugged flint]

Wait, I prolly don't even know what the cosine formula is tbh

And that's the problem

[rugged flint]

I just calculated sine 56,4 degrees and then multiplied the result with 24

[ancient kindle]

ohh my bad i thought 18 was this length

okay so

[ancient kindle]

you know that H is 20, and AB (the hypotenuse) is 24. what does that make the other leg of the larger right triangle?

the red one im showing here^^

[rugged flint]

Yeah, that one I also calculated, but I used pythagoras' theorem, is that cheating?

[ancient kindle]

im not really sure exactly what your teacher or textbook want

[rugged flint]

13,26 approximately

Me neither ngl

[ancient kindle]

do you know the c^2 = a^2 + b^2 -2abcos(C) formula?

if so, i think that one is probably what they wanted, but im not sure

[rugged flint]

Nope

What's the name of it?

[ancient kindle]

damn

[rugged flint]

Sorry, me very stupid

[ancient kindle]

the law of cosines

[rugged flint]

Yeah, I'm gonna watch a video on it lol

[ancient kindle]

nonono

wait

dont 😭 it might be too advanced

[rugged flint]

I thought it was something different

Okay, what's my alternative then

[ancient kindle]

well

okay if it's not too much to ask do you know what level of math you're at

so i can properly do this bc

i dont want to accidentally give you something too advanced 😭

[rugged flint]

So the textbook obviously wants me to use the cosine formula

So it seems rather logical that I learn it

[ancient kindle]

do they have the explanation on some other page

[rugged flint]

Yeah, I kinda skipped that chapter

Anyways, I did learn the sine formula right now, which made sense

[ancient kindle]

the sin(a)/a = sin(b)/b = sin(c)/c?

or

is it

sin = opposite/hypotenuse

[rugged flint]

The former

[ancient kindle]

okay yeah if you learned that one you should definitely learn the cosine one

[rugged flint]

Okay, I'm gonna try to learn it and then practice on some question

I'll be back

[ancient kindle]

aigh

[rugged flint]

Thanks for the guidance 🌸

[ancient kindle]

np!

[austere mortar]

how could I get the values of any given angles in trig functions such as cos(36°) or tg(18°)?

[pseudo raft]

If you just want an approximation, you can probably just use the first couple terms of their macularin series

[austere mortar]

Im rather looking for fractions with roots and so on

how could I obtain them?

[faint pasture]

Tan 18 is ugly

For cos 36 nd sin18

Use sin 90 = 1 = sin(5*18)

[austere mortar]

will try, thank you

[drifting cipher]

@light dock

[light dock]

just ping the whole mod role next time

but thanks for letting me know.

[drifting cipher]

Me afraid to

[light dock]

we get them all the time :P no need to worry.

[ivory flax]

you can use herons

it would be much simple

[graceful talon]

how about law of cosines

[ivory flax]

similar triangles

just dm me if you want solution

[sand patrol]

Anyone can help. Show PQ^2-RS^2=4OY^2-4OX^2

[upper karma]

use the property the line from centre to midpoint of chord is perpendicular to the chor

chord

[minor marten]

how much is 2+2

i need help solving that

[upper karma]

what do you think it is

[minor marten]

hmm

let me think

[upper karma]

with pythogoreas theorem .

[minor marten]

i think i skipped a math class

that wasnt on the adding topic

but i got it

it was 4

[upper karma]

nice

its right

[royal citrus]

this took me so long

so happ

[upper karma]

Nvm

It's broken

[dark sparrow]

anyone know the anwsers?

[trail tendon]

do you know how to start?

[dark sparrow]

no I dont i just need the anwser though

[hoary prism]

all angles in a triangle sum to 180

and once u figure out the last angle the sides are gonna be easy cause its a common type of triangle

but in general you can do it with law of sines even if it were a weird triangle

[dark sparrow]

I know b is 45 degrees

but beyond that know clue just trying to get this exam turned in

[hoary prism]

ok so you have a 45-45-90 right triangle

how are the sides related in those

[dark sparrow]

s=4 right?

[hoary prism]

yeah

and then it's a right triangle so to find the hypotenuse what do u do

[dark sparrow]

is it 2.828?

[hoary prism]

no

use the pythagorean theorem

[dark sparrow]

can you please just tell me

[hoary prism]

Look up the Pythagorean theorem and apply it

[dark sparrow]

already failed the question

[brazen ridge]

yo

[upper karma]

Yes

And h is 4 underoot2

[upper karma]

proove that angle r is 90 degree

<@&286206848099549185>

[flint marlin]

Use mid point theorem

[flint marlin]

Dont pin

[upper karma]

isnt this server for helping

[flint marlin]

Yes

But

There are rules

[upper karma]

can u mentions me the rules so i can read them

[flint marlin]

"With rules , we live in a world with animals" - Winston Churchill

Idr the exact phrase lol

[flint marlin]

Just check #rules man

[flint marlin]

Anyone got some ideas regarding THIS?

[upper karma]

if someone comes here then send me a proof of it

[upper karma]

Ok i dont need it now i have figured it out by myself

[dire furnace]

Im sorry but i cant read like half of this

[flint marlin]

I solved it though

[dire furnace]

ABC is an acute angle triangle such that angle A= 60°

Thats as far as i could understand it

[dire furnace]

Oh sweet

[flint marlin]

But umm isnt it too obvious to understand what it means ? (If you do alot of Geometry then yes)

[dire furnace]

No like i cant read the writing

[flint marlin]

H is orthocente and I is the incentre

Also C > B

Find AHI/B

[dire furnace]

Angle HI/B?

[flint marlin]

Angle AHI / Angle B

[dire furnace]

Or angle AHI/ angle B

Oooooh okay

[elder halo]

how do i find x

[frank plinth]

use pythagore

[wooden island]

It can be anything

Since it will change all the values

[frank plinth]

it is way easier to use pythagore

[wooden island]

And what will u get by doing that? 0 = 0

[frank plinth]

the length of x lol

[wooden island]

Pls show how

[frank plinth]

we have the hypotenus

then

[obsidian harness]

any 30-60-90 triangle will have side lengths in that ratio

[wooden island]

U won't get anything

[obsidian harness]

the x tells you the ratio of the sides is 1 : sqrt3 : 2

[wooden island]

x can take any positive value

[frank plinth]

you will just wait

[wooden island]

4x² = x²+3x²
4x² = 4x²
Then what??

It won't help

[frank plinth]

this

[wooden island]

Huh

[frank plinth]

then solve idc

[wooden island]

= x²??

[frank plinth]

idk wait

[wooden island]

What ru doing?? 😭

[frank plinth]

whats wrong with it???

[wooden island]

4x²-3x²=x²

[elder halo]

isnt it supposed to be x²

[frank plinth]

Ah yes srry

yes yes

[wooden island]

But What r u going to do with that

I'm telling u X can take any positive value

[frank plinth]

I learned in belgium

the national olympiad

[wooden island]

So maths in Belgium is different?

[frank plinth]

no???

wtf

[wooden island]

So wtf were u doing ??

[frank plinth]

it was an error

[wooden island]

Ok

Nvm

[frank plinth]

so

[wooden island]

But still Pythagoras won't help

[frank plinth]

yeah I just realised lmao

this is so awkward

[wooden island]

Let's just act like nothing happened

[frank plinth]

yeah

imma die lol

[wooden island]

Everyone make silly mistake

[frank plinth]

yeah

[wooden island]

Just saw someone doing 12 x 0 = 12

So don't worry u r not alone

[frank plinth]

LMAO

where r u from?

[wooden island]

India

[frank plinth]

yo its so cool

[wooden island]

What is so cool?

[frank plinth]

India

[wooden island]

Oh yess

Too much competition

Too many kids

[frank plinth]

yeah

here in Belgium math is so fucking hard

[obsidian harness]

India would be harder by a mile I feel

[frank plinth]

idk

[obsidian harness]

though yeah I heard someone from Belgium say they found maths hard too

[frank plinth]

yeah it is

and the weather

[faint pasture]

Hell nah

More or less every other country has a better curriclum

[frank plinth]

AHAHHAHHAH

[dire furnace]

Do any of yall need any help?

[frank plinth]

no thanks

[faint pasture]

With life sure

!da2h

[frank plinth]

lol

[faint pasture]

We really need a dont ask to help page too

[frank plinth]

yeah fr

[wooden island]

Tru

[faint pasture]

Also coming back to that people in here are just good rant-ers so that

[wooden island]

So hardde

[frank plinth]

but

[wooden island]

I'm preparing for engeneering

[frank plinth]

yo me too

[faint pasture]

Jee =/= indian avg guy math

[wooden island]

With 20 lacks competitors

[frank plinth]

where r u from

[faint pasture]

Id rather not

[frank plinth]

why