precalculus

[languid crane]

lmao

algebREEEE

[tawny nacelle]

$(a + b)(c + d)=ab + ad + bc+ bd$

[obsidian monolith]

soap:

[tawny nacelle]

tru men never FOIL

[languid crane]

ye

o

All Students Take Calculus is bad mnemonic

wait no actually its good one

[tawny nacelle]

hey cotton

[languid crane]

what

[tawny nacelle]

you're guilty of being too cute

[languid crane]

sorry

[tawny nacelle]

[languid crane]

c: uwu

[tawny nacelle]

oWo

now u have to call me cute

[languid crane]

cute

[tawny nacelle]

[idle dust]

HJOW DO U TURN ARCTAN into something u can simplify with sine ;CCC

[languid crane]

9+x < 12+x

cry

[tawny nacelle]

IS IT RIGHT THO ??? @idle dust

[idle dust]

WELLI MEAN I Put the thing in wolfram alpha and its right but like i dont get hjow u did it

[tawny nacelle]

damn

im still samrt

hold on lemme draw the triangle

[idle dust]

lolis are smart

[tawny nacelle]

https://prnt.sc/nkou8o

arctan(sqrt(x)/(1 + x))

gives u the angle theta

and sin(theta) = opp/hyp

ez

done

🍻

🍰

[idle dust]

WAT

wait¸

ket me think

[tawny nacelle]

k dude

take ur time

[languid crane]

oh no

i failed my exam bc no use arctan

[idle dust]

OH

OK

[languid crane]

:(

[idle dust]

OK

IM DUMB

YES

[languid crane]

i try use tan instead

[tawny nacelle]

[languid crane]

jk

its just that I forgot arctan

damn arctan it is

arctan i try to remember!

[idle dust]

omg its been like 4 months i didnt do trig now i completely forgtot it damn

FWFJWFJWJ

[tawny nacelle]

ajksdhkasjhdjka

[idle dust]

FRANKS FOR U HELP KIND sire

[languid crane]

@tawny nacelle is 9 + x < 12 + x just 0 < 3

[tawny nacelle]

np dude

bRUh

@languid crane

??

[languid crane]

lmao

[tawny nacelle]

what on earth did u do ??

[languid crane]

?

I did -x from both sides

[tawny nacelle]

just subtract 9 from both sides

[languid crane]

i did that, too

[tawny nacelle]

even if u did -x on both sides wheres the -x on the right ??

[languid crane]

+x-x

0

[tawny nacelle]

$9 + x < 12 \implies x < 3$

[obsidian monolith]

soap:

[languid crane]

yes

[tawny nacelle]

why DID U WRITE 0 < 3

asdasdasdas

askldjkjgperjpg

[languid crane]

idk lmao

idk what did I do

im tired

[languid crane]

:(

[tawny nacelle]

hahha

jk

[languid crane]

dude

[tawny nacelle]

nothing personal kid

[languid crane]

you need to subtract the x

[tawny nacelle]

[languid crane]

from both sides

[tawny nacelle]

yeah

[languid crane]

REEEEE

so its 9 < 12

[tawny nacelle]

bRUH

[idle dust]

a

[tawny nacelle]

where the x go on the right ??

[languid crane]

bruh what is this

[tawny nacelle]

[idle dust]

ok i will do the same for the other questions and ill tell u if i need help

[tawny nacelle]

k sure buddy

[languid crane]

9+x < 12+ x //-x, -9

0 < 3

[tawny nacelle]

ok

[languid crane]

what

dude

you can show me

if this aint right

[tawny nacelle]

its right ig

[languid crane]

imo x can be any value lmao

dude what is this!

how i write it

[tawny nacelle]

write wha t ?

[languid crane]

how do I solve this :( if that aint right

im cry

[tawny nacelle]

oh no

dont cry

[languid crane]

imo this is right cause idk what else to do

or x < 3+x

[tawny nacelle]

whats the question to being with ?

[languid crane]

solve inequality 9 + x < 12 + x

[tawny nacelle]

its like ur starting with something true and then ur proving that its ture

9 + x < 9 + x + 3

so ofc rhs is bigger

[languid crane]

yes

there is no answer for inequality?

[tawny nacelle]

the answer is it works for any x

any real

[languid crane]

ok but

[tawny nacelle]

🍑

[languid crane]

how to write it in terms of math

[tawny nacelle]

idk use logic notation or something

[languid crane]

ok

[idle dust]

wait

im stupid

oK

[tawny nacelle]

ok

[languid crane]

so x works for any real number

ok

[idle dust]

cos(2x)= 2(cos(x))??

NANI

ok

i will think

NOWP IM WRONG OK

[tawny nacelle]

bRO

cos(2x) = cos^2(x) - sin^2(x)

[idle dust]

aaaaaaaaaaaaa

[tawny nacelle]

bbbbbbbbbbbbb

[idle dust]

RRRRRRRRRRR

[tawny nacelle]

EEEEEEEEEEEE

[idle dust]

OK

HOW about you help me and i stop being dumb

[tawny nacelle]

[idle dust]

cos(2arcsin(x)

)

So

i drew a triangle already

and

ok i will show u

[echo plaza]

use double angle

[tawny nacelle]

welp

that smeels wrong

sin(theta) = 1/x but it should be x

[idle dust]

oh wait

i drew it wrong

lol

[tawny nacelle]

[idle dust]

oops

OK THERE

i drew that on my paper

and then i got confusion

[echo plaza]

$\cos(2\arcsin(x)) \equiv 1-2\sin^2(\arcsin(x))$

[obsidian monolith]

CaptainLightning:

[tawny nacelle]

now use le double angle identity

hmm

[idle dust]

wooot

ok

[tawny nacelle]

,w cos(2arcsin(x)) = 1 - 2x^2 true or false ?

[idle dust]

let me try

[obsidian monolith]

Display results online and refine query

Requested by soap#7503

[tawny nacelle]

ok u remember the triangle we drew ?

[idle dust]

yea

[tawny nacelle]

throw that to the garbage

[idle dust]

NANI

ok

[tawny nacelle]

now use cos(2x) = cos^2(x) - sin^2(x)

also we know that cos^2(x) = 1 - sin^2(x)

so overall we get that cos(2x) = 1- 2sin^2(x)

[idle dust]

ye

ok

i undestanjddd

so far

[tawny nacelle]

[idle dust]

so that means

1-2x^2

wait

BRUH

[tawny nacelle]

yeah

[idle dust]

BRUhruRBurURHUrbuIRBrh MOMENT

[tawny nacelle]

[idle dust]

thankS VERY MUICHO SIR

[tawny nacelle]

no problemo

with this i shall take my leave

[idle dust]

OK YEET LAST QUESTION HOPEFULLY I WONT NEED UR HELP BECAUSE IM look like a total idiot

YES

[tawny nacelle]

sayonara

[idle dust]

ok baiii

[torn swift]

conveniently chosen points

the parent function is sqrt(x)

so you can graph that first using easy to graph points, then you apply the transformation to each of those points

[vestal plaza]

Can someone explain what exactly is this question asking for or what exactly is it looking for?

[short sorrel]

it wants you to rearrange the equation to isolate x

ie, get x by itself, with k on the other side

[vestal plaza]

so log_3(x(x+5))=k

than log_3(x^2+5x)=k

@short sorrel something like that?

[short sorrel]

yes, then what?

[vestal plaza]

uhh

I forgot whats the reverse of log

[short sorrel]

you can convert from logarithmic form to exponential form.

[vestal plaza]

like e^k?

[short sorrel]

$a = b^c \iff \log_{b}{a} = c$

[obsidian monolith]

Namington:

[vestal plaza]

which will drop the log leaving x^2+5x=3^k

like that?

[short sorrel]

yes

[vestal plaza]

mm what can I do next?

[short sorrel]

if you prefer, you can also conceptualizing this as making everything raised above a base of 3, ie
\
$3^{\log_{3}{x^2 + 5x}} = 3^k$

[obsidian monolith]

Namington:

[short sorrel]

and then the 3^log_3(stuff) just becomes that stuff

[fringe stream]

you can use the quadratic formula after that

[short sorrel]

anyway, next you want to solve $x^2 + 5x = 3^k$ for x

[obsidian monolith]

Namington:

[vestal plaza]

how would I solve for x with the K in there?

[short sorrel]

you can complete the square

[vestal plaza]

uh explain

I havent done math in 6 years

[short sorrel]

uh

maybe better to use the quad formula then

so, we know for ax^2 + bx + c = 0

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

[obsidian monolith]

Namington:

[short sorrel]

to use the quad formula, we need to get one side to 0

so, subtract 3^k from both sides

$x^2 +5x - 3^k = 0$

[obsidian monolith]

Namington:

[short sorrel]

then lets plug that into the quad. formula. our a value is 1, b is 5, and c is -3^k

$x = \frac{-5 \pm \sqrt{5^2 - 4(1)(-3^k)}}{2}$

[obsidian monolith]

Namington:

[vestal plaza]

-5+-sqrt(25-4(-3^k))/2

[short sorrel]

simplifying, we get $x = \frac{-5 \pm \sqrt{25 + 4(3^k)}}{2}$

[obsidian monolith]

Namington:

[short sorrel]

and thats the solution in terms of k

[vestal plaza]

Im guessing they want us to type the whole solution like that

[short sorrel]

probably something like that, yeah

[vestal plaza]

I hate Lumen, worst math program ever

[short sorrel]

however, do note

[vestal plaza]

doesnt even tell you the answer after you get it wrong, so how are you supposed to learn?

[short sorrel]

can we take the log of a negative number?

[vestal plaza]

no

[short sorrel]

not over the reals

[fringe stream]

remember to reject the negative value

[short sorrel]

so we can usually eliminate a solution

[fringe stream]

since x > 0

[short sorrel]

^^

[fringe stream]

oh whoops

way ahead of me

[short sorrel]

we're only gonna care about the + case, not the - case

because we'll always get a negative x value if we use the - case

sorry, mental mathed wrong, nbd

[fringe stream]

its all good

[vestal plaza]

rip still took it wrong

Yea this program is garbage, next semester is going to cost money in order to be using it 😃

[swift glacier]

So if I have to find the limit as x approaches zero

and it's a rational function with x as the denominator

would it be possible to multiply by a conjugate?

[willow bear]

what conjugate

if it's a rational function then you never need to do that

[swift glacier]

like that

the three methods I know of is subsitution, which doesn't work, factoring or multiplying by a conjugate

and I don't think the rest really help

[echo plaza]

that looks like conjugate time

[swift glacier]

unless it's conjugate the top one?

[echo plaza]

yes

[swift glacier]

oh wait I'm so dumb thank you

[echo plaza]

np

[swift glacier]

I can't believe I didn't notice

I literally just thought about it

I love your pfp btw

[willow bear]

this isn't a rational function

rational functions cannot have radicals like that

@swift glacier

[swift glacier]

wait it isn't?

What would it be then?

I sort of categorized fraction looking function as a rational oops

[willow bear]

a rational function is specifically polynomial/polynomial

[hybrid charm]

√(16+x) is irrational

[half axle]

pls help

idk how to answer this

[willow bear]

what have you tried?

[echo plaza]

$

[swift wagon]

Let $x=\cos(\theta)$

[obsidian monolith]

BigPhatdog:

[swift wagon]

Hint^

[willow bear]

is that synthetic division that the author of this work used?

if so, then they knocked the polynomial down two degrees

by dividing out the factors responsible for the roots already found (1 and -1)

[3, -10, 8] <-> 3x^2 - 10x + 8

the rightmost entry in a list like this is always the constant term

yes

what

[echo plaza]

multiply it out and see

[willow bear]

^

the constant term is wrong

(3x-4)(x-2) = 3x^2 - 10x + 8

what?

...i mean are you not able to expand this thing out and see for yourself what goes into the constant term?

[torn swift]

well what happens when x=-2?

well they wanted to plot a convenient point, so they want to get 3^0 which would be 1

but if you plug in 0 for x you end up getting 3^2 which is 9

so to get 3^0 you need to plug in x=-2

yes, you would get (0,9) using that equation

but we want to have 1 as the y coordinate

so we have to set x=-2 to so you get 3^0 which is 1

it's just an easy point to plot

you can plot anything else, but (x, 1) is pretty easy to graph

no, you get the same graph of course because you are using the same function, it's just that graphing other points may be harder since the numbers may not be as "nice" (they could be very big because this is an exponential function)

why did you plot (9,0)?

it was (0,9)...

but you said you plotted (9,0)

so you reversed the coordinates

[sleek pawn]

Hi, I need some help. I have to fill these two tables but I'm very confused on the first one, also I'd like to know if I don't have any errors on the second one.

<@&286206848099549185>

[torn swift]

only been a few seconds and you pinged us already...

[sleek pawn]

sorry

[sleek pawn]

thanks, and what is #28 about?

[willow bear]

wdym

your thing for #28 is weird

[sleek pawn]

I don't know where to start filling it

[willow bear]

you put the sin values in the rad row so idk

[sleek pawn]

that's what the textbook had, I just checked

so that's bothering me

well then I'll have to ask my teacher

[viscid thistle]

Is showing these problems fulfill the pythagorean identity enough as I did in the problem below?

[royal gull]

above*

[viscid thistle]

thanks

[royal gull]

but yeah, sin^2 a + cos^2 a = 1 is very known, its enough

[viscid thistle]

am I making it to easy on my self?

seems way too essy

[royal gull]

I mean you can prove it if you want, but it is very well known

[viscid thistle]

factor, multiply by sin or cos and expand tangent, cotangt etc.

it just feels like I'm cheating

[royal gull]

yup

[viscid thistle]

like I'm missing the harder part

[viscid thistle]

problem 37 is messing with me

am I on the right track? Feel like I missed a trick somewhere

[eternal folio]

seems to me that you overcomplicate it

just multiply top and bottom of the left hand side by 1 - sin(t)

[viscid thistle]

ahh

and then pythagorean identity time

[eternal folio]

yes

and then it's just the definitions

[viscid thistle]

ok let me try again

still having trouble with the left side

thought by putting everything on the left side in terms of tangent I could simplify but still not making it work

[torn swift]

,rotate

[viscid thistle]

<@&286206848099549185>

[eternal folio]

Don't manipulate the two sides at once

just concentrate on one side

start with the left hand side (of the original) and then work your way from there until you reach the same thing that's on the right hand side

I suggested that you multiply top and bottom by 1-sin(t)

[frozen needle]

[eternal folio]

to get $ \frac{\left(1 - \sin(t)\right)^2}{1 - \sin(t)^2} $

[frozen needle]

[obsidian monolith]

uno20001:

[eternal folio]

Yeah, this is it

[frozen needle]

[eternal folio]

Problem 37

[frozen needle]

Ah ok

[viscid thistle]

I got it, thanks guys!

[viscid thistle]

I have a maath revision question, I am able to voice chat, it wont let me upload a photo of the question unfortunately

math*

[serene heath]

type.it @viscid thistle

[hallow flicker]

do you guys know how to represent sin like a formula

or something

[serene heath]

wym

[viscid thistle]

@serene heath Root(9y+1) = 3 root(4y-2)

@hallow flicker From my knowledge you could show sin as sin=opp/hyp but if your talking about specific angles it can vary.

[serene heath]

u want to solve for y?

square both sides

[viscid thistle]

Thanks, I got some help, thank you so much!

[ruby otter]

Let $f:[1,10] \to Q$ be a continuous function and $f(1) = 10$ then$f(10) = $ ??

[obsidian monolith]

Radical Ninja:

[ruby otter]

how can we determine f(10) from this given info?

[frozen needle]

Could be anything

[ruby otter]

but its given 10 how??

[frozen needle]

Oh wait

$f:[1,10]\to\bbQ$

[obsidian monolith]

Tuong:

[ruby otter]

rational numbers

[frozen needle]

Now that's more interesting

[ruby otter]

still it could be anything right?

[frozen needle]

My guess is that f is constant on [0,10]

[ruby otter]

even i thought of that but we can't be sure about that can we??

coz the options are
1/10 , 1 , 10 , can't be obtained

[frozen needle]

By contradiction, if f wasn't constant on [0,10] there would exist some c in [0,10] such that f(c)≠f(1)

however there do exist some irrational number in between f(1) and f(c), and since f is continuous, the intermediate value theorem holds so, f outputs some irrationals

and here's the contradiction

[tawny nacelle]

if its not a constant fucntion then it might output some irrational number on the way ?

[frozen needle]

so f really is constant on [0,10]

[tawny nacelle]

nice maff

[ruby otter]

thanks

[frozen needle]

You're welcome

[ruby otter]

mean value theorem = LMVT??

[frozen needle]

Oips

[tawny nacelle]

intermediate value theorem no ?

[frozen needle]

Fixed xd

yeah

[ruby otter]

there's also CMVT

[ruby otter]

can find differentiability of a function by differentiating it ??

and subs a value for x

[willow bear]

[ruby otter]

doubt

$\cos|x| + (x^2 -4)|x^2 -5x+6|$

[obsidian monolith]

Radical Ninja:

[ruby otter]

how do i check differentiability for this function

[willow bear]

consider instead checking where differentiability may fail

also, \cos

[ruby otter]

can I find derivative of this

and find where differentiability may fail ??

[obsidian monolith]

Radical Ninja:

[willow bear]

no

[ruby otter]

why?

[willow bear]

because it's a bit weird trying to find something which you don't even know if it exists, don't you think

[ruby otter]

but it does work for many functions like

|x|

[willow bear]

the absolute value function is differentiable everywhere besides zero, and is not differentiable at zero.

use this fact to establish where your function may fail to be differentiable - there will only be finitely many points like that - and then examine each of those points individually

[ruby otter]

ya it failed

thats why i'm asking

whats the fastest way to find this

[willow bear]

what failed?

[ruby otter]

leave that just help me with how to find differentiability

[willow bear]

i just told you exactly what to do

[ruby otter]

how do i find where differentiability might fail??

[willow bear]

the points which are suspect for non-differentability are the points where one of the expressions inside an absolute value is equal to zero.

or more generally, where any input to a non-differentiable function equals one of said function's points of non-differentiability.

[ruby otter]

so for my func it must be 0, 2, 3

[willow bear]

yes. everywhere besides that, your function is guaranteed differentiable.

[ruby otter]

im sorry but my original question was this i think

how do i find which point among 0, 2, 3 is the culprit

[willow bear]

you know the limit definition of the derivative, right

[ruby otter]

is that the only way??

[willow bear]

it's the only foolproof way

[ruby otter]

no other way??

[willow bear]

no other way

[ruby otter]

[willow bear]

i mean like

what

do you want me to give you ad-hoc strategies for each of those points for this function in particular or what

bc that's equivalent to just doing the thing for you

[ruby otter]

what is bc

[willow bear]

because

[ruby otter]

👍

[thick raptor]

[viscid thistle]

Why does multiplying an odd function by an even function = an odd function?

is it because multiplying is the same as adding the powers so functions follow the same rule or even/ odd as adding/ subtracting numbers?

For example x^2 * x^3 = x^5 becuase 2 +3 = 5

Is that making sense?

[wild trout]

Well as an intuition, maybe

But you need to be rigouros

Let $f$ and $g$ be two functions defined on $D_{f}$ and $D_{g}$ respectively, and to be odd and even respectively. $ \ $The function $h:=fg$ is defined on $A:=D_{f} \cap D_{g}$, and is that, for all $x \in A$: $\ h(-x)=(fg)(-x)=[f(-x)][g(-x)]=[-f(x)][g(x)]=-f(x)g(x)=-(fg)(x)=-h(x) \$
So $h$ is odd

[obsidian monolith]

Patrick Salhany:

[wild trout]

Here I am supposing you're talking about functions with one variable on R let's say, and both domains are centered at the origin

[viscid thistle]

So you're method serves as a useful heuristic but it doesn't speak to the deeper explanation

?

[wild trout]

?

[viscid thistle]

The whol idea of thinking of them following the same property as adding even / odd numbers

[wild trout]

Well I said maybe they called it even odd functions by analogy of what were you saying

I thought you were asking why multiplying an even and an odd function gives an odd one, like prove why

Maybe someone can think like this
Negative to odd power is negative
Negative to even power is positive
So it is an analogy to call even and odd function, because in the integrable case, if you're calculating algebraic area under a curve on let's say a segmet [-a;a], you'll get 0 for odd functions and double area from 0 to a for an even one, cause the odd function have positive and negative sign, but even has either, supposing O is a center of symmetry an (Oy) is an axe of symmetry for each case properly

[honest igloo]

Can someone help me with this? If sin(x) = − 3 and x is an angle in Quadrant 3, find the value of tan x

[hard hornet]

isn't sin(x) range -1<sin(x)<1?

[honest igloo]

If so how would I apply that?

[hard hornet]

are you sure thats the right question?

[honest igloo]

I’ll double check

I apologize it was an error

[hard hornet]

ok yea, scared me

[honest igloo]

It’s sin(x) = -3/5

[hard hornet]

sin(x) will always spit out a value between -1 and 1, never less or more than that

ok so you're given that

sin(x) = -3/5

[honest igloo]

Ok

I think I’m following

[hard hornet]

the key here, is NOT to solve for angle x

ok so draw a picture with the 4 quadrants

[honest igloo]

Ok done

[hard hornet]

now draw an angle of x, that leads to quadrant 3

[honest igloo]

Ok done

Wait any angle?

[hard hornet]

its arbitrary

the key is not to solve for the angle x

[honest igloo]

Ah ok

[hard hornet]

shud look something like this, right?

[honest igloo]

Yes exactly

[hard hornet]

ok so when it says sin(x) = -3/5

that means

hmmm lemme try explain it intuitively

[torn swift]

I like how he had a question but now you are asking questions about it

[honest igloo]

Go for it

[hard hornet]

lol

heheheeheh

ok lets go back to the basics

when you are given some ratio for example, sin(x) = something

it usually measures the ratio of two parts of the triangle

opposite leg to the angle, adjacent leg to the angle, or the hypotenuse right?

[honest igloo]

Yes I follow

[torn swift]

it usually measures? you mean is that

[hard hornet]

i mean not measures, it gives

SORRY LOL

damn stressful teaching calc when u got amphy spying on u

[torn swift]

just be confident in your answers

if you don't sound confident then how is the other guy going to trust you lol?

[hard hornet]

what should you add to the diagram to make it a right triangle, that allows you to measure the ratio of how much the x, y, changes

[honest igloo]

Woah uh

[hard hornet]

recall that the trig functionsi really a measure of the ratio between the horizontal component, vertical component, and the hypotenuse (or the line you drawn with angle x)

[honest igloo]

I’m honestly not sure

Ok

[hard hornet]

did you learn in class, that when you usually draw a right triangle in the quadrants to help you compute some trig functions/

gimme a sec

[honest igloo]

Yes

I remember now

[hard hornet]

very shotty drawing

[honest igloo]

However that’s the part I got stuck on

[honest igloo]

Yes I recall the professor drawing this

[hard hornet]

in your question, sin(x) = -3/5, what does sin(x) = -3/5 tell you about its ratio of the triangle

[honest igloo]

The ratio is 3/5?

[hard hornet]

more specifically, if i told you sin(x) or sin of some angle, gave you 3/5

which specific part of the triangle

is it taking the ratio of

[honest igloo]

I’m not sure sorry

[hard hornet]

sin measures the ratio between the opposite leg and the hypotenuse of the triangle

[honest igloo]

Ahh

[hard hornet]

sin = opposite / hypotenuse
cos = adjacent / hypotenuse
tan = opposite / adjacent

remember?

[honest igloo]

Yes lol damn it

[hard hornet]

now, again, what is sin(x) = -3/5 telling you

[honest igloo]

That the -3 is the opposite and 5 is the adj side?

[hard hornet]

yup

[honest igloo]

Hyp*

Sorry

[hard hornet]

the ratio of the opposite side of the triangle with the hypotenuse is -3/5

[honest igloo]

Alright

[hard hornet]

ok, heres something i want u to remember, im not sure how to put it into words yet, but just remember. WHen you draw your right triangles, always draw the line PERPENDICULAR to the x axis

so remember how we drew some line in quadrant 3?

[honest igloo]

Oh like you did in the drawings too

[hard hornet]

build a right triangle, and assign the proper sides its ratio

with the angle x

[honest igloo]

Ok

[hard hornet]

mmmm now show me ur drawing

if u dont mind

[honest igloo]

Wait almost done

Ok

[hard hornet]

mmmmmm not quite not quite

ok this is kinda like the confusing part

when you draw your right triangle, think about it as having like two angles

[honest igloo]

Ok

[hard hornet]

one angle like you drew in ur picture, and another angle that always comes from the nearest x axis (negative or postive) to the terminal line

i know, its a bit weird, i really dont know how to put it to words

but i think you'll c it later on

[honest igloo]

Isn’t that similar?

[hard hornet]

when we refer to the angle x, we also refer to the inner angle of the triangle

yes and no

the original angle x is the question, and what we've drawn is more of a "short cut". When you practice more of these, you'll soon see why

for now, lets continue

[honest igloo]

Ok

[hard hornet]

now what sin(x) refers to, is that triangle

where should your 5 and -3 be?

(it was wrong in your original picture

[honest igloo]

Should I switch the values?

[hard hornet]

mmm recall, that sin is the measure of the OPPOSITE over the HYPOTENUSE

[torn swift]

caps lock life

[honest igloo]

Is that correct?

[hard hornet]

yes

exactly

awesome awesome

[honest igloo]

Nice

[hard hornet]

is this making sense?

if not, dw about it, you'll understand it the more u practice

[honest igloo]

A lot of sense

[hard hornet]

i didnt understand this at all, but when i did, idk how to explain it in words lol

ok so now

from geometry, do you think you can solve for the final leg of the triangle

[honest igloo]

Yes

[hard hornet]

wut is it

[honest igloo]

Hold up

4?

[hard hornet]

yes and no

another thing i forgot to mention

[honest igloo]

-4

[hard hornet]

ahhh

u got it

[honest igloo]

Heh

[hard hornet]

im sure u know why

[honest igloo]

Yeah my apologies

[hard hornet]

cuz u wanna think of it as a graph

now

do you think you can compute tan(x)?

[honest igloo]

Is tan sin/cos?

[hard hornet]

yuo

yup, or if ur lazy like me, opposite / adjacent

[honest igloo]

Ok cool I’ll try

[hard hornet]

lmk what u get

[honest igloo]

-3/4?

[hard hornet]

close

be a bit careful with your negatives and positives

[honest igloo]

Hmm I don’t see where I messed up

Oh nvm

.75

Or 3/4

[hard hornet]

yups

exactly

ok for these types of problems, i recommend doing a lot and a lot of practice

its one of those things where you're like asked to memorize your multiplication table in elementary

and then figure out later on why it all works

[honest igloo]

Yes I know my brain just farted

Apologies

[hard hornet]

ye np

[honest igloo]

Oh that’s the answer?

[hard hornet]

yep lol

[honest igloo]

No way

[hard hornet]

in a nutshell

or lemme summarize

[honest igloo]

That’s so easy

Damn you explained it well

[hard hornet]

you're given sin(x) is some ratio. Based off of that, you have to draw the line in the correct quadrant, build a proper right triangle, and compute for the other trig values

i cant really explain why you do it

[honest igloo]

Nice I’ll keep this in mind with the next problem

[hard hornet]

liek i said, its one of those things you just learn how to do

then later down the line, u figure out why you do it

but even then, idk lmao

[honest igloo]

Thank you for the help!

[hard hornet]

ya anytime

[torn swift]

should have mentioned that x coordinate is cosine value and y coordinate is sine value

much easier to see that tan(x) in quadrant 3 would be a positive quantity

[hard hornet]

ah tru

damn

[torn swift]

that's assuming he knows about unit circle though

[honest igloo]

I do but I don’t remember it

Like all the values

Can you elaborate on that? The comment you said earlier

[torn swift]

the applicable part here was helping you determine the signs

[honest igloo]

Well isn’t the sign tied to the actual computation

[torn swift]

in quadrant 3, the x and y values are all negative so cosine and sine will have negative values

[honest igloo]

Alright I follow

[torn swift]

then it easily follow that tan(x) will be positive

can you see why that is?

[honest igloo]

Oh because two negatives = a positive?

[torn swift]

correct

[honest igloo]

That makes this much more clear

[torn swift]

tan(x)=sin(x)/cos(x) and if both are negative values then the answer is positive

again you should review that fact that cosine will take on the sign of the x value and sine will take on the sign of the y value

so the sine of an angle in quadrant 2 will always be? because?

[honest igloo]

Positive

No

[torn swift]

and the cosine of an angle in quadrant 2 will always be? because?

[honest igloo]

Negative

[torn swift]

think of quadrant 2

[honest igloo]

Not negative?

[honest igloo]

Yeah negative

[torn swift]

look at the signs of the numbers on each axis in quadrant 2

sine takes the sign of the y coordinate so in quadrant 2 sine will always be...?

[honest igloo]

Oh sin will always be positive in quadrant 2

[torn swift]

and cosine will always be?

[honest igloo]

And cos will always be negative

Making tan negative as well

[torn swift]

correct

[honest igloo]

Wow I literally have grasped this topic

[torn swift]

do the same thing for quadrant 4, sine will always be? cosine will always be? therefore tangent is always?

[honest igloo]

Quadrant 4 sin will be negative, cos will be positive which makes tan negative again

[torn swift]

perfect

[honest igloo]

Thanks for the help

I appreciate it

[torn swift]

[hard hornet]

damn explained it so much better than me

[torn swift]

all a matter of practice

[tropic crown]

dont really understand how this works

[torn swift]

was cot(x) restricted to be between pi and 2pi initially?

[tropic crown]

let me give the original question

[torn swift]

ok makes sense then

did you read the posts above? I go over the same thing

[tropic crown]

no i did not, let me go over it

first

[torn swift]

I explain the signs of cos(x) and sin(x) for the various quadrants

[tropic crown]

okay... so since cot is positive, that means quad 3 will

be i suppose the focus?

[torn swift]

the focus? what?

[tropic crown]

this is what i understood i suppose. even tho it says pi < x < 2pi, since cot is positive, its really more like pi < x < 3/2pi

[torn swift]

the restriction is just telling us that beta is in quadrant 3 or 4

if there was no restriction then it could have been in quadrant 1 or 3

[tropic crown]

ok i think im following

without the boundaries, there would no distinction between quad 1 or 3,

so if we were to write out the trig ratios, we would have to consider both quad 1 and 3, but since there are restrictions, we only care about 3

[torn swift]

yes

[hard jacinth]

Can I have some help with a problem

Please

Number 6

[velvet ether]

multiplying by 2 on both sides does not cancel out the 2 on the right hand side...

[wet hare]

Yep, just figured it out.

[velvet ether]

oh lol okay

[wet hare]

Thanks either way!

[valid vector]

How do I make the denominators the same in #5?

[limber bone]

u dont

u just a/b - c/d = ad-bc/bd

[valid vector]

I had no idea that was a thing. Thanks!

[prime prawn]

I need help evaluating $$\sum_{n=1}^{\infty} \frac{1}{2^{n}} \tan \left( \frac{\pi}{2^{n+1}} \right)$$

[obsidian monolith]

Linguist = Trash:

[prime prawn]

idk, but 2/pi seems a good approximation

but i need help on how to get there, 2/pi was just a guess

Please ping if you can help

[limber bone]

i mean

jusut a suggestion

try just computing partial ssums?

see if anything just cancels out

[thick raptor]

posting in 3 places smh

[torn swift]

being offline all the time even though you are clearly online smh

[hard hornet]

i forgot but in summations can you split terms

[thick raptor]

[echo plaza]

what do you mean by split terms

[thick raptor]

cutting them in half

🔪

[hard hornet]

find 1/2^n and then find tan(pi/2^n+1) then multiply em

i forgot my seires

[echo plaza]

[hard hornet]

im assuming thats illegal

[echo plaza]

you can't multiply the two different series together like that

[hard hornet]

aah ok ima commit sudoku

🔪

[limber bone]

u mean

seppuku

XD

[hard hornet]

shuddup nerd

let me die in peace

[echo plaza]

$14=\sum_{n=1}^3 n^2\neq \left(\sum_{n=1}^3 n\right)\left(\sum_{n=1}^3 n\right)=36$

[obsidian monolith]

CaptainLightning:

[echo plaza]

need to use \left and \right more

[thick raptor]

need to use \bigg too tbh

[echo plaza]

no I decided to stop that

[thick raptor]

[fossil kiln]

Im taking precalculus over the summer. What can I expect and should I focus my review on anything?

[willow bear]

algebra, algebra, algebra

[tawny nacelle]

algeBRUH

[torn swift]

precalcuLOST

[thick raptor]

Is this loss?

[viscid thistle]

This looks more like calculus than pre-cal because I dont reqcongise a thing.

[hard jacinth]

Hey can I have some help with this

Number 3

[willow bear]

what have you tried so far

[willow bear]

$z^{\star}$ most likely refers to the conjugate of $z$

[obsidian monolith]

Ann:

[hard jacinth]

Oh hey Ann I remember you

And wow yes it does

You're right

I thought it meant multiplication

Thanks 😂

[serene heath]

true

[coral jay]

You're not wrong

[torn swift]

All three of the above are incorrect

[serene heath]

[serene heath]

who pong

[sleek pawn]

hi

any tips on verifying trigonometric identities?

I'm stuck and need to hand in a set of exercises in a few hours

$cos^4\alpha+sin^4\alpha=1-2sin^2\alpha cos^2\alpha$

[obsidian monolith]

a-regular-kugelblitz:

[sleek pawn]

this is one of them

one of the simplest in appearance

any help from the helping gods?

[willow bear]

from the left hand side, add & subtract 2sin^2(α) cos^2(α)

[sleek pawn]

I don't understand quite well what that would do

[willow bear]

you can make (cos^2+sin^2)^2

[sleek pawn]

ok I didn't really understand how to do the first thing, but what change would the second thing do?

ok so the first step would eliminate 2sin^2(α) cos^2(α) from the right side of the equation and would be added to the left is that correct?

I'm left with all this nonsense equaling 1 and there's no identity there that can be simplified more

$2sin^2\alpha cos^2\alpha+cos^4\alpha+sin^4\alpha=1$

[tawny nacelle]

[obsidian monolith]

a-regular-kugelblitz:

[tawny nacelle]

do you know the binomial formula ?

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

[sleek pawn]

mhm

[tawny nacelle]

ring any bells ?

[sleek pawn]

ok let me see...

yeah

ok

lol I'm so dumb

but then what happens to the 1?

[tawny nacelle]

it stays as it is

[sleek pawn]

but I need to verify it

were did the 1 come from?

[tawny nacelle]

???

IT WAS THERE TO BEGIN WITH

[sleek pawn]

yes

but

I'm supposed to say why it's there no?

[tawny nacelle]

bRuH

[sleek pawn]

omg

[tawny nacelle]

im just gonna write it out

[sleek pawn]

yes please

[tawny nacelle]

much easier that way

[sleek pawn]

because afaik the right side of the equal sign has to be empty so that the original thing can be compared

[tawny nacelle]

$(\cos^2(x))^2 + + 2 \cdot (\cos^2(x)) \cdot (\sin^2(x)) + (\sin^2(x))^2 = 1 \ (\sin^2(x) + \cos^2(x))^2 = 1 \ 1^2 = 1 \ \qed$

[obsidian monolith]

soap:

[sleek pawn]

God I am really dumb

thank you

[tawny nacelle]

np

[sleek pawn]

this is still problematic, if there's anyone who could give me any universal advice on this stuff I'd be enormously grateful

now I have another equation:
$1-2sin^2\alpha = \frac{1-tan^2\alpha}{1+tan^2\alpha}$
this is what I've got:
$2+2tan^2\alpha-2sin^2\alpha=1$

[obsidian monolith]

a-regular-kugelblitz:

[sleek pawn]

I feel like I'm very close but I'm not sure what to do

$2+((2\tan \alpha)^2+2(2\tan \alpha)(2\sin \alpha)-(2\sin \alpha)^2)=1$

am I going in the right direction?

I feel like the quotients for tan and sin should go somewhere

<@&286206848099549185>

(I'm sorry for tagging you all again)

[willow bear]

okay first off

can you please use \sin, \cos and \tan

it's gonna look much less ugly

[sleek pawn]

wait that can be done?

nicee

so...

any help?

what to do with the quotients and integers

[obsidian monolith]

a-regular-kugelblitz:

[sleek pawn]

:(

[tawny nacelle]

are you trying to solve this equation ? @sleek pawn

[sleek pawn]

it's the same as before, verifying

also thanks for listening to my cry for help

[tawny nacelle]

ok i proved it

lemme send you the solution

$1 - 2 \sin^2(x) = \frac{1 - \tan^2(x)}{1 + \tan^2(x)}$

[obsidian monolith]

soap:

[tawny nacelle]

$\frac{1 - 2\sin^2(x) + 1}{1 - 2\sin^2(x) - 1} = \frac{1 - \tan^2(x) + 1 + \tan^2(x)}{1 - \tan^2(x) - 1 - \tan^2(x)}$

[obsidian monolith]

soap:

[tawny nacelle]

$\frac{1 - \sin^2(x)}{ - \sin^2(x)} = - \cot^2(x)$

[obsidian monolith]

soap:

[tawny nacelle]

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

and then cos/sin = cot

QED

[sleek pawn]

I'm gonna try and decipher this now, thanks

[steady aspen]

Can someone help me with trig equations

[tawny nacelle]

sure

[steady aspen]

I just started it so I dont know like anything of the concept rn

how would I solve something like this, 5sinx+12cosx=1

[sleek pawn]

@tawny nacelle could you describe to me what all this means please?

[tawny nacelle]

google "componendo dividendo"

it basically says

if u have

$\frac{a}b = \frac{c}d$

[obsidian monolith]

soap:

[tawny nacelle]

then $\frac{a + b}{a - b} = \frac{c +d}{c - d}$

[obsidian monolith]

soap:

[sleek pawn]

ohhhhh

interesting

but I still have a few questions

like:

shouldn't this cancel sin^2(x)?

$\frac{1 - \sin^2(x)}{ - \sin^2(x)} = - \cot^2(x)$

[obsidian monolith]

a-regular-kugelblitz:

[sleek pawn]

@tawny nacelle

[tawny nacelle]

why would this cancel the sin ?

[sleek pawn]

well Ig it would become 1

[tawny nacelle]

$\frac{1 - \cancel{\sin^2(x)}}{ - \cancel{\sin^2(x)}} = - \cot^2(x)$

[obsidian monolith]

soap:

[tawny nacelle]

are you saying that ?

[sleek pawn]

yeah, but now that you point it out, I'm not sure it's correct because of the 1 there

[tawny nacelle]

yeah u cant do tha t

the best u can do is to split up the numerator

like $\frac{1 - \sin^2(x)}{-\sin^2(x)} = \frac{1}{-\sin^2(x)}- \frac{\sin^2(x)}{ - \sin^2(x)}$

[obsidian monolith]

soap:

[tawny nacelle]

then u end up with 1 - csc^2(x)

[sleek pawn]

uhhhh

but that's not what happened

[tawny nacelle]

huh ?

[sleek pawn]

how did you get the 1 - sin^2 = cos^2(x) in the end?

[tawny nacelle]

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

[sleek pawn]

but where did that come from?

[tawny nacelle]

thats the fundamental theorem of trigonometry

[sleek pawn]

yeah not that

lol

how did you go from 1 - csc^2(x) to sin^2(x) + cos^2(x) = 1

[tawny nacelle]

i didnt ?

which part are you talking about ?

cuz i applied sin^2(x) + cos^2 = 1 when i didnt break up the numerator

and then u asked if we could break up the numerator

so i did break it up

an THEN i applied the other identity csc^2 - cot^2 = 1

[sleek pawn]

I only asked for the explanation you are giving me right now

ok so can I see that?

[tawny nacelle]

see what ?

[sleek pawn]

$\frac{1 - \sin^2(x)}{ - \sin^2(x)} = - \cot^2(x) and then \csc^2 - \cot^2 = 1$

[obsidian monolith]

a-regular-kugelblitz:

[tawny nacelle]

ok lets do that part again

$\frac{1 - \sin^2(x)}{-\sin^2(x)} \ = \frac{ \cancel{-1}(\sin^2(x) - 1)}{\cancel{-1}(\sin^2(x))} \ = \frac{\sin^2(x)}{\sin^2(x)} - \frac{1}{\sin^2(x)} \ = 1 - \csc^2(x)$

[sleek pawn]

the two last parts I understand

I'm genuinely trying my best but this doesn't make much sense to me

[obsidian monolith]

soap:

[sleek pawn]

ok

you changed their signs

[tawny nacelle]

yeah

[sleek pawn]

but what about the -1 that used to be a positive 1?

[tawny nacelle]

as u said "changed their sign"

that +1 on the top became -1

cuz i factored out a -1

[sleek pawn]

lmao I'm dumb

ok but then it seems as if an extra sin^2(x) appeared out of nowhere

[tawny nacelle]

WHAT ???

[sleek pawn]

in the third step, when you separate the fractions and start to show the co-secant

[tawny nacelle]

bRuH

[sleek pawn]

wait

[tawny nacelle]

$\frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}$

[obsidian monolith]

soap:

[sleek pawn]

oh shoot I did it again

self-bamboozling 100

have to practice it like crazy

I'm an idiot hahaha

well thank you so much for wasting your time on me

it really helps and I appreciate it

bye

[tropic crown]

So I just proved this question by using conjugate

But tbh, I don't really know the definition of conjugate and why were allowed to use them on one side without having to multiply the other

Can anyone help explain?

[timid basin]

@tropic crown Can't really explain in detail rn, but if you look you'll notice that 1-cos/1-cos is the same thing as 1.

[tropic crown]

ahh thanks, make sense

but when u can, i would appreciate a full detailed explanation

[timid basin]

Yeah, sure.

[hard hornet]

i can tag in a bit

often, when you have things in the form of a+b or something like a-b in a very VERY nasty fraction

as above

what we often do is multiply by the conjugate (aka a+b is the conjugate of a-b, and a-b is the conjugate of a+b). reason why we do this is it makes the fraction a lot cleaner to work with

its also one of those tricks you develop

think about it in this case

lets say i wanted to simplify

1 / ( sqrt(3) + sqrt(5) )

without any square roots on the bottom

@tropic crown

what would you do?

[tropic crown]

multiply by the conjugate

so we have no sqrt on the bottom

[hard hornet]

exactly

why we do that in trigonometry?

because we have things called pythagorean triples

and in this case

1 + cos(theta)

it looks like we can get a pythagorean triple

or something in the form

if we multipled that with 1 - (cos theta)