geometry-and-trigonometry

[fallen ivy]

not really?

I'm confused as to how it acts as such

[upper karma]

=tex |n^i|^2 = 1

[upper karma]

n = -2

you said "only positive integers"

[fallen ivy]

yeah...you didn't compute (-2)^i

[upper karma]

k

so

don't have to

[fallen ivy]

=tex (-2)^i = (2 * e^{pi * i})^i = 2^i * e^{-pi}

^ has magnitude > 1

[upper karma]

=tex |(-2)^i|^2

[upper karma]

is what I computed

not sure what you're talking about

[upper karma]

=tex |(-2)^i|^2 = (-2)^{-i} (-2)^i

[upper karma]

can we agree this is correct

[fallen ivy]

god why am I so bad at tex today

no?

that's not how the norm function works

[upper karma]

can we agree this is correct

=tex |z|^2 = z^* z

[fallen ivy]

yes

[upper karma]

what is z^* when z= (-2)^i

[fallen ivy]

it's not (-2)^(-i)

[upper karma]

(-2)^{-i}

do you know what complex conjugate is

[fallen ivy]

yes

[upper karma]

then why are you saying it's wrong

[fallen ivy]

because it is...

[upper karma]

you just agreed with everything I said

[fallen ivy]

no

I didn't agree that (-2)^i has conjugate (-2)^(-i)...

[upper karma]

what is the complex conjugate of (-2)^i

[fallen ivy]

https://www.wolframalpha.com/input/?i=complex+conjugate+of+((-2)^i)

[surreal bolt]

Um you can only find the complex conjugate after you put it in a + bi form

[fallen ivy]

https://www.wolframalpha.com/input/?i=(-2)^(-i)

different things

QED :P

[surreal bolt]

The "I" in a+ bi is multiplied. The "I" in (-2)^(i) is exponential.

[upper karma]

oh ok

[stiff heart]

See

While I know what I^I equals

[upper karma]

in that case, can't it equal just infinitely many things then

[fallen ivy]

wah?

[haughty prawn]

wat happing?

[stiff heart]

I didn't expect it to go backwards

[fallen ivy]

oh yeah

i^i is not equal to an individual thing

[stiff heart]

Or become transcendant

[fallen ivy]

complex exponentials aren't uniquely defined

(technically you can say the same of (-n)^i)

[stiff heart]

And if it did go backwards

[fallen ivy]

(-n)^i is not uniquely defined

[upper karma]

=tex i^i 😦 e^{i (\pi/2 + 2\pi n)})^i = e^{-(\pi/2 + 2\pi n)}

[stiff heart]

Why isn't it one or negative one

[upper karma]

lmao the emoji broke my latex

[fallen ivy]

ye, you get similar problems with (-n)^i

[stiff heart]

It's strange

So why is it not uniquely defined

[upper karma]

so is there anything stopping it from me defining my version to be correct then lol

[fallen ivy]

because i can be either e^(i * pi/2) or e^(i * 5pi/2)

nah @upper karma, your version's shot either way

[stiff heart]

Are they the same value

[fallen ivy]

yes, but not if you apply your method to them

[upper karma]

the correct way?

[fallen ivy]

one will give e^(-pi/2) one will give e^(-5pi/2)

complex exponentials aren't uniquely defined

[stiff heart]

Why?

[upper karma]

i can be infinitely many choices

[fallen ivy]

unless you give them meaning by picking the lowest exponent nicely...

[upper karma]

not just those two

[fallen ivy]

(branch cuts of ln ftw)

yeah @upper karma

but just two choices is nicer for illustration

[stiff heart]

When i mean i

[fallen ivy]

=tex (-n)^{i} = \left(n * e^{i * (2k + 1)\pi}\right)^{i} = n^i * e^{-(2k+1)\pi}

[fallen ivy]

technically even (n^i) is shot

because n = n * e^(i * 2pi)

[stiff heart]

I mean something that is one positive unit in a number line perpendicular to real numbers

[fallen ivy]

yeah

so do we

but that can be expressed as an exponential of e in a number of ways

ok, here's how we "usually" define exponentials

=tex x^y = e^{y * \ln x}

[fallen ivy]

the reason for that is that e^something has a nice series representation

the problem is

\ln i can equal many different things

it can be i * pi/2, or i * 5pi/2 or i * 9pi/2...

[stiff heart]

It doesn't have a natural log then

[fallen ivy]

yeah, and because it doesn't have natural log

and powers are defined using natural logs

it doesn't have a unique power

[stiff heart]

also

About the way the exponential is made

Can x and y be switched with eachother

[fallen ivy]

y^x = e^(x * ln y) is valid yes, but that's not anything new, it's the same as the first definition

[stiff heart]

I was just wondering

[fallen ivy]

but x^y ≠ y^x

(not in general anyway)

[stiff heart]

Very few cases

[fallen ivy]

the reason that exponentials are defined in terms of e^something (otherwise abbreviated exp(something)), is because it has a nice series representation that only uses integer powers

and integer powers are just repeated multiplication

[stiff heart]

See

I was going to try repeated multiplication to try and figure it out

Or in this case repeated exponentials

But it really behaves erratically

[fallen ivy]

wdym?

[stiff heart]

I view i as turning a number 90 degrees

[fallen ivy]

sure, that's what it does under multiplication

[stiff heart]

And since i^i basically went backwards

I expected it to do it again or the opposite

[fallen ivy]

how does it go backwards?

[stiff heart]

It became a real number

[fallen ivy]

oh yeah, well "became"

really it doesn't become anything

it becomes a set of real numbers

e^(-pi/2 - 2kpi)

k ranging over the integers

[stiff heart]

I'm using 0.20787957635

[fallen ivy]

okay, branch cuts on ln it is then ^^

[upper karma]

neo the main thing is

=tex 1= e^{i 2\pi}

[upper karma]

so you can raise this to any integer power and still get 1 on the left

so you can get any integer multiple of i2pi on the right

[fallen ivy]

complex exponentials are only valid if you choose a place for ln to be discontinuous and then define ln by that place

[stiff heart]

Okay

[fallen ivy]

the most common is that ln is discontinuous on the line emanating from the origin and going left along the real numbers

and the place where it's discontinuous shifts the i component of the ln(z) function from -pi to pi

[stiff heart]

Well it was a nice thought experiment

Thanks everyone

[fallen ivy]

so, yes i^i = e^(-pi/2)

[upper karma]

lol

[fallen ivy]

but there's a lot of assumptions and choices that go into that

[upper karma]

like one choice, the same choice you're making when you decide sqrt(4)=2 instead of -2 basically

[fallen ivy]

yeah, but this choice has much more far-reaching applications

BTW, the graph of ln(z) is neat https://www.wolframalpha.com/input/?i=ln(z)

https://www.wolframalpha.com/input/?i=i^z

specifically the image graphs

(Complex Map)

https://www.wolframalpha.com/input/?i=z^i

[stiff heart]

Why choose z

[upper karma]

don't look at the riemann surface, otherwise all the stuff he said about cutting up the function is wrong

[fallen ivy]

because z is a common variable name for complex numbers

shhhh

riemann surface is nice but special

[upper karma]

far reaching consequences

but let's not go that far

heh

[stiff heart]

Z^i is nice

But shouldn't we look at sqrt(-z)^i

https://www.wolframalpha.com/input/?i=Sqrt(-z)^i

I just realized the complex plans is transformed the same

Nevermind

[fallen ivy]

oh another thing

this is the reason that

=tex \int_{\gamma} \frac{\mathrm dz}{z} = 2\pi i \text{ where } \gamma \text{ denotes the unit circle}

[tranquil yew]

Since when can you use a shape as a bound for an integral

[forest dove]

It's parametrized

So you define it as follows

=tex \int_{\gamma} \omega = \int_a^b \omega(\gamma(t))\gamma'(t) dt

Where gamma is the parametrization mapping [a,b] to wherever

[upper karma]

you forgot gamma'

[forest dove]

Whoops

[dark sparrow]

@tranquil yew complex analysis!

line integrals!

[final prairie]

aren't you missing a dt?

[golden crater]

Wow that's a cool bot! (Hi, I'm new lol)

[forest dove]

Orangalo: Were you asking me?

[final prairie]

Yes

On the left side

[forest dove]

I used omega as a differential form, so it includes the d stuff

[final prairie]

Oh ok

coolio

[forest dove]

Yeah, I'd recommend learning a bit about differential forms, they're pretty nice

[trim belfry]

is it possible to have a hyperbolic space which wraps into itself like a spherical space as to be hyperbolic on the small scale but spherical on the large scale?

[dusk snow]

Oo

well you can quotient hyperbolic space to get a compact space that is locally hyperbolic

quotient by a suitable subgroup of transformations

if you have ever seen tilings of hyperbolic space you can get an idea from there

[polar moon]

ok my trig is really bad

but

does the cosine similarity fulfill the triangle inequality assuming no 2 vectors are linearly dependent and 0 doesn't count

[dusk snow]

cosine similarity ?

[polar moon]

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

[violet nest]

oh well im here

anyone?

[dark sparrow]

yes?

[violet nest]

anyways to solve cos(5x)

i had to find out what sin^4(x) equalled to

[dark sparrow]

solve cos(5x)

[violet nest]

luckily i found a very elegant identity which i advertised a lot and got ignored

@dark sparrow oops wrong term

[azure storm]

so you havent learnt Euler identities for cos and sin, right?

[violet nest]

no

[azure storm]

RIP

[violet nest]

its not in da curriculum

but i watch youtube so yeah i know the famous formula

[rare talon]

Once you learn that, deriving that kind of stuff will be much easier

I heard

[azure storm]

you could do cos(3x+2x) maybe instead of one by one

[rare talon]

I am not familiar with that though

[violet nest]

@rare talon we're not allowed to use advanced techniques i think

[azure storm]

and use that sin^2(3x)+cos^2(3x)=1 as you already have sin(3x)

[rare talon]

Yeah I know, you can just learn that for your self study

I think I have a handout about that

[violet nest]

@azure storm thats now i did it, you end up with cos3xcos2x - sin3xsin2x

[azure storm]

i have NO idea how it should end up... Wolfram alpha it (to verify!)

[violet nest]

my question was show that cos(5x) = 16cos^5(x) - 20cos^3(x) -5cos(x)

@azure storm thats now i did it, you end up with cos3xcos2x - sin3xsin2x

[azure storm]

=wolf cos(5x) = 16cos^5(x) - 20cos^3(x) -5cos(x)

[charred spear]

Query made by @azure storm
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=cos(5x)+%3D+16cos^5(x)+-+20cos^3(x)+-5cos(x)
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

[rare talon]

@azure storm tell me if I refer something wrong, but is it this? https://artofproblemsolving.com/community/c6t96f6h609795s1_updated_2015complex_numbers_in_trigonometry

[violet nest]

copied the answer wrong

its +5cosx

btw i have an idea

[azure storm]

for what, Nya?

[violet nest]

can we have perms to remove mathbots pictures

cause it gets spammy if you do something wrong

[rare talon]

Euler identity

The 2nd page?

I didn't learn complex number in my high school

[violet nest]

=wolf cos(5x) = 16cos^5(x) - 20cos^3(x) +5cos(x)

[charred spear]

Query made by @violet nest
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=cos(5x)+%3D+16cos^5(x)+-+20cos^3(x)+%2B5cos(x)
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

[violet nest]

yep

thats equal

[dark sparrow]

gimme a moment

[azure storm]

https://en.wikipedia.org/wiki/Euler's_formula#Relationship_to_trigonometry
Was talking about that, maybe my english translation was wrong as its formula not identitities

[rare talon]

Oh yeah

That's exactly the same that you mentioned

[violet nest]

=wolf cos(3x)cos(2x) - sin(3x)sin(2x)

[violet nest]

lets see if wolfy can do it

[charred spear]

Timeouts
FunctionProperties, GlobalExtrema, InterestingDefiniteIntegrals, InterestingMultiDimensionalDefiniteIntegrals, MathematicalFunctionData

Query made by @violet nest
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=cos(3x)cos(2x)+-+sin(3x)sin(2x)
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

[violet nest]

ha lol it failed

at least they didnt put it in polynomial terms

[azure storm]

when there's such a "complex" trig query, there's just to many possibilities to simplify

[violet nest]

this is not complex this is literally high scool maths

[azure storm]

i'm almost sure you can find at least 3 to use for different things (polynomial, factorization...)

I meant complex in length, not difficulty

[violet nest]

anyways whats really sad is that i didnt figure out i had to use the factor formula to prove that cos(5x)+cos(3x) = 2cos4xcosx

and instead jumped on finding cos(5x)

[dark sparrow]

you can also do cos(5x) = cos(4x+x)

[violet nest]

not sure how that would help we dont know what cos(4x) is

[dark sparrow]

and then you can use the double angle formulae twice

cos(4x) = 2cos^2(2x) - 1

you know cos(2x) = 2cos^2(x) - 1, right?

[violet nest]

.. yes

[dark sparrow]

okay i'm going to use c_n and s_n to denote cos(nx) and sin(nx) respectively for clarity, with an absent subscript being equivalent to 1

=tex c_5 &= c_4c - s_4s \ &= c(2c_2^2 - 1) - 2ss_2c_2 \ &= c(2(2c^2 - 1)^2 - 1) - 4s^2cc_2 \ &= c(2(4c^2 - 4c + 1) - 1) - 4(1-c^2)c(2c^2 - 1)

[dark sparrow]

and then you can simplify that polynomial

and eventually you'll get what you need

hang on a minute

...

fuck now i can't see the fuckup in my work

oh

right

4c^4 - 4c^2 + 1

🤦

=tex c_5 &= c_4c - s_4s \ &= c(2c_2^2 - 1) - 2ss_2c_2 \ &= c(2(2c^2 - 1)^2 - 1) - 4s^2cc_2 \ &= c(2(4c^4 - 4c^2 + 1) - 1) - 4(1-c^2)c(2c^2 - 1)

[dark sparrow]

there we go

@violet nest that polynomial in c will eventually simplify down to 16c^5 - 20c^3 + 5c

[violet nest]

@dark sparrow thats a very newton way of writing cos and sine

[dark sparrow]

it's ad-hoc

[final prairie]

Is it common?

c()=c_1()?

[dark sparrow]

no

it's ad-hoc

i literally made it up on the spot

[final prairie]

Mhm

[onyx cypress]

Could I get a bit of help with this?

[dark sparrow]

have you tried drawing the triangle in question?

[onyx cypress]

yeah I have it drawn out

[dark sparrow]

i've denoted BC with x since that's what you wanna find out

is it clear that cos(B) = 3/x?

[onyx cypress]

yeah

[dark sparrow]

=tex \frac{3}{x} = \frac{4}{5}

[dark sparrow]

can you solve this eq for x?

[onyx cypress]

3/(4/5)

or just 15/4

[dark sparrow]

👏

[onyx cypress]

OH that's the answer

thanks very much

[onyx cypress]

for this question, would finding AD help with anything?

[dark sparrow]

probably wouldn't

i suggest examining the ratios DB/AB and AB/BC

what are they both equal to?

[onyx cypress]

DB/AB is 1/3 and AB/BC can be 6/x

[dark sparrow]

no no no

i don't want you to think about their numerical values

i want you to see something they have in common

[onyx cypress]

that the denominators are hypotenuses?

[dark sparrow]

you're getting close

what are their numerators?

[onyx cypress]

the legs

[dark sparrow]

admittedly though you didn't mention the thing i had in mind

[onyx cypress]

lol

[dark sparrow]

the lines involved in both of those ratios frame the same angle

😉

[onyx cypress]

frame

as in angle B?

[dark sparrow]

yup!

now

if you think about it

both ratios are cos(B)

😛

[onyx cypress]

ok yes I see

[dark sparrow]

cos(B) = DB/AB = AB/BC

[onyx cypress]

ohhhhh

thank you

[upper karma]

How to calculate cos(135deg)?

[dark sparrow]

can you draw 135° on the unit circle?

@upper karma

[upper karma]

yeah it's like 3pi/4

-3pi/4*

[dark sparrow]

no

[upper karma]

wwhat

[dark sparrow]

135° is 3π/4

no minus sign

[upper karma]

????????

No what

wait what

wait noooo

im dumb

I thouught lower y axis was 180 xDD

I thought first half was only 90 deg

lmfao

is it same for sin?

[dark sparrow]

what?

[upper karma]

sin(135)

[dark sparrow]

all angles on the unit circle start out on the right and are counted counterclockwise

that point where the red ray hits the circle has coordinates (cos(3π/4), sin(3π/4))

by definition

(the axes here are unlabeled, so sorry for that)

do you follow?

@upper karma?

[upper karma]

trying to squeeze this into my nonfunctional brain

So that just means it's the same

??

Or not?

Dont get it

[dark sparrow]

that point is in the second quadrant

its x-coordinate (and thus cos(135°)) is negative, while its y-coordinate (and thus sin(135°)) is positive

cos(135°) = **-**sqrt(2)/2

sin(135°) = **+**sqrt(2)/2

[upper karma]

Wait what

You just said

cos(135) is positive

[dark sparrow]

no i didn't

[upper karma]

3pi/4

hows that negative

[dark sparrow]

3π/4 itself isn't negative

cos(3π/4) is negative

[upper karma]

Yeah cause it's on left side of coordinate system

Where do you get sqrt(2)/2 from?

[dark sparrow]

do i need to explain further?

and explain, perhaps, why the blue and purple lines both have length sqrt(2)/2?

[upper karma]

forgot about this server

@dark sparrow no need explanido

I think I missed a khan video

I don't know how to do it for 30/60 angles

[dark sparrow]

honestly, values of trig functions for 0, 30, 45, 60 and 90 degrees are really worth committing to memory

[fallen ivy]

I know the values for 30 and 60 have sqrt(3), 1, and 2

I arrange them on the fly based on angle-order theorem

[upper karma]

@dark sparrow that don't explain much, as I said I haven't seen khan video about it so I don't know about that, only the 45 angle

https://www.khanacademy.org/math/algebra2/trig-functions/trig-values-special-angles-alg2/v/solving-triangle-unit-circle

@fallen ivy Well done you earned sauce

[fallen ivy]

heh

I just think it's easier than memorizing the values

[haughty prawn]

This might help explain https://qph.ec.quoracdn.net/main-qimg-a18451afffa7b9e269c9189fae642502

[upper karma]

Nope it doesn't explain at all

[haughty prawn]

Oh :(

[upper karma]

sorry im dumb

I know how Angles work

But not measurements

[haughty prawn]

Nah nah, my fault

So you can split up the 30-60-90 triangle like shown above right?

That leaves you with an equilateral triangle and an isoceles triangle, so if you know the side on "the left" is x, you'll know one of the identical faces on the isoceles triangle is x as well.

x+x=2x, so you know that if the base is x on a 30-60-90 then the hypotenuse will be 2x. You can get the third side via Pythagoras theorem.

Geometry is all about knowing where to draw the lines.

[upper karma]

so for 30 it's sqrt3

60 it's 0,5??

and 2 for 90

[fallen ivy]

cos(60) = 1/2, cos(30) = sqrt(3)/2

[haughty prawn]

Gotta remember the hypotenuse will equal 1 on the unit circle, so 2x=1

[vast fern]

So what exactly is it asking for here?

x in terms of each of all the other variables, like x in terms of r, x in terms of h, etc?

[eager pendant]

the length of x in terms of r, h and t

[violet nest]

can someone tell me what is demoivre's theorem? (misspelled it)

[vast fern]

I got x=r(t/h)

Wait that's not right

Brb

[brazen roost]

=tex x = (r \cdot \frac {t}{h} ) -t

[brazen roost]

Gotcha buddy

[vast fern]

Ya that's what I did

I put x as the triangle's side instead of x+t

@brazen roost could I put it as (rt/h)?

[brazen roost]

Sure

[vast fern]

Ty

[brazen roost]

Hell

=tex x = \frac {t}{h} (r-h)

[brazen roost]

Works too

[foggy oxide]

@onyx cypress i have to tell you that this method only works because AD is perpendicular do CB and CAB is a right triangle

just saying as a advice

[brazen roost]

Generalization includes the use of the cosine rule

[foggy oxide]

also he could just use pithagorean theorem in this one no need for trigonometry

[onyx cypress]

Yeah I see what you mean

Thanks for tip

[restive moat]

BC = 18

iamso smwart 😄

yeah, u don't need trigonometry for this exercice

[onyx cypress]

I was gonna use trig but then root2 told me to do otherwise lol

definitely was quicker and more simple. Helps too since i'm not allowed to use calculators

[tranquil yew]

@violet nest ik I'm a lil late, but it says if you take a complex number and put it in polar form (i.e. rewrite a and b in terms of an angle and the magnitude of the number) you can use this neat identity to calculate powers of it

where k is an integer

[charred spear]

Support the bot on Patreon: https://www.patreon.com/dxsmiley

[tranquil yew]

wait

I messed that up thats for roots

=tex (r(\cos \theta + ir \sin \theta))^n=r^n(\cos(n(\theta+ 2\pi k))+\sin(n(\theta+2\pi k))

[upper karma]

You can just rewrite as

=tex (r(cos \theta + ir \sin \theta))^n=r^n(cos(n(\theta))+sin(n(\theta))

[tranquil yew]

You won't get every possible solution with that

[upper karma]

If you add 2kpi then cos and sin dont change

[tranquil yew]

err yeah I guess n has to be fractional for it to matter

[upper karma]

Then yeah I guess

[tranquil yew]

but that formula only works for integers

[hardy shore]

I have one thing

that is

driving me crazu

My prof uses these stupid numbers

that I can't read

If one of u could read it for me

is that a 5, a 6, an 8!?

I have no idea

I tried the equation with 7

6*

with 6 it was wrong

with 8 it was wrong

I'll let u know how it goes w/ 5

But the prof shouldn't do that in the first place

[azure storm]

whats the equation?

[hardy shore]

can't even freakin read it

It's a measurement

Using that to find circumference

which is easy

IF I COULD READ THE FREAKIN NUMBERS HE POSTS

FINALLY

YES

IT WAS A 5

OMG YESSSSS

[minor quest]

Can someone help me with this?

I don't get the steps after they written sinA = 9/22

[rare talon]

Do you know why sin A = 9/22?

[minor quest]

Yea cause we want find that angle using sine

[rare talon]

Okay

Then do you know A = sin^-1(9/22)

[minor quest]

That's all that they taught me._.

[rare talon]

I think it called the inverse function of sn

[minor quest]

I think u minus sin? And get sin inverse?

[rare talon]

No

=tex A = \sin^{-1}(\frac{9}{22})

[minor quest]

So how did they get from sinA= 9/22 to that?

Clearly they moved sin?

[rare talon]

It's just an inverse

I mean:
If sin A = x , then A = sin^-1 (x)

I hope you have been taught that

[minor quest]

Oh right

[rare talon]

Make sense?

[minor quest]

Yes so far

[rare talon]

Then just put sin^-1 (9/22) into your calculator

Then you will know what is angle A measure

[minor quest]

Uh

I just learned how to put the sin cos in degrees for answer

[rare talon]

Oh I see

Damn try to look at the button

See if there's a shift there, and find sin^-1 button

Usually it's also on the sin button? Could you show me your calculator?

[minor quest]

I'm away rn..

[rare talon]

Ah I see

Yeah you might be taught later

Then by putting that value in your calculator, you would get 24.15 degrees

[minor quest]

Aight thanks anyway!

[rare talon]

After that part, everything else should make sense to you

It's just simple geometry problem

(don't forget try how to input sin^-1 function into your calculator)

I myself never use calculator on math, so idk

[minor quest]

Then how did you calculate?

[cedar prawn]

With simple problems one can do it in one's head but I'm not sure about that with sines O.o

[restive moat]

Cosines would be simpler

[fallen ivy]

Small angle approximations are fun

[foggy oxide]

@minor quest

since ABD is a right triangle

u could discover the lenght of the AB using pythagorean theorem

i highly recommend you to use tan^-1 (x) instead of sin^-1 (x) obv, both work as the guy told you but tan^1 would be better since the radius was a bisector from D

and those functions are frequently called arcsin and arctan too

you can read them as

tan^-1 (x) = arc that the tangent gives x

[rare talon]

@minor quest ask someone who is familiar with calculator

Anyway I use sin, because the solution suggests it, so I just go along with it

[minor quest]

Alright I think I got the basic down now thanks!

[rare talon]

Okay!

[chrome ruin]

Anybody up willing to give me a hand on this question?

Hmm I'm not too sure about figure 8. I believe I need to calculate as if its 1/2 cylinder and a rectangular?

[haughty prawn]

Basically

Aka literally

[foggy oxide]

i can help u

just noticed you already knew how to do it

well yes first the semi cylinder

then the rect that is 10*16*400

well

the second question you should calculate the surface area (of both figures) and multiply by the price that is given

[chrome ruin]

thanks all

[foggy oxide]

np

@chrome ruin don't forget about the unit measure

idk what is yd

maybe you'll have to convert the surface area that is in meters to yd

[chrome ruin]

yards

[foggy oxide]

idk how to convert meters to yards, i`ll let it 4 u

we dont use this unit in my country so im not used to it

[chrome ruin]

all good apreciate the help expect now i think ive confused my self for the cylinder the formula would be or Height + raidus?

[foggy oxide]

what do you mean

you`re saying about the second question?

the cilyinder heigh is actually 400m

8 is the radius

[chrome ruin]

oh

[foggy oxide]

that`s your semicylinder

[chrome ruin]

how did u got the rest away so quick

[foggy oxide]

i just used paint

xD

[chrome ruin]

lmao

[foggy oxide]

xDD

if u need a hint or something just call me here xD

[chrome ruin]

ms paint wizard on the job!

[foggy oxide]

xD hahaha

[chrome ruin]

help on number 7? https://cdn.discordapp.com/attachments/333461112576278538/337069072988504064/Capture5.JPG

idk where to even start tbh

[foggy oxide]

i'm here

well let me see

if i got it right

i'll draw it

that's our figure right

my english make things difficult

[chrome ruin]

paint WIZARD

i understand u

[foggy oxide]

uhauhauhahu

[chrome ruin]

an looks fine

and*

[foggy oxide]

well

[chrome ruin]

well its has a volume of 116 inches cubed

[foggy oxide]

first

you have

to find the triangle side lenght

or maybe no

i think u can do this without the triangle side lenght

[cunning zinc]

it's a minimization problem, no?

[foggy oxide]

yeds

you can

[cunning zinc]

minimizing one value while having the other one defined

[foggy oxide]

you can use calculus

[cunning zinc]

both of which are based on parameters that control the size

of the prisim

[foggy oxide]

but don't

[cunning zinc]

prism*

[foggy oxide]

don't do it with calculus

[cunning zinc]

why no?

not*

[foggy oxide]

well

this is a highschool problem

he is supposed to don't use calc i think

[cunning zinc]

because the total surface area is a function of both the base triangular side length

and the hieght of the prism

[foggy oxide]

i know ...

[cunning zinc]

I took calc in high school

and we did these sorts of problems, but idk what is expected to be used to solve it

so calc is the first thing I thought ot

of*

[foggy oxide]

are you in high school @chrome ruin

?

[chrome ruin]

correct

[foggy oxide]

did u learn calculus ?

[chrome ruin]

this is my last course rn

[foggy oxide]

show 2 solutions

[cunning zinc]

is this a pure geometry problem or are you expected to use calc?

[chrome ruin]

grade 12 college

[foggy oxide]

show 2 solutions

[chrome ruin]

pure geometry

[foggy oxide]

well

whatever

just do like that

let h be the height

h is constant

in both cases

so u can set h= 1 or 2

it'll make things easy i think

the volume is given by the area of the figure multiplied by the height

sqrt3 /4 * a^2 * h = volume, the surface area would be the sum of lateral areas and base areas

3ah + 2* sqrt3/4 a^2

that's the surface area in function of a and h

do this helps ?

[chrome ruin]

quite a bit

[foggy oxide]

did u got what i did ?

3 rectangles with sides a and h

• the area of two triangles that are the base

now you can just set h= 1 and solve for a

this won't change the answer

[chrome ruin]

how did u get the area?

[foggy oxide]

well

wich area ?

those laterals are rectangles of side h and a

they appear 3 times in the figure

got it ?

[chrome ruin]

oh

[foggy oxide]

here

all rectangles are equal

triangles are equal too

this is the figure decomposed

[chrome ruin]

gotcha

[foggy oxide]

here

better image

looking at the expressions i gave you, when the surface area will be the < possible

[chrome ruin]

i can't understand the equation for the life of me

[foggy oxide]

well

how do you calculate equilateral triangles areas ?

[chrome ruin]

u said the surface area would be the sum of all lateral bases right?

[foggy oxide]

yes

two equal and triangles triangles areas

[chrome ruin]

but what are the numbers for the bases

[foggy oxide]

there are 2 bases

two triangles

and 3 laterals

3 rectangles

don't think i've drawn this haha,

[chrome ruin]

wikihow!

[foggy oxide]

yes yes

got it ?

3 laterals and two bases

[chrome ruin]

im sorry i really do not

i know it's late

[foggy oxide]

how many sides a triangular prism does have

?

[chrome ruin]

3?

[foggy oxide]

yes

wich figures are those ?

if you drawn a side of a triangular prism what it would looks like

[chrome ruin]

pyramid and trinagular prisim

[foggy oxide]

don't think about the pyramid

if you drawn a side of a triangular prism what it would looks like

[chrome ruin]

the fiqure you drew?

[foggy oxide]

yes the rectangles

it would actually looks like a parallelogram deppending on how you draw it but focus on the rectangles

how do you calculate rectangle areas ?

[chrome ruin]

lw

[foggy oxide]

yes

forget about the images i sent you

[chrome ruin]

ok

[foggy oxide]

what is the height of those rectangles

[chrome ruin]

whihc ones

which*

[foggy oxide]

the 3 area equal

because

the base is an equilateral triangle

forget about the problem, the focus it's about understanding this

[chrome ruin]

1 or 2?

[foggy oxide]

you have to understand the intuitive way of calculating surface areas

[cunning zinc]

are you understanding that if you kind of "unfold" the prism, you get something like this?

http://puu.sh/wObST/e1cdb96045.png

[foggy oxide]

^^

thanks

[chrome ruin]

okay now i get it

[foggy oxide]

this is a good picture

[chrome ruin]

but how do we gtaher the height

[foggy oxide]

well

actually you don't need to get anything

the surface are will be <

when the height of the figure is 0

so you solve the a with the given volume as h=0 or 1

[cunning zinc]

wait what?

[foggy oxide]

1*

[chrome ruin]

okay is the height alwyas 0 when we are trying to minimize the SA

[foggy oxide]

sorry

it's 1

sorry

[chrome ruin]

is it always 1?

[foggy oxide]

well, yes

[cunning zinc]

I did the problem with optimization, and I didn't get h=1 as giving the smallest SA

[foggy oxide]

what did u get

[cunning zinc]

I got SA was about 155.7

h came out to be

[foggy oxide]

your units are m^2 right ?

[cunning zinc]

about 4.4

[foggy oxide]

and

triangle side

?

[cunning zinc]

whatever the problem gave as the units

cm

I think

[foggy oxide]

and the triangle side is ?

[cunning zinc]

7.7

[foggy oxide]

wait a sec

[cunning zinc]

see if you can find if I made a mistake

[foggy oxide]

i think you did xD

(not really sure)

[chrome ruin]

idk whats going on rn lol

[cunning zinc]

when I plug them into the volume equation, I got 116

but, I had an interesting and tiring few days so it's possible I made a mistake

@chrome ruin I used calculus to get the answer in order to compare when you guys did it the other way

[foggy oxide]

https://www.wolframalpha.com/input/?i=1%2F4*sqrt3+*+a%5E2+%3D+116 triangle side https://www.wolframalpha.com/input/?i=3*+(4+sqrt(29))%2F3%5E(1%2F4)+%2B+2*+(4+sqrt(29))%2F3%5E(1%2F4) surface area

[cunning zinc]

where does the sqrt(29) come from?

[cunning zinc]

yea, how did you get that

[foggy oxide]

well

solved this

1/4*sqrt3 * a^2 * h = 116, (h=1)

[cunning zinc]

oh true

your SA euqation is wrong in wolfram

you forgot the sqrt(3)/4 and to ^2 the a

in the second term

[foggy oxide]

fuck

didn't saw it

[chrome ruin]

okay can soebody just show me how i would wirte this on paper

somebody*

[foggy oxide]

i'll

but

did u understand the intuition ?

[chrome ruin]

not really cuz you started talking about the diffrent wasy to get the equation

[foggy oxide]

.-

[cunning zinc]

sorry...didn't mean to confuse you

[foggy oxide]

well

[chrome ruin]

its all good man

[foggy oxide]

i'll try to explain you the intuition

[chrome ruin]

or could we pick it back up were we left off? @foggy oxide

[chrome ruin]

yes

[foggy oxide]

this

look at this

how many rectangles are in the figure ?

[chrome ruin]

the h is 1

3

[foggy oxide]

and what is the area of those 3 rectangles

(don't give the numerical value yet)

just say how you wold calculate them

[chrome ruin]

lw

[foggy oxide]

so

3* a*1

right ?

[chrome ruin]

tyes

[foggy oxide]

and what about the triangles ?

[chrome ruin]

what about them?

[foggy oxide]

their areas

and how many triangles

[chrome ruin]

3

[foggy oxide]

there are only two triangles

[chrome ruin]

bh/2

[foggy oxide]

look at the figure

yes

[chrome ruin]

i2

2

2

[foggy oxide]

those triangles are equilateral, so you don't need 2 find their heights to calculate their areas

just because they are equilaterals

the problem gave you the formula

look back at the problem

[chrome ruin]

sqrt3/4

?

[foggy oxide]

sqrt3/4 * a

how do you calculate a ?

[chrome ruin]

• means?

[foggy oxide]

multiplication

[cunning zinc]

sqrt(3)/4*a^2

don't for the ^2

[foggy oxide]

thanks fuck always forget about the square haha

[chrome ruin]

a is the legnth of each side?

[foggy oxide]

"by ma is the lgenth of each side"

yes

it is

[chrome ruin]

ya sorry about that lol

[foggy oxide]

he said that for me it's my fault

here

it's the lenght of each side

now you have to find a

how do you find a ?

[chrome ruin]

idk

[foggy oxide]

you have to solve the equation for a knowing that h=1

[chrome ruin]

are we doing bh/2

[foggy oxide]

nah

here

[chrome ruin]

i saw that but i do not have anumber for a so how would i do that equation

[foggy oxide]

but you have the value for the volume

the volume it's 116

and you know the equation that involves volume and a

[chrome ruin]

i just don't get it im sorry man

[foggy oxide]

😦

i failed as a teacher

sorry

but

you can just think about the relation that involves the volume, and the triangle side

knowing that u already have the height and volume you can calculate the triangle side

[cunning zinc]

it might not be your abilities as a teacher, it might just be tough without being in person with maybe a whiteboard or something to illlustrate it more effectively

[chrome ruin]

i may understad better if u were to write it on paper where im stuck at

understand*

[foggy oxide]

you guys may be right

i think it would be easier in person to explain that

look at your pm @chrome ruin

i sent you a pic

[chrome ruin]

Will do

[lost hinge]

Is trigonometry geometry?

[dark sparrow]

trig usually goes in this channel, yes

[lofty python]

Do any of you have any good resources on verifying identities? I'm getting butt kicked. I'm totally lost lol.

For example I need to verify sin^2(a)sec^2(a)+sin^2(a)csc^2(a)=sec^2(a)

I know it's true but I need to transform the right side into the left.

[hazy field]

well first off

sin^2(a)csc^2(a) = 1

[lofty python]

I think I may have figured it out. sec^2(a) is the same thing as 1/cos^2(a)

I didn't think you could move it to the bottom when it has an exponent.

[hazy field]

I don't like the notation sec^2(a) because it makes that not obvious

is it obvious that (1/x)^2 = 1/(x^2)?

[naive scarab]

yes

in the sense that you shouldn't ever need to explain why unless directly asked

but to see why formally, (1/x)^2 = (1/x)(1/x) = 1/(x^2)

[hazy field]

that's not really a formal proof :P

(x^2) (1/x)^2 = x x (1/x) (1/x) = x (x1/x)(1/x) = x11/x = x*1/x = 1

[naive scarab]

ok fair enough

[foggy oxide]

wow

i thought i wouldn't solve this one

well

it's surprisingly easy to solve this

the areas of similiar triangles are related to the square of their sides ratio

[lofty python]

Hey guys, do any of you have any tips on how to solve sec(x)=csc(2pi/3). I need to find one value that satisfies it.

I'm assuming I need to use a cofunction identitiy but I'm not sure how to use it with pi.

[dark sparrow]

csc(2π/3) is just a number

do you know what that number is?

@lofty python

[lofty python]

120 degrees?

[dark sparrow]

no, that's just 2π/3 converted to degrees

i'm asking you for this angle's cosecant

[final prairie]

Find 2π/3 and find it's inverse (i think)

[dark sparrow]

no

[lofty python]

Im not sure

[dark sparrow]

well

[final prairie]

You could though no?

[dark sparrow]

you know that csc(x) = 1/sin(x), right? @lofty python

[lofty python]

yes

[dark sparrow]

do you know what sin(2π/3) is?

[lofty python]

like the decimal value?

[dark sparrow]

no decimals.

you ought to know the exact expression

[lofty python]

3rad2?

[dark sparrow]

=tex 3\sqrt{2}

[dark sparrow]

did you mean this?

[lofty python]

yep

[dark sparrow]

no, that is not sin(2π/3)

it's greater than 1, for one!

[lofty python]

hmm

[dark sparrow]

can you locate 2π/3 on the unit circle?

[lofty python]

(-1/2, 3sqrt2)

sqrt3/2*

[dark sparrow]

yes, that's the point corresponding to it

so sin(2π/3) = ?

[final prairie]

How come :O

[lofty python]

sqrt3/2?

[dark sparrow]

bravo

csc(2π/3) = ?

[lofty python]

1/(sqrt3/2)

[dark sparrow]

can you simplify that?

[quiet orbit]

what do you need it's value for? the eq reduces to sin(pi/2 - x) = sin(2pi/3) so you can see right off from here which are the solutions

[dark sparrow]

that requires a potentially tricky step immediately afterwards though @quiet orbit

[lofty python]

2sqrt3/3

[dark sparrow]

great

so now

=tex \sec(x) = \frac{2\sqrt{3}}{3}

[dark sparrow]

you know that sec(x) = 1/cos(x), right?

[quiet orbit]

it requires no tricky step imo, see which are the fundamental solutions in [0,2pi) then add 2pi * k to them

one of them is obvsly 2pi/3

[lofty python]

yes

[dark sparrow]

so if sec(x) = 2sqrt(3)/3, what must cos(x) equal?

[quiet orbit]

the other one by looking at unit circle is pi - 2pi/3

[dark sparrow]

also raesh i'm pretty sure x = 2π/3 isn't a solution to the eq

:p

[lofty python]

-2?

[dark sparrow]

no

[lofty python]

Man I am so lost lol

[dark sparrow]

1. how did you get that?
2. -2 < -1!

[quiet orbit]

yeah, you get pi/2 - x from that. obvsly, but there's no point in computing sin( ) then finding the inverse

[dark sparrow]

=tex \frac{1}{\cos(x)} = \frac{2\sqrt{3}}{3} \ \cos(x) = ~ ?

[lofty python]

I know the answer is -pi/6

oh

[final prairie]

=tex \frac{1}{a}=\frac{b}{c}\implies a=\frac{c}{b}

[dark sparrow]

@lofty python so?

[lofty python]

3/2sqrt3?

[dark sparrow]

yes

[final prairie]

Yas

[lofty python]

nice

[dark sparrow]

i take it that it doesn't need explaining that this simplifies to sqrt(3)/2

now... can you locate 2 angles on the unit circle that have sqrt(3)/2 as their cosine?

[lofty python]

30, 150,

[dark sparrow]

no

does cos(150°) equal sqrt(3)/2?

(also, radians. get used to stating angles in radians.)

[lofty python]

oh its equal to -sqrt(3)/2

[dark sparrow]

so

you've said - correctly - that cos(π/6) = sqrt(3)/2

but there is another angle on the unit circle that also has a cosine of sqrt(3)/2

[lofty python]

330

[dark sparrow]

radians

what is 330° in radians?

[lofty python]

11pi/6

[dark sparrow]

bravo

so now

the solutions to your equation are all angles that can be written in one of these two forms:

=tex \frac{\pi}{6} + 2 \pi k, \frac{11\pi}{6} + 2\pi k

[dark sparrow]

for arbitrary integer k

does that need explaining?

[lofty python]

Never seen it before

How come the answer in the back of the book is -pi/6

Not just pi/6

[dark sparrow]

were you given a range to solve the equation over?

[lofty python]

no just 1 value

[dark sparrow]

...then the question is ill-posed

there are many angles that satisfy the equation

-π/6 does

as does +π/6

as does 11π/6

[lofty python]

oh ok

[dark sparrow]

as does 13π/6

as does 25π/6

[lofty python]

because you can go around the unit circle an infinite amount of times?

[dark sparrow]

i'd rather say you can make as many full turns as you like

[lofty python]

Because the question doesn't have a range?

[dark sparrow]

but yes

[lofty python]

oh ok

That makes sense.

Thanks. 😀

[dark sparrow]

@white wolf

[dark sparrow]

ok

[white wolf]

I don't even know what shape that is

[dark sparrow]

it's not a simple shape

as in, it's not one of those shapes that you can make a pre-computed area formula for

so

the first question i'm going to ask you is

how long is the side i've marked in red?

[white wolf]

4 units

[dark sparrow]

4 cm, specifically, but yes

now

imagine slotting a 2 by 4 rectangle into that dent

won't you agree that the shape you get is another rectangle?

@white wolf

[white wolf]

yes

[dark sparrow]

(i have to disappear for a moment, but this should give you an idea on how to find the area of the shape!)

[white wolf]

thanks!

[dark sparrow]

ok i'm back