geometry-and-trigonometry

[rapid tartan]

anyone?

[abstract saffron]

complete the square in x and y

i.e. (x-h)^2+(y-k)^2=r^2

use cylinder volume formula, r^2*pih

[abstract saffron]

write in (x-h)^2+(y-k)^2=r^2, where r is the radius and (h,k) is the center

[rapid tartan]

😅

[abstract saffron]

did it help?

[rapid tartan]

sorta

hard to learn this topic through a computer class tho lol

[abstract saffron]

ok

lel true

you could read up the wikipedia article on conic sections

its a big help

[rapid tartan]

conicsections?

they help with geometry?? @abstract saffron

[abstract saffron]

conic sections

are like

ellipse hyperbola parabola

its like

useful stuff abt circles n stuff

[untold cosmos]

is this acute?

[night hedge]

I'm trying to find the name of a shape of something not terribly common. I tried asking the math help general question area but it doesn't feel like the right place to be asking. Would this be a better space for such inquiries?

[abstract saffron]

isosceles

[untold cosmos]

thx bro

[abstract saffron]

lel np

[abstract saffron]

you cannot assume that it is acute because the measure of ADC cannot be deduced

however, you can confirm by CPCTC that AD is congruent to DC

which makes it isosceles

[untold cosmos]

welcome bro

[blissful obsidian]

Isn't it's a kite

[abstract saffron]

yesh

[haughty eagle]

solve for x

pls help me

[dark sparrow]

@haughty eagle how familiar are you with the basics of trigonometry?

[haughty eagle]

im ok

i know a little

like cos

tan

[dark sparrow]

are you familiar with the SOH-CAH-TOA mnemonic?

[haughty eagle]

sin

yes

[dark sparrow]

okay

[haughty eagle]

i did tan(22)x24 and it looked way wrong

[dark sparrow]

so you have two sides here that you care about: the side labeled 24 (because it is known) and the side labeled x (because it is sought)

...oh, so you did make an attempt

why didn't you say so in the first place

[haughty eagle]

my bad

[dark sparrow]

did you do it on a calculator?

[haughty eagle]

yes

[dark sparrow]

did you, perhaps, get a ridiculously small number like 0.212?

[haughty eagle]

yep

[dark sparrow]

switch your calculator from radian to degree mode

[haughty eagle]

ok

i got 9.7 when rounded to the nearest tenth

is that right

[dark sparrow]

yes, assuming rounding to the nearest tenth is what you're asked for

[haughty eagle]

yep

thank you so much

[abstract saffron]

360/8 sides

so it’s 45

[dark sparrow]

you should not give out answers, wtqpped

[dawn mango]

someone please message me if they want to teach me how to draw a rectangular hyperbola

in the form of:

i have spent 3 housr of this and cant do anymore

i know how to find the asymptote, intercept

but i do not know how to sketch it

at my wit's end rn

ping me or msg me pls if willing

[abstract saffron]

sorry

[night hedge]

Aren't kites a rhombus?

[abstract saffron]

lol what do you need help on

[blissful obsidian]

No

They are not

[night hedge]

You're taking about this?

[tepid yacht]

if someone can help me with my test ill pay just dm me!

[abstract saffron]

bro

did u like

not read the rules

...

[sick torrent]

lol

[abstract saffron]

lol

[lofty oxide]

can someone help cause i got no clue

[lofty oxide]

plsss pinggg if u answer plssss

[abstract saffron]

a changes the amplitude

if a is 0 it has 3 x intercepts in that interval

so it must be tangent

to the x axis

which means d=+-a

cuz |a|=d is necessary for it to be tangent to x axis

@lofty oxide

[lofty oxide]

ahhh yeahh

i didnt think about the tangent part

[abstract saffron]

when its in the form y=a*sin(x)+d and d=+-a it will be tangent to the x axis at 3pi/2

if d is positive

[lofty oxide]

ok thanks so much!

[abstract saffron]

if its negative

it will be tangent at pi/2

[abstract saffron]

np

it looks like this if a is negative

this if a is positive

yeah only one intersection with x axis in interval 0,2pi

just remember that d shifts the graph up and down and a changes the amplitude

[lofty oxide]

ok!

thanks again lol

mr kind stranger

[abstract saffron]

lol

np

[chrome cradle]

How this need to be solved? I've searched shearing on internet and couldn't find exemples with angles and other points excepting the origin. Can someone help?

[night hedge]

This is what I'm talking about. I don't think this is a kite.

[untold cosmos]

yo could someone help me

[trim breach]

Do you know the formula to determine the total degrees in a polygon?

[abstract saffron]

what do the exterior angles in a polygon add up to?

hint: ||360||

[abstract saffron]

for 2, use the polygon interior angle formula i.e. (n-2)*180/n

[vagrant ore]

helppp

[trim breach]

I’m assuming that implied scaling down.

In which case, do you know how scale factors work?

[trim breach]

Which part is giving you the problem?

[sick torrent]

the 2nd

and 3rd

[trim breach]

For angles 1 & 5, can you eliminate any of the choices?

[sick torrent]

complementary

would it be supplementary?

[trim breach]

Yes, and one more applies here.

[sick torrent]

vertical

[trim breach]

Not for 1 and 5.

[sick torrent]

i mean adjust

[trim breach]

Yes.

They share a common side and vertex.

[sick torrent]

oh

[trim breach]

Do you have an idea about 2 & 5?

[sick torrent]

is that just a line

if so would it be complementray

[trim breach]

We don’t know what measure angles 2 & 5 are.

So we cannot determine if they add up to 90 degrees.

And for the same reason, we cannot determine if they are supplementary.

So they are either vertical or adjacentZ

[sick torrent]

vertical?

[trim breach]

Yes!

[sick torrent]

oh thank you so much

[trim breach]

Glad to help!

[sick torrent]

wait what would 2 and 4 be

supplementary?

[trim breach]

Not quite. 2, 3, and 4 together add to 180 degrees.

[sick torrent]

oh so complementray

[trim breach]

Yes. 3 is a right angle, so the sum measures of 2 and 4 must be 90 to give 180 degrees total.

[sick torrent]

Thank you again

[sweet jetty]

Y= 44
Z= 43?

[trim breach]

Why z=43?

A hypotenuse cannot be shorter than either of the legs.

[upper karma]

does anyone understand this

its 45 45 90

[keen lion]

thats really simple. use pythagoras theorem

from the pattern of the solutions you can understand the general solution for 45 45 90 trianlge

[upper karma]

can you do one to see how you do it ?

[keen lion]

first one- it is an isosceles triangle so the other side is also 17. from the pythagoras theorem $17^{2} + 17^{2} = x^2$ that means $x=17\sqrt(2)$

[somber coyote]

ömer

[upper karma]

for 2. is it 29? making sure cause i tired it on my own and i wanna know if i got it correct

tried

[keen lion]

its $20 \sqrt2$

[somber coyote]

ömer

[keen lion]

you should be able to do the algebra

[upper karma]

um idk im not that smart ngl

[keen lion]

I am not trying to be rude you can do it.

[upper karma]

so 3 is 12 and 4 is 13 ?

with the sqrt 2

[keen lion]

let me do the math in first one more clearly. $17^2 + 17^2 = 2 (17^2)$ but thats also equals to $x^2$ that means $\sqrt(x^2) = \sqrt(2 (17^2))$ from the rules of the square root $x=17\sqrt(2)$

[somber coyote]

ömer

[upper karma]

is 5 just sqrt(2)

?

[keen lion]

yes

[upper karma]

mm ok

wait can you do one of the other ones so i could understand how you do it and i could go off of that yk? that have sqrt

[keen lion]

no sorry. I helped enough you should work through the rest

[upper karma]

so the math im doing is pre university ?

im confused lmao

[keen lion]

it is

[upper karma]

what?

[keen lion]

it is pre university

[narrow kraken]

Am I allowed to post pictures here for questions?

[upper karma]

im just in 10th grade

lmao

[keen lion]

that is pre university

[upper karma]

im hela dumb

hella

[keen lion]

be more confident about math

it doesn't care if you are dumb or not

learning math at highschool level is doable by everyone

[narrow kraken]

"Find the exact value of each expression "

How do you find the tan A or Tan B/ Sin A , Sin B

[upper karma]

alright, thanks

[narrow kraken]

My teacher taught me to use significant angles but I see a different method used by other people

[keen lion]

it seems like you have to use this guys

[narrow kraken]

I understand that part but idk where they get these numbers from

4pie over 12 and 3pie over 12

[keen lion]

because you want to use significant angles.

1pie over 12 and 11pie over 12 wouldn't work

[upper karma]

@keen lion yo i did question 6 and is the answer 6? i tired it on my own and i just wanna know if i did it correct so i could know if im going on the right path

[narrow kraken]

This method will work to find it right?

[keen lion]

yes

[narrow kraken]

I get 105 degrees

[upper karma]

one more thing before i go do i always need to add the sqrt next to it ?

[narrow kraken]

Sin 60 + 45

[keen lion]

ok try to write 105 degrees in the form of addition of the significant angles what do you get?

now transform 60 and 45 to the pi form

[keen lion]

I didn't understand

[upper karma]

the answer is 6 so do i write down sqrt(6) or just 6

[keen lion]

6

[narrow kraken]

How do I do that? I usally use this

[upper karma]

oohhhh i understand it now thanks @keen lion

[keen lion]

60 = 180/3 so 60 = pi/3
45 = 180/4, so 45 = pi/4

sin (pi/3 + pi/4) = sin (4pi/12 + 3pi/12)

then the addition rule

[narrow kraken]

Wow

I dont understand why my teacher didn't include this

[keen lion]

you are welcom :)

[narrow kraken]

Thanks

What would 30 be ?

Just pie?

[keen lion]

solve the equation 30 x = 180

[narrow kraken]

Oh ok

[keen lion]

30 = 180/x so 30 = pi/x

[narrow kraken]

Is there a name for this equation?

Because I need to put this on my notes

[keen lion]

I don't think so

[narrow kraken]

Ok thanks again

[keen lion]

no problem

[narrow kraken]

im using the pie circle

[keen lion]

thats also works

[narrow kraken]

i forgot we learned that a while back

[keen lion]

👍

as long as it feels natural

[narrow kraken]

yup

also how would i know what order to do it

for example

sin 60 + 45 or sin 45 +60

[keen lion]

sin 60 + 45 = sin 45 + 60

[narrow kraken]

that confuses me

[keen lion]

it doesn't matter

[narrow kraken]

ahh ok

[dark sparrow]

y'all need parentheses

yes

sin x + y reads as sin(x)+y, not sin(x+y) as you intended

[keen lion]

true

[dark sparrow]

then why this "no"?

[keen lion]

i was on the same page with him so i didn't feel like it was necessary but you are right

this is pre-university

someone might miss understood

[hard vale]

this is the only one i dont understand out of 14 problems

@dark sparrow

[dark sparrow]

???? why ping me

[hard vale]

help

[dark sparrow]

no, why ping me specifically instead of waiting for someone to come along like you're supposed to?

[hard vale]

i am sorry i wont do it again i am sorry

[dark sparrow]

anyway, okay, whatever... have you made any progress so far?

[hard vale]

i tried to put in the triangular prism formula i found the area for the base of the trapezoid but then i got height of 504

i cant figure ito ut

[dark sparrow]

but this is not a triangular prism. why would you use the formula for a triangular prism on an object that isn't a triangular prism?

[hard vale]

someone told me it was a triangular prism

wait it is a triangular prism

[dark sparrow]

the problem literally says "A trapezoidal prism..."

at the very beginning

the only way to miss it is to not read the problem at all lmao

[hard vale]

oh wait i just typed the wrong thing

im so tired rn im sorry

[dark sparrow]

in any case

the volume of any prism is just the product of its height and its base area.

[hard vale]

what do i write down

[dark sparrow]

i mean, idk, you can write $V = Ah$ or something and say what stands for what

[somber coyote]

Ann

[hard vale]

like through the whole problem what should i write down

i just dont get it at all

[dark sparrow]

i was going to prompt you to calculate the area of the base as an intermediate step

[hard vale]

what should i start with

[dark sparrow]

i literally just said it

[hard vale]

oh ok

[dark sparrow]

or would you like me to repeat it but more explicitly?

[hard vale]

theres 2 bases

which base do i write down

[dark sparrow]

eh?

no, i'm not talking about the bases of the trapezoid

i'm talking about the bases of the prism

which are trapezoids themselves

[hard vale]

so you want me to find the base of the prism as the first step of thje problem

[dark sparrow]

i want you to find the area of the trapezoid serving as your prism's base, yes.

[hard vale]

thats 84

i already did that

[dark sparrow]

84 what

[hard vale]

is the area oif the trapezoid

[dark sparrow]

84 cubic miles?

84 dollars per hour?

[hard vale]

inches

[dark sparrow]

just inches?

[hard vale]

i think

idk

[dark sparrow]

what do you measure area in?

[hard vale]

in^2?

[dark sparrow]

square inches, that's right.

okay, so you got the area of your base as 84 in^2

now, do you remember how to find the area of a trapezoid in terms of its height & bases?

[hard vale]

wait i wanna know if i did it right'

for the area

i did 6+8/2 x 12

and got 84

[dark sparrow]

$6 + \frac{8}{2} \cdot 12$?

[somber coyote]

Ann

[dark sparrow]

this gives 54, not 84.

[hard vale]

https://gyazo.com/4f73dcd2b000d5db5ce6dbdd07fa0184

[dark sparrow]

but 12 inches is the height of the prism, not of the trapezoid!

[hard vale]

oh

so whats the height of the trapezoid

[dark sparrow]

that's what you are asked for

[hard vale]

so youre telling me

i just put

wait

so do i start with 588 =

[dark sparrow]

if you insist on phrasing it that way, sure

(prism height) * (trapezoid area) = 588

[hard vale]

what is the trapezoid area

[dark sparrow]

that's what i suggested you find

as an intermediate step

[hard vale]

but i cant find that

without the height

[dark sparrow]

yes you can

[hard vale]

how

[dark sparrow]

you are given the height of the prism

[hard vale]

i used that and u just said it was wrong

12

[dark sparrow]

read this again

and try to read something more true to reality in my messages than just a flat "it's wrong"

[hard vale]

im just horrible at this

im sorry

[dark sparrow]

horrible at reading comprehension?

[hard vale]

that too yeah

im in critical reading

[dark sparrow]

i literally said numerous times, 12 inches is the height of the prism

12 inches is the height of the prism. 12 inches is the height of the prism. 12 inches is the height of the prism.

[hard vale]

ok

so for the numbers

is it

the two bases as 6 and 8

then

height 12

right?

[dark sparrow]

no

[hard vale]

what is it

[dark sparrow]

refer to this again

the prism height is 12

12 * (trapezoid area) = 588

[hard vale]

but i thought iw as trying to find height

sp the trapezoid area is 576?

[dark sparrow]

need i repeat myself a fourth time?

our goal is to find the trapezoid height, yes.
the intermediate step i'm suggesting, and the one we're going through now, is to find the area of the trapezoid.

no, the trapezoid area is not 576.

[hard vale]

so we got it

how]

[dark sparrow]

how did you get 576? did you subtract 12 from 588?

[hard vale]

yes

[dark sparrow]

look at this again. what symbol is between the 12 and the (trapezoid area)?

[hard vale]

multiplication

oh

so i divide by 12

[dark sparrow]

and how do we undo multiplication?

[hard vale]

???

[dark sparrow]

that's right, you divide.

[hard vale]

alright

so 49

whats the next step

[dark sparrow]

the area of the trapezoid is 49 in^2.

[hard vale]

yes

[dark sparrow]

now we come back to a question i asked you but never got the answer to: do you know how to find the area of a trapezoid in terms of its bases and height?

without using an online calculator to do so, that is.

[hard vale]

no

[dark sparrow]

so the formula $A = \frac12 (a+b)h$ is foreign to you?

[somber coyote]

Ann

[hard vale]

no

not completelty

[dark sparrow]

this is the formula for the area of a trapezoid

the one i asked you to recall, and the one you said you did not know.

[hard vale]

so now i plug in everything i know

[dark sparrow]

here a and b are the bases, and h is the height.

yes, sure.

[hard vale]

for A do i put 588 or 49

i put 588 right

[dark sparrow]

no

588 cubic inches was the volume

49 square inches is the area of the trapezoid

[dark sparrow]

as i said here again

though this probably slipped past your apparently horrible reading comprehension!

[hard vale]

so i put 49 = 1/2(8+6)h

[dark sparrow]

yes

[hard vale]

and then i do the algebra and thats my answer

[dark sparrow]

yes, assuming you are able to do the algebra correctly.

[hard vale]

i got 7

is that right

[dark sparrow]

yes

7 inches

[hard vale]

and thats my answer

[dark sparrow]

yes, because that's what they ask you for

[tawny plaza]

hi I have a simple question

sin(2theta) times cos(4theta)
the next step was that
sin(-2theta) plus sin(6 theta)
anyone got an Idea how the former is equal to the later?

[dark sparrow]

i think you're missing a factor of 2 somewhere

[tawny plaza]

maybe

but what is the idea

nvm about the factor

[dark sparrow]

product-to-sum identities

$\sin(x) \cos(y) = \frac12 (\sin(x+y) + \sin(x-y))$

[somber coyote]

Ann

[dark sparrow]

(also do you not have a way to type + and * on your kb?)

[tawny plaza]

what does the dollar sign mean?

and also the meaning of the backward slash ?

[tawny plaza]

I forgot to use them lol

[dark sparrow]

those are a LaTeX thing. this bot over here can make your equations look nice if you type them using the LaTeX markup language

i summoned it just now

[dark sparrow]

look at this

[tawny plaza]

oh yeah I see

do you know a name for this rule?

or no wait

I think I saw it before

just give me a minute to soak it in

[tawny plaza]

@dark sparrow thank you this rule you just gave me is a wonderful key to a bunch of my problems

specially concerning laplace stuff

btw, if I have a question on laplace where should I ask

I mean which branch of math does laplace belong to

[dark sparrow]

calculus i guess

[keen bobcat]

Diff eq?

[dark sparrow]

or that

[tawny plaza]

ok thanks

[dull thorn]

If I have a triangle UVW, is its midpoint triangle essentially a scaling and a rotation about the centroid?

Like it kind of looks to me like the midpoint triangle is obtained by scaling the triangle UVW by 1/2, then rotating it by pi about the centroid

[tawny plaza]

yes

[tawny plaza]

yeah you are right

[serene obsidian]

yo

i got 54.3 am i rigght

A regular polygon is inscribed inside a circle. Find the area of the shaded region to the nearest tenth.

[rapid umbra]

geometry kinda sussy ngl

[dark sparrow]

@serene obsidian you rounded incorrectly at some point

[serene obsidian]

yea i eneded up fixing

i put 54.4

[sonic needle]

Do this I double dare ya

[coarse shoal]

could anyone help me with these? i’m terrible at it lol

[idle bobcat]

can someone help me with this problem?

this is my work but i know i'm doing something wrong

im not even sure if I modelled it correctly

[idle bobcat]

oops i forgot to take the square root when i did the cosine law

but even then i still think its wrong

[upper karma]

x^2 + (mx)^2 = 1 -> x = +- sqrt(1-(mx)^2), y = mx ?

[upper karma]

<@&286206848099549185>

[dreamy ridge]

So this line has equation y=mx

Unit circle has equation x^2 +y^2 =1

Sub them into each other

I’m assuming it wants it in terms of m

[upper karma]

didn't i do that?

[dreamy ridge]

Yea you did

So you’re right, I was just saying

[dark sparrow]

yeah but then you did not solve for x as you were supposed to

[upper karma]

what do you mean

i was supposed to solve for points

these are: (sqrt(1-(mx)^2), mx) (- sqrt(1-(mx)^2), mx ?

[dark sparrow]

no, this is nonsense

you are making it seem like any pair of values for x and m yields a valid pair of points, when it does not

your points should be expressed in terms of m only

you need to solve the equation x^2 + m^2 x^2 = 1 for x

properly

instead of this bullshit you seem so deathly insistent on pulling

[upper karma]

thank you

[upper karma]

how?

this means that either of the coordinates has a negative sign

so if 1. is in quadrant 1, then 2. is in quadrant 2 or 4

the x coordinate is 5 times bigger/y coordinate 5 times smaller

since if 2 were in quadrant 2, x coordinate would be smaller, this must mean that it is in quadrant 4

[dark sparrow]

is this supposed to be an explanation for why the two radii are perpendicular?

[upper karma]

so it makes intuitive sense that they are perpendicular

dunno how to prove it mathematically

[dark sparrow]

this explanation doesnt hold any water

so what that the two radii occupy adjacent quads, this alone doesnt mean they must be at a 90 degree angle to each other

they CAN be at a 90 degree angle to each other but like, why not 89 degrees

[upper karma]

sec

i'm trying to write it

ok i give up

[dark sparrow]

after all of 3 minutes

have you made a diagram?

[upper karma]

ok my best bet is
coordinates of 1 are
(a, 5a) -> 5a/a = 5
coordinates of 2 are
(a, -a/5) -> (-a/5)/a = -1/5

1: a^2 + 25a^2 = 1 -> cos x = sqrt(1/26), sin = 5sqrt(1/26)
2: cos y = sqrt(1/26) = cos x (opposite degree)
@dark sparrow

i think it proves it sufficiently

actually i don't think it proves it at all lol

[dark sparrow]

coordinates of 1 are
(a, 5a) -> 5a/a = 5
coordinates of 2 are
(a, -a/5) -> (-a/5)/a = -1/5

you are introducing the variable a twice here

this is bad

is a the x coord of the first point or the second point? if both, then how can you say they have the same x coordinate?

[upper karma]

i don't know

so if 1. is in quadrant 1, then 2. is in quadrant 2 or 4
the x coordinate is 5 times bigger/y coordinate 5 times smaller
since if 2 were in quadrant 2, x coordinate would be smaller, this must mean that it is in quadrant 4

i based than on tan

[livid moss]

You seem to be overcomplicating it. But you need to know 2 things, or you basically have to prove them

1. If a line makes an angle α (not equal to 90 degrees or the line will be vertical) with the x axis then it has slope tan(α)

[dark sparrow]

have you made a diagram?

[upper karma]

no

what do i need a diagram for

[livid moss]

2. Two lines are perpendicular when the product of their slopes is -1

[dark sparrow]

why not?

do you hate diagrams? are you allergic to diagrams?

[upper karma]

i know in which quadrants they are

[dark sparrow]

you can make a diagram like this, and everything will become crystal clear

[upper karma]

proving things algebraically is funnier anyways

algebra shows relationships nicely and geometry does not

[upper karma]

nice reminder, thanks

[dark sparrow]

subjective

geometry is by its very nature more visual

[upper karma]

idk how

ye idk whatsoever what to do with it except
tan (theta + pi/2) = -cos theta/sin theta

[livid moss]

If you have shown that then you are done...

[upper karma]

why?

[livid moss]

Because 1/tan theta = cos theta / sin theta

[upper karma]

and?

(sin (theta + pi/2)) / (cos (theta+pi/2)) = -cos theta/sin theta
-cos theta/sin theta * cos (theta+pi/2) = sin (theta + pi/2) * sin theta

how do i evaluate it

[livid moss]

I don't understand

[livid moss]

Here, replace cos(theta)/sin(theta) with 1/tan(theta) and you are done.

[upper karma]

i don't see why -1/tan(theta) = tan (theta + pi/2)

[livid moss]

You showed this right?

[upper karma]

i have just rewritten it

-1/tan(theta) = tan (theta + pi/2) follows from tan (theta + pi/2) = -cos theta/sin theta
because they are the exact same thing

but how is that a proof

[dark sparrow]

1/(a/b) = b/a

[upper karma]

where do you even have a/b

[livid moss]

So you understand that:
$$\tan( \theta + \frac{\pi}{2}) = - \frac{\cos(\theta)}{\sin(\theta)}$$
and
$$ \frac{1}{\tan(\theta)} = \frac{\cos(\theta)}{\sin(\theta)}$$
But you cannot conclude that
$$\tan(\theta + \frac{\pi}{2}) = - \frac{1}{\tan(\theta)}?$$

[somber coyote]

Lunasong the Supergay

[dark sparrow]

i was trying to remind you of a property of fractions that you seem to have forgotten

[upper karma]

i don't understand why is it negative

[livid moss]

Where?

[upper karma]

1st, 3rd

same thing but rewritten

[livid moss]

Well, if the first one is negative, then it follows that the third one should be, right?

[upper karma]

why is the first one negative

[livid moss]

You said this.

So you already showed it's negative...

[upper karma]

i just rewrote that

didn't show anything

[livid moss]

Show me what you did

I don't understand

[upper karma]

i saw -1/tan theta

so i just used the definition of tan

to rewrite it

i left -1 as is

[livid moss]

😭😭😭😭😭

Don't say tan(theta + pi/2) = -cos(theta)/sin(theta)

You made me think you started at the LHS and got to that by simplifying.

[upper karma]

i already said that idk how to ismplify
sin(theta + pi/2)/cos(theta+pi/2)

[livid moss]

The way you wrote it is confusing. You keep saying things are equal when you haven't shown they are equal, so then I assume you have.

[upper karma]

sry

[livid moss]

There are two trig identities:
$$\sin( \theta + \frac{\pi}{2}) = \cos( \theta)$$
$$\cos(\theta + \frac{\pi}{2}) = - \sin( \theta)$$

[somber coyote]

Lunasong the Supergay

[livid moss]

You use these

[upper karma]

they hadn't been introduced

i don't know where they come from

axler sometimes asks you to prove results from the next section in the exercises of the previous one

[livid moss]

Do you know that sin(pi/2 - theta) = cos(theta)?

[upper karma]

no

[livid moss]

That you can see from a triangle tbh

[upper karma]

he only explained what sin, cos, tan, the opposites of these are

[livid moss]

Okay, he looks like he just wants an explanation and not like a trig proof.

So idk what he is looking for, sorry

[upper karma]

so am i not supposed to know where does the - come from?

and he just wants me to think about it as i read the next sections?

[livid moss]

You are supposed to know, but you have to give some kind of explanation not a proof.

I don't know what explanation he wants

Because I don't know what you have and what knowledge you can use

[upper karma]

doesn't it work

[livid moss]

Yes it does

But you prove that using identities you said you don't know

[upper karma]

no, why?

i just used the definition of the tangent and it worked

dunno how to do these

or this

[livid moss]

For the first equation I sent you need the identities

[upper karma]

no, bcuz it's in the question

[livid moss]

I feel like we are totally talking past each other. Sorry, if you have to explain why what they give you is true, then you can't just use it as if it's true.

[upper karma]

why not

i'm supposed to explain that it's true

so i just show that the two sides are equal to each other

dunno how to do this

[fervent imp]

Hi guys! How do I find the value of k?

[upper karma]

go to a different channel

[upper karma]

.

i have asked 4 questions here, last one was 6 minutes before your post <@&286206848099549185>

[fervent imp]

Ok. I am just going to ask my Mom for help.

[upper karma]

5th and the last q from this section:
pi/4 is in the specified interval, therefore pi/4 is irrational
irrational number * rational number is still an irrational number which means that pi is irrational
?

[humble pulsar]

something something something contrapositive

[upper karma]

i don't know what that is

never used that

i only know proof by contradiction @humble pulsar

[humble pulsar]

Ok well assume $\pi\in\mathbb{Q}\implies \frac{\pi}{4}\in\mathbb{Q}$

[somber coyote]

Mosh

[upper karma]

and

what's wrong with my proof

[humble pulsar]

you say "pi/4 is in this interval, therefore it's irrational"

being in the interval doesnt make it irrational

[upper karma]

'use ... to explain why this result implies ...'

[humble pulsar]

yes, but what you said makes no sense

[upper karma]

tan theta is irrational in that interval

according to that proof of his

[humble pulsar]

no

[upper karma]

i mean wherein theta is rational

[humble pulsar]

IF theta is in Q and theta is on (0,pi/2), THEN tan(theta) is in not Q

[upper karma]

oh wait

[humble pulsar]

if you assume pi is in Q, then you show that 1 is in not Q

which is false

therefore pi cant be in Q

[upper karma]

how?

[upper karma]

?

i mean now that i look at it i don't really udnerstand the question

theta is a rational number

tan theta is irrational

and we use an irrational theta

pi/4

so this result of his doesn't apply

[humble pulsar]

Assume $\pi\in\mathbb{Q}$, this means that $\frac{\pi}{4}\in\mathbb{Q}$ since division of rational numbers remain in the rational numbers. Applying tan to $\frac{\pi}{4}$ will get us an irrational number by Lambert's lemma, so $\tan(\frac{\pi}{4})\in\bar{\mathbb{Q}}$

[upper karma]

oh wait i see

[somber coyote]

Mosh

[upper karma]

isn't there a mistake

tan(pip/4) not included in q

at the end

[humble pulsar]

pi/4 is rational, so tan(pi/4) is irrational cause we assume pi is rational

[upper karma]

ye so if it's irrational then it isn't included in q

because Q is the set of rational numbers

[humble pulsar]

yes, so it's in Q bar

[upper karma]

what does Q bar mean

[humble pulsar]

irrationals

[upper karma]

haven't seen it before

oh thanks

[humble pulsar]

$\mathbb{Q}\cup\bar{\mathbb{Q}}=\mathbb{R}$

[upper karma]

ok and since tan pi/4 = 1 then that means 1 is irrational

[somber coyote]

Mosh

[upper karma]

which is false

[humble pulsar]

yes

[upper karma]

thanks

could you help me with the other questions'

[upper karma]

first 2

[upper karma]

<@&286206848099549185>

[silent plank]

are you allowed to use a calculator?

[upper karma]

ye

[silent plank]

have you tried using arctan?

[upper karma]

don't know it yet

it's introduced in the next chapter

[silent plank]

do you know what you get when you take the tan of something very close to 90° from each side?

[upper karma]

ye

sin x/cos x -> big numbers

[silent plank]

from the left side you'd get a really big positive number,
from the right side you'd get a really big negative number

so consider numbers slightly under and above 90°

[upper karma]

not the other way around ?

sin, cos are positive in the 1st quadrant

in the 2nd cos is negative

[silent plank]

from the left side, i.e. under 90°, i.e. Q1, where sin and cos are pos
pos/pos is pos

[upper karma]

oh, you mean from the left side of the horizontal axis?

[silent plank]

yes

[upper karma]

ok thanks

[upper karma]

what about this one?

[silent plank]

consider the definition of tan and that cos is bounded

[upper karma]

bounded?

[silent plank]

-1 <= cos(x) <= 1

[upper karma]

but cos can be bigger than sin, no?

|sin x| <= |tan x| = |sin x / cos x | , if cos x >= sin x then |sin x| >= |tan x|, no?

[silent plank]

doesn't matter if cos is bigger

in your tan(x)

sin(x) is being divided by something between -1 and 1

which results in something of equal or greater magnitude

[upper karma]

thans

[upper karma]

this is the last one

could you also help me with it?

[dark sparrow]

write down the equation of the line connecting the point (sin(θ), cos(θ)) and the origin

and verify that the point (tan(θ), 1) satisfies that equation

[upper karma]

i'm not sure how

[dark sparrow]

do you know how to write down the equation of a line passing through two points with known coordinates?

[upper karma]

y = cos a/sin a *x = sin a/cos a * cos a/sin a = 1

is this it

[dark sparrow]

$y = \frac{\cos(\theta)}{\sin(\theta)}x$ is the equation of that line

[somber coyote]

Ann

[dark sparrow]

that's what i was asking you to produce

[upper karma]

which is what i did?

[dark sparrow]

and then plugging in x = tan(θ) yields y = 1 as needed

you mangled those steps together

[upper karma]

not sure what you mean

[dark sparrow]

you tried to write those steps in one line

apparently in an effort to save time or keystrokes

[upper karma]

haven't you done the same thing

how did you want me to do it? write
y = cos a/sin a *x
then
x = sin a/cos a
y = 1
1 = sin a/cos a * cos a/sin a
?

[dark sparrow]

i wanted you to use more words

[upper karma]

so like this?

[dark sparrow]

that's... one (1) word in your entire line of work

[upper karma]

what do you need words for

[dark sparrow]

"the equation of the line through (0,0) and (sin(θ), cos(θ)) is y = cos(θ)/sin(θ) * x.
plugging in x = sin(θ)/cos(θ) yields y = cos(θ)/sin(θ) * sin(θ)/cos(θ) = 1.
therefore the point (sin(θ)/cos(θ), 1) lies on the line."

[dark sparrow]

to read your work? lol

[upper karma]

isn't my work readable

[dark sparrow]

and to like, not get lost in seas of symbols

i'm complaining about its quality, not its readability

[upper karma]

quality of readability?

lol

[dark sparrow]

was that an intentional misinterpretation

[upper karma]

i don't understand

you are saying that words are needed to make the work more readable

due to its nature it's readable as is

and then

i'm complaining about its quality, not its readability

so i don't know what you mean by quality

[dark sparrow]

i'm complaining about the quality of your work

your sea-of-symbols work is readable but it is not a joy to read at all

so unless you're sadistic i would advise against drowning your reader in seas of notation

[upper karma]

ok

thanks

[upper karma]

perpendicular line - one that forms a straight angle with something?

[dark sparrow]

a right angle

[upper karma]

thanks

right triangle
find all lengths for:
perimeter 29 angle 42
straight angle is to the right
sides read clockwise are
a b h
sin 42 = b/h -> h sin 42 = b
cos 42 = a/h -> h cos 42 = a
h = a/cos 42
perimeter: h sin 42 + h cos 42 + a/cos 42 = 29
what now

[dark sparrow]

you could have just... not replaced h with anything, and kept it as-is

h sin(42) + h cos(42) + h = 29

also, right angle. it's right angle. in english, "straight angle" means 180°.

[dreamy ridge]

Well now this is just a normal equation, remember sin 42 and cos 42 are just numbers, so can just get h on its own

It’s like saying 2x + 4x + x = 29

Obvs cos and sin 42 isn’t 2 or 4

As a example just to make it clearer

[inland mountain]

Can someone help explain how this would be done

[pure cape]

,rotate

[pure cape]

So what have you tried

[inland mountain]

I've tried to just do the normal reflection of keeping the x the same and the Y's turn negative and then minus one from x

[pure cape]

Ok you have to imagine the picture in your head

So basically you have a triangle, and then there is this axis x=-1 which is vertical

Channel is busy, please move

[upper karma]

oh ok sorry

[pure cape]

So that means you are reflecting thw point across the axis, meaning moving it on a horizontal line

So that means the y should be the thing thats kept the same and the x should be the one thats reflected

So now comes to the reflecting part

Are you clear on the above?

[inland mountain]

So y stays the same while x changes?

[pure cape]

Yes

[inland mountain]

Then yes im clear

[pure cape]

So now, we know it reflects across an axis, but dont know how much it changes

So that means we have to find the distance between the point and the axis

Lets take B for example

Can you find the distance from B to the axis x=-1?

||Hint: subtract one x from another||

[inland mountain]

Uhm for some reason im stumped wouldn't it be -2?

[pure cape]

So | x_1 - x_2 |

Is the distance

Plug in the coordinates we get | -2 - (-1)| = 1

So the distance is 1 unit

Are you clear on this?

[inland mountain]

Yes

[pure cape]

Great

And since x=-2 is to the left of x=-1

We need a point that is to the right of x=-1

and it must be 1 unit away

How do you think we can find this?

[inland mountain]

By adding - 1 and - 2?

[pure cape]

No its 1 unit away from x=-1 and is to the right of it

Which means we just need to add 1 to x=-1

So our coordinate of x would be?

[inland mountain]

0

[pure cape]

Yes

So thats our reflect of B

(0,3)

For E, its a bit different

E is to the right of x=-1

So you need to find a point to the left of x=-1, which means yiu have to subtract by the distance from E to x=-1

Thats just all

[inland mountain]

Ok so would E be after reflection -2,4?

[pure cape]

No, the distance from E to x=-1 is not 1 unit

[pure cape]

Use this and plug in the coordinates to find the distance

[azure pebble]

how tf do u solve this question without a height

[silent plank]

the triangle is indicated to be equilateral

[haughty spire]

it has to be an equilateral prism

[silent plank]

consider trig and/or formula for area of an equilateral triangle

[azure pebble]

@silent plankok i solved that one

can u help me with this

how do i find the volume if i dont know the height of the cylinder

[proven matrix]

You know the radius of the sphere

[azure pebble]

ok

the hemisphere*

how do i find the height of the cylinder

theres no way ur gonna solve this problem without the height of the cylinder

[proven matrix]

Subtract radius from total height

[azure pebble]

and how is that gonna get u the correct answer

[proven matrix]

Because that gives you the height of the cylinder

[upper karma]

if cos is a function (and it is)
why can you do
cos(theta) = sin theta + pi/2 ?

[pure cape]

why not, and you are missing a pair of important brackets

x^2 is a function too

and i can do x^2 = 1

this is when it becomes an equation with unique solution(s)

[upper karma]

but you are not changing the x itself

[pure cape]

sometimes my equation can have infinite solutions, for example x^2 = x^2

were you supposed to write cos(theta)=sin(theta+pi/2)?

[upper karma]

sin (theta) + pi/2 ?

[pure cape]

well you could write so too

[upper karma]

with sin (theta + pi/2) i'd be okay

[pure cape]

i mean it doesnt really matter, because both are equations

they can have solutions or dont

[upper karma]

wait

[pure cape]

in equations, they arent considered function anymore

[upper karma]

cos(theta - pi/2) = sin theta

oh

nvm

just noticed something

sry

[pure cape]

so anyway, whats still troubling you?

[upper karma]

could someone please explain

[silent plank]

complementary property of tan and cot

[upper karma]

could you elaborate

[silent plank]

tan(pi/2 - x) = cot(x)
tan(x) = 1/cot(x)

(where defined)

[upper karma]

i don't get it

tan 5pi/12 = tan (pi/2 - x) = cot(x)

meaning that i have to calculate x

but here x is assumed to be 5pi/12

which makes no sense

but still gives the right result

[silent plank]

tan(pi/2 - x) = cot(x) also implies
cot(pi/2 - x) = tan(x)

[upper karma]

i don't see why

5pi/12 = pi/2 - x -> x = y/12 -> 6pi/12 -y/12 -> y = pi -> x = pi/12

[silent plank]

tan(5pi/12) = cot(pi/2 - 5pi/12) = 1/tan(pi/2 - 5pi/12)

[upper karma]

ok but how do you know cot(pi/2 - 5pi/12)

you have to first calculate x

[upper karma]

i just don't see why is this the case

[silent plank]

trig identites

its pretty much the same identity

[upper karma]

?

this is what i know

[silent plank]

so yeh the one with the tan

consider what happens if you set theta = pi/2 - k

[upper karma]

tan(k)

[silent plank]

no

what?

wtf is theta(k)

[upper karma]

pi/2 -pi/2 + k

corrected

[silent plank]

actually even simpler is just rearranging that equation

multiply both sides by tan(theta)/tan(pi/2 - theta)

[upper karma]

both sides of what

[silent plank]

tan(pi/2 - theta) = 1/tan(theta)

to get:
$$\tan(\theta) = \frac{1}{\tan(\frac{\pi}{2} - \theta)}$$

[somber coyote]

ℝamonov

[upper karma]

got it

[upper karma]

what did you want to do with this?

[silent plank]

you can ignore that

was initially going for another route

[upper karma]

what route

[silent plank]

working with tan and cot

instead of fractions with only tan

that would lead to
tan(k) = cot(pi/2 - k)

which is pretty much the same thing

[upper karma]

where does this come from?
I just know that sin u = cos(pi/2 - u)

<@&286206848099549185>

[silent plank]

assuming sin(u) is non-negative,
pythagorean trig identity

[peak flower]

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

[somber coyote]

𝔙eryhappyperson

[upper karma]

right, thanks

[upper karma]

how?

cos u = pi/2 - sin u = (2pi - sqrt(7) -1)/4 and don't know what to do now

<@&286206848099549185>

don't know how to do these either

aren't there only two distinct angles satisfying some cosine/sine ?

and these two distinct angles are always theta and -theta

[humble pulsar]

reflex angles

and any angle with a multiple of 360 added on will have the same trig values

[upper karma]

right, thanks

could you help me with the other questions?

[humble pulsar]

I mean just use compound angle for the proofs

and the explanation you can split into cases, n is even and n is odd

[upper karma]

i'm not sure how

[humble pulsar]

just use compound angle

$\sin\left(t+\frac{\pi}{2}\right)=\sin(t)\cos(\frac{\pi}{2})+\sin(\frac{\pi}{2})\cos(t)$

[somber coyote]

Mosh

[upper karma]

i'm not sure where does that come from

[humble pulsar]

compound angle formula

[upper karma]

don't know it, this is what i know

[humble pulsar]

Ok we can talk about that different question, sure

if n is even, then you're adding an even multiple of pi, which is effective to doing nothing due to the 2pi-period of cosine

[upper karma]

ik

[humble pulsar]

so $\cos(x+2k\pi)=\cos(x)\implies \abs{\cos(x+2k\pi)}=\abs{\cos(x)},k\in\mathbb{Z}$

[somber coyote]

Mosh

[humble pulsar]

if k is odd, then the graph has been translated half a period, so it's negative cosine, and their abs value are the same

[upper karma]

ik all that

[humble pulsar]

so why are you asking that question if you know the answer?

[upper karma]

i don't see how does this

follow from all of this

[humble pulsar]

ok but that had nothing to do with the question you had just asked about

[upper karma]

what

don't know it, this is what i know

[humble pulsar]

yeah but then you posted something not related to that question, and the other question

[humble pulsar]

.

that picture has nothing to do with t+pi/2

since 1/2 isnt in Z

[upper karma]

don't know it (the compound angle formula), this is what I know (trigonometric identities with theta +pi * n)

@humble pulsar so how am i supposed to deduce the compound angle formula from it

[humble pulsar]

you can do the proof w/ the unit circle, but i cba to draw out the proof

[upper karma]

here's all i know about trig:

[upper karma]

could you help me with other questions then

[bright cairn]

<@&286206848099549185>

[steady shoal]

How many times do you have to be reminded that don’t ping before 15 min

And to not post same q In different server

[bright cairn]

What do you mean

[bright cairn]

Can you help

[upper karma]

read the rules, you've broken two:

1. this channel is occupied (by me) and yet you have asked a question here
2. you've pinged helpers before waiting a sufficient amount of time

[upper karma]

how can i assess whether the inverse trigonometric function will result in a positive number?

[upper karma]

why? pi/3 is within the range of cos^-1

[pure cape]

Pi/3 is indeed within the range of cos^(-1)(x)

But the domain is what defines cos^(-1)(x)

[upper karma]

thanks

could you help me with the other questions?

[pure cape]

Just post it

[upper karma]

already did

[upper karma]

first ones

[pure cape]

Can you factor cos^2(x) for me?

Oh wait

Looked at wrong one

Ok so looks at the first two terms

[upper karma]

cos^2 (x) = (cos x + 1)(cox x +1) -2cos x -1 = cos x^2 + 2 cos x +1 - 2cox x - 1 = cos x^2

[pure cape]

cos^3(x) + cos^2(x)sin(x)

[upper karma]

i see

[pure cape]

Factor out cos^2(x)

[upper karma]

do you want me to use
cos^2 x + sin^2 x = 1

[pure cape]

Yes

And then just foil

[upper karma]

foil?

[pure cape]

Factor cos^2(x) from this

[pure cape]

Apply this to cos^2(x)

And then foil for me

[upper karma]

cos^3 (x) + (sin^2 x sin(x)) = cos^3 (x) + sin^3 x

[pure cape]

You will see that something will cancel out with the 3rd and 4th term

[upper karma]

i don't know what foiling is

[pure cape]

Oh, expand it then

I thought thats how you usually say it

[upper karma]

?

[pure cape]

No no,

First

Factor out cos^2(x)

[pure cape]

From this

What do you get?

[upper karma]

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

[pure cape]

Yes

Now apply the pythag identity

What do you get?

[upper karma]

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

[pure cape]

Forgot a pair of brackets there

(1-sin^2(x))(cosx+sinx)

So now expand it

Tell me what you get

[upper karma]

(1-sin^2(x))(cosx+sinx) = cosx + sinx - cosx * sin^2 (x) - sin^3 (x)

[pure cape]

Yes very good

See how -cos(x)sin^2(x) - sin^3(x) cancels out with the last two terms

[upper karma]

ok ty

[upper karma]

what about this one

same thing?

factor it out?

[pure cape]

Remember (a+b)^2?

Can you tell me what you after expanding it?

[upper karma]

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

[pure cape]

Yes

[upper karma]

expanding what

(cos^2 (u) + sin^2 (u))(cos^2 (u) + sin^2 (u)) = 1 ?

[pure cape]

Yes

[pure cape]

Expanding this

(a+b)^2= a^2 +2ab +b^2

So if a=sin^2(u) and b=cos^2(u)

[pure cape]

This turns into a^2 + 2ab +b^2 =1

And since we know what (a+b)^2=

No

[upper karma]

idk where you are going with this

[pure cape]

This?

[pure cape]

See how this is related?

I mean its just blatantly there

[upper karma]

yes

the expression is always positive

what about it

[pure cape]

No not that the expression is always positive

a^2 + 2ab + b^2 is (a+b)^2

[upper karma]

yes ik

[pure cape]

Hence this is equivalent of (a+b)^2