geometry-and-trigonometry

[cyan violet]

@storm rover

[storm rover]

Well the video still is there

Idk how to do it tbh

[cyan violet]

Oh, peculiar.

It's on the interenet?

[storm rover]

Yeah

[cyan violet]

Odd.

[storm rover]

It’s a rlly weird solution imo

[cyan violet]

You know how I solved it.

Pressed on the graph.

Examined the asy.

Found the angle.

Lol.

I want to actually solve it this time.

Without.

Cheating.

Essentially, that's what that was.

[storm rover]

Wym presses on the graph

[cyan violet]

One second.

I solved it by pressing the graph and examing the asy.

[storm rover]

Cool ig

How old r u

[cyan violet]

Meh, AoPS' solution was way simpler and normaler.

6th grade.

[storm rover]

Lol

[cyan violet]

Going into 7th.

[storm rover]

I’m in 7th

I hate geometry

[cyan violet]

Wow.

[storm rover]

It’s to shaded solution wise imo

[cyan violet]

Hmk.

Are you going into 7th.

Or you're actually in 7th.

[storm rover]

No 8th

[cyan violet]

Oh, I see.

[storm rover]

Going into

[cyan violet]

I'm like one chapter away from learning rudimentary trig.

Like basic laws od sins and cos.

[storm rover]

Who would have the idea of drawings triangle from the center

To the smaller square

I don’t like it lmao

[cyan violet]

Meh.

better than the video solution.

To be frank.

[storm rover]

Yeah ig

[cyan violet]

Mhm.

[storm rover]

Lol what are these moms and Mehs lol

Moms*

Mhms *

Lol

[cyan violet]

What?

This stumped a stanford math major???

What???????

It's not even that hard.

[storm rover]

It’s not hard when you know the solution lmao

It’s a weird way to get there

like I said

[cyan violet]

True.

[storm rover]

How would have the idea of drawing a radius of the circle that intersects the smaller squares vertex

[cyan violet]

You'd have to make a weird approach to derive the answer.

True.

Oh.

[storm rover]

That’s why I don’t like geometry

[cyan violet]

@storm rover

That's where it comes from.

[storm rover]

Oh nice

I did mathcounts

My team sucked though lmao

[cyan violet]

You know what's hard.

Geo proofs.

🤦

Hell.

[storm rover]

Depends

[cyan violet]

Mostly.

[storm rover]

On the proof

[cyan violet]

I think I proved power of the point though in very little time.

From what I recall.

Or it was another.

[storm rover]

Like my favorite thing to do in class is to prove

[cyan violet]

There was this proof that I proved in like one minute.

Yeah.

[storm rover]

Elementary theorems

Of shapes and stuff

[cyan violet]

Cool.

Have you proved heron's formula.

It's a meme according to most people on this.

[storm rover]

What’s that

[cyan violet]

Yeah.

That's what I mean.

Nobody remembers it.

[upper karma]

Crazy algebraic proofs with just pure plug and chug are pretty annoying

[cyan violet]

It's like a meme.

[upper karma]

Heron’s formula is a good example

[cyan violet]

@storm rover

It was like.

3 pages of bash.

And confusion.

[upper karma]

Just a bunch of variables you plug in

[storm rover]

Yeah lol

[cyan violet]

Exactly.

[storm rover]

Like deriving the

Formula for

[cyan violet]

Cool.

[storm rover]

Like 3rd degree polynoms

[cyan violet]

Do you study with AoPS?

[storm rover]

Sometimes

I mostly just ask around here

[cyan violet]

Basic polynomials, the ones discovered in algebra, are really easy.

[storm rover]

I haven’t done work with polynomials yet

[cyan violet]

Not real work.

[storm rover]

Haven’t felt like it tbh

Wait

Have you derived the quadratic formula

[upper karma]

Yes

[cyan violet]

Like.

A long time ago.

Jesus, it's been a while.

I miss algebra.

[storm rover]

it’s a nice one

[cyan violet]

Geometry is hell.

Yeah.

Fun.

I proved it on a board.

[storm rover]

Lol geometry is just a bunch of rules

[cyan violet]

In front of my dad and mom 🤣 .

I was so proud.

[upper karma]

I remember my first proof being proving Pythagorean’s theorem using these squares geometrically

That was so fun

[storm rover]

I hate visual proofs

[cyan violet]

I'm saying like.

these polynomials.

Easy.

With basic solutions like such.

[storm rover]

Lol I never learned that

[upper karma]

That’s not real polynomials

[cyan violet]

I know.

It's polynomials that you get introduced into in Algebra 2 with AoPS.

Really easy stuff.

Not real polynomials work, though.

[storm rover]

I don’t like polynomials

Even the basic ones

Don’t care about them

[cyan violet]

And then there's.

Real polynomial stuff.

[storm rover]

Do you do math comps lol

[cyan violet]

Not really.

Not much.

[storm rover]

I don’t like it too much

It’s too experienced based especially for math

[cyan violet]

I get most of these problems from the AoPS website.

Oh, okay.

Well the problem with AMC's is the little time given/

It makes it less feasible.

[storm rover]

I failed all my amc’s

[cyan violet]

Even though the problems aren't too crazy, you need to solve them in like two minutes.

Which is insane.

Meh, it's fine.

It's probably caused by the little time given.

Which can be annoying.

So, yeah.

[storm rover]

Yeah

Wanna go into like serious or something

Cuz we are going off topic

[cyan violet]

Uh.

Sue.

Sure*.

[devout shell]

visual proofs have their place

a picture is worth a thousand words as the saying goes

[storm rover]

Yeah but

I just don’t like them as much of numerical proofs

[hallow smelt]

@quartz edge when I apply the double angle I get sin of 18/41

[cosmic trellis]

can someone help me with cos^3x - cos^2x = 0

[trail minnow]

um

solve for x lol?

wait

$\cos^{2x} - \cos^{3x} = 0$

[somber coyote]

hegel:

[trail minnow]

uhhh

[spark stag]

$\cos^3x - \cos^2x = 0$

[somber coyote]

Namington:

[trail minnow]

ohhh lmao

[spark stag]

what do you realize about the left side right away?

@cosmic trellis

[cosmic trellis]

well cos^2x = 0

[spark stag]

what do you mean?

did you factor out a cos^2 x?

[cosmic trellis]

yeah

[spark stag]

yeah, so then we get

$\cos^2x (\cos{x} - 1) = 0$

[somber coyote]

Namington:

[spark stag]

right?

[cosmic trellis]

yup

[spark stag]

so our classes of solutions are where:

$\cos^2x = 0$ \
$\cos{x} - 1 = 0$

[somber coyote]

Namington:

[spark stag]

we can rearrange the second line to $\cos{x} = 1$

[somber coyote]

Namington:

[spark stag]

and the first, well, a number squared is only 0 if that number is 0

so its the same as $\cos{x} = 0$

[somber coyote]

Namington:

[spark stag]

in other words, we

are looking for when cos(x) = 0

or when cos(x) = 1

when is that true?

[cosmic trellis]

do I solve

for x?

[spark stag]

im assuming thats what the question was asking for, yes

what x values make cos(x) = 0?

what x values make cos(x) = 1?

[cosmic trellis]

cos(0) = 1

[upper karma]

yaas

and what about 1

[spark stag]

thats one solution, yes

did the question ask for answers in a certain domain/interval?

or just all the solutions?

[cosmic trellis]

all solutions

so pi/2

for 0

i think

[spark stag]

yep, but theres more

remember, cosine is periodic

which means it repeats every 2pi

[upper karma]

just try to visualize it

[spark stag]

,w graph cosx

[somber coyote]

Display results online and refine query

Requested by Namington#1983

[upper karma]

each time you make a 360degree rotation you end up in the same place

[cosmic trellis]

ohh

[upper karma]

90 + 360 = 450

[spark stag]

so if we want a general solution

[upper karma]

cos 450 is also 0

[cosmic trellis]

the question was in the range [0,2pi]

[spark stag]

cos(1) = 0, yes

[upper karma]

oh ok

[spark stag]

oh, ok

then we dont need a general solution

so your answers are

x = 0, x = pi/2

theres one more

er actually, two more

[cosmic trellis]

3pi/2

[spark stag]

yes

and one last one

[cosmic trellis]

ermm

oh

-pi/2

[spark stag]

thats not in our domain

,w plot y=cosx from 0 to 2pi

[somber coyote]

Display results online and refine query

Requested by Namington#1983

[spark stag]

cosx=1 when x=0 (left edge)

where else does cosx=1?

[upper karma]

easy enough

visualize it

might not be the best pic but you can see

cosine is in the x-axis

and at 0degrees its 1

||just add 180deg||

[spark stag]

erm

180?

cos(pi) = -1

[upper karma]

oooh right

[spark stag]

the last solution is

x = 2pi

which is on the very edge of our domain, but is apparently included

[upper karma]

kinda forgot about that

[spark stag]

[0,2pi]

so our solution set is

x = 0, x = pi/2, x = 3pi/2, x = 2pi

we can test each of these individually if you wanna check the answer

[upper karma]

😃 I guess I can say I learned too from this

[cosmic trellis]

is it 2pi because it happens every period?

idk im confused

[spark stag]

we have a solution at x = 0

[upper karma]

if you rotate by 2pi you still get back to the starting position

[spark stag]

which means we also have one at

x = 2pi

[upper karma]

2pi is 360 degrees

[spark stag]

x = 4pi

x = 6pi

x = -2pi

etc

every multiple of 2pi

[upper karma]

Each time you rotate by 2pi, you get back to the starting position

[spark stag]

buuut our domain is from [0,2pi]

so we dont care about most of those

we do care about x=0 and x=2pi, though

[cosmic trellis]

oh ok thanks alot

[sly belfry]

hello everyone

I need some help in a very particular assignment I've been working on

i basically have to comment on a 18th century book written on conic sections

in this case, they're defined as the points whose distance to the directrix is p, and whose distance to a focus is q; and such, depending on the ratio p/q, the three different conic sections (ellipse when p/q > 1, parabola when p = q, hyperbole when p/q < 1) arise

i'm working on the p/q = 1 case, but the author makes two claims

the first, is that the curve never meets at the center (thus becoming an ellipse)

the other claim is that the distance between the two sides of the parabola stabilises

almost everything is proved based on euclidean geometry and the book of elements, but the teacher has given us liberty to use whatever tools we wish

how could I, for example, show that the tangent of the parabola tends to 1 over time?

[hallow smelt]

Not sure if this is right

for D when I try Tan-1 (3/5) - Tan-1 (2/-6) I do not get the same answer I did doing the U*V = |U| |V| Cos(x)

[quartz edge]

@hallow smelt let me see the work

[hallow smelt]

that is all my work

[quartz edge]

same problem?

I meant the one from earlier

[hallow smelt]

oh no this is a different one

[quartz edge]

I guess you figured that one out then

[hallow smelt]

from earlier I did 2 * 9/40 * 40/41

and got a different answer it was 720/1681

9/41**

[quartz edge]

So after you find inverse tan you use the indentity for sin2x

which should be 2sinx*cosx

then you just multiply

[hallow smelt]

yup thanks man. Do you happen to know how to do the second one I posted?

[quartz edge]

I forgot how to do vectors completely

[devout shell]

[quartz edge]

[hallow smelt]

My trig final is tomorrow

[west tapir]

cot x sec^4x = cot x + 2 tan x + tan^3x

i need to prove the left to the right

[cinder portal]

you mean

[west tapir]

i dont know what to do past

[cinder portal]

cot(x) + (sec(x)) ^ 4

?

[west tapir]

(sec^2x)(sec^2x)

yeah

[cinder portal]

wait lemme rewrite

$cot(x) * (sec(x))^4$

is this what you mean?

[west tapir]

yeah

they wrote it differently but it means the same as that

[cinder portal]

o wait no

they're multiplicative

sry

[somber coyote]

MemesPlease:

[cinder portal]

there

is that right now?

[west tapir]

oh yeah that

[supple abyss]

this looks right

[cinder portal]

ok heres a hint

do you know

your trig identity for sec, tan?

pythag identity

[west tapir]

yeah

[cinder portal]

use that

foil

[west tapir]

i got it to sec^2

[cinder portal]

and you should get your answer

do it twice

sec^4(x) = (tan^2(x) + 1) * (tan^2(x) + 1)

foil

and you should have your answer

[west tapir]

i should multiply the cot after right?

[cinder portal]

mhm

[west tapir]

because on the right its a plus

[cinder portal]

just foil it

and see how far you get

[west tapir]

ok

i got

tan^4(x) + 2tan^2(x) + 1

[cinder portal]

cot(x) * (1 + 2tan^2(x) + tan^4(x))

perfect

and then distribute

[west tapir]

oh i see

thanks

[cinder portal]

ye np

[onyx basin]

how was x found

my mind's blanking rn

[trail minnow]

um

divide both sides by tan 35

then divide by (x+1500)

for

$\frac{x}{x + 1500} = \frac{\tan 30}{\tan 35}$

[somber coyote]

hegel:

[onyx basin]

ok

[trail minnow]

you know how to simplify this right

[onyx basin]

would (x + 1500) still be in its parentheses

i'd assume it would, right

[devout shell]

what do you mean by that?

[trail minnow]

?

[onyx basin]

when we divide by (x+1500), would (x+1500) still be in parentheses when it's the denominator

[trail minnow]

uh

[devout shell]

$\frac{x}{(x + 1500)} = \frac{\tan (30^{\circ})}{\tan (35^{\circ})}$

[somber coyote]

⚡Amphy⚡:

[onyx basin]

yeah ok

[trail minnow]

yes and no. i mean whether you put in parenthesis isn't really relevant

[devout shell]

you can write it that way if you want to

[onyx basin]

tan30/tan35 ~= .824542

[trail minnow]

don't simply fy

*simplify that yet

[onyx basin]

x / x + 1500 = 1 / 1500

[devout shell]

no

[trail minnow]

no

it is equal to

[devout shell]

you cannot do that

[trail minnow]

$\frac{x}{x} + \frac{x}{1500}$

[somber coyote]

hegel:

[trail minnow]

so $\frac{x}{1500} + 1$

[somber coyote]

hegel:

[devout shell]

what?

no

[supple abyss]

[devout shell]

you cannot break the denominator like that

[onyx basin]

i have no recollection of doing that method in school

[supple abyss]

demented algebra

[devout shell]

it really be like that sometimes

@trail minnow you will have to explain your case lol

[trail minnow]

wait what

fuck im dying lkhsfdlkjsdf

now my brain is collapsing

[devout shell]

you cannot break the denominator

[trail minnow]

yes that is dumb and wrong

im dying internall fuckin help

[devout shell]

just move x+1500 to the right and distribute it out

[trail minnow]

conjugate

[onyx basin]

move x+1500 to the tan30/tan35?

[devout shell]

yes I would move x+1500 and then distribute the tan30/tan35

[trail minnow]

honestly easiest method is :
$x = \frac{(x + 1500)\tan 30}{\tan 35}$

[somber coyote]

hegel:

[trail minnow]

then distribute tan 30 and divide

$x = \frac{x \tan 30 + 1500 \tan 30}{\tan 35}$

[somber coyote]

hegel:

[devout shell]

$x\tan(35^{\circ})=(x+1500)\tan(30^{\circ}) \to x=(x+1500)\frac{\tan(30^{\circ})}{\tan(35^{\circ})}\
\to x=\left(x\cdot \frac{\tan(30^{\circ})}{\tan(35^{\circ})}\right) + \left(1500 \cdot \frac{\tan(30^{\circ})}{\tan(35^{\circ})}\right) $

[trail minnow]

ugh

[supple abyss]

[trail minnow]

l a t e x

[supple abyss]

ew

\tan pls

\\

[devout shell]

I missed one?

[trail minnow]

can anything fix this

[supple abyss]

y brackets tho

[devout shell]

so the person won't get confused, he was asking about parentheses earlier

[onyx basin]

and then i can solve the last term right

1500 * tan30/tan35

[devout shell]

let me fix mistake first

[somber coyote]

⚡Amphy⚡:

[devout shell]

so move the x(tan(30)/tan(35)) to the left

and then you can factor out an x to get:
x * (1-(tan(30)/tan(35))

[onyx basin]

and then i can divide 1500 * tan30/tan35 by (1-(tan(30)/tan(35))

[devout shell]

correct

[onyx basin]

ok cool ty

[thorn zinc]

question 18

[oblique spire]

Do i just have to find the projection of (3i - j) in the direction of vector (5i-2j)

[olive solar]

yeah

[oblique spire]

saying 3i-j is vector a and 5i-2j is vector b

is it (a dot b)/|a|

or (a dot b)/ |b|

to find the scalar resolute

[olive solar]

projection of a onto b

which do you think it would be?

[oblique spire]

uh

(a dot b)/ |b|

[olive solar]

yep

[ornate lodge]

Why does sin(x) = 16/sqrt(65) become sin(x) = 8/sqrt(65) if we know x is in (0, pi/2)?

[dark sparrow]

what?

sin(x) cannot be equal to 16/sqrt(65). that's greater than 1.

where did you get this from

[ornate lodge]

I had worked out cos(2x) = -63/65 to cos(x) = 1/sqrt(65). From that sin(x) = 16/sqrt(65)

[dark sparrow]

how did you get sin(x) = 16/sqrt(65) from cos(x) = 1/sqrt(65)?

[ornate lodge]

I made a mistake

I see now

[stoic nova]

Hi I need help with a problem

Find the area of this prism

[devout shell]

,rotate

[devout shell]

You seem to be on the right track

Add the areas of all the faces to get the total surface area

[stoic nova]

I got the wrong answer

I’m trying to correct

[thorn zinc]

@dark sparrow

k

[dark sparrow]

okay

we're back on topic now

so the big question now is, how familiar are you with the very basics of trigonometry?

@thorn zinc?

[thorn zinc]

Decently familiar

but on a very basic level

Im not very confident in this area

[dark sparrow]

you don't need anything above the very very basics here

[thorn zinc]

okay

[dark sparrow]

all you need to know is the definitions of the three basic trig ratios

i.e. sin, cos, and tan

[thorn zinc]

oh k

[dark sparrow]

calculate the length of every side you see in terms of n and y, until you find a side that you can express in terms of n and, independently of that, in terms of y

[thorn zinc]

What are the sin, cos and tan definitions

[dark sparrow]

...okay let's put it this way

do you even know what sin, cos and tan ARE

... @thorn zinc?

[thorn zinc]

nope

you just use them for SOHCAHTOA

thats all i know

and sine and cosine rule

[dark sparrow]

i mean if you don't know what sin, cos and tan are in the first place, you can't really say you're "decently familiar" with the basics of trig

all you're doing is shuffling symbols around without much regard for their meaning

and repeating what may very well be magic spells (liek "SOHCAHTOA") after your mentors

[thorn zinc]

Lol

ty

isnt it

[dark sparrow]

maybe just maybe it's high time you actually went ahead, got a textbook, and familiarized yourself with this shit

[thorn zinc]

sin o/h

and tan o/a

[dark sparrow]

becuse clearly there's something bigger at play than just the fact that you don't know this one basic thing

[thorn zinc]

cos a/h

[dark sparrow]

i feel like it would be utterly unproductive if i tried to assist you but all you did was parrot back mnemonics for things you don't understand

[thorn zinc]

It seems that way i guess

[deep swift]

What is the most reduced form of 3x/2x?

[opaque gate]

how do i solve -2cos(f3) = -1

?

[trail minnow]

@deep swift 3/2

@opaque gate solve for what

variable f?

[opaque gate]

I've figured it out, nvm

[trail minnow]

Lmao

99% of qs asked on this server end up not needing any help and already knowing the answer

[fleet lake]

Lmao

[whole lynx]

I need some help for my problem,
so I got tan x = 3/7 and the problem asks us the exact answer of sec x - sin x?
Thanks in advance ^^

[upper moat]

If tanx = 3/7, then the opposite side of x is 3n, and the adjacent side is 7n for some common ratio n. The hypotenuse is then sqrt((3n)^2 + (7n)^2) = sqrt(58n^2) = sqrt(58)n. Dropping the ns really doesn't matter as they'll get cancelled out in the trig ratios anyways.
secx - sinx = hyp/adj - opp/hyp = sqrt(58)/7 - 3/sqrt(58)
And from there just do some fraction math

[whole lynx]

Thanks so much :)

[upper moat]

np 👍

[zinc grail]

I see how to solve this

1/4

but arent we assuming that line BC is the diameter

and also assuming that the left side is similar to the right?

[upper karma]

Yea by the definition of a diameter we see BC to be just that. Then we can draw a line from A to D and find the angle. Someone correct me if I’m wrong, but I believe that the angle found as pi/4 for ABD would equal ACD

would x = 75 degrees?

Mandatory: not homework

That's what I got

Woop.

@upper karma this ones being harder for me...

Im thinking y = 30 but I dont trust meself

Could you move the page over a bit more so I can see the right hand more clearly

Yeah mt bad

All good

Thanks

thank u* hahaha

Yea I got y=30 as well

OwO

thats interssting

than what is x? @upper karma

90?

Yea

I got that for x

very cool. @upper karma

ahahahah

But the thing is

Hold up

?

oh nvm

Im dumb :c

Sorry

All good

[upper karma]

@upper karma hi

So, the book says that C = w (and w = 60)

I know why w = 60

But I do not get why w = C

actually

wait

what the fuck

🤔

yes I di

do

,rotate

[somber coyote]

Couldn't find an attached image in the last 10 messages

[upper karma]

nvm wait

I dont

So like

Why would C = 60? Theyre not corresponding?

This is my thought process...🤷 I thought it made sense

C and W are equal because of the transversal

https://www.mathplanet.com/education/pre-algebra/introducing-geometry/angles-and-parallel-lines

And x =70 because they are opposite

So you do 180-50-70 = 60

Yeah

I get that

But look at my process

Why wouldnt C = 70?

By vertical angles

In the bottom left where you wrote 70 for the top left I don't think that that is equal to 70

but doesnt that correspond

Wouldn't that be 70+50?

with the 70 from above

oh fuck

yeah

LOL

lol

I initially put 120 on another paper uwu

anyway, C != 120 though

so what gives

?

@devout shell

Why isnt C = 120?

Oh

Actually C is 120

[devout shell]

huh?

[upper karma]

w and the angle above c are equal

I think

But the book says c = 60

[devout shell]

I have not been looking at any of these questions lol, so I have no idea what you mean at all

[upper karma]

Ill send a pic @devout shell

gimme a sec

Because, 70+50=120

Corresponding angles and then vertical angles, so C = 120

[devout shell]

which is C? I don't see it labeled

must be mistake then lol

[outer frost]

how do you find x

[upper karma]

Uh

@devout shell C is bottom, next to the left w

where it says 120

[dark sparrow]

@outer frost what have you done so far and where are you stuck?

[upper karma]

Ann, can u help?

[devout shell]

yea, your book is suspect...it's wrong then, because no way it is 60

[upper karma]

hmmm

lemme search their forums to see

if anyone has noticed this too

[devout shell]

good idea, that should clear it up

[outer frost]

we've covered most theorems but im stuck in a few places

[upper karma]

This is the original problem.

@upper karma 😜

hey cutie:3

bro

just noticed

thsy might mean inside the triangle

which yeah itd be 60

LOL

@devout shell

[outer frost]

@dark sparrow we covered the theorems about arcs, chords, and secants in circles. Im just stuck on that problem (its probably an easy solution but i was probably sleeping in class lol)

[dark sparrow]

i'll give you a hint: draw radii from the center to those points of tangency

[outer frost]

🤦

thats embarrassing lol

[jolly stream]

2 * sin(4x) * cos(2x) - sin(6x)

One of the rules to be used is
sin(2a) = 2sin(a) * cos(a)

But that only got me so far

4 * sin(2x) * (cos(2a))^2 - sin(6x)

Doing this to sin(6x) doesn’t bring me anywhere, any tips?

I’m supposed to simplify this by the way

[dark sparrow]

try writing sin(6x) as sin(4x+2x)

[jolly stream]

I have

What rule is this by the way?

Don’t have the formulas on hand

[dark sparrow]

angle sum formula for sine

sin(x+y) = sin(x) cos(y) + cos(x) sin(y)

[jolly stream]

Yeah, that’s the one. Thanks, I’ll look around

sin4xcos2x + cos4xsin2x

Got to here using that

and then I assume I use double angle rules for sine and cosine for sin4x and cos4x

Right?

[dark sparrow]

no, you don't need to.

[jolly stream]

HMM

Oh wait

This is the sum rule

Or the one you showd me I mean

and then we go backwards

and I get sin(6x)

[dark sparrow]

no, why would you undo what you just did?

[jolly stream]

Oh no, I made a mistake after doing the first step

Because I had

a * c - b

I thought I got a * c - a + b

nvm I wrote it all incorrectly

But I see my mistake

[dark sparrow]

2sin(4x)cos(2x) - sin(6x) = 2sin(4x) cos(2x) - sin(4x)cos(2x) - cos(4x)sin(2x)

= sin(4x)cos(2x) - cos(4x)sin(2x)

= sin(4x-2x) = sin(2x)

[jolly stream]

Oh nvm then I did it right

OHHHHHHHH

Minus 😃

not plus!

Okay. Thank you once again!

[rancid mantle]

anyone know the answer? r = 2/(4 - (5cosθ))

[plucky marlin]

yeah nice polar equation

[upper karma]

hey

wondering if there is anyone online who can help me with some geometry EOC review that i'm doijg

the questions is this

[spare ridge]

help

[snow fable]

Could anyone help me find measure of angle E, and length y and x

[storm rover]

If you know the base area of a pyramid, can you just multiply that by the height of the pyramid and divide by 3

[devout shell]

ED is tangent to the circle, there is a theorem you can use to find that length. Segment FC is part of a chord that is cut by another chord. There is yet another theorem, probably named intersecting chord theorem, that you can use the find the length of FC

[spark stag]

@storm rover yes, thats essentially how the formula is derived

assuming you're finding volume.

[storm rover]

Oh okay so I don’t need complicated forms

Thanks

[balmy sparrow]

How would I solve for T in terms of W for this question : sin(Wx)=sin(Wx+T)?

[dark sparrow]

is that an equality meant to hold for all x

[balmy sparrow]

yes

I wanted to solve for it formally answer should be 2pi/W

[dark sparrow]

huh what

no, this looks like it's satisfied by T = 2kπ for k ∈ Z

[balmy sparrow]

right, can you show me how you got that from sin(Wx)=sin(Wx+T)?

[spare ridge]

someone help me with this please

how do u do it

[spare ridge]

YY

TY

[devout shell]

Every week someone has a question about the length of a tangent segment lol

Power of a Point

Like every week there is at least one question

I’ve posted the formula so many times now lol

Few seem to remember them though

A lot of my classmates would forget thms when I took geometry in hs

People seem to just remember for the test then forget lol

[fossil lotus]

People seem to just remember for the test then forget lol

I feel like this is true for most things you learn in school

[stoic nova]

^^

[tired terrace]

Can someone talk me through some work with Trig functions and the unit circle? I have a test on it and im confused. 😦

[fossil lotus]

@tired terrace Are there specific problems you need help with?

[tired terrace]

yes

well

kinda

but i feel like if I get help solving one ill catch on and be able to solve the others

[fossil lotus]

I can try and help you

What’s a problem you’re stuck on?

[tired terrace]

2-2sin^20/cos0

0=theta

its simplification

[fossil lotus]

So for these kinds of problems you want to look for trig identities

[tired terrace]

ok

[fossil lotus]

In this case I think a useful one would be tanθ = sinθ/cosθ

I’m not sure if there’s anything else to do besides that

[tired terrace]

so

2-(2sin)(tan) ?

[fossil lotus]

Yeah

I don’t think you can simplify it any further than that but I could be missing something

[tired terrace]

ok

what ab

csc = (cot/sec) + sin

this one is just proving

[fossil lotus]

What do you know about cot and sec?

[tired terrace]

that cot = cos/sin

and sec = 1/cos

[fossil lotus]

See if you can do anything with that

[tired terrace]

csc = (cos * 1 / sin * cos) + sin?

[fossil lotus]

Not quite

Since sec is in the denominator it would initially be (cos/sin)/(1/cos)

[tired terrace]

so

csc = (cos^2)/(sin) + sin

[fossil lotus]

Mhm

[tired terrace]

and bc

csc= 1/sin

then we can multiply sin on each side to get 1

so (cos^2 + sin^2) = 1

which is a pythagorean identity

[fossil lotus]

Yep

[tired terrace]

WOOHOO

we did it!

Im one step closer to passing that class

Can you do a couple more with me?

[fossil lotus]

Sure

[tired terrace]

cos^2 = (csc*cos)/(tan+cot)

ok uh this hurts my head

[cinder portal]

work on the right side

convert everything to sin and cos

tan = sin/cos
cot = cos/sin

is what I would do first

[tired terrace]

ok

so

uh

[fossil lotus]

The first thing I would do is divide both sides by cos

[tired terrace]

cos^2= cos^2/sin^2+cot?

[cinder portal]

$cos^2(x) = \frac{\frac{1}{sin(x)} * cos(x)}{\frac{sin(x)}{cos(x)} + \frac{cox(x)}{sin(x)}}}$

[somber coyote]

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

[cinder portal]

i would multiply top and bottom by cos(x) * sin(x)

[tired terrace]

ok

well

when I do that

[dark sparrow]

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

[tired terrace]

I get to cos^2= 1/sin

[somber coyote]

Ann:

[cinder portal]

thank you Ann

when you multiply top and bottom by cos(x)*sin(x)

the denominators in the bottom become a bit nicer

[tired terrace]

right

[cinder portal]

you'll notice it

multiplying top and bottom by cos(x)*sin(x) gives you the answer

[tired terrace]

they turn to sin^2+cos^2

[cinder portal]

thats an identitty

[tired terrace]

and then the top turns to cos^2 * sin

yea thats uhh 1

[fossil lotus]

The sines cancel out on the top

[tired terrace]

righttt

so then

cos^2 = cos^2/1

which is truh

true

[fossil lotus]

I solved it a different way but this also works

[cinder portal]

yup

that works

[tired terrace]

ok

next one

it gets harder 😭

[cinder portal]

post it

send a picture

its easier that way

[tired terrace]

cot+1 = csc *(cos+sin)

I cant

I wish i could

[cinder portal]

o

[tired terrace]

my phone is broken to shit

[cinder portal]

i would say

work on the right hand side again

convert csc to 1/sin(x)

and distribute

and you should have your answer

[tired terrace]

cos+sin is just 1 right?

[fossil lotus]

No

[cinder portal]

no

cos^2 + sin^2 is 1

[fossil lotus]

That’s cos^2+sin^2

Yeah

[cinder portal]

lemme rephrase

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

x has to be the same in both

[tired terrace]

oh

ok

[fossil lotus]

Yes

[cinder portal]

u dont need that identity for this one though

just do what i said

and you should be fine

[fossil lotus]

But yeah for any of these problems your first step should just be to try and simplify things as much as possible

Then try rearranging things and looking for identities

[tired terrace]

oh ok

so when i multiply the 1/sin by the stuff in the par

then i get cos/sin

which is just cot

and sin/sin

which is 1

[cinder portal]

lol yup

[tired terrace]

ok last one

[cinder portal]

and of course, these problems, the more you do, the better you get at it

[tired terrace]

(cot-tan)/(sin*cos) = csc^2 - sec^2

[cinder portal]

work on left hand side

convert everything to sin and cos

[tired terrace]

ok

(cos/sin)-(sin/cos)/ sin *cos

and then multiply the top and bottom by sin-cos?

[cinder portal]

lemme c

yup

[fossil lotus]

Yeah

[cinder portal]

sin(x)*cos(x)

[fossil lotus]

Or split it into two fractions and simplify

Same result

[tired terrace]

ok one sec

[cinder portal]

are u sure ur right hand side is csc^2 minus sec^2?

[sleek thistle]

GrandPiano you play piano?

[fossil lotus]

I do

[tired terrace]

holy shit wait

how do i do

(sin*cos)(sin-cos)

[sleek thistle]

How long have you been playing

[fossil lotus]

About 14 years

[cinder portal]

did you copy the problem correctly?

[sleek thistle]

Oh wow, I just started about 3 months ago or so

[cinder portal]

something doesnt seem to match, or im just mad dumb

play around with it

lemme c

[fossil lotus]

I think it’s right

[tired terrace]

ye im sure

[fossil lotus]

So

[cinder portal]

ok im prob just high

[fossil lotus]

(sin*cos)((cos/sin)-(sin/cos))

(In the numerator)

Should be what you’re simplifying

[tired terrace]

holy shit wtf

im not getting that at all

i got -cos^2+sin^2

in the numerator

[fossil lotus]

Wait

No you don’t multiply the top and bottom by sin*cos

[tired terrace]

sin-cos

[fossil lotus]

So you have ((cos/sin)-(sin/cos))/(sin *cos)

Split that into (cos/sin)/(sin*cos) - (sin/cos)/(sin*cos)

Then simplify from there

[cinder portal]

yup

[tired terrace]

ok I see

thank you!

[fossil lotus]

Do you see how to solve the problem?

[tired terrace]

yes

then once you simplify

you get cos/sin - sin/cos

then you multipy by sin*cos

then

you get it down to 1

which is exactly was csc^2-sec^2 is right?

[fossil lotus]

How did you get cos/sin - sin/cos?

[tired terrace]

by subtraction

they had the same denomenator

so i just subtracted them like the sign indicated

[fossil lotus]

But then you’re just back where you started

That gets you ((cos/sin)-(sin/cos))/(sin *cos)

The denominator doesn’t disappear

[tired terrace]

youre right

Holy crap this is hard

I missed this entire unit

[fossil lotus]

What I’m saying is split it up into two fractions

(cos/sin)/(sin*cos) - (sin/cos)/(sin*cos)

And then simplify each fraction

[tired terrace]

so

you multiply them both by sin*cos?

[fossil lotus]

No

[tired terrace]

wait

no

[fossil lotus]

Look at each one individually

[tired terrace]

ok

they are all either cot or tan

\cdot

maybe that can help

[fossil lotus]

$\frac{\frac{\cos{x}}{\sin{x}}}{\sin{x} \cos{x}} - \frac{\frac{\sin{x}}{\cos{x}}}{\sin{x} \cos{x}}$

[somber coyote]

GrandPiano:

[fossil lotus]

Might be easier to think about in this format

[tired terrace]

ohhhhhhhhhhhhhh

sin*cos

not sin/cos

[fossil lotus]

In the denominators yes

[tired terrace]

so you can just move the cos and the sin down to the denominators

so you get sin/ sin * cos^2 - cos/sin^2* cos

then you simplify

which gives you cos^2-sin^2

right?

[fossil lotus]

Not quite

Maybe you’re thinking of how a/(b/c) = ac/b

The reverse doesn’t really work

But think about how you can simplify each fraction

[tired terrace]

oh

well then how do you get it simplified?

[fossil lotus]

Is there anything you can cancel out?

[tired terrace]

the sin in the first term and the cos in the second?

wait no

flip that

cos in first

sin in 2nd

[fossil lotus]

Yes!

[tired terrace]

so

that gives us

1-1

[fossil lotus]

?

[tired terrace]

bc sin/sin - cos/cos

[fossil lotus]

If you divide cos/sin by cos you get 1/sin

[tired terrace]

right

so 1/sin^2- 1/cos^2?

[fossil lotus]

There you go

[tired terrace]

which gives me

drumroll please

[fossil lotus]

drumroll sounds

[tired terrace]

csc^2-sec^2!!!!

[fossil lotus]

Ding ding ding!

[tired terrace]

and just like that

the problem's solved

Ill make sure to practice, and can you help me if get stuck around this time tmr?

[fossil lotus]

Sure

I actually need to go to bed now

[tired terrace]

same

gn ❤

[fossil lotus]

Night

[ornate lodge]

In the triangle made from a circle chord and two radii, if the central angle is θ, will the other two angles be (180 - θ)/2?

[dark sparrow]

yes

[upper karma]

A polygon has 8cm for its given side a similar larger has 24 that corresponds to the 8cm side the 8cm has an area of 100cm what is the area for the larger

I need major help

[brave kayak]

give me a moment

a similar larger?

[upper karma]

It's similar and it's larger @brave kayak

[brave kayak]

900 cm^2, i think

my geometry is rusty but here's my intuition:

imagine a right triangle whose legs are both 1

it has an area of 1/2, right?

and now there's a similar right triangle whose legs are 2

the area would be 2

the ratio between the sides is 2

the ratio between the area is 4

it's like the ratio between the areas is just the ratios of the sides squared

mate you there lol

[upper karma]

Yeah thanks lol I just needed the area because it doesn't say what polygon it is

[brave kayak]

this rule applies to any kind of similar polygons

[upper karma]

I come from a town where there's a small population so we don't have many people to teach math so our teacher is some fresh grad that doesnt know any teaching guideline

[brave kayak]

you can experiment with different kinds of shapes if you're bored lol

(good lord this is so un-mathematical)

[upper karma]

What about this?

Solve for X and it's a polygon with 115 degrees and 120 degrees and on the bottom it has 3x and x + 25

@brave kayak do you know this one

[brave kayak]

the 115 and 120 are where?

outside?

[upper karma]

Inside

All of them are inside

[brave kayak]

"polygon"