geometry-and-trigonometry

[upper karma]

how do you go from 1 to 2?

[keen tangle]

expand the fraction inside the sqrt by 2

[upper karma]

expand?

you mean times 2?

oh ok

[keen tangle]

$\frac{a}{b}=\frac{2a}{2b}$

[somber coyote]

Gonzo17:

[upper karma]

ok thx

now I get it

[keen tangle]

glad to help 😁

[upper karma]

latex is really useful lol

I should learn how to use it better

or else I always get : $}$

[somber coyote]

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

[cold plaza]

@sick veldt the answer is (a), and the way I checked it was by using the distance formula from (6.5,8.5) to (4,8), and verifying that that was indeed a third of the distance from (6.5,8.5) to (14,10)

i dont think repeatedly finding the midpoint works

[sick veldt]

No

Gotta use the section formula

[cold plaza]

what is the section formula

[sick veldt]

so

the ratios

lets say 2:3

2 would be m

3 would be n

and the formula is

mx2 + nx1/ m + n

[cold plaza]

so when u say 2:3, it means that AC = 2/5 AB?

[sick veldt]

my2 + ny1/ m + n

uhh

its hard to explain haha

But you just plug in the x and y's

[cold plaza]

yeah im not too sure about it lol

and im in uni studying math

smh

[sick veldt]

oo lol

[somber coyote]

luka:

[upper karma]

Random words: I'd start with sum and difference formulas

ok, what are these formulas?

$cos(3x)+cos(x) = 0$

[somber coyote]

luka:

[upper karma]

http://www.sosmath.com/trig/Trig5/trig5/trig5.html

ok

hmm how does this help?

can I simplify this? $cos(3x)+cos(x) = 0$

[somber coyote]

luka:

[upper karma]

<@&286206848099549185> any ideas?

[steady sleet]

$$\cos(x+x+x)=~?$$

[somber coyote]

Simple_Art:

[upper karma]

idk

I never learned a formula for this

$cos(3x)$ 😃

[somber coyote]

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

[timber hinge]

cos(x+2x)

[steady sleet]

Also:
#❓how-to-get-help, rule #4: If your question has not been answered for a minimum of 15 minutes, you may use the Helpers tag once.
Please read the rules: #❓how-to-get-help!

[upper karma]

right

[steady sleet]

And you have all the formulas above

[upper karma]

oh ok didn't know that

from the link?

[steady sleet]

yes, the image you posted to be precise...

[upper karma]

ok

how does that help though?

https://media.discordapp.net/attachments/326138757474680852/521134206697603094/404533869845544971.png?width=500&height=65

the solutions belong to [-2pi; pi]

[timber hinge]

rewrite it as

$4\cos\left(x\right)\sin\left(\frac{\pi}{4}-x\right)\sin\left(\frac{\pi}{4}+x\right)$

[somber coyote]

Colen:

[timber hinge]

Then its trivial to solve for = 0

[upper karma]

ok

@timber hinge how do you rewrite it like that?

I tried using the formula from the image but it doesn't give me your equation

<@&286206848099549185> any ideas?

[timber hinge]

$ \cos\left(3x\right)+\cos\left(x\right) = 2\cos\left(x\right)\cos\left(2x\right) \ \cos\left(2x\right)=\left(\cos\left(x\right)-\sin\left(x\right)\right)\left(\cos\left(x\right)+\sin\left(x\right)\right) \ \cos\left(x\right)-\sin\left(x\right)=\sqrt{2}\sin\left(\frac{\pi}{4}-x\right) \ \cos\left(x\right)+\sin\left(x\right)=\sqrt{2}\sin\left(\frac{\pi}{4}+x\right) \ \cos\left(2x\right)=2\sin\left(\frac{\pi}{4}-x\right)\sin\left(\frac{\pi}{4}+x\right) \ \therefore 4\cos\left(x\right)\sin\left(\frac{\pi}{4}-x\right)\sin\left(\frac{\pi}{4}+x\right)=\cos\left(3x\right)+\cos\left(x\right) $

[somber coyote]

Colen:

[timber hinge]

You can probably deduce that first bit on your own

$ \cos\left(3x\right)+\cos\left(x\right) \ \cos\left(x+2x\right)=\cos\left(x\right)\cos\left(2x\right)-\sin\left(x\right)\sin\left(2x\right)+\cos\left(x\right) \ \cos\left(x\right)\left(\cos\left(x\right)^2-\sin\left(x\right)^2\right)-2\sin\left(x\right)\sin\left(x\right)\cos\left(x\right)+\cos\left(x\right) \ \cos\left(x\right)\left(2\cos\left(x\right)^2-1\right)-2\sin\left(x\right)^2\cos\left(x\right)+\cos\left(x\right) \ \cos\left(x\right)\left(2\cos\left(x\right)^2-1\right)-2\left(1-\cos\left(x\right)^2\right)\cos\left(x\right)+\cos\left(x\right) \ \cos\left(x\right)\left(\left(2\cos\left(x\right)^2-1\right)-2\left(1-\cos\left(x\right)^2\right)+1\right) \ \cos\left(x\right)\left(4\cos\left(x\right)^2-2\right) \ 2\cos\left(x\right)\left(2\cos\left(x\right)^2-1\right) \ 2\cos\left(x\right)\cos\left(2x\right) $

[somber coyote]

Colen:

[timber hinge]

I feel like i'm overkilling this problem but it would technically work

[keen tangle]

yeah Colen there's a pretty easy way

[timber hinge]

Enlighten

Also @upper karma

[keen tangle]

wait how do you break lines here?

[timber hinge]

double backslash

[upper karma]

ok thanks

I've never learned these formulas (cos(3x))

(this is school homework I don't get)

I've only learned the duplication one with cos(2x) or sin(2x)

is there a method in which we can use these?

[keen tangle]

$\cos(3x)=\cos(2x)(\cos x)-\sin x\sin(2x)=\cos^3 x-\sin^2 x \cos x-2\sin^2 x\cos x=\=\cos^3 x-3(1-\cos^2 x)(\cos x)=4\cos^3 x-3\cos x$

[somber coyote]

Gonzo17:

[keen tangle]

so $0=\cos^3 x +\cos x=\cos x(4\cos^2 x-2)=4\cos x(\cos x-\frac{\sqrt 2}{2})(\cos x+\frac{\sqrt 2}{2})$

[somber coyote]

Gonzo17:

[upper karma]

ok get it now

[nova jewel]

I need to justify that cos^2(x/2)=(1+cos(x))/2

@everyone How can i do that?

[oblique plank]

It's a double angle formula derived from cos (2x).

[nova jewel]

What do you mean ?

How can i explain it?

@oblique plank

[oblique plank]

I mean that follows from this.
@nova jewel

[nova jewel]

I don’t get the point , How can you only find cos^2(x/2) with ^2 meaning square

[oblique plank]

Replace sin^2 (x) with (1 - cos^2 (x)).

[nova jewel]

Wait... it give me cos(2x)=cos^2(x)-(1-cos^2))

So cos(2x)=2cos^2(x)-1

That doesn’t help me

Or i’m juste the dummest person on earth

Pls @oblique plank help

[oblique plank]

You're almost there. Just transpose the 1 and the 2 to the RHS and make the squared term the subject.
Then substitute x/2 for x throughout.

[nova jewel]

Srry i’m not english what does rhs mean??

[oblique plank]

Also, don't @ everyone. People don't like it. Some severs will even warn or ban you for it.
RHS = Right Hand Side of the equation.

[nova jewel]

Ok understood

[oblique plank]

Is your pfp Discord chan?

[nova jewel]

Pfp ? Srry again

thanx MaDDogx for your help and your time

[oblique plank]

Profile picture. You new to the internet or something @nova jewel?

[nova jewel]

I’m not english so all of this is new for me and i dunno who is she i just found her funny

[oblique plank]

👌

[eager plume]

what is pythagoras' theorem

[astral hornet]

In a right triangle, The square of the hypotenuse length equals the sum of the squares of the 2 legs lengths

[eager plume]

ok thanks

[silent sparrow]

Need to find b and c

I’m stuck

how would i find c

@limber lance

[serene field]

Any of the trigonometric ratios, or the law of sines.

[silent sparrow]

i found it ty for the help 😃

youtube works wonders sometimes

[hot flax]

Write 26 in terms of tan?

[limpid basin]

do you know what identity comes from 1 + tan^2 x?

[hot flax]

sec^2 x

[limpid basin]

k so sec^2x/csc^2x =?

[silent sparrow]

Is this right

Number 37

[chrome fiber]

you need to square both terms

[real sparrow]

No they're squared

[chrome fiber]

oof

[real sparrow]

Jinxed.

[silent sparrow]

I think this is wrong

Its sec squared theta

[real sparrow]

Yep.

[silent sparrow]

How tho

Idk what I did wrong

[chrome fiber]

$\tan^2\theta = (\tan\theta)^2$

[somber coyote]

june:

[real sparrow]

tan ^2 x= (sinx/cosx) ^2

[chrome fiber]

same for the other one too

[silent sparrow]

Can I see a visual

[real sparrow]

cosec^2x = (1/sinx)^2

[silent sparrow]

I forgot the squares

[real sparrow]

Multiplying will yield sec^2 x

Yeah

[silent sparrow]

I don’t cancel anything

[real sparrow]

Well.

[silent sparrow]

So it’s sin^2/cos^2 times 1/sin^2

[chrome fiber]

yep

[real sparrow]

^

[silent sparrow]

Ok then

Multiply across ?

Sin times 1 is sin

Idk lol

I need help

Lol

[real sparrow]

Lol what ?

[silent sparrow]

Dude idk

I’m confused

Idk how it’s sec

[chrome fiber]

you have this

[real sparrow]

@silent sparrow https://socratic.org/questions/how-do-you-simplify-tan-2x-csc-2x

Go live your life now.

[chrome fiber]

$\frac{\sin^2\theta}{\cos^2\theta}\cdot\frac1{\sin^2\theta}$

[somber coyote]

june:

[chrome fiber]

simplify it

[silent sparrow]

Yes that’s what I have

[chrome fiber]

simplify it?

[silent sparrow]

Idk how

I’m stuck

[real sparrow]

LOL

[silent sparrow]

Lol I’m not good at math

Or this section

I’ve been on this prob for like 30 mins

Its sad

Nvm

I found out how to do it

Lol

Ty

Sorry for the fuss

Not cool to lol either

[tawdry monolith]

Hi could anyone help me with a trigonometry question

[serene field]

[tawdry monolith]

So basically I have to find CEA

Is it clear

[upper karma]

Not really

[tawdry monolith]

Hmm

CE||BD

[upper karma]

As in, take a better picture

[tawdry monolith]

ok

Yea thats what I meant too

Is it better

[upper karma]

🆗

[tawdry monolith]

Cool

Now I have to find CEA

anything up?

[upper karma]

CED = CBD right?

[tawdry monolith]

right

[upper karma]

CBD = CEA

[tawdry monolith]

true

[upper karma]

There ya go

[tawdry monolith]

I have to find it in pure numbers

[upper karma]

Are you given the angle?

[tawdry monolith]

no

but im given that BD=1.8AC

what's hannen?

did anything pop up

[real sparrow]

What's the question again ?

@tawdry monolith?

[tawdry monolith]

to find the angle CEA

[real sparrow]

Well in the question it says angle CAD is twice of angle DBC but in the figer angle CAB is twice that of angle DBC

Is the correct or a typo ?

[tawdry monolith]

ohhh myyy goddd

I thought its CAB

[real sparrow]

Well

[tawdry monolith]

I'll try to solve it now

[real sparrow]

Yeah its done

The Essence is just use the fact sum of the internal angles of a traingle is 180 and follow the route till you get angle BCE and then you'll get angle AEC

Thanks.

[tawdry monolith]

its been a long time since i've solved trigo

is Sin(a)=1.8Sin(0.5a) solvable?

dont think so

[real sparrow]

Ya it is

[tawdry monolith]

how so

[real sparrow]

sin(a) = 2sina/2Cosa/2

Half angle formula

Dividing both side by sina/2

[tawdry monolith]

is that slash dividing?

ok?

[real sparrow]

sin(a) = 2sin(a/2)cos(a/2)

[tawdry monolith]

I dont think we've learned that yet

[real sparrow]

Well its an identity ?

[tawdry monolith]

aha

[real sparrow]

Still sloving that will give your the value of cos(a/2)

and you can continue with whatever you're doing

[tawdry monolith]

Thnx for ur help

[real sparrow]

Welcome.

[upper karma]

hi could anyone help me with my analytical geometry

[rugged moat]

What property says that AH²=GH×HE

I guess similar triangles?

$\frac{AH}{GH}=\frac{HE}{AH}$

[wild hamlet]

pythag theorems

maybe

[rugged moat]

Eh

I will remember this

Fuck however one derives that, should be easy if I tried

Thanks anwyay

[wild hamlet]

np

[bitter viper]

anyone know the answer to this?

https://gyazo.com/ed63ffa0321ed0dd4fa1727655f06293

[keen aspen]

Split the log as $ \log_{5}(35)-\log_{5}(3) $ first..

[somber coyote]

PJS:

[keen aspen]

Now you need to turn log_5(35) into something with log_5(7)

See what you can do?

[bitter viper]

5log_5(7) - log_5(3)?

@keen aspen

[keen aspen]

Nope right idea though

log_5(7*5)

-> log_5(7)+log_5(5)

log_5(5) is 1

so just 1+log_5(7)

[bitter viper]

ahh yes

thank you for the help

[keen aspen]

Np

[bitter viper]

Hi guys just wondering if "-1" is the right answer here. Just want to double check because i am on my possible entry for this question

https://gyazo.com/5fe41a80268f254b432ea2007218c296

[serene field]

Why would there be -1 zeroes? 🤔

Or rather, how?

[bitter viper]

now that i think of it u right lol

the answer isnt 0 though, i tried that

[keen aspen]

Factor out the x^4

And set = 0

you will see that it id not -1 for a zero

[bitter viper]

ohhh is it 5?

x4(1-x)

[keen aspen]

Yeah

so x^4=0 and 1-x=0

[bitter viper]

Hi yall, sorry to bother just wondering if this is the right answer before i submit

https://gyazo.com/e1e21f0c36e7bb4f87bd8dd0950d9966

[keen aspen]

$ \left(\frac{f}{g}\right)(x))=\frac{x+3}{x^2-2} $

[somber coyote]

PJS:

[keen aspen]

so plug -1 in

you get -2

-g(3) -> -7

so -2-7 -> -9

[bitter viper]

ooo ok thank you that will help with my next questions

[meager acorn]

arcsec is inverse of cos right>

?

[jaunty plume]

no

sec is 1/cos

arcsec is 1/arccos

[meager acorn]

os fo rthis than

how do you solve this

[jaunty plume]

wait no

idk

[meager acorn]

nvm

i gotit

2sqrt3/3 = pi/6

30deg

[meager acorn]

angle from radians to degrees
-3.3

-3.3 x 180/pi ?

-189.08 deg?

[jaunty plume]

yup

[meager acorn]

whats the arc on the circle of radius formula

1/2 pi r^2?

S= r theta
A = 1/2 r^2 theta

i m guessing S = R theta >.>

[umbral snow]

s = rθ
Arc of circle = radius of circle × subtended angle

In radians of course

[upper karma]

what

i dont understand anything here I feel lonely

[meager acorn]

cot^2(x/2)
= 1/tan^2(x/2)
= cos x/ 1-cos x

[upper karma]

w

[meager acorn]

how can 1+cos x can become cscx+cotx

i tried with cscx + cotx/cscx-cotx didnt work out well so i went for cot^2(x/2)

[dire rampart]

multiply the top and bottom of the rhs by sinx

then use double angle formula for cos> cos(x)=cos(x/2+x/2)

[meager acorn]

by the way

i still do not understand how can 1/2tan ^2(x/2)
becomes cosx / 1-cos x

[dire rampart]

same question?

[meager acorn]

yep

i m stuck at 1/tan^2(x/2)

[dire rampart]

yea multiply top and bottom by sinx

what do you get?

[meager acorn]

wait what do you mean

wait

tan ^(x/2) = sin/cos?

nononono

[dire rampart]

no i meant the first question you posted

one with the fraction

[meager acorn]

its identies

1/tan^2 (x/2)

[dire rampart]

its way easier to start from the rhs imo

[meager acorn]

ok

let me try that

wait i got a question

[dire rampart]

always start with the more complicated side, cuz you have more to work with

[meager acorn]

somthin like that right

1+cosx/1-cosx

@dire rampart

do i still have to use double angle formula

[dire rampart]

ya you do

[tulip yarrow]

Question Guys!

Not sure if I did this correctly.

[dire rampart]

you mean the circled one?

8 is correct

[tulip yarrow]

Alright! Thank you!

[languid pecan]

Hello can someone help me get started on solving a triangle given 2 medians and a vertex?

[pseudo pecan]

@languid pecan
What do you mean solving?

[upper karma]

<@&286206848099549185>

[chilly flower]

[steady sleet]

#❓how-to-get-help, rule #4: If your question has not been answered for a minimum of 15 minutes, you may use the Helpers tag once.
Please read the rules: #❓how-to-get-help!

[upper karma]

At least he posted question before pinging that's really the worst type of pings

half the time I ping helpers 15 minutes later no one still helps me reeeeeeee

[chilly flower]

Well just to double check with myself, that symbol means they're congruent right? @upper karma

[upper karma]

I already answered it thank though, sorry for pinging early.

[lofty scaffold]

How did u do

Im new too

Common side?

@upper karma

Did u use an ang?

[chilly flower]

To verify that two triangles are congruent, both triangles have to share certain traits

In this case, both of them have equal sides (lines JS and JF are marked, lines SK and KF are both 3, and the line JK is shared)

Because all the sides are congruent, there is no way for the angles to not be congruent as well, so the two triangles are congruent

I think it's like

SSS, SAS, and ASA

Dunno

[lofty scaffold]

Ah yes so its by sss

Comon side , 3cm by 3cm and also equ segment

[chilly flower]

Yeah, just a roundabout way of showing all those legs' lengths are equal lol

[tawdry monolith]

Guys real quick is there any website that has all of the geometry (circle related) "sentences"

Or rules idk what do u call it in english

[left ginkgo]

postulates and theorems?

[vocal minnow]

circles as in unit circle or

[upper karma]

perhaps

http://mathworld.wolfram.com/EuclidsPostulates.html

[cursive spire]

Is it viable to find the distance between two points without y?

(y equating to 0 here)

[glass fjord]

Hi! I need help answering these four questions and also, can someone explain the congruency tests to me please?

[outer aspen]

Hi can someone help me with this problem

So the question was finding the measure of angle X and I thought it was 140 until my friend said it was 130

I’ve gtg @ me if you can help

Thanks!!

[paper glacier]

yep that's correct

x = 140

[outer aspen]

@paper glacier Awesome thanks!

[lament bay]

I’m straight up stuck on this one

[lament bay]

<@&286206848099549185>

[supple haven]

whats pi/pi/4

[lament bay]

1/4

[supple haven]

$\frac{pi\pi}\frac{4\1}

oh so it is 1/4

thanks dude

[lament bay]

It should be opposite away around

Thanks dude

[sacred cradle]

can someone help me on this one

[lament bay]

@supple haven what about B?

[supple haven]

um

ohh uhh

I think I got it

gimme a sec

[lament bay]

Ok

[supple haven]

is it 37?

[lament bay]

I don’t know

[supple haven]

oh I thought you were quizzing me

erm

<@&286206848099549185> I need you rn y'all

[lament bay]

Nah I just don’t know how to do it

[supple haven]

ohh shit I think I got it

I think its 56

dont trust me for sure though

[plucky marlin]

it's 34

[supple haven]

can you explain it?

[plucky marlin]

yeah sure

https://prnt.sc/luatx2

ADC = 180 - (71 + 90)

= 19

so in the triangle BDC

BCD = 180 - (127 + 19)

[supple haven]

damn I'm hella dumb, I should've seen that

[plucky marlin]

[sacred cradle]

ty

[lament bay]

Any idea on B @supple haven

[supple haven]

lemme see

I think you gotta add them together then factor

[winged fulcrum]

yo can someone help me with trig plz?

simplifying trig expressions

i dont get how it got thAt

and when i use an online calculator it says the original cant be simplified

[plucky marlin]

online calculator has the dumb

$\frac{\cos x}{\frac{1}{\cos x} + \frac{\sin x}{\cos x}}$

[somber coyote]

Salty Boi:

[plucky marlin]

$\frac{\cos x}{\frac{1 + \sin x}{\cos x}} $

[somber coyote]

Salty Boi:

[winged fulcrum]

then i got cos(x)/(1+sin(x))/(cos(x)

[plucky marlin]

$\frac{\cos^2 x}{1 + \sin x} $

[somber coyote]

Salty Boi:

[plucky marlin]

$\frac{1 - \sin^2 x}{1 + \sin x} $

[somber coyote]

Salty Boi:

[plucky marlin]

$\frac{(1+ \sin x)(1 - \sin x)}{1 + \sin x}$

[somber coyote]

Salty Boi:

[plucky marlin]

$1 - \sin x$

[somber coyote]

Salty Boi:

[plucky marlin]

DONE

[winged fulcrum]

OH

TY

[plucky marlin]

[winged fulcrum]

OML

you saved me ❤

[plucky marlin]

[long sapphire]

i felt that mic drop

[winged fulcrum]

i got to like 3rd step then got stuck ty ty

xd

[plucky marlin]

np

👍

[shut swift]

hey guys, whats a good resource to practice proofs?

[little osprey]

What kind of proofs

[shut swift]

aaa, aas, sas, hl

triangle congruence mostly

@little osprey

[little osprey]

AAA as in Angle Angle Angle?

[shut swift]

yeah

[plucky marlin]

what in the world ???

if all the angles of a triangle is the same that doesn't mean they are congruent

they would be similar triangle @shut swift

[shut swift]

meant sss sorry

brain is kinda fried right now

[plucky marlin]

my brain is fried too

i shouldn't know geometry

[shut swift]

so any good resources?

[winged arch]

I am pretty confused on how to do number 1 on this study guide

can any one show me the method to solve this problem?

[upper karma]

Its just proofs

[little osprey]

@plucky marlin Yeah, thanks for clearing the point

[plucky marlin]

wut point ?

[little osprey]

Scroll up a bit

[plucky marlin]

i'm too lazy

nvm

[real sparrow]

Ooof

[keen tangle]

geometry-trigonometry?

[void osprey]

this is geometry

would it be #probability-statistics

[keen tangle]

#probability-statistics

yeah

[oblique ibex]

show that the equation sin^2x + 3 sinxcosx = 4cos^2x can be written as a quadratic equation in tanx

so basically i should make the sin and cos in tan form?

[plucky marlin]

$\sin^2 x + 3 \sin x \cos x - 4 \cos^2 x = 0$

[somber coyote]

Salty Boi:

[plucky marlin]

i would divide by cos^2(x)

$\frac{\sin^2 x}{\cos^2 x} + \frac{3 \sin x \cos x}{\cos^2 x} - \frac{4 \cos^2 x}{\cos^2 x} = 0$

[somber coyote]

Salty Boi:

[plucky marlin]

$\tan^2 x + 3 \tan x - 4 = 0$

[somber coyote]

Salty Boi:

[plucky marlin]

DONE

🥞

@oblique ibex

[oblique ibex]

oh thanks lol

[plucky marlin]

[exotic plover]

i'm really stuck on this equation, i know i can use compound angle formulae to change it but i end up overcomplicating questions like these... could somebody help me?

[upper karma]

whats the issue when you use them

show your working

[exotic plover]

I just don’t know where to take it to get it to look like that

I got up to there

[upper karma]

uh

you could use the fact that sin(x) = cos(90-x)

[plucky marlin]

damn ur smart

[upper karma]

no you are mom

[plucky marlin]

actually on second thoughts, i'm not ur mom

[exotic plover]

So where the sinx is, could I write cos20 then?

[plucky marlin]

umm

no

[exotic plover]

I'm just super confused by all the sins and cos bc I'm not good at trigonometry

[upper karma]

bud.

sin20 = cos70

[plucky marlin]

uwu

[upper karma]

do you see why

dont memorise it

[plucky marlin]

yes i do

[upper karma]

ok

[plucky marlin]

but wait

[upper karma]

stfu

[plucky marlin]

dats wrong tho

[upper karma]

what?

[plucky marlin]

sin(20) = cos(70)

smh

[upper karma]

what?

[plucky marlin]

wut ?

[upper karma]

go slow buddy

check my post again

[plucky marlin]

wut ?

u fucking cheater

u changed it

not cool

[upper karma]

wtf is this a game

[exotic plover]

Okay, so cos70 then where the sin20 is

[upper karma]

@exotic plover why though..

[plucky marlin]

it seems like i have to take matters into my hands

[upper karma]

@plucky marlin go ahead

[plucky marlin]

😤

[upper karma]

i exit

[plucky marlin]

but i give u the honor @upper karma

go ahead

[exotic plover]

Because at first I thought you could collect all the terms together or something, I just don't really see it

[upper karma]

no use collecting if there is no possibiliy of having common terms

using sinx= cos(-x) gives you that

use it and come back

[exotic plover]

Okay, thanks

[plucky marlin]

alright

i solved it

if u need the steps ping me @exotic plover

[exotic plover]

Thank you I appreciate the help

[uneven ruin]

cos(x^2-2x)=0
How can I solve this trigo function ? I need to find points with the X line (Y=0)

[gritty siren]

You need to use the known characterisation of zeroes of the cos function

[umbral snow]

Could you solve cos(u) = 0?

[uneven ruin]

I know how to solve things like
cos(u)=0
...
u=1 obviously

[gritty siren]

=pup cos(1)

[charred spear]

Do more with Wolfram|Alpha pro

Query made by

@gritty siren

[gritty siren]

cos(1) doesn't look like it's 0

[uneven ruin]

It is in my calculator, at least I think it is but nvm that.

This is the function right ?
cos(x^2-2x)=0
I know that no matter what, 90 degrees should be inside the cos ( ) to make it equal zero
But how do I solve it, I don’t know how to get rid of the cosine, can I just write it over like this and claim it’s supposed to be equal to 0.5Pi (cuz half pi is 90 degrees right)? ==> x^2-2x=0.5π

[gritty siren]

π/2 isn't the only solution to cos(u)=0

anything of the form π/2 + kπ with k in Z also work

[sinful isle]

hi guys

cointerior angles

whats the diffrence between cointerior and coextirior

[uneven ruin]

cos(u) = 0
So there’s 2 possible solutions for this ?
u = 1/2π +2πκ
u = -1/2π +2πκ
I can just write it as:
u = πκ
right ?

[serene field]

If you were to test that

Let k = 1

If u = pi

cos(pi) = 0

Is this true?

[gritty siren]

there are infinitely many solutions to the equation of unknown u: cos(u) = 0

[sinful isle]

5x=a what is a

[uneven ruin]

Mb I missed that π/2

[sinful isle]

uh

360

and then it goe sup to like

540?

it goes up by 180 if its 5 lines ye

[uneven ruin]

Is the same principle works for higher degrees ?
Say that I have something like
Cos(X^9-8X^5)=0
To solve this I just realized I need to use cos^-1 on the right side in order to convert it into π
And one time the solution will be +1/2π and one time -1/2π ?
So a general solution will be like this ?
X^9-8X^5=1/2π + 2πκ
X^9-8X^5=-1/2π + 2πκ

[gritty siren]

the exercise is about finding the set of all Xs that work

[uneven ruin]

Well in those kind of exercises I will have to solve X between a certain set of values by going over all the K’s so it shouldn’t be a problem.
But is my assumption right about the general solution (doing what I did there) ?
Ofc it needs some more algebra into it to make it in a form of X = something but I will do that later.

[queen socket]

So for the graph of tanx, pie/2 is essentially 1/0 which makes it undefined? Because tan = sin/cos

And the coords for 90 degrees is (0,1) on the unit circle

[keen aspen]

Yes

Which is why there is asymptotes for pi/2+pik

[lament bay]

Help on B

<@&286206848099549185>

[dawn stone]

@lament bay Since I is centroid, BI/IF=2/1. Solve for y from there.

[rare ravine]

k so i got no help in #geometry-and-manifolds so i'll ask here i think :/

can u plz go take a look ? i've to do it before Monday

that's important

[upper karma]

guys

can someone show me the incenter theorem pls?

i mean, a demostration. I dont know why this is true

https://gyazo.com/7e3460a7df9073e7c321a4a66f2b7a7e

why m/n is proporcional to a+c / b

[upper karma]

can someone help me with this ?

https://gyazo.com/7178e065cbcd0d833af235e657fff3da

[upper karma]

The ratio of the sides BF:FC will be equal to AE:EC which is 2:3 @upper karma

Yeah I got it right.

I did it already, haha.

That's great!

I'm struggling with something else though.

[upper karma]

ty for answering

[keen tangle]

do you still care for that answer?

[upper karma]

what makes u think i dont??

[keen tangle]

idk

do you know anything about mass point geometry?

[upper karma]

no

[keen tangle]

basically it's about putting some mass at certain points and seeing where the center of that mass is

like if you put equal masses at the vertices A, B, C you'd get that the center of them is the centroid of ABC

i guess you could just read sth about here

https://en.wikipedia.org/wiki/Mass_point_geometry

it's the easiest way I know of to solve stuff like this

[upper karma]

i wanna know why that is proportional

[keen tangle]

yeah I can tell you when you understand what that is about

[upper karma]

i know what those mass points are

[keen tangle]

ok then

do you know the angle bisector theorem then?

[upper karma]

nop, i know what bisectors are but not a theorem

[keen tangle]

it basically says that in your pic AD/DC is equal to c/a

because the altitudes from D to AB and CB are equal

[upper karma]

ah okey

[keen tangle]

and then areas manipulation

do you see it?

[upper karma]

i think (?)

[keen tangle]

[X] - area of X

$\frac{[ADB]}{[DCB]}=\frac{c}{a}$

when you take sides AB and BC and altitudes dropped on them

but $\frac{[ADB]}{[DCB]}=\frac{AD}{CD}$

when you take sides AD and CD and the altitude (it's one line) dropped on them

[upper karma]

https://gyazo.com/285c0a7540185909b86ff6d3f0530656

would that be B?

[keen tangle]

that would be A

also Aza do you get it so far?

[left folio]

How do i graph y=5sin(pix/2)-4

[upper karma]

https://gyazo.com/7425b88bf0b93cf01d55aaaa06126b63

I've been stuck on this homework problem

for 45 minutes.

UGH.

??

anyone?

[upper karma]

not D

not A

what'd you think now

[bitter stratus]

how do i prove this

[dire rampart]

use double angle formula

for both sin2x and cos 2x

[bitter stratus]

idk what else to use i keep getting back to the same thing

[dire rampart]

you see the second line?

divide top and bottom by cos^2

then use the identity $\tan^{2}(x)+1=\sec^{2}(x)$

[somber coyote]

lemon catto:

[bitter stratus]

i can just divide the top and bottom by cos^2?

is that allowed?

actually i have a question

where did the 2sinxcosx go?

[dire rampart]

ye you can do that since its just dividing by 1 over all

what they did is much quicker

(cosx+sinx)=cos^2+sin^2+2sinxcosx

now cancel a factor of cosx+sinx from the top and bottom

and divide top and bottom by cosx

[bitter stratus]

i dont get it, they just took out the 2sinxcosx?

[dire rampart]

no

expand (cosx+sinx)^2

what do you get?

[bitter stratus]

ohhh (cos^2+2sinxcosx+sin^2) ?

[dire rampart]

yup

they just compressed it into a bracket

[bitter stratus]

so the 1 the used Pythagorean identities?

and for the sin2x they used the double angle formula?

[dire rampart]

yea

[bitter stratus]

oh that makes more sense

thank you so much

[bitter stratus]

can i simplify sin4x to sin 2(2x) ?

[upper karma]

yes

it becomes 2sin2xcos2x

which becomes 4(sinxcosx)cos2x

though idk howll that help you

[bitter stratus]

okay thank you

and another question

is this possible "1 - cos(4x) = 2sin^2 (2x) "

[gritty siren]

Yes

[bitter stratus]

how though

shouldnt you have a 1-

infront of the 2sin^2(2x) ?

[gritty siren]

tbh I plotted the two and the curves were the same, there's nothing too deep backing up my reply

[bitter stratus]

but wouldnt it make a difference if i was proving trig identities?

[gritty siren]

It's surely doable using the identities you Know

[bitter stratus]

does this even equal tanx?

[gritty siren]

by plotting the thing, it looks like it works

[bitter stratus]

tf

i been doing this for an hour and im not getting it

[gritty siren]

$\sin 4x=2\sin 2x\cos 2x$\
$1-\cos 2x=1-\cos^2x+\sin^2x=2\sin^2x$\
$1-\cos 4x = 2\sin^22x$

[somber coyote]

Tuong:

[gritty siren]

$\frac {2\sin 2x\cos 2x ×2\sin^2x}{2\sin^22x×\cos 2x}$

[somber coyote]

Tuong:

[gritty siren]

$\frac{2\sin^2x}{\sin 2x}$

[somber coyote]

Tuong:

[gritty siren]

$\frac{2\sin^2x}{2\sin x\cos x}$

[somber coyote]

Tuong:

[bitter stratus]

hmmm okayy thank you!

[old warren]

kinda need a tutor. I live in Hawaii, my school offers no math help, our "teacher" assigns trig work from chiacago, essentially we're doing an online trig course. We have a permanent sub unqaulified to teach math. So if anyone wants to help me, that'd be greatly appriciated. Half the grades in my school dropped trig for that reason. I'm still enrolled in the course to learn the subject, for college, and to recieve an honors certificate for completing highschool with 5 credits of math, 5 creds of english, & 5 creds for science. If anyone can assist me on the work I've been given or reccomend me tutor sites/helpful math stuff, I'd greatly appriciate it for me and for my fellow clasmates that strugle.

[upper karma]

Hello there, I am struggling with is this inequality, cos(sin(x)) =< sin(cos(x)), any hint please ?

[upper karma]

not sure

but probably along the lines of sin(90-sin(x))<= sin(cos(x))

now you can eliminate the sin since sin is an increasing function from [0,π/2]

90-sin(x) <= cosx

[dire rampart]

are you trying to solve for x?

[smoky violet]

cos(sin(x)) doesn't equal sin(cos(x)) for any real x

also sin(cos(x)) is not greater than cos(sin(x)), it's always less

[upper karma]

[dire rampart]

roses are red, violets are blue, cos(sinx)>sin(cosx) for all real x, that much is true

[upper karma]

violets are violet

[plucky marlin]

violets are purple not fucking blue

smh

[surreal patrol]

Violets are cyan

[warm pine]

Violets are violet

Because violet is a strong independent colour who don't need no blue

[clear haven]

violets at least are

[bitter stratus]

how would you find important points for this?

to graph it

[opal blaze]

@bitter stratus you need to make it into a sine function (if you don't know how, I can show you), then find the amplitude, "line of symetry", frequancy and the "shift".

[bitter stratus]

yes please

i dont know how to do that

@opal blaze

[cinder brook]

Boys last 2

Last 3 if possible

[opal blaze]

@bitter stratus Ok, just let me write it out

[bitter stratus]

okayy thank you so much

[cinder brook]

Somebody help me as well plz

[opal blaze]

@cinder brook I might look at it after this

@bitter stratus

[bitter stratus]

hmm okay, i see what you're doing :p

thank you so much!!

[opal blaze]

@bitter stratus and btw, the reason for me multiplying with sqrt(3^2 + 4^2) is so that cos^2(phi) + sin^2(phi) = 1

such that phi is the same number in both sine and cosine

[bitter stratus]

ohh that calrifies things a bit more, thank you again!

[cinder brook]

Any1 had a go at my problema

[opal blaze]

I am looking at it now

@cinder brook idk, it is quite complex

I do not know how to solve it

Im looking at power of a point theorem, angle bisector theorem and isosceles triangles. But there is nothing that really jumps out at me

[keen tangle]

i got the 10th one

idk if you want hints or full solutions

@cinder brook

[cinder brook]

Ur choice full solution if possible

[keen tangle]

ATDB is cyclic

just angle chasing

[cinder brook]

Did you make one circle or 2

[keen tangle]

wdym make circles?

[cinder brook]

For tangents

[keen tangle]

there's one circle

the circumcircle of ABC

[cinder brook]

This correct

@keen tangle

[keen tangle]

yeah the picture's correct

[cinder brook]

How's atdb cyclic

[keen tangle]

angle BDT=angle ACB

bc TD||AC

and ACB=TAB

[cinder brook]

Still don't get it

[keen tangle]

do you get why BDT=ACB?

[cinder brook]

Yes

Alternate angle

[keen tangle]

yeah

and ACB=TAB

[cinder brook]

Tangent triangle property

[keen tangle]

it's a pretty well known fact in olympic geometry

yeah that

[fallow edge]

Olympic geometry?

[keen tangle]

yeah what?

[fallow edge]

Never heard that term

[keen tangle]

idk I'm not native english

[fallow edge]

Oh

maybe planar geometry?

[cinder brook]

So what next @keen tangle

[keen tangle]

so BDT=BAT

[cinder brook]

How

[keen tangle]

BDT=ACB=BAT

[cinder brook]

Ok

Ic

And we now BTD

We know BTD

[keen tangle]

but that's not what we want

we want to use ADB=180-ATB

[cinder brook]

Ok

[keen tangle]

Do you see it?

[cinder brook]

No

[keen tangle]

We want to prove AD=CD

[cinder brook]

Ya

[keen tangle]

so we just want to prove that ADC is iscosceles

[cinder brook]

Ya

[keen tangle]

so if we get the ADC=180-2ACD we're done

ok let's take a step back

[cinder brook]

I mean DAC =180-2ACD

U*

[keen tangle]

we know that ADB=180-ATB right?

[cinder brook]

Thts wht I'm not getting

[keen tangle]

cause ATBD is cyclic

[cinder brook]

T and D are not on circumference

[keen tangle]

the opposite angles sum up to 180 in a cyclic quadrilateral

[cinder brook]

So howr they cyclic

[keen tangle]

and we know that ATBD is cyclic

[cinder brook]

For cyclic all points must be on circumference

[keen tangle]

but they don't lie on the same circle as C

we found that BDT=BAT

[cinder brook]

Yes

[keen tangle]

which proves that the quadrilateral ATDB is cyclic

the points ATDB all lie on some circle

[cinder brook]

Ok

I see

[keen tangle]

and since ATDB is cyclic

its opposite angles sum up to 180 degrees

so ADB+ATB=180

[cinder brook]

Yes

Got it

Thanks

[keen tangle]

np

I hope it made sense

[halcyon hollow]

Can i run a proof by someone real quick? I need to write a paper essentially based off it and i need to make sure it's right

[umbral snow]

Go for it!

[halcyon hollow]

Given 3 circles, a large circle with a radius R and two identical small circles inside, with a radius r. All circles are tangent to each other, find the arclength of the large circle between the two points of tangency to the small circles.

I got 2R*arcsin(r/(R-r))

[normal elbow]

yo

how can I turn $$((sinx-1)/(cosx))$$*$$(1/(1-sinx))$$ into -1/cosx?

[somber coyote]

Zoomin in the foreign:

[normal elbow]

meh good enough 😂

[hard gale]

$\frac{\sin(x)-1}{\cos(x)}\cdot\frac{1}{1-\sin(x)}$

bettre enough

[normal elbow]

yea that ❤

[somber coyote]

emeric75:

[normal elbow]

its sinx-1 on top

yea

[gritty siren]

Good bot

[hard gale]

so yeah first you could multiply your two fracs $$\frac{\sin(x)-1}{\cos(x)}\cdot\frac{1}{1-\sin(x)} = \frac{\sin(x)-1}{(1-\sin(x))\cos(x)}$$

[somber coyote]

emeric75:

[hard gale]

now notice that $\sin(x)-1 = -(1-\sin(x)$)

[somber coyote]

emeric75:

[normal elbow]

yup

[hard gale]

hence we get $$\frac{\sin(x)-1}{\cos(x)}\cdot\frac{1}{1-\sin(x)} = \frac{-(1-\sin(x))}{(1-\sin(x))\cos(x)}$$

[somber coyote]

emeric75:

[normal elbow]

ahhh

[hard gale]

good bot ^^

[normal elbow]

beautiful, thankyou so much

[hard gale]

👌

[normal elbow]

uhhhh

what dis

[hard gale]

waht what? the sec thing?

[gritty siren]

1/cos

[normal elbow]

how do you prove it

I know wut sec is

[warm pine]

Is this an A Level paper?

[normal elbow]

I mean, how do you prove it using the triangle

[hard gale]

real q : why do they even give a figure?

lel

[normal elbow]

its maths a level yeah

[hard gale]

ik

[normal elbow]

cus if there is no triangle its only worth 4 marks, 5 max

😄

[gritty siren]

Rewrite 1 as cos²/cos² and tan² as sin²/cos² and it all goes smoothly

[warm pine]

You can just do it from the identity

I assume they want you to show to geometrically

Yeah just label AD as cos(t) + BD as sin(t) and BC as tan(t)

Then use Pythagoras

Although using the identity for sin and cos is faster

[normal elbow]

ive got tit

it

thanks

[feral fractal]

Hey guys. I have an impossible exercise I'd need some help with. A cube and a ball have the same surface area. What's the coefficient between their the volumes of the objects? Let it be V1/V2 where V1 is the cube volume and V2 is the ball volume

[chrome fiber]

if the radius of the sphere is r, and the side length of the cube is a and their surface areas are equal, then whats the r/a?

[feral fractal]

i dont have a clue

wqit

wait

6a^2 = 4 pi r^2

[dire rampart]

yea now whats r/a

[feral fractal]

could you give me a hand? I think I messed it up.

[normal elbow]

ey bois

I've found both the x-values for the intersects

but idk where to substitute the x values for y

excuse the retardation

nvm

im a stinky nonce

I just thought it through

I forgot to minus 4 cause I was just looking at mod[2x+3]

got the right answer

[dire rampart]

@feral fractal so 6a^2=4pi r^2, meaning a^2/r^2=2/(3pi) meaning $\frac{a}{r}=\sqrt{\frac{2}{3\pi}}$

[somber coyote]

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

[dire rampart]

reeeee

$\frac{a}{r}=\frac{\sqrt{2}}{\sqrt{3\pi}}$

[somber coyote]

lemon catto:

[dire rampart]

now you want volume of cube/volume of sphere right?

so $\frac{a^{3}}{4/3 \pi r^{3}}$

[somber coyote]

lemon catto:

[dire rampart]

or $\frac{3a^{3}}{4\pi r^{3}}$

[somber coyote]

lemon catto:

[dire rampart]

so $\frac{3}{4\pi} \cdot \bigg(\frac{a}{r}\bigg)^{3}$

[somber coyote]

lemon catto:

[vague pier]

is anyone available to help with geometry?

[hard gale]

just post your things someone will answer

[vague pier]

ok,

i need help with 4 midpoint problems

[serene field]

Recall the midpoint formula.

[vague pier]

M= x1+x2/2 and y1+y2/2 right?

[somber coyote]

Yue:

[vague pier]

i got 1/2, 0/2

am i doing something wrong?

[serene field]

Your x coordinate of the midpoint looks off.

-3 + 2 is -1

[vague pier]

so -1/2, 0/2?

[serene field]

0/2 is the same as 0

[vague pier]

ok,

this is my next problem,

i got 4, -2.75

[vague pier]

@serene field are you still there?

[halcyon hollow]

Your x cordinate is right

the y cordinate is wrong @vague pier

[vague pier]

where did i go wrong

[serene field]

Your two y-coords are -2.5 and -8

They should not average to -2.75

[halcyon hollow]

@vague pier you when you add -2.5+(-8) you get -10.5 not -5.5

which is where i assumed you messed up

[vague pier]

oh! yeah

[kind mist]

https://cdn.discordapp.com/attachments/367528357614845952/523970434614558730/IMG_20181216_131106.jpg

Can some1 help me on #7

[dire rampart]

factor out a cosx

[kind mist]

yeah got that, confused on whats after that

[dire rampart]

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

[somber coyote]

lemon catto:

[kind mist]

Ah thanks I see it now

[wraith pumice]

I'm not understanding the special right triangles: is the 45-45-45 sqrt2 for the two legs, and 2 for the hyp. or is it 1/sqrt2 for the two legs and 1 for the hyp?

which one should I use when looking for exact values of all 6 trig funtions?

[serene field]

Both are valid for isosceles right triangles.

If you apply the Pythagorean Theorem, you'll see that the legs:hypotenuse ratio of sqrt(2):2 and 1/sqrt(2):1 is true.

[hollow token]

Can someone explain #7A

[serene field]

Plug -1 into the function.

i.e. look at x = -1

[hollow token]

Do I plug y=f(-1). ?

[left ginkgo]

the line on the graph represents the function f(x)

the x axis is where you want to look.
Particularly at where x = -1

at that point, you determine the y value based off of the y axis

and that is your f(-1)

[hollow token]

I don’t understand that lol.

[left ginkgo]

uhhh

[hollow token]

Is #24 in the 3rd quardrent?

[keen aspen]

no

cosine is positive

so either I or IV

tangent is negative

so either II or IV

so that limits to IV

[hollow token]

Btw #27 is in 4 quardrent right?

[keen aspen]

Correct

[tardy plinth]

I have 12 pages of geometry idk how to do due by tomorrow for a final study guide

Thought you guys could have a bit of a chuckle with that

[dire rampart]

oof

[tardy plinth]

🤠 🤠

Okay good talking gamers

[sick veldt]

I think this is right

also for the last one it has to be CPCTC since HL theroem is only for right triagnles

[upper karma]

yes

https://gyazo.com/576cd5d394065e464d965edb32f83e4d

Is this right

[sick veldt]

Looks good