geometry-and-trigonometry

[lime dune]

no

ask a better question

one that isn't so broad to the point of being useless

[obsidian harness]

can you send the problem in your textbook / online?

[silent plank]

being helped atm at #help-47

[pure knoll]

hey is it possible someone can help me in geometry so i wont get summer school on it this is mad hard

i got like 57 / 148 and i gotta complete it all before may 20 and like its not possible for me bc IDK HOW TO DO GEOMETRY

so my dms are open to anyone who wants to help

[dark sparrow]

then maybe you in fact need summer school.

[manic swan]

basic geometry should be super doable in summer school

• fun

learn your pythagorean and your thales and stuff

[upper karma]

i need help 💔

forgive me if these are dumb questions i wasnt in school for 3 weeks due to surgery

[bright zephyr]

for the first question, extend the PO line till it touches line WE at say point B and also make a line, say RA which touches the line PO at A

this way youll get a 2 triangles, OBW and PAR and we'll also get a square RABE

you can find the area of them individually and add them up

[daring frigate]

Anyone here to help me ;-;

[bright zephyr]

here we are given the focus and the equation of directrix. we know that the distance between directrix and any point on parabola is equal to the distance between the focus and the point

so let there be a point P(x,y) that lies on the parabola

[daring frigate]

Can U check my working ?

[bright zephyr]

sure

[daring frigate]

I assume I did right

@bright zephyr is the color fine or should I change it ?

[bright zephyr]

nah its fine

but we need to find the distance between them

[daring frigate]

x-1 ^2 + y^2 = ??

what's the radius ?

then

[bright zephyr]

uhh do you know how to find the distance between a point and a line??

[daring frigate]

we are getting a as zero if we try to solve using perpencidular method

[daring frigate]

no clue what's that

[bright zephyr]

uhh there's a formula to find the distance between a point and a line

where the equation of a line is given

so let's say your equation of line is ax+by+c=0

[daring frigate]

well My clg taught me like this
distance from the directrix to focus is 2a
2a = |directix|/ sqrt[((coeff of x) ^2 + (coeff of y)^2)]

[daring frigate]

@bright zephyr ^^

[bright zephyr]

and you want to find the distance from point P(p,q)

[bright zephyr]

what is |directrix| here??

[daring frigate]

the line parallel to the axis

[bright zephyr]

no i meant like |directrix|

oh got it

nvm

[bright zephyr]

but this wont help us here

[daring frigate]

Hmm

[bright zephyr]

we need to find the distance of a point P(x,y) which lies on the parabola from directrix and from focus

[fading dove]

easy problem, A is the correct answer

let p be some (h,k) (parabola is a locus )

then take root ((h-1)^2 +k^2 = h +2k -1 / root(1^2 +2^2)

at the rhs denominator you get root(5)

then square both the sides

you will get it in the form of

4h^2 4hk + k^2 -8h +4k +4 = 0

then put h = x and k = y

which is option A

[azure helm]

how to find bcd?

if AD = DC
8α + x = 180
x1 = 180 - 8α

oh I think I see where this is going

8α + x1 + x2 = 180

8α + (180-8α) + x2 = 180

no wait

[daring frigate]

Thanks @fading dove

[fading dove]

ur welcome

[next grotto]

how did you say bd=dc?

[obsidian harness]

DCA = 4alpha and from sum of angles in triangle ABC, angle BAD = 180 - 12alpha
hence angle ADB = 6alpha, angle DBC = alpha

but then angle DBC = angle DCB implies DB = DC

so BA = AD = DC = BD

[whole sequoia]

need help with 4

all i know is there is a vertical angle meaning they are congruent

but idk how to prove they are similar

[summer cradle]

2 triangles are similar if either all their angles are congruent or their sides ratio is constant

[whole sequoia]

ye ik

i have a problem to both angle e and h

idk how to find their angles

[summer cradle]

you dont have to

you just need to know if their angles are the same

regardless of their measure

[whole sequoia]

how do i two column proof it tho

[whole sequoia]

ik the F angle can be proved congruent by VAT, and 60 is definitoin of congruent angles, but idk what i use to the rest

[summer cradle]

sum of angles in a triangle is always 180 right?

[whole sequoia]

ye

how i put it

[whole sequoia]

i already know the rest are congruent but idk how i put it on two column proof

[summer cradle]

you can just write sth like:
"Both triangles have 2 congruent angles. Since the sum of angles in a triangle is 180°, the third angle must be congruent too. Therefore the both triangles have congruent angles. Hence they are similar."

that's just an example

you can use your own words

[whole sequoia]

ye ig so

never knew they accepted those

ill try

[slender bronze]

Am i correct ?

That's maybe a bit harder compared to use 3 isosceles triangles ..

[azure helm]

you need to realize that angle obd = alpha,

[visual flume]

help pls

[summer cradle]

try to find similar triangles

[next grotto]

what?

[next grotto]

yes how?

[obsidian harness]

what about it didn't you understand?

[upper karma]

Guys u know u can geometrically Derivate sinex

[next grotto]

@obsidian harness

[ripe helm]

prove that triangle HKT is a square triangle
thanks for your help!

[onyx brook]

Messiest diagram i hv ever seen

[ripe helm]

yeah thats the geometry question from the entrance exam for grade 10 specialized school

[onyx brook]

Oh alr jesus christ

[obsidian harness]

Man if you just said something rather than waiting for me to come back

We could have got it over now

Oh right that, OBD = alpha

You should agree that angle ABD = 6alpha then, BAD is isosceles and we know angle BAD

Then ABC is already 5alpha (given)...

[signal vapor]

There is no information about the diagram

[sage radish]

thanks for the problem

[sour scaffold]

use the cge triangle to fint the small part

or oe

use pythagoras theorem

but need to find the base first

[azure helm]

AD = DC:

angle ACD=4alpha then angle DAC = 4alpha,

now look at triangle AOC;

we know angle DAC = 4alpha, we know angle OCA = 3alpha
thus because the sum of interior angles of a triangle is 180, then angle AOC is 180 - 7alpha

now look at the triangle BOA

because (180 - 7alpha) + angle BOA must sum to 180
(180 -7alpha) + angle BOA = 180
angle BOA = 7alpha

we know angle ABO is 5alpha, thats given

the interior angles of triangle BOA must sum to 180

angle ABO + angle BOA + angle BAO = 180
5alpha + 7alpha + angle BAO = 180
angle BAO = 180 -12alpha

now look at triangle BOD

because angle BOA + angle BOD = 180
angle BOA = 7alpha
7alpha + angle BOD = 180
angle BOD = 180-7alpha

from knowing angle BOD we know (angle OBD + angle ODB = 7alpha )

now, since we know angle BAD is 180-12alpha

then, since BA = AD, we know angle ABD = angle BDA

so

if angle BAD = 180 -12alpha, that must mean angle ABD + angle BDA = 12alpha
but since angle ABD = angle BDA, and knowing that angle ABD = 5alpha + angle OBD then we get this

12alpha = 2(5alpha + angle OBD)
6alpha = 5alpha + angle OBD
and then angle OBD = alpha

since we know angle OBD = alpha, we know triangle BDC is iso

[sage radish]

I got like 6 or something

[next grotto]

thnx @azure helm and @obsidian harness

[chrome token]

what do we call similar with same orientation?
i mean a thing which homothety + rotation can make equal

[visual flume]

ohhh my god i kept tryna use pythagorean theorem

tysm

cuz the lesson was based off of that

my glourious king

[upper echo]

If AE=a, DF=b, CG=c then R=2abc/(ab+bc+ac).

[visual flume]

lol can yall help me with this one

[bright zephyr]

where are you stuck at??

[sage radish]

i feel like its possible for there to be multiple solutions to this but i guess ill check

[knotty jacinth]

guys

someone help wit ts

1,4,3,and 7

see how the x is on end of the side instead of in the middle

i am so confused

[bright zephyr]

tangent is the line which touches the circle at only one point

in first, XY is a tangent because it only touches the circle at one point which is X

[sage radish]

maybe if its on the side you just double it cause its a radius

tangency is basically just a right angle so you can just check if it fulfills a^2+b^2=c^2

[knotty jacinth]

yea i get that

oh wait

your right mb

i was just overthinking

thanks tho

so this would be x=6

am I wrong?

[sage radish]

unsure for this one

[sage radish]

solution is 9+10

(21)

[knotty jacinth]

Is the radius outside of the circle the same with the one inside?

[bright zephyr]

i think that here x is the radius so by applying pythagorus: $x^2+12^2=(6+x)^2$

[somber coyote]

Aria

[bright zephyr]

the whole hypo is 6+x ig

[knotty jacinth]

chatgpt gave me x=9

[sage radish]

lmao that also makes sense

[knotty jacinth]

so yeah ur right

[bright zephyr]

dont use chatgpt, its sometimes unreliable

[knotty jacinth]

ur right

[bright zephyr]

that doesnt mean chatgpt is always right bruh, it makes mistakes

[sage radish]

After solving the polynomial related to this question the two solutions are 21 and 12 so you can express the final solution as 33/2 +- 9/2
but looking at the orientation in the problem its more likely to be 21

[sage radish]

true I tried to verify my answer was correct for the last one and it got the answer totally wrong

[signal vapor]

O3 ???

It is surprisingly accurate

[bright zephyr]

yeah happened to me a lot of time so now i avoid it

[sage radish]

idk you try

I highly doubt it could

[signal vapor]

I tested it with my local competition level problems got all integer types right

[signal vapor]

I tested 11 questions

[sage radish]

this problem is hard as to solve it most methods lead to a quartic equation

[knotty jacinth]

how to ts pls 💔

[signal vapor]

<@&268886789983436800>

[knotty jacinth]

164/360 times pi(11)^2

is this correct

[signal vapor]

Why do people even post scam links here

[knotty jacinth]

who?

[signal vapor]

They deleted it

[sage radish]

is 16 the arc length?

[signal vapor]

Someone sent some steam gift card

[knotty jacinth]

no

pretty sure

its the angle

[sage radish]

like degrees?

or radians

[knotty jacinth]

yes

like inside the traingle

16

[sage radish]

??

degrees or radians

[knotty jacinth]

degrees

[sage radish]

oh okay thanks

[sage radish]

thats sector area

it would be 2pi*11

[knotty jacinth]

oh wait

ur right

mb

but wouldnt be

so it would

164/360 times pi times 2 times 11?

[knotty jacinth]

so wouth this be 180/360 times 2pi times 5?

[sage radish]

yeah

[silent plank]

careful, that doesn't give you the full arc

using 180 only gives you the length of arcMK, not what they asked

[sage radish]

wait what is MKL then is it like arcMK+ arcKL

[silent plank]

yes

it's the arc from M to L passing through K

with two letters the shorter arc is implied
with three the middle letter indicates the path/direction

[azure helm]

np, let me know if you still didnt understood why alpha = 10

[knotty jacinth]

guys

how would u do this

would it be

(12.6)^2+(2x)^2=(21)^2

guys

guys i need this please

i am watching your stream why are you laughing

this is not funny 😭

[silent plank]

yes

[knotty jacinth]

alright thanks

btw

I think you are a " 💔

also

who masticates daily bro

thats weird

[silent plank]

most people chew daily

[trail tendon]

you've underestimated the power of elderly people

[meager sinew]

48

[meager sinew]

it's just calculating circumferences

circumference = 2 * pi * radius

the radius of the outer circle is obviously 16

the radius of the inner circle is 16 - 4.5 = 11.5

[summer cradle]

mark unknown sides with variables
use triangle similarity and Pythagoras
form equations and solve

[summer cradle]

you can do area of whole rectangle minus area of cutout part (which is a trapezium)

[dark sparrow]

!nosols

[lime crown]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[dark sparrow]

dunno if that was intended as "oh yeah here is the answer to the area problem just above" but if it was then the reminder stands

[lapis moon]

!showwork

[lime crown]

Show your work, and if possible, explain where you are stuck.

[echo sigil]

Prove that X and Y lie on the circle with diameter BC.BX and CY are bisectors ,F,D,E are tangent points

[lapis moon]

!showwork

[lime crown]

Show your work, and if possible, explain where you are stuck.

[lapis moon]

!xy

[lime crown]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[summer cradle]

I am curious about the solution to this problem

[ruby oasis]

@summer cradle

[summer cradle]

wrong problem

[summer cradle]

I mean this one

[ruby oasis]

Sure

[summer cradle]

oh using trigonometry

nice

but I was wondering if there is a pure geometry way

[robust jacinth]

hi

[summer cradle]

^ this is my own proof check it out

[wanton jay]

i hate geometry so much

[ruby oasis]

I guess i sent wrong solution

[summer cradle]

oh

[ruby oasis]

Is this question correct?

[summer cradle]

this question?

[ruby oasis]

Yes it is correct

[obsidian hornet]

This is a famous lemma:

[echo sigil]

Thaanks!!

[ruby oasis]

@summer cradle

[summer cradle]

ohh I didn't know that theorem

what does cevians mean?

[obsidian harness]

calling $BA = p$ and $BC = q$

$$(-p + q/2) \cdot (p/2 + q/2) = 0$$
$$-2|p|^2 - 2p \cdot q + q \cdot p + |q|^2 = 0$$
$$|q|^2 - 2|p|^2 - p \cdot q = 0$$

but $CF = -q + p/2$ and hence:

$$|CF|^2 = |q|^2 - p \cdot q + |p|^2 / 4$$
$$= |q|^2 - (|q|^2 - 2|p|^2) + |p|^2 / 4$$
$$= 9/4 \cdot |p|^2$$

now recognising $|p|^2 = (2/3)^2 (AD^2 + BE^2)$ it follows that $CF = \sqrt{AD^2 + BE^2}$

[somber coyote]

south

[plain storm]

Is the definition of a tangent line of a graph connected to the trigonometric definition of tangent?

Or is it just called that because it matches the behavior on the geometric properties of a unit circle's tangent

[dreamy hound]

there are various ways to define a tangent line of a graph.

could you tell what you mean by trigonometric definition of a tangent

[plain storm]

Opp/adj

Graph tangent as in the slope at a infinitely small rate of change of a point on a curve

Badly worded but yk what I mean

[dreamy hound]

well once you defined your tangent line somehow then yea, of course they are connected.
the slope of this line is precisely equal to the tangent of the angle between it and the x axis

[plain storm]

Ok I think I get it, I thought tan being slope was only for unit circles

Since then hyp=1

Ok wait I'm dumb nvm tan doesn't use hyp

[dreamy hound]

even though it is very convenient to think of trig functions as coordinates of a point on a unit circle they still describe relations between side lengths and angles of any right triangle

and using one of the definitions of a tangent line (if the limit of the ratio delta y / delta x as 1 point on the graph moves closer to the other exists then we can call the line at that point with the slope equal to the limit a tangent) the construction does imply you have a right triangle in there

[plain storm]

Ok I get it now

Thanks

[brazen mason]

hey, can anybody help me out with these 2 questions? i know that i need to split each into 2 rectangles, but i'm confused which way

[cunning lion]

it doesn't matter which way you split it

as long as you are consistent about it and don't double count

[brazen mason]

for example if i did 2 * x as one rectangle, what would the other be for a?

oh wait (x+3)-2 * (x-2) ?

[cunning lion]

needs more parentheses but that's the idea, yes

[brazen mason]

thx idk why it took me typing it in discord to realize

[lapis moon]

This interpretation is problematic at the rightmost vertex (x₀, y₀) on any smooth curve F(x, y) = 0, where
∂`F (x₀, y₀) = 0.
∂ y

For example, the point E in the diagram below is the rightmost vertex of the unit circle x² + y² = 1

[somber coyote]

vin100

[lapis moon]

The segment ET is the tangent of the unit circle (from Euclid's Elements, in which the "tangent of a circle" is defined without calculus), but you can't use the "Δy / Δx" interpretation for this vertical tangent

[dreamy hound]

By the way, our calculus prof modified the basic definition to account for this case funnily enough

The tangent exists if the limit of alpha := arctan(y/x) exists

But i dont wanna introduce these cursed ass definitions in a normal conversation

[grand path]

,rrcw

How do i rotate?

So i am on an exercice where ABCD is a parallelogram, O is the center of that parallelogram, F is the middle of [AO], G is the centroid of ADC(in french it is called the center of gravity, i don't know if i have the right name in english) And H is the intersection of (EG) and (DC)

The problem arives when after a serie of operation, we obtain the vector AH in two forms that are in function of vector AB and vector AD
And after that I have to show that vector AB and vector AD are colinear

Which doesn't make sense to me, just from the figure , they have different direction so shouldn't they not be colinear?

[grand path]

Plus, all the steps make sense so am i missing something?

[grand path]

Do i have to post all the steps used?

[delicate horizon]

Preferably

[lapis moon]

That's to let helpers know what you've already known, so that they can

• give you a response that adapts to your level
• respond to you more quickly by avoiding repeating what you've already known

[grand path]

Okay, i'll try to not include unnecessary things

[upper karma]

how do i use the inscribed formula to solve this? X would = 44 i think, but how do i solve for y and z then?

[signal vapor]

X is 88

The angle theorem will be used for y

[upper karma]

wait why would it not be used for x, since it is the midpoint? So y=44, what do i do to get z

[grand path]

Is there a problem with this ? Because for me it doesn't make sense for AB and AD to be colinear since they have different directions

[obsidian harness]

$GH = k' DC$ is false

[somber coyote]

south

[grand path]

Sorry. I made a mistake while writing, GH= k'EG

[obsidian harness]

no but you also assumed that EG is parallel to AD

when that's not true in fact

[grand path]

Where?

[obsidian harness]

why is it $\frac{2k'}{3} AD$?

[somber coyote]

south

[grand path]

Because GH = k'EG=k'(2/15 AB + 2/3 AD)

[obsidian harness]

ah I see

[obsidian harness]

well from what you wrote, that doesn't imply AB and AD are collinear

because k and k' are free to vary, so you'll never get a scalar multiple except in some contrived cases

might I propose a simpler method if you want to find the vector AH?

[grand path]

?

[obsidian harness]

if one vector is a scalar multiple of the other, then those vectors are parallel, right?

[grand path]

Yeah

[obsidian harness]

okay so it's like if you want (1/3 + 4x + 5y) / (3x + 4y), or something like that algebraic fraction

to be a constant, sorry

that function is never constant

[grand path]

Well that isn't the same case

Those two aren't linked

[obsidian harness]

I just don't understand

what's the original question? you haven't told me

I feel you're overcomplicating this massively

[grand path]

In the end it is written that AD and AB are colinear, and i just don't understand how

It doesn't make sense, it would mean that A,B and D are aligned ,and thus since this is any parrallelogram, tht a parrallelogram has all of it's point aligned

[obsidian harness]

I don't understand either

I bet there's some kind of mistake, maybe not necessarily in the derivation, but in the conclusion

[grand path]

I found this in a math textbook cpreection btw

So i'm even more conflicted

[grand path]

I bet too, i just can't find what

[obsidian harness]

sorry about that... yeah

[mystic umbra]

weird problem 😔

[grand path]

I'm pretty sure the problem is hiding in the last three lines

[mystic umbra]

just saying, $\vec{AB}$ and $\vec{AD}$ being collinear means they're parallel but that means that ABCD is a "degnerate parallelogram" or not a parallelogram in the usual definition 😔

[somber coyote]

parabolicinsanity

[mystic umbra]

so i feel like there's some issue in your answer

because that's never true for a non-degen parallelogram 🤔

[grand path]

We 1ll feel like there is one

[grand path]

Yeah, 1nd since this is any parrallelogram, it would mean all parallelogram are degenerate

[mystic umbra]

which is... absurd so disproven by nuh uh that's not true... or something like taht

[grand path]

Real

[proper hornet]

@vivid schooner ULDM for aperture still not out 😕

[topaz jewel]

why does the line "normal" always bisect the angle between the 2 foci?

[visual flume]

thx lol

[quaint chasm]

Anyone have a better way to create this shape (ideally parametrically and not implicitly)? I'm not sure what it's called (might be an astroid, but from what I've seen the curvature isnt quite right)

It's essentially a 'anti-circle', I'm constructing it by flipping the circle's halves on the vertical and horizontal axes

This equation that I threw together works but it's not continuous, I rendered a gif of colors going in a path along it, and you can see it has sudden jumps

[visual flume]

help pls

[solid needle]

complete off the cuff intuition with zero justification, so might not work

since there is are points that arent differentiable, maybe try something with absolute value

take the diamond, subtract the circle so the result is always negative, then absolute value would flip it

[quaint chasm]

mk

by continuous btw i dont mean like for derivatives, i mean like for a t that is increasing, it should sort of keep going the same direction

[solid needle]

yeah

[solid needle]

try coordinates?

[quaint chasm]

does desmos have a thing for tracing a path, actually?

like for directionality

ig i can use the domain limits lol

[sage radish]

4root2

[quaint chasm]

hm if i piecewise it maybe

[visual flume]

coord bashing rahhh

last resort

[visual flume]

ty

[solid needle]

!noans

[lime crown]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[visual flume]

mb

[sage radish]

mbmb

[safe stratus]

how do i do this

using this

[covert quarry]

For qn 3 and 4, you can use sine rule
Qn 5 you need to use cosine rule

[safe stratus]

do i use the pythagorean theorem to find the sides

[covert quarry]

No you don't have to

In fact you don't even have to find the sides

[upper echo]

rotate ACB around C clockwise by 90 degrees to get an isosceles right triangle.

[mystic umbra]

late but notice how for $\triangle DCB,\ \angle DCB=90^{\circ}$ and $\overline{DC}=\overline{BC}$ so use these to find $\overline{BD}$

[somber coyote]

parabolicinsanity

[sly venture]

Multiple of root 2

[signal vapor]

I tried that didn't get any answer

So i used coordinate geometry

Coordinate geometry>>>euclidean geometry

[sly venture]

But its easy right

[signal vapor]

Which one

[sly venture]

U talking bout this

[signal vapor]

I was unable to find ac

Through euclidean geometry

[sly venture]

In which question?

[signal vapor]

This

[sly venture]

Oh lemme try

[signal vapor]

Heron's formula might work in this

I will also give it a try

[sly venture]

Mate it's a cyclic quadrilateral and it got the property where if u construct DB the angle CAD = angle DBC the ac becomes angle bisector

[signal vapor]

I am just terrible at spotting cyclic quads it has been 2 years since I did those

[sly venture]

And after that u can use ptolemy theorem

[sly venture]

Well u just have to see that opposite angle sum is 180 degree

In quadrilateral

Man answer is absurd its 40 by root 50 dem

Well it come out nice when u simplify mbmb

4root2

[signal vapor]

I would never have thought of using cyclic quads

So i used heron's formula

And got a long eqn

[sly venture]

Fr that must be tiring

[visual flume]

law of sines

[visual flume]

ty

[visual flume]

ty

[dense carbon]

Let me try something uhhhh
D B is sqrt(50) which is 5(sqrt(2))
Triangle DBC is icoseles or whatever it’s spelled as so 50/2 =25 so DC and BC is 5
Other than that I have no idea how to solve it probably

[stiff orchid]

I am so lost

Trigonometric identities make no sense to my brain

[lime dune]

!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/

[lime dune]

show us a specific example you’re having trouble with

[stiff orchid]

It is "Simplify sec x cos^3 x - 1"

I replaced the sec x with 1/cosx

and then from there i have no idea what to do

[lime dune]

is there anything you can cancel?

[stiff orchid]

My first thought is to cancel the denominator cos x and the 1 in the numerator

[lime dune]

uhhhhhhhh

what

that’s not how that works

substituting sec(x)=1/cos(x) should give you

cos^3(x)/cos(x)-1

notice I’ve written it in a suggestive way, is there a common factor you see in the first term’s numerator and denominator that we can cancel?

[stiff orchid]

Wait so I can put the cos^3 as the numerator?

[lime dune]

for the first term yeah

so cancel that cos(x) and what do you get for the first term?

[stiff orchid]

1?

[lime dune]

cos^2(x)

the numerator is cos^3(x), cosine cubed

[stiff orchid]

Wait so if I replace the numerator 1 with cos^3 x - 1

And the denominator is now cos x

[lime dune]

if you divide by cos(x) you get cos^2(x), cosine squared

i think you’re getting confused about what actually goes in the fraction

gimme a few minutes and I’ll write it out

[stiff orchid]

Oh the numerator doesnt include the -1

Thats a seperate term

[lime dune]

yep

[stiff orchid]

So cos^3 x / cos x

and then it's cos^2 x - 1

but how do i simplify it further

or is that the simplest form

[lime dune]

ye

[stiff orchid]

Amazing, thank you for your help

[graceful talon]

Simplify $\sec x\cos^3x-1$

[somber coyote]

0_א

[stiff orchid]

Then how does

Verifying trig identities work

[lime dune]

turn one side of the equation into the other

[stiff orchid]

So you always have two sides

[graceful talon]

,align \cos^2x-1&=1-\sin^2x-1\
&=-\sin^2x

[stiff orchid]

oh so you just want them to be the same exact terms

[lime dune]

alternately you could use Pythagoras to simplify further, to -sin^2(x)

[somber coyote]

0_א

[lime dune]

but not super necessary

[stiff orchid]

wait what identity is that

i dont think i have it noted down

[lime dune]

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

it’s called the Pythagorean identity bc you can derive it using Pythagorean theorem

[stiff orchid]

But we dont have the sin^2 and the 1 is a negative no?

[lime dune]

the key is that you can write cos^2(x)=1-sin^2(x)

and in the example above that extra 1 you introduce cancels out the -1

[stiff orchid]

Is that another identity?

[lime dune]

no, i just rearranged the Pythagorean identity

subtracted sin^2(x) from both sides

there are two more Pythagorean identities that are useful to know

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

cot^2(x)+1=csc^2(x)

[stiff orchid]

Wait so basiclly as long as you have either sin^2 or cos^2 you can just subtract the term you don't have from both sides?

[lime dune]

you can introduce it into the expression by writing 1-[whatever]^2

[stiff orchid]

My brain isn't braining I can't quite understand it

What would be the requirement to use that identity

[lime dune]

if you see squares of sines or cosines

think Pythagoras

[stiff orchid]

does it matter if I have a 1 or -1?

[lime dune]

can you find another example for us to work through

not really

[stiff orchid]

Ok there is "Verify (Cos x - 1) (1 + cos(-x)) = -sin^2 x"

[lime dune]

ok what’s the first thing that comes to mind

the LHS looks a lot less nice than the RHS

so which side you want to turn into the other

[stiff orchid]

How can you tell what's the most effecient way to make them equal to each other

Would you start with the bigger side

[lime dune]

usually take the less nice one

so i would turn the left into the right

[stiff orchid]

Ok so always go with the complicated side

[lime dune]

it depends but generally yes

ok so what’s the first step we should take

[stiff orchid]

first thing i think is converting the cos into sin

[lime dune]

for the left side?

[stiff orchid]

yea

[lime dune]

what’s your reasoning for that

not to be harsh, i just want to know your thought process

[stiff orchid]

To end up with the same trig function as the RHS

[lime dune]

do we know any identities to convert sines into cosines

or vice versa

[stiff orchid]

Yea

[graceful talon]

pattern recognition

[stiff orchid]

Cos theta = 1/ sin theta

[lime dune]

?

nope

[stiff orchid]

hmm

sin^2 + cos^2 = 1?

[lime dune]

what if we tried to expand the left side?

[lime dune]

yup

[stiff orchid]

Wait no we dont have

[lime dune]

how can we get there

[lime dune]

^^

[stiff orchid]

sin and cos in the LHS

[lime dune]

don’t need both, just get the square of one of them

[stiff orchid]

Do we convert one of the cosines and leave the other?

Wait so

[lime dune]

what happens if you just expand

[stiff orchid]

but how can I get the cos to become one term if they don't have the same value inside them

One is x and the other is -x

[lime dune]

do we know any identities to resolve that?

(have you learned unit circle definitions for the trig functions)

[stiff orchid]

Wait is it the odd even identities?

[lime dune]

yup

[stiff orchid]

Ok I'm not sure how to use those but I wrote them down

So cos(-x) = cos x

Does that mean I can literally just flip a negative value cos into a positive whenever I want?

[lime dune]

yea

[stiff orchid]

No way

[lime dune]

ok now what happens when we expand

[stiff orchid]

So now we have both of thjem as cos x

but wouldnt they being in parenthesis not allow us to combine them

[lime dune]

multiply it out

[stiff orchid]

multiply the -1 in and the positive 1?

[lime dune]

did they not teach distributive laws 💀

hold on lemme write it out

does “FOIL” or “distributive property” ring a bell

[stiff orchid]

yea

[lime dune]

ok use that

[stiff orchid]

so first term with first term

first term with second term

second with first

second with second

[lime dune]

what do you get when you do that

[stiff orchid]

Cos + cos x ^2 - 1 - cos x

Wait so whenever I want to get rid of parenthesis I just foil?

[lime dune]

if you’re trying to expand a product sure

anyway is there anything in there that cancels?

[stiff orchid]

And btw what do you mean by expand

As in like I have terms that I want to combine from different parenthesis?

[lime dune]

oh boy did they not teach you the vocab 💀

expanding basically means “apply distributive property” to write the big product as a sum of smaller products

[stiff orchid]

ok that makes sense, mb I have a hard time remembering vocab for math rules.

So now I can cancel the cos x and the - cos x

left with cos x ^2 - 1

And then use the sin^2 cos^2 = 1 identity

[lime dune]

yep

[stiff orchid]

wait so like

the 1 cancels

and then the sin^2 I just can't wrap my head around introducing it without it being there in the first palce

place

is there like a rule if i want to use a function that wasn't their originally to use an identity

[stiff orchid]

Btw do you think Ill have a hard time in calc if I don't remember them by name? 😭

[lime dune]

…its just algebra?

ultimately most of the identities you’re applying on the way are just substitutions

[stiff orchid]

so lets say I have a 1

Can I use it

To get the cos^2 and sin^2

[lime dune]

you can write that as sin^2(x)+cos^2(x) yes

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

[stiff orchid]

This is crazy

Maybe im just crazy

Ok so it becomes

Wait but in the case that I only have cos, it wouldn't equal to 1 tho right

since the sin is missing

so how would you set up the equation

[lime dune]

easiest way is to factor out -1

then the stuff inside reduces by Pythagoras

[stiff orchid]

ok let me try to factor the cos^2 into two

[lime dune]

no factoring needed at this point

[stiff orchid]

(cos(x)−1)(cos(x)+1)

[lime dune]

just directly apply Pythagoras

see the last two lines of the solution I sent

[stiff orchid]

Yea

Did you set up

cos^2 = sin^2 +1

[lime dune]

?

[stiff orchid]

wait no the 1 is already

[lime dune]

I substitute sin^2(x)=1-cos^2(x)

[stiff orchid]

the cos ^ 2 is equal to sin^2

[lime dune]

and thats all we need to finish

[lime dune]

?

[stiff orchid]

i have to be brain dead or something

Ok so between the cos ^ 2 x - 1 to the -(1-cos^2 x) what happend

[lime dune]

I factored out a -1

-1 is -1*1

cos^2(x) is -1*-cos^2(x)

[stiff orchid]

So you multiplied the -1 by the 1 in the identity, and the sin ^ 2 by

[lime dune]

now as shown above we can use Pythagoras on the stuff inside the parentheses

there’s no multiplication involved at this point

look at the stuff at the blue arrow

[stiff orchid]

So we are

expressing the cos^2 -1 in terms of sine, cos^2(x) - 1 = -sin^2(x)

[lime dune]

yep

[sterile tiger]

need help with my Trig Study guide for a test tmr anyone help

@lime dune u seem smart

[lime dune]

what specifically in the study guide

are there exercises on there you need help with

[sterile tiger]

i have screen shots of the study guide if u want me to send them

yes

some problems i need help with if thats what u asked

[lime dune]

sure go ahead

[sterile tiger]

here is the first one

now i understand some of the things like raddii and chord

[lime dune]

point me to which ones you don’t understand

[sterile tiger]

but some make no sense i think for number 3 its DC but for diameter you have to go across or something so im just confused how its DC if you have to go across point A

all them besides thos few lol

[sterile tiger]

maybe im wrong with the diameter thing idk

[paper spear]

even if theres a point in the middle you can still name the points right

so df would be the diameter

?or no

[lime dune]

yeah

[sterile tiger]

thats what i was think chatgbt got me wrong

[lime dune]

this is just definition chasing

[lime dune]

!nogpt

[lime crown]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[paper spear]

do nottt use chatgpt for maths

its awful

[lime dune]

it does not actually have any “intelligence”

[paper spear]

i dont think its helped me a single time

[sterile tiger]

ok i was just trying something then i thought of discord

anyways the others i need help on

[lime dune]

all it does is word vomit out whatever is statistically most likely to come next

[paper spear]

exactly

[sterile tiger]

how do i find tangent secant etc

[paper spear]

waiti have a good image for that one second

[lime dune]

it’s just definition pushing

what’s the definition of a tangent

[paper spear]

is it this ?

[lime dune]

nah

[sterile tiger]

no idea math is not my strong suit

[paper spear]

😞

[lime dune]

they want “lines tangent/secant to a circle” not trig ratios

[paper spear]

i see

[lime dune]

a tangent line hits the circle at exactly one point

[sterile tiger]

so F

[lime dune]

nope

[sterile tiger]

A

[lime dune]

that one also hits the circle at D

A is a point, and it doesn’t even intersect the circle

[sterile tiger]

oh

[lime dune]

look outside the circle

[sterile tiger]

I see but why could it not be C or B but D is correct

B has its own line and touche sthe circle but outside same with D

[lime dune]

??

the line through B is the tangent line

bc it hits the circle at B and B only

line DE is the secant line, because it intersects the circle at two points D and E

[sterile tiger]

oh

and for the problem at the bottom you would use A square B sqaure

which i got that

[lime dune]

basically that question is asking equivalently

“does that triangle satisfy the Pythagorean theorem, ie is it a right triangle”

so does the square of the longest side equal the sum of the squares of the shorter sides

and this is because

a line tangent to a circle is perpendicular to the radius drawn to the point of tangency

[sterile tiger]

need to move it down so i dont gotta scroll

[lime dune]

do you want me to draw an example of this

[sterile tiger]

yes pls

[lime dune]

ok

[sterile tiger]

oh

got a question

too solve this what formal would you use

[lime dune]

^

ie

[sterile tiger]

ok

[lime dune]

does 9^2+15^2 equal 18^2?

[sterile tiger]

i see

im so lost on these ones

[lime dune]

||answer is no, as 306 != 324||

[sterile tiger]

yup

wait

where did you got 324 from

[lime dune]

18^2

[sterile tiger]

i didnt do the math but do they equal that when added uop

oh ok

and 306 is from the 15 and 9

added

[lime dune]

yep

[sterile tiger]

i see

[paper spear]

sorry to interrupt but what doe sthe symbol between AC and BE mean under the first circle?

[lime dune]

ok for this one, you have that AC is a diameter and perpendicular to BE

[lime dune]

perpendicular

[sterile tiger]

Yup

[graceful talon]

,,\perp

[somber coyote]

0_א

[paper spear]

Ooo

[lime dune]

what does this tell you about the angles?

central angles

[sterile tiger]

i dont know lol

[lime dune]

AC perpendicular to BE

[sterile tiger]

tells me what they are

[sterile tiger]

yea

[lime dune]

what do we know about the angles between perpendicular lines?

[sterile tiger]

umm

that they go to E?

there AC perpendicular

[lime dune]

how many degrees are in these angles?

[sterile tiger]

40

90?

[lime dune]

90 yep

[sterile tiger]

so would AB be 90

and ABC would be 180?

[lime dune]

so AEB and CEB are 90 degrees

[upper karma]

I have a question, why is proving triangle inequality relationships so hard?

[lime dune]

proving the triangle inequality?

[sterile tiger]

what would ABD be

[lime dune]

you know arc ABC now

[upper karma]

No, proving triangle inequality relationships

[lime dune]

and you’re given arc CD

[sterile tiger]

And CD is 40

[lime dune]

we can add them

to get ABD

[lime dune]

what kinds of problems

[sterile tiger]

Is AD 140

180-40=140

[lime dune]

like “what’s the longest side in this diagram?” type stuff?

[sterile tiger]

u talking to me

[upper karma]

i don't have any difficulties with proving triangle congruency

[sterile tiger]

nvm

[upper karma]

Nope, i don't have any difficulties with that, it's like this: You're given a figure and some relationships like this side is greater than this side, and this angle is congruent to this angle, and then you'll have to prove a statement using the triangle inequality theorems such as the hinge theorem, etc

They mostly involve two column proofs

[sterile tiger]

@upper karma ur taking my tutor lol

[upper karma]

Sorry 💀

[lime dune]

me when multitasking (I’m also practicing piano rn)

lmao all cool

[sterile tiger]

i got a test i must pass dont want summer school

[upper karma]

What are u having difficulties with anyway

[sterile tiger]

the things i screen shoted

[upper karma]

What number do you need help with this one?

[lime dune]

oh going from triangle inequality to hinge theorems

got it

[sterile tiger]

@lime dune idk if u wanna check but the ones i typed like the signs are they correct

[lime dune]

the ones you’ve answered so far look good

[sterile tiger]

ok

[lime dune]

yeah uh id personally just trig bash those instead

[sterile tiger]

and how

the bottom one

same thing

[lime dune]

notice AD is a diameter

what does that tell us about AED?

[sterile tiger]

yup

180 degrees

[lime dune]

and you’re given AE

so what’s DE?

[sterile tiger]

125

[lime dune]

nice

[sterile tiger]

YESS BROTHER

[lime dune]

ok and the last one

[sterile tiger]

If you save me from my math test which u are im pay palling u

ok

gotta lock in

[lime dune]

that’s a full circle right there

[sterile tiger]

yup

[lime dune]

what’s its arc measure

[sterile tiger]

idk

[lime dune]

360

[sterile tiger]

would you

oh yea

[lime dune]

and how can we solve for x?

[sterile tiger]

put the things in the ( ) multiplyed by 5

[lime dune]

uh not quite

all those arcs that are given

sum to the full circle

which is 360 degrees

[sterile tiger]

yup

so

got a question

[lime dune]

so ||5x+40+70+75+2x=360||

[sterile tiger]

oh

[lime dune]

can you finish from here?

[sterile tiger]

yup

[lime dune]

nice

[sterile tiger]

thank you brother

[lime dune]

np

[sterile tiger]

https://tenor.com/view/tears-of-joy-tears-happiness-kevin-hart-crying-gif-18835009

[lime dune]

it’s been many years since I took geometry lol

[sterile tiger]

u in college

[lime dune]

but ik it can be kinda overwhelming when they throw so much new vocab at you at once

[lime dune]

yep I’m 3rd year

[paper spear]

wow

[sterile tiger]

my teacher sucks she does like new units so often so when test come im still trying to figure out the last unit

[lime dune]

oof the pacing

I remember when i took geo we didn’t get to circles until like

3-4(?) weeks before finals

[paper spear]

do you think a slow unit or a faster unit would be better

as in like

a bad teaching style

would you say one that speeds through it is better than an unneccesarily slow one

[lime dune]

id rather have foundations be more solid

even if that means less material is covered

[paper spear]

i see

[proud glen]

Hi. Can someone help me understand why my book says that

5 cos(Theta) = 3 ?

How can I visualize this differently?

[mystic umbra]

uhm, there's not really much to visualise differently all things considered

you have a force x acting at an angle theta relative to the area vector

[proud glen]

Yeah I know that, what I don't get is why 5cos(Theta)=3

I mean, looking at the FRONT right triangle I can assume Theta is the angle above the right angle

But how are that Theta and the Theta labeled in the picture related?

[mystic umbra]

what do you mean 5cos theta is 3? 🤔

[proud glen]

The book says that

[mystic umbra]

yeah, that happens

what about it?

[proud glen]

I don't get why

[mystic umbra]

@proud glen here's a visualisation how

[proud glen]

Ok the two Theta in green are equal because they are opposite angles

[mystic umbra]

$\overline{\theta}$

[somber coyote]

parabolicinsanity

[mystic umbra]

it's supposed to be this 😔

grrr, i'll label again clearly just wait

it's a straight line so A+B+90=180 i.e. A+B=90

then if the top angle is K, K+B+90=180 i.e. K+B=90 or... K=A

then from that you get 5cosA = 3

@proud glen

[proud glen]

You mean K the one on the leff?

[mystic umbra]

The top triangle

there's no vertices points so its a bit harder to think about it

but it is how you obtain the value

[proud glen]

@mystic umbra

So this is it

[mystic umbra]

yes

😃

so you see how that came to be?

[proud glen]

Ok but I needed to do some sketches and reason about it... how can I immediately see things like these without going through all these sketches?@mystic umbra

I mean, how could I immediately see that?

[mystic umbra]

probably deal with a few more of them, most of it is just intuition from practise 😔

angle chasing and stuff

[proud glen]

Thank you for your help 💙@mystic umbra

[mystic umbra]

happy mathing

[cerulean cipher]

CHAT

[stuck anchor]

dead chat

[cerulean cipher]

Can u find the side lengths of a square given only a diagonal line across

[stuck anchor]

ngl

[cerulean cipher]

💀

[graceful talon]

yes

||Pythagorean theorem||

,texsp||Let $d$ be the length of the diagonal, let $x$ be the length of a side of the square.
\begin{align*}
x^2+x^2&=d^2\
2x^2&=d^2\
2x^2\textcolor{red}{\cdot\frac12}&=d^2\textcolor{red}{\cdot\frac12}\
\cancel2x^2\textcolor{red}{\cdot\frac1{\cancel2}}&=d^2\textcolor{red}{\cdot\frac12}\
x^2&=d^2\cdot\frac12\
\sqrt{x^2}&=\sqrt{d^2\cdot\frac12}\
x&=d\sqrt{\frac12}
\end{align*}||

[somber coyote]

0_א

[mystic umbra]

white bg jumpscare

[graceful talon]

or just remember the diagonal length of a square is √2 times of the side length

[mystic umbra]

the latter is easier

[whole sequoia]

Help on the check ur progress part cuz idk how

[obsidian harness]

similar triangles

[whole sequoia]

Yes

Help me im lost

[obsidian harness]

if you understand the previous examples you should be able to do this easily

and practiced through them

[whole sequoia]

Nah i dont

All i know are the proportion thingy

[obsidian harness]

okay

[whole sequoia]

My teacher aint even teach us this

Just went straight to assignment

Help me pzlzz

[obsidian harness]

so for example 2

you can do (large long side) / (large hypotenuse) = (small long side) / (small hypotenuse)

can you sub into each of those brackets?

[whole sequoia]

Ye ig

[obsidian harness]

you do it then

[whole sequoia]

Ight thx

[obsidian harness]

no worries, you can always come back and ask

[whole sequoia]

Same thing for the check ur progress thing?

[obsidian harness]

the steps are:

1. set up an equation like this
   so you have to be consistent with the order - notice how I did large/large = small/small
2. sub in given the values in the question
3. use algebra to solve, so the crisscross multiplication way or whatever

[obsidian harness]

yes

[whole sequoia]

Yeye ty

But how do i prove this

[obsidian harness]

ah, so there's a very good reason why similar right triangles are AA similar

if you call blue angle = x, then red angle = 90 - x

then 90 - red angle = 90 - (90 - x) = x

that's how it goes

so there are three similar triangles in a setup like this

smallest, medium, and the big one are all similar

[cerulean cipher]

Thats pretty cool

[paper spear]

greetings!

i have a quick question!

when solving for an angle using the law of sines why do you multiply by arcsin?

i know you have to but im curious how it works and why it's like that

[cunning lion]

you don't "multiply" by arcsin, you apply the arcsin function

the arcsine function is an inverse function to sine (on a limited domain), so if you apply one and then the other, they cancel out

similar to squaring and square rooting

[paper spear]

ohhh

i see

thank you

[somber coyote]

The Mathematics Guy

[dark sparrow]

that link looks like a scam

you would be surprised to see the number of scam links people post here

[upper karma]

,, \tan^{i}{(x)}

[somber coyote]

! ! Ｎ Ｏ Ｒ Ｔ Ｈ ⭐ ^w^

[upper karma]

why

[mystic umbra]

$\tan{^i}(x)=\cos{(\ln\tan(x))}+i\sin({\ln\tan(x)})$

[somber coyote]

parabolicinsanity

[mystic umbra]

this is just a specific form of $a^i=\cos\ln{a}+i\sin\ln{a}$

[somber coyote]

parabolicinsanity

[mystic umbra]

is E the centre of the circle or the intersection of AC and BD?

[short panther]

I have no idea

I would assume the intersection

[mystic umbra]

sure let's go with that

notice how since AC is the diameter and DE=EB

that would mean AC is orthogonal to BD

as all chords are bisected and orthogonal to a diameter

[short panther]

Idk if I learned that ima see rq

[mystic umbra]

😔 sure

[short panther]

No😭

[mystic umbra]

well, that's the only way to solve it

but

[mystic umbra]

uh

i guess you can use

chord - chord theorem

or whateer it's called

[short panther]

Yeah I am using chord chord theorem

[mystic umbra]

$\overline{DE}\cdot\overline{EB}=\overline{AE}\cdot\overline{EC}$

[somber coyote]

parabolicinsanity

[mystic umbra]

let $\overline{EC}=x \implies \overline{AE}=\overline{AC}-x$

[somber coyote]

parabolicinsanity

[mystic umbra]

figured out now?

[short panther]

I did this but like it’s not a whole number so I’m cooked

[mystic umbra]

what are you doing 😭

[short panther]

Idk bro I saw😭

A organic chemistry video and I just did what he did but with my problem😭

[mystic umbra]

$6\cdot6 =(13-x)\cdot x$

[somber coyote]

parabolicinsanity

[mystic umbra]

evaluate this isntead

$x^2-13x+12=0$

[somber coyote]

parabolicinsanity

[short panther]

I’m too lazy to factor so I’m just going to use desmos

[mystic umbra]

lazy ass

[short panther]

(X-1)(x-12)

[mystic umbra]

now use that to obtain the two values of AE

[short panther]

Um

When substituting it’s just 12

😭

[mystic umbra]

yeah?

$\overline{AE}=1,12$

[somber coyote]

parabolicinsanity

[short panther]

But that doesn’t equal 36 though

[mystic umbra]

Why would it need to be 36?

Why?

[short panther]

6 * 6 ≠ (13-12)12

[mystic umbra]

😔 slight multiplication error

it's supposed to be

$x^2-13x+36=0$

[short panther]

😭

[mystic umbra]

i accidentally did $6^2=12$

[somber coyote]

parabolicinsanity

[short panther]

It’s okay

[mystic umbra]

happens

[obsidian harness]

$1 + 2 + 3 = 1 \cdot 2 \cdot 3$

[somber coyote]

south

[short panther]

Um😭

[mystic umbra]

ugh

[somber coyote]

parabolicinsanity

[mystic umbra]

evaluate this instead

[mystic umbra]

okay then $\overline{AE}=4,9$

[somber coyote]

parabolicinsanity

[short panther]

What was the problem when getting the quadratic

[mystic umbra]

silliness 😔

[short panther]

😭

Thx

[trail trellis]

How do you find angle measures to the nearest tenth of a degree (for example, if it’s cos-1(5/9) it will be 56.251 then when you change it to degrees and minutes it’s 56degrees 15minutes but how do u find it to the nearest tenth degree???)

[silent plank]

if you want it to the nearest tenth of a degree,
that means 1dp while in decimal form,
i.e.
round 56.251 to 1dp

[trail trellis]

So 56.3

Degree?

Or like just 56.3

[silent plank]

56.3 degrees

56.3°

[trail trellis]

Thank you!!!!!!!

[dark sparrow]

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