geometry-and-trigonometry

[upper karma]

<@&286206848099549185>

[rich wolf]

Then you can use double angle

[upper karma]

Buddy I know the formulas

[onyx cloud]

sass

[upper karma]

That's not the part I can't get past

[onyx cloud]

damn

we're just trynna help my guy

[rich wolf]

Ok lol no need to get cranky

[upper karma]

Sorry just felt like it was getting repetitive

I can show you the progress I've made so far

I tried getting to the question from what we had to prove

Cos^2(theta)+cos(alpha)=0

That's what that led to

I mean that's what I need to prove

Or that Alpha=2n*pi

This is what I got after trying to convert it into a quadratic and do some stuff

If you know any relevant formulas to this then it'd help

[onyx cloud]

if you convert it to quadratic there shouldnt be sines and cosines

[upper karma]

There were two different arguements

So you can at most eliminate one

[onyx cloud]

pulse use the law of sines

[tiny halo]

Not good at that

[onyx cloud]

sin(A)/a = sin(B)/b

[acoustic jungle]

Why do you need law of sines

the question is what is the angle z

[onyx cloud]

oh

[acoustic jungle]

you just use the sum of triangle = 180 and right triangle = 90 and 28 degrees is 28

[onyx cloud]

i thought it asked for side length

my bad

[acoustic jungle]

oh max, do you know combinations stuff?

I need some help on that

[onyx cloud]

a little

sure go to a question channel

theres a ton going on here

[acoustic jungle]

I have it in #combinatorial-structures

[upper karma]

This channel is taken @acoustic jungle

Oh shoot that message is late

Sorry for that

[acoustic jungle]

no worries

[onyx cloud]

same stuff pulse

[upper karma]

Hey max,think of anything?

I'm stuck and can't even derail myself to think of something else

[onyx cloud]

im working through it

i isolated sin(theta) and i can see how to get a root 2

[upper karma]

Alright take your time

[onyx cloud]

but i dont see how they got further

[upper karma]

Same

[onyx cloud]

freaks

https://media.discordapp.net/attachments/326138757474680852/694354459928690749/IMG_20200331_064631.jpg?width=773&height=702

[tiny halo]

Is A 60

[onyx cloud]

the first line is wrong

yes pulse

oh wait nvm

i thought it was cos(2theta)

[upper karma]

Ye

[tiny halo]

So what do you do

[upper karma]

Use the law of sines and cosines

Pulse

[onyx cloud]

pulse yours is fine

its 60

[upper karma]

No I meant to find the missing length

Use those laws

[onyx cloud]

he only has to find the angle

[upper karma]

Oh okay

[tiny halo]

Idk wth to do in this

[upper karma]

Use Pythagorean theorem to find y

And then you can find the angles of that smaller triangle

Also if you are struggling with basic trigonometry you might want to brush up your theory

It would save a lot of time

[tiny halo]

Idk Wht Pythagorean Theron is

Theorem

[upper karma]

Hmm

I could tell you but that would destroy the whole purpose

Of learning lmao

Alright so id suggest watching a few videos from khan academy

[high zephyr]

maybe just give him the formula?

[upper karma]

Then he'll get stuck at the next problem

And the next

[tiny halo]

😐 khan academy

[high zephyr]

all of them involve pythag?

[upper karma]

He's doing basic trigonometry,so I reckon yeah

[tiny halo]

Oh ok

[high zephyr]

we can give him the formula and let him work out the answer by himsef

[swift walrus]

wasn't there that (10sqrt5)^2 = 20(20+x)

thing

[onyx cloud]

freaks

im confused

https://media.discordapp.net/attachments/326138757474680852/694347949790199838/IMG_20200331_025309.jpg?width=1442&height=289

if you multiply both sides by sin(theta) and sin(theta-alpha)sin(theta+alpha)

you get that sin(theta) = 2sin(theta-alpha)sin(theta+alpha)

[upper karma]

Yeah

[onyx cloud]

and under the assumption that alpha = 2npi

sin(theta) = 2sin(theta)sin(theta)

and i just realized that i added instead of multiplied

i was gonna say sin(theta) = 4sin(theta)

but math doesn't work like that

[upper karma]

I don't get your doubt

And also

If the alpha=2n(pi) condition doesn't hold the second one will

[onyx cloud]

no but it has to hold

cause its the end of the proof, but its not 4sin(theta)

its sin(theta) = 2sin^2(theta)

1 = 2sin(theta)

[upper karma]

Share the proof

[onyx cloud]

1-2sin(theta) = 0

[upper karma]

We skipped over a bit of stuff I think

[onyx cloud]

yeah i did

give me a sec

freaks i cant get it sorry

[upper karma]

No worries

[onyx cloud]

i might come back to it

[upper karma]

I appreciate you trying to help

Yeah let me know when you do

[onyx cloud]

law of cosines

[tiny halo]

Idk what that is

[upper karma]

Yeah or if you don't want to use trigonometry you can draw a perpendicular

It will give a positive X and a negative one

Lengths aren't negative

Lmao nvm that's way too lengthy

Use loc

[upper karma]

@onyx cloud that question is wrong

[acoustic jungle]

which question

[upper karma]

The one i asked

[onyx cloud]

what was wrong about it

[acoustic jungle]

the textbook one?

[upper karma]

@onyx cloudcosec(theta + alpha) + cosec(theta - alpha) = 2 cosec(theta)

This is what the question should have been

[onyx cloud]

ah i see

[upper karma]

It was stupid of me to not consider the possibility of the problem being wrong

[acoustic jungle]

I've been trying to do it for so long

[runic monolith]

If the image of point J under a 180° rotation about the origin is (7,-3), what are the coordinates of point J?——- am I post to apply the rotation to the coordinate point (7,-3)

I’m confused cause it doesn’t give me a picture for reference

[onyx cloud]

image and preimage are geometric terms

[runic monolith]

Oh I’m just finding the pre image

[upper karma]

Lmao i didn't know that too

Curious

[summer spire]

that's a tangent line

[eternal crag]

...

[acoustic jungle]

that is not the diameter unless you are given more information

[acoustic jungle]

wait this is solvable??

You can't solve this without knowing angle DAB

if you visualize this, the height can be anything without any restrictions

[vale nimbus]

wtf

[idle bloom]

Yea wtf this has infinite solutions essentially

[iron rain]

DAB is a right angle I think, if you look very closely

there a small right angle symbol

[idle bloom]

Ahhh yes that'll do it

Ok so the strategy is to do a bunch of systems of equations using Pythagorean theorem

Label the unknown sides with variable names for easiness of writing

And then you get 3 unknowns with 3 equations

[silent sequoia]

how do i figure out the changes to a parent function from the equation

i just completely forgot everything from functions from last year

[onyx cloud]

do you have a specific question in mind

[silent sequoia]

yeah but its from an assignment

isnt that bannable

[onyx cloud]

nah

as long as you don't bet for the answer

i can walk you through it

[idle bloom]

Assignments are good

Tests and quizzes are bannable

[silent sequoia]

gkve me a sec

lemme screenshot

[onyx cloud]

test and quizzes are elements inside a subset of assignments

therefore if assigmnets and allowed, so are tests and quizzes

checkmate

[idle bloom]

🤔

[onyx cloud]

sorry i watched a video on introduction to set theory and i thought i was cool

[silent sequoia]

So i can’t post it?

[onyx cloud]

yeah go ahead

[onyx cloud]

ok so lets work through the first one

problem a that is

are there any functions you can't recognize?

[silent sequoia]

the last two

[onyx cloud]

so if you look at the green text

it'll show you which function it is

[silent sequoia]

i honestly didnt notice that

[onyx cloud]

the more you know 😄

well either way

now that you know all the parent functions we can work on part b

so looking at the first one

what is the parent function

[silent sequoia]

a quadratic so f(x)=x^2

[onyx cloud]

yep

ok

do you know what domain is

[silent sequoia]

i dont remember entirely

[onyx cloud]

ok

any guesses?

[silent sequoia]

x = real numbers things

but i dont remember how to phrase it

[onyx cloud]

yeah i gotchu

its a good start

so what domain is

is the possible values of x to satisfy the function

so if our function is say
f(x) = 1/x

then x can be all real numbers

except for a single number

that number being 0, as 1/0 is undefined

so if we apply the same logic to x^2, what is the domain

[silent sequoia]

all numbers can be real

i think

[onyx cloud]

yep

thats right

now

that's for f(x)

what about for 2f(0.5(x-1))+3

(i got that function from the first picture btw)

[silent sequoia]

im not sure

[onyx cloud]

so lets do it looking at the graph

are there any x values for the red curve that don't work?

[silent sequoia]

0

wait

nvm

no

its only for y

that it doesnt work

[onyx cloud]

um im not sure what you mean but you're right

its no

therefore the domain of f(x) is x is an element of all real numbers

or

x ∈ R

[tiny halo]

Sry

[onyx cloud]

np

johnny do you get what i said

[silent sequoia]

yeah i get a bit of it

since x can be any number

all are real

[onyx cloud]

since x can be any number
@silent sequoia

thats right

but what we say is that x can be all real numbers

not that it matters too much rn but yeah

now you wanna do range?

[silent sequoia]

range cant be below 3

[onyx cloud]

what can't be below 3

[silent sequoia]

the y

[onyx cloud]

how do you write that mathematically

[silent sequoia]

y is greater than or equal to 3

[onyx cloud]

good

now im assuming you did that by observation

[silent sequoia]

yea

[onyx cloud]

ok

yup ok its right

[silent sequoia]

but i think thats what the +3 means

right

[onyx cloud]

pretty much

like actually i think its 100% what it means

but ok you're good with range and domain

now intervals of increase and decrease

do you know what that means

[silent sequoia]

horizontal phase shift and vertical

[onyx cloud]

nono

at least i don't think so

[silent sequoia]

oh

[onyx cloud]

before we go on

do you know derivatives?

[silent sequoia]

no

[onyx cloud]

ok

so interval of increase

this is asking for what values of x

the y is going up

so lets work it out with f(x) = x^2

looking at the green graph, for what values of x does the y increase

[silent sequoia]

0.5

or 1

yeah 1

[onyx cloud]

so im gonna say no

but i wanna know why you said 1

[silent sequoia]

thats where the centre of the red line is

[onyx cloud]

ah ok i see

lets do it with the green line first

[silent sequoia]

0

[onyx cloud]

so is the y increasing at 0?

[silent sequoia]

yea

or 0.1

[onyx cloud]

so i notice what you're thinkning

but what about .01

or .0001

or .0002

or 1

or 10

is it a single value of x or is it multiple values of x?

[silent sequoia]

single

i mean

multple

because its increasing the same from other side

aswell

[onyx cloud]

ah ok

let me clarify

[tiny halo]

Max can I ask my question when u guys done

[onyx cloud]

increasing means the y goes up

as the x goes up

pulse head over to one of the #❓how-to-get-help channels

decreasing means the y goes down

as the x goes up

did i lose you

[silent sequoia]

No

[onyx cloud]

ok

so looking at the graph one more time

where in the graph does the y go up as the x goes right

[silent sequoia]

1

[onyx cloud]

1 what

[silent sequoia]

It’s going up at any point on the right isn’t it

[onyx cloud]

thats true

and when is it decreasing

[silent sequoia]

On left

[onyx cloud]

so how can we express the left and right of the graph

in math terms

[silent sequoia]

-x

For left

+x for right

[onyx cloud]

how do you express that as an inequality

sort of what you did with the range

[silent sequoia]

I’m not sure

[onyx cloud]

so what does -x mean

[silent sequoia]

X is below 0

So x is less than 0

[onyx cloud]

👍

what about the increasing

[silent sequoia]

X is above 0

[onyx cloud]

or

in math terms

[silent sequoia]

X is greater than 0

[onyx cloud]

perfect

so increasing for x^2 is x>0

and decreasing is x<0

what about the red function?

[silent sequoia]

X is increasing at 2

At x is greater than 1

And decreasing at x is below 1

[onyx cloud]

perfect

now the last one

turning points

know what that means?

[silent sequoia]

The point where the graph changes

To the opposite

[onyx cloud]

wdym?

[silent sequoia]

Where the line turns

[onyx cloud]

from doing what to what

[silent sequoia]

Not sure

[onyx cloud]

ok well you're right

its turning from decreasing to increasing

or increasing to decreasing

so for the green function where is that at

[silent sequoia]

There is none

[onyx cloud]

but the function increases for some x

and decreases for some other

so it has to turn at some point right?

[silent sequoia]

0

The origin

[onyx cloud]

yep

what about the red function

[silent sequoia]

1

[onyx cloud]

yep

so now

can you try the other ones

and see how you do?

i'll be here if you need hel

p

[silent sequoia]

K

[tiny halo]

Can u help me plz

I’m trying to get help but they confusing me

[onyx cloud]

sure

let me see

[onyx cloud]

pulse do you know pythagoras theorem

[tiny halo]

Kinda

I’ve been on this question for pretty much an hour

[onyx cloud]

ok

so lets break it down

[tiny halo]

Ok

[onyx cloud]

can you give me 2 secs while i draw something

so this is pretty much what we have right?

[tiny halo]

Yeah

[onyx cloud]

so theres a few things we know

[tiny halo]

Theirs 3 triangles?

[onyx cloud]

yep

so i labeled the sides to make this clearer

now

do you know what the hypotenuse of a right triangle is

[tiny halo]

I heard of hypotenuse

[onyx cloud]

ok

if you look at this triangle here

the hypotenuse is the longest side

now

[tiny halo]

So lone AB

Line*

[onyx cloud]

on the first image yes

[tiny halo]

Yeah

[onyx cloud]

but focus on this second one for now

what pythagoras theorem tells us

is that the hypotenuse

(we'll call it c)

has a length that if we square (c^2) will be equal to the sum of the other two side lengths squared

we'll call them a and b

so this looks like:
a^2 + b^2 = c^2

[tiny halo]

Ohhh I’ve seen this

[onyx cloud]

and this is true for all right triangles

[tiny halo]

A square plus B square equals C square

[onyx cloud]

yep

so now going back to this picture

https://cdn.discordapp.com/attachments/326138757474680852/694597612136824832/unknown.png

[tiny halo]

Ok

[onyx cloud]

https://cdn.discordapp.com/attachments/326138757474680852/694597837916340355/unknown.png

this one has the labeled sides sorry

theres 3 different hypotenuses

can you tell me which ones?

[tiny halo]

A^2+C^2=B^2

[onyx cloud]

i don't see how you got there

but focus on the hypotenuses first

here i

i'll highlight one

the hypotenuse is in red

and the legs are in yellow

thats 1 of them

https://cdn.discordapp.com/attachments/326138757474680852/694599182337441803/unknown.png

there are 2 more in the image

can you name them

[tiny halo]

Y

Line

[onyx cloud]

yes

[tiny halo]

X?

[onyx cloud]

no

[tiny halo]

Line CB?

[onyx cloud]

yep

so now we know that y, z, and cb are hypotenuses

and let's call cb a for simplicity

so a, y, and z are the hypotenuses

we good thus far?

[tiny halo]

Yeah

[onyx cloud]

alright

now what do we know about the side lengths of these sides based on pythagoras theorem?

let me make the question a bit simpler

[tiny halo]

That they are the same

[onyx cloud]

they aren't the same length

[tiny halo]

Value? I

[onyx cloud]

nono

[tiny halo]

Ok

[onyx cloud]

so i'll give you the z one for example

we know that $z^2=x^2+20^2

$z^2=x^2+20^2$

[somber coyote]

Maximo:

[onyx cloud]

do you see how i got here?

[tiny halo]

😐no

[onyx cloud]

so z was one of the hypotenuses right?

[tiny halo]

Yea

[onyx cloud]

and

what does pythagoras theorem tell us about the length of the hypotenuse?

[tiny halo]

The length?

[onyx cloud]

do you remember pythagoras theorem?

[tiny halo]

A square plus B square equals C square

[onyx cloud]

and what is c in that equation

[tiny halo]

20

[onyx cloud]

no

just for any right triangle

what does c stand in for

https://cdn.discordapp.com/attachments/326138757474680852/694598785107492874/unknown.png

[tiny halo]

The longest length

[onyx cloud]

which is the hypotenuse yeah

so

if we take the triangle again

can you use pythagoras theorem for the triangle i highlited?

[tiny halo]

Yes

[onyx cloud]

how would that end up looking?

[tiny halo]

A+B=Z

[acoustic jungle]

You can also use similar triangles

[onyx cloud]

so pyhtagoras theorem is $a^2+b^2=c^2$

[somber coyote]

Maximo:

[onyx cloud]

fishraider you think tht

that'd be easier

to explain

[acoustic jungle]

I dunno

[onyx cloud]

wanna give it a shot

[acoustic jungle]

ok

bpa is similar to bac

that would give us $20/z=z/29$

[somber coyote]

Fishraider:

[tiny halo]

??

U talkin to me?

[acoustic jungle]

yes

[tiny halo]

Oh

That just confused the heck out of me

[acoustic jungle]

oh

well you can also do a more tedious way

write out 3 equations with pathetic theorem

[strange laurel]

I would just make these 3 equations add up to 360 and then solve right?

[devout shell]

right

[strange laurel]

ok ty

[tiny halo]

I’m confused

I just need this question to turn it in

[acoustic jungle]

did you get the three equations

[tiny halo]

No

[acoustic jungle]

use the pathageorean theorem and get them then

do you see the three right triangles

[tiny halo]

Yeah

[acoustic jungle]

get the equations

use pathagorean

[tiny halo]

Like this

[acoustic jungle]

it should be

[tiny halo]

Ok

[ancient jacinth]

help>?

[devout shell]

should be theorem for that

[acoustic jungle]

I think I got it

[ancient jacinth]

oh yea

but like my teahcer cant teach

so i have no idea how to apply the theorem

[acoustic jungle]

all you need to know is isoceles triangle

and from the end of the circle to a chord is have the angle to chord subtends

I actually like this problem a bit because it requires some steps

also whenever you are given a problem use every piece of information, have you used the 50 degree info

@ancient jacinth

[ancient jacinth]

yea

i got this

i got the problem!

[acoustic jungle]

Nice!

[ancient jacinth]

but thanks

[acoustic jungle]

did you do this?

[prime ledge]

your welcome

[tiny halo]

?

[prime ledge]

i zoomed in

[tiny halo]

Ooh do u know how to do this?

@prime ledge

[prime ledge]

no

[tiny halo]

Ok

[dark sparrow]

there was supposed to be an image there but it ain't loading

on your screen

[upper karma]

How do I calculate only the black area? This is a roundabout.

[onyx cloud]

do you know how to calculate the area of the middle circle

@upper karma

by middle i mean between the black and the white

[silver radish]

Please help

[onyx cloud]

which one

[silver radish]

Both

[onyx cloud]

lets start with the top

what have you tried

[silver radish]

I tried using the formula for radius and got 4.37 and got it wrong

[onyx cloud]

what is the formula for the volume of a cylinder

[silver radish]

V= pie r2 h

[onyx cloud]

perfect

now what do you already know

[silver radish]

I know the height

And how much it can hold

Which is 480 cubic centimeters

[onyx cloud]

so substitute that in

[silver radish]

What would be r though?

[onyx cloud]

that's what they want you to find

[silver radish]

Oh yeah

V= pie r2 8

[onyx cloud]

what's v

[silver radish]

Would it be 480?

[onyx cloud]

yep

[silver radish]

I get 4.3702

[onyx cloud]

show me your steps

[silver radish]

Well I just put it into a calculator but let me try by hand

[onyx cloud]

ok if you want my help show me your steps

[silver radish]

Ok

[onyx cloud]

that looks right

did you write cm in your answer?

or just the number

[silver radish]

Wym? My old answer was 4.37

So 19.1 is the answer ?

[onyx cloud]

no

its $\sqrt(19.1)$

[somber coyote]

Maximo:

[dark sparrow]

$\sqrt{19.1}$

[somber coyote]

Ann:

[dark sparrow]

@onyx cloud

[silver radish]

4.3703

[onyx cloud]

mb

4.3703 cm

[silver radish]

I think I got it wrong because I didn’t round to the nearest tenth

[onyx cloud]

and it says rounded to the nearest tenth

[silver radish]

So 4.4

[onyx cloud]

cm

[silver radish]

Ok

For the next one I did 4 x 1.2

And got 4.8

But that’s wrong so it told me “if the scale factor is r=4, the volume would be affected by what factor”

[onyx cloud]

so the volume is V

what is the volume of a rectangular prism

[silver radish]

L x w x h

[onyx cloud]

perfect

now what if we multiply the length width and height by 4

what would that look like

$V = l\cdot w\cdot h$

[somber coyote]

Maximo:

[silver radish]

V•4 ?

[onyx cloud]

do the multiplications on the right side first

[silver radish]

V= 4l•4w•4h

[onyx cloud]

so then what will the factor be?

[silver radish]

4?

[onyx cloud]

$4l\cdot 4w\cdot 4h=4\cdot 4\cdot 4\cdot l\cdot w\cdot h$

[somber coyote]

Maximo:

[silver radish]

64?

[onyx cloud]

👍

thats the factor

[silver radish]

Ty I have one last question

[tiny halo]

Max

[onyx cloud]

yeah

[tiny halo]

Are u done ?

[tiny halo]

Rip I have canvas too

[silver radish]

I did 18.75 divided by 1.2

Lol

[onyx cloud]

so now you have that the volume will be v*64

so plug in your original V and you'll have what you're looking for

whats up pulse

i'm boutta go to bed so hope its quick

[tiny halo]

Bed at 4?

Pm

[silver radish]

64 divided by 18.75?

[onyx cloud]

i woke up at 5 pm yesterday and haven't slept since

and also remember i might be in a different time zone

yk why are you dividing

oh lmfao

[silver radish]

Idk

[onyx cloud]

i was thinking it was question 4

hahahaha no you're right

[silver radish]

Lol

[onyx cloud]

i think you got it

but pulse whats up if its quick i can give you a hand

[silver radish]

Question 4 isn’t 64 and 5 isn’t 3.41

[onyx cloud]

nono i meant the picture was right

18.75/1.2

4 should be 64*1.2

[silver radish]

Then 5 is 18.75/1.2?

[onyx cloud]

yep

@tiny halo last chance

[silver radish]

That’s wrong tho

[onyx cloud]

which one

[silver radish]

15.625

Question 5

[onyx cloud]

ah shit

do the cube root of that

let me explain why though

let me explain why though

[silver radish]

2.53

Ok

[onyx cloud]

you start with V times S the scaling factor alrgith?

or no

you have $V_1 and V_2$

[somber coyote]

Maximo:

[onyx cloud]

$V_1 = l\cdot w\cdot h$

[somber coyote]

Maximo:

[onyx cloud]

and that is the small cereal box of 1.2

then we have that

$V_2 = sl\cdot sw\cdot sh$

[somber coyote]

Maximo:

[onyx cloud]

where s is the scaling factor

so then if we plug in the numbers we know

we get

$1.2 = l\cdot w\cdot h$

[somber coyote]

Maximo:

[onyx cloud]

and $18.75 = sl\cdot sw\cdot sh$

[somber coyote]

Maximo:

[onyx cloud]

but this formula right here $18.75 = sl\cdot sw\cdot sh$

[somber coyote]

Maximo:

[onyx cloud]

can be rearranged

$18.75 = s\cdot s\cdot s\cdot l\cdot w\cdot h$

[somber coyote]

Maximo:

[onyx cloud]

and then we see the lhw

[silver radish]

Mhm

[onyx cloud]

and we know that

$1.2 = l\cdot w\cdot h$

[somber coyote]

Maximo:

[onyx cloud]

from the other equation

so

$18.75 = s\cdot s\cdot s\cdot l\cdot w\cdot h$

[somber coyote]

Maximo:

[onyx cloud]

$18.75 = s\cdot s\cdot s\cdot 1.2$

[somber coyote]

Maximo:

[onyx cloud]

$\frac{18.75}{1.2}=s\cdot s\cdot s$

[somber coyote]

Maximo:

[onyx cloud]

whic means that $15.625 = s^3$

and thus

[somber coyote]

Maximo:

[onyx cloud]

$s=2.53$

[somber coyote]

Maximo:

[onyx cloud]

did you get confused by anything at all

[tiny halo]

Ok

I was washing dishes

[silver radish]

No

[onyx cloud]

alright

as long as you understand how we got there

ight pulse if its quick ill help if not i gtg

[tiny halo]

It’s too big to fit in one pick

[onyx cloud]

yeah thats a lil much for me

[tiny halo]

Dang ok

[onyx cloud]

gl though

[brittle condor]

pretty easy vector Q but i am a little confused
(imagine below)
so i gotta find the tangent vecotr of this
which is simply c'()
but i also have to find the equation of the tangent line to the given path at the point c(1)
how would i do that ?

[weary drift]

the expression for a point on the tangent line is c(1)+c'(1)t

[brittle condor]

this is what i get from the book for the tang line to a path

(what is t -t0) ?

[visual mist]

can someone check a and b for me? Also for part c im not really sure how to do it but i drew a diagram, can someone help me please?

[acoustic jungle]

how do I rotate an image

[visual mist]

oh i will retake it then

[high zephyr]

,rotate

[visual mist]

nvm

[acoustic jungle]

oh for the last question you need to know the angle bisecting theorem thing

[visual mist]

huh

[acoustic jungle]

basically AB/AD=CB/CD

I think**

let me search it up

[visual mist]

ok

[acoustic jungle]

yep it's right

[visual mist]

wot is

wait is part a) and b) right tho???

[acoustic jungle]

a is right let me check b

[visual mist]

okay

[vital frost]

Hey so I have a question for my homework: (cos^(4)x-sin^(4)x)/(cos^(2)x)=1-tan^(2)x ; here is the problem but how would the left side of this problem look like to verify this is true?

[acoustic jungle]

yes it's right

I like your proof

please use the $$ thing

[somber coyote]

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

[acoustic jungle]

that gives me big brain damage

[weary drift]

the expression for a point on the tangent line is c(1)+c'(1)t
@brittle condor there is no need to compare what i said to the book

[brittle condor]

there is if i wanna understand what i have in front of me with what you are saying 😛

[visual mist]

okay thankq

what about part c? idk how to do it

[weary drift]

give it some thought and you'll see they'll get the same points on the line

[brittle condor]

probably. i just wanted to see. i got l(t) = (t-1, e^2)

could you check or .. ? LOL

[acoustic jungle]

@vital frost I got it

[weary drift]

i won't check your computation but i will say that c(1) is the point of tangency, c'(1) is a vector that gives the direction of the tangent line, so c(1)+c'(1)t where t is real will pass through all of the points on the tangent line @brittle condor

[vital frost]

Thank you for taking the time to help me

[acoustic jungle]

do you still need help

[vital frost]

yeah

[acoustic jungle]

ok do you know how to factor

for example a^2-b^2

[vital frost]

i understand the factoring on the top

but i dont know where to go beyond that

[acoustic jungle]

do you know what $sin^2x+cos^2x=$

[somber coyote]

Fishraider:

[vital frost]

that would be 1 correct?

[acoustic jungle]

yes

did you do that before?

if you didn't then you will solve the problem knowing this

[vital frost]

ok I believe i understand

[fallow karma]

can somebody help me with a quiz?

[vital frost]

thank you

[acoustic jungle]

yes what is the question jett-

[weary drift]

asking for help during a quiz is a bannable offense

[high zephyr]

i thought it's only for exam

[brittle condor]

i got a different answer with ur way (a neg sign isntead of pos)

[acoustic jungle]

jett what is the question

area or surface area

[fallow karma]

volume

[acoustic jungle]

sorry I meant volume

[weary drift]

jett, you're taking a quiz right now in class?

[acoustic jungle]

well area of circle*height=volume

[fallow karma]

no, its a homework assignment, im doing online but I dont understand

[weary drift]

so what did you mean by "quiz"?

[acoustic jungle]

if you know the area of the circle then you will find the answer

[fallow karma]

sorry I didnt mean that

is this it? 11.6619in^3

thats what I got

[visual mist]

area of circle*height

[acoustic jungle]

tbh if he wanted to cheat he literally could've searched up volume of cylinder calculator

[visual mist]

lol

can someone help me with c)?

[acoustic jungle]

did you use the theorem I told you

angle bisector theorem

[visual mist]

i dont get it

[acoustic jungle]

let me try the question

[visual mist]

alright

[acoustic jungle]

so far I got AE = BD with congruent triangles

[visual mist]

o how

[acoustic jungle]

OH

I got it

ok

so you actually don't need the angle bisector theorm after all I was small brain

label the angle abd theta and try to come up with as many angles as you can

@visual mist

[visual mist]

okay

lemme draw it out

[acoustic jungle]

you should be able to see many similar triangles

(that are most likely congruent but I haven't proved them yet)

[visual mist]

i c some too

but then idk how to prove it

[acoustic jungle]

can you show me your drawing

[visual mist]

k

[high zephyr]

neat

[acoustic jungle]

do you see ACE and BCD

[visual mist]

ye

[acoustic jungle]

how are they related

[visual mist]

well uh CB and AC is equal bc isosceles triangle

all share angle c

[acoustic jungle]

about the triangle

[visual mist]

wdym

[acoustic jungle]

Can you prove Ace = bcd with SAA

[visual mist]

i think so

wait let me draw it

well ok AC and BC is the same because ABC is an isosceles triangle, they share angle c and angle DFA and EFB is the same because they are vertically opposite

[acoustic jungle]

yes

prove afd and bfe are similar first

[visual mist]

okay

wait

im not really sure, but angle DFA and BFE is the same because vertically opposite, DF and EF is the same, and FA and FB is the same???

[acoustic jungle]

you don't know df and ef are the same

[visual mist]

tru

[acoustic jungle]

just prove that afd and bfe are similar and after we will find a side length which will prove they are congruent

and you do need the angle bisector theorem

[visual mist]

wots the bisector theorem

[acoustic jungle]

you have to search that up

[visual mist]

ok

its kiinda confusing i dont get it

[acoustic jungle]

ah

I am not a good explainer

maybe nightgale can help

[visual mist]

is he online

[acoustic jungle]

I mean you can also ask him how to do it

maybe he doesn't need to use that theorem

[visual mist]

can i say that DA is the same as EB

probs not

[acoustic jungle]

oo

yes you can

adb is equal to bea

because SAA

now you have da=eb

[visual mist]

oo

lol

[acoustic jungle]

now you can prove those are congruent with SAA angle

[visual mist]

ye

[acoustic jungle]

again*

then you get your answer

[visual mist]

wait isnt it SAS because angle DFA is the same as EFB and then FA is the same as FB and AD is the same as EB

[acoustic jungle]

yes you can say that too

[visual mist]

and also u said abd is equal to bea is it because they share AB and DAF is the same as EFB and then they share AFB

ok

[acoustic jungle]

my brain is going to explode but I hope I what you wrote matches up with what I think you wrote

it's just about proving similar triangles to congruent triangles with Angle Side angle

[visual mist]

hahahahaha same tbh

thanks for ur help anyways

[silent plank]

overcomplicating the problem

[visual mist]

u have better way?

[silent plank]

look at the info given.
the base angles of the isosceles triangle are bisected right?

[visual mist]

yes

[acoustic jungle]

it shouldn't be too bad all we did was we proved EB=DA with ASA and then proved ADF = EBF with ASA

[visual mist]

hahahahaha ye

[silent plank]

which means those 4 angles formed at the base are equal right?

[visual mist]

yes

[silent plank]

specifically ABD and BAE

[visual mist]

ye

[acoustic jungle]

hmm

[silent plank]

and what type of triangle would that make (Tri)ABF

[acoustic jungle]

jesus

my brain is so small

that is a great way

marvolous

fantastic

[visual mist]

wait what

isosceles?

[silent plank]

yes

[visual mist]

then wot

[silent plank]

and then apply a theorem related to isosceles triangles

[visual mist]

wot theorem

isosceles triangle theorem??

[silent plank]

something relating sides to angles which you had already used when constructing your diagram

[visual mist]

ummmm

im not sure

[silent plank]

in short: the sides opposite equal angles of an isosceles triangle are what?

[visual mist]

equal

?

[silent plank]

yes

and what do you get when you apply that?

[visual mist]

wdym

[silent plank]

specifically ABD and BAE (are equal)

in your triangle ABF,

which side is opposite angle ABD

[visual mist]

DBA

oh

wait no i thought u meant angle

AE?

[silent plank]

no

in your triangle ABF

perhaps if I labelled it as angle ABF instead (which is still the same angle)

[visual mist]

idk

[silent plank]

try not to overthink it

focus only on the triangle ABF

and you can pretty much ignore everything else now

[visual mist]

and?

[silent plank]

specifically ABD and BAE (are equal)

[visual mist]

mhm

[silent plank]

those are the base angles of that triangle right?

[visual mist]

yes

[silent plank]

that triangle has 3 sides:
AB, BF, AB

which of those sides is opposite angle ABD?

[visual mist]

2 ab?

[silent plank]

no? also why's there a 2 there

[visual mist]

u repeated AB twice

[silent plank]

oh sry

autocorrect

[visual mist]

lol

wdym by opposite

[silent plank]

side that isn't touching

is what it means for a side to be opposite an angle in a triangle

[visual mist]

well AB and BF are touching ABD

[silent plank]

so the side that isn't touching is:

[visual mist]

u gave me the sides AB, BF and BF.

[silent plank]

fml

last one should be AF

[visual mist]

AF then

[silent plank]

similarly which side is opposite angle BAE

[visual mist]

BF

[silent plank]

and from the definition/properties of isosceles triangle, would they be equal?

[visual mist]

the angles or the sides

[silent plank]

well we already established that the angles were equal earlier

because the angles are equal, would those specific sides also be equal?

[visual mist]

ye

s

[silent plank]

and that's it

[visual mist]

Oh

omg thanks

[dull remnant]

https://cdn.discordapp.com/attachments/694070956875972639/694758598361874432/large_photo974922_7625169.png

what is csc of that theta

i got -17/8

because i drew a triangle from origin to -8,15 to the y axis

[silent plank]

what's the definition of csc?

[dull remnant]

@silent plank its cosecant

its the reciprocal of sin

[silent plank]

in terms of add, opp, hyp

and/or what is sin(theta) here

[dull remnant]

obvi hyp/opp

i think

[silent plank]

and what is the length of the opp side

[dull remnant]

that sthe problem

idk where the triangle is formed

is it from origin to 8 15 to x axis

or origin to 8 15 to y axis

[silent plank]

draw a vertical line from (-8,15) to the x axis

[dull remnant]

wait why there

isnt the angle on the other side?

if thats the real angle, then the csc should be 17/15

i got it from the angle of the x axis to y axis and then opposite over adj

[silent plank]

that look ok

[dull remnant]

then reciprocal

so u think it is 17/15

thats what i got at first

[silent plank]

wait

[dull remnant]

but it doesnt seem right

[silent plank]

yeh

[dull remnant]

@silent plank but

can u explain

why its not the other angle

because its asking for the theta

[silent plank]

look up stuff like reference angles

[dull remnant]

yea

so its the reference angle?

ok ty

[eager kraken]

Is this system check right in saying that my work in wrong?

I did the equation
cos8x-cos2x = -2sin((8x+2)/2)sin((8x-2x)/2)

[quiet mason]

6/2=3

not 5

[eager kraken]

oh it's minus

ooooof

[quiet mason]

lol rip

[eager kraken]

i kept double checking every single sign making sure they were right but didn't stop to do the signs right

thanks

[upper karma]

could anyone help me with these 2

[summer spire]

@upper karma what extra measurements can you determine from those diagrams?

[cold oriole]

can someone help me

sin(x)-cos(x)=1.4

with goniometric relations

in G [0;2pi[

[rich wolf]

Goniometric

[summer spire]

it's an actual word, but I have no clue how it comes into play here

[rich wolf]

Oh it just means angles and shit

Heres a useful tip

$\sin(x) = \cos(x-\frac{\pi}{2})$