precalculus

[viscid thistle]

hm that graph looks off

for 2 reasons

at least

one is that it should be going through the origin just from looking at the volume equation

[void quiver]

yeah

i drew it wrong

[viscid thistle]

and another is we have a domain of (0,5)

[viscid thistle]

so i think you only need to draw it for (0,5)

is that what it means by window?

[viscid thistle]

oh

[void quiver]

yeah

but i think she won't take off to many points for graphing

just the max volume

[viscid thistle]

honestly id only graph it for the domain

[void quiver]

which is 216.913

do you have a graphing calc on u?

[viscid thistle]

how do you know its 216.913

yes

[void quiver]

cause you do

2nd trace

then find max

you guess max and you go off

the y vavule it gives you

then for d you go off the x vauvle

to see how much you need to cut

which then is 2.15 x 2.15

units are in^3

[viscid thistle]

ok didnt realize this was a graphing calculator activity

[void quiver]

oh

yeah sorry

so am i right or wrong?

[viscid thistle]

yea it looks right

[void quiver]

then e is 1.15 in

cause it says at least

[viscid thistle]

how are you getting it from calculator

oh calc value?

[void quiver]

2nd trace

then you go to intersec

intersect*

[viscid thistle]

what is that doing?

[void quiver]

findin where it intersects

and shows you ho wmuch you shoud cut

i think i got that down

[viscid thistle]

oh did you put another equation in

[void quiver]

yeah

[viscid thistle]

ok i see

[void quiver]

so for y1

you pu t

put*

the (10-2x)(22-2x)(x)

then y2 you put 175

[viscid thistle]

yea i see

[void quiver]

ok

i need more help

if you don't mind

[viscid thistle]

sure

[viscid thistle]

ok for the first one

start with a side length of x

ok jeez i dont know how to explain without just giving the answer

[viscid thistle]

well that wont quite work

but ty now we can work with that instead

[void quiver]

wait so it would be the other way

[viscid thistle]

you can find the height of a triangle like that right?

[void quiver]

wouldn't it?

[viscid thistle]

not sure what you mean by other way

[void quiver]

so it would be like flip flop

the x woud be the base

[viscid thistle]

that wouldnt help

[void quiver]

and the x-3 would be the height

[viscid thistle]

doesnt actually change the height

just rotates it

[void quiver]

oh yeah

freak

ok so lets work with that

[viscid thistle]

couldnt you label the height of that triangle?

by drawing an altitude?

[void quiver]

see

i dont' think i've learn this

[viscid thistle]

you probably know what i mean if you have any geometry experience at all

[void quiver]

see

i had a perverted geo teacher

that didn't teach

are you supposed to use hypothiense thingy

[viscid thistle]

its a line that is perpendicular to a side of a triangle and crosses through a vertex of the triangle

a side and vertex that are opposite of each other

[void quiver]

yeah

im lost

like lost lost

[viscid thistle]

ok well we want something kind of like the middle one

lets call the bottom side of the triangle x

[void quiver]

?

[viscid thistle]

not quite but almost

that height is 3 greater than its base

h=x
b=x-3 there

[void quiver]

wait so the basse is good?

or the base half would be

b=x-1.5?

[viscid thistle]

the base is supposed to be x

or...

it could b x+3 i suppose but i think that will make computations slightly more annoying

[void quiver]

so what do i do next?

[viscid thistle]

or the base could be x-3 and height x-6

well one other thing is that side length on the right isnt x

that side length wont matter actually

we just need to fix the base and height

[void quiver]

could you draw a diagram?

[viscid thistle]

https://i.imgur.com/rENRIXe.png

[void quiver]

AH

ok so what next?

1/2x?

[viscid thistle]

actually this can maybe be simplified

[void quiver]

well

if you are trying to do the

wat can we call?

or no go

[viscid thistle]

https://i.imgur.com/PvCJSCA.png i think we can just do this instead lol

[void quiver]

yeah that works

then wut

[viscid thistle]

but the altitude approach would work as well if we assumed the triangle to be isosceles

its just a triangle area calculation

$\frac{bh}{2}$

[obsidian monolith]

Botnuke:

[void quiver]

so A= x*x-3

?

[viscid thistle]

over 2

[void quiver]

and thats the equation?

[viscid thistle]

yes

[void quiver]

so

[viscid thistle]

that diagram doesnt lead to that equation in one step

because it would be 2 separate triangle areas added together

each with a base of x/2 and h of x-3

[void quiver]

yeah

i see now

[viscid thistle]

and also probably better to replace x with b in the final equation

[void quiver]

wait so

A= (x)(x-3) / 2

[viscid thistle]

since b=x and it kind of asks the question in terms of b

[void quiver]

so

A = (x)(b-3) / 2

?

[viscid thistle]

replace x with b

[void quiver]

oh

ok so

a + (b)(b-3) /2

a=*

[viscid thistle]

yea

sorry i made that more complicated than needed when we started

[void quiver]

now to b LMFAO

its fine

I NEED TO LEARN

FOR THIS TEST

ITS WORTH 40%

I THANK YOU SO MUCH

so

A triagnle whose height is the square root of its base

what now

[viscid thistle]

we can just construct this in our head really

no need to draw anything

lets call the base b

what do we want h to equal?

[void quiver]

the sqaure root of the base

[viscid thistle]

yes

and the area is $\frac{bh}{2}$

[obsidian monolith]

Botnuke:

[viscid thistle]

so you can just substitute that in for h

[void quiver]

so

A= (B)(sqaurerootB) over 2

[viscid thistle]

yes

[void quiver]

OH

I

M

SMART

sorry for bothering u

theres so much sutff

[viscid thistle]

alright lemme google cone volumes first

[void quiver]

uh

[viscid thistle]

dont need?

[void quiver]

thats the problem i go off of

so it would be

(1/3)(pie)(1.5)^2(1)

right

[viscid thistle]

what would be?

[void quiver]

the volume

wait i think i got it

its fine

but i need help on this

Write an equation for your new salary after receiving a 2.5% raise.

[viscid thistle]

if x is your salary, 1.025x is your new salary?

[void quiver]

so

is it?

[viscid thistle]

if thats the whole question then id think so yea

[void quiver]

lemme show u

[viscid thistle]

ya i think thats all it wants

[void quiver]

so like

[viscid thistle]

let x be your salary, 1.025x is your new salary

[void quiver]

but lke

wait

why is it 1

isn't it supposed to be

0.025

[viscid thistle]

thats the increase

you have 1x salary right

and we add .025x salary

[void quiver]

OH

[viscid thistle]

1x+0.025x = 1.025x

[void quiver]

ic

okok

i think

im done

besides the grain problem

if you can do it and tell me the asnwer and compare

then sure

[viscid thistle]

ya sure lmk when you get an answer

i did it out earlier

[void quiver]

OH

while ur doin git

can you check my asnwers?

[viscid thistle]

if my fortitude lasts long enough

[void quiver]

thats it

on confident on the rest

ok what you get for the grain?

[viscid thistle]

you first

[void quiver]

66.356

ftin3

i think i missed a step

[viscid thistle]

how can the final answer be a volume?

it asks for the height of the cone

[void quiver]

wait

so i did miss a step

OOK

ait so is

66.356 on a good track?

[viscid thistle]

not sure i dont remember seeing one but it could be

[viscid thistle]

uh

i think you might have missed this

[void quiver]

oh

you you put

4 for h?

[viscid thistle]

the initial volume should use 4 as h yea

[void quiver]

ok

so

[viscid thistle]

keep in mind there are 2 h's in the equation not just 1

[void quiver]

9.424.

[viscid thistle]

that seems way too low

[void quiver]

is that wut you got for the first one?

well

you add the

60?

[viscid thistle]

i think you missed putting in a 4 somewhere

[viscid thistle]

what does the radius equal?

[void quiver]

wait

so r would be

so r would be

6

?

[viscid thistle]

yes, r=1.5h and h=4

[void quiver]

AHHAHA

IM SMART

150.796

?

[viscid thistle]

i think that is what i got yea

[void quiver]

so

210.796

[viscid thistle]

where did you get that?

[void quiver]

you add the

the rate

after 90 mins

so

[viscid thistle]

how are you getting 60 from that?

[void quiver]

(rate)(time)

the rate is

2 feet per min

and its after 80 mins

[viscid thistle]

so, 160?

[void quiver]

so 2 x 80

oh fuck

im stupikd LMFAO

310.796

[viscid thistle]

yep

[void quiver]

ok then

since ur solving for height

r/h

which is

[viscid thistle]

we want to find the height when the volume is 310.796

[void quiver]

r=6/4

[viscid thistle]

well

r=1.5h

important detail

not just 1.5

[void quiver]

310.796=1/3pie(6/4h)^2(h)

right

then solve for h

[viscid thistle]

r gets squared for cone formula doesnt it?

[void quiver]

yeah

i forgot to hadd

add*

[viscid thistle]

ok that looks good i believe

solve for h from there and you should be done

[void quiver]

so

h=5.827

?

sorry

forgot the 5

[viscid thistle]

im not sure actually i didnt throw the final form into calculator lol

[void quiver]

wait

so the

pile will be

5.827ft^3 in 80 mins?

[viscid thistle]

if this comes out to h=5.827

then the pile is 5.827 feet after 80 minutes

[void quiver]

thats

it

IM DONE

HOLY SHIT

[viscid thistle]

(not ft^3 btw)

[void quiver]

oh

jsut feet

[viscid thistle]

yes its only a height

not a volume

[void quiver]

oh ok

watch i get every question

wrong on this hw

[viscid thistle]

thats so funny to think about

[void quiver]

its ad

i have a 89

and i odnt wanna take finals

[viscid thistle]

i mean thinking about how it could be possible i helped in all the wrong ways leading you to all the wrong answers

[void quiver]

wow

rude

DISHONORABLE

[viscid thistle]

given i think i have helped you arrive at correct answers

its funny to think about how its possible they could be all wrong

[void quiver]

checking on my notes

it makes sense

but did you check 3

[viscid thistle]

this is one that can be easily checked by yourself

just multiply out your final answers

and see if they add to 150

or 149.999 for rounding

[void quiver]

OK

THANKS

IM

DONE WITH EVERYTHING

HOLY

TOOK

2 30 MINS

[viscid thistle]

ok im off to do math for 5 hours cya and ty for the ti84 help

[void quiver]

oh

well you helped me more than I did

but thanks tho

and your welcome

[sick lava]

How do i find solution set for sin x = .31

[gilded prawn]

Can someone help me out with number 4

Genuinely stuck

[uncut mulch]

draw a triangle

[gilded prawn]

Oh frick dude thank you

I forgot about that

[fluid shore]

Lmao ‘draw a triangle’ is probably the most commonly given advice and it’s possibly the most commonly ignored advice for such problems

[trim fable]

hii

@uncut mulch remember mee

i have a question how would u do this

coz u cant change 48 to a common base of 2

im confused

[uncut mulch]

didnt we do something similar or this exact question last time?

start with manipulating the left side

are you able to factorise it?

[trim fable]

ye i think but then i stopped

i think

uh

😦

[uncut mulch]

iirc the hint I provided was
a^(m+n) = a^m * a^n

[trim fable]

oh

(2^x)(2^2)

@uncut mulch right?

[uncut mulch]

yeh
2^(x+2)= (2^x) * (2^2)

can you continue?

[trim fable]

uh

(2^x)*(2^2)-2^x=48

but what do i do with the 48?

[uncut mulch]

nothing yet, try factorisingbthe left side

after splitting the exponent like that, the next step should be clear

[trim fable]

factor the 2?

[uncut mulch]

not quite

you can factor more than just 2

[trim fable]

2^x

[uncut mulch]

yes

[trim fable]

but

2^2

doesn't have x

[uncut mulch]

the thing it's being multiplied to does

[trim fable]

oh true

so then why didn't

we just leave it like the

original question

and just factor in

the first place?

[uncut mulch]

you could, but if you were struggling to see stuff, you need this step

[trim fable]

oh

yeah

true

thank uuuu

so the thing with

these types of questions

is that u have to try to work with one side

then

isolate?

ok

so can i do this one with u

so u can

see if

[uncut mulch]

sure

[trim fable]

i get it

ok so

(4^x)(4^3)+4^x=260
4^x(4^3+1)=460
4^x=4
x=1

[uncut mulch]

yep

[trim fable]

yayyyy

[trim fable]

@uncut mulch 😛

[uncut mulch]

what have you thought about?

[trim fable]

what i did so far is

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

then u know how its 3x so isn't that just

(3^3)^x

[uncut mulch]

put parentheses in appropriate places

[trim fable]

uh

[uncut mulch]

$3^{3x} \cdot 3^{4x -2} = 3^{x+4}$

[trim fable]

oof

[obsidian monolith]

ramonov:

[trim fable]

ye

so factor a 3?

so like

3x+4x-2=x+4

then solve for x

[uncut mulch]

yeh that works

[trim fable]

ok yay

[upbeat prairie]

what does it mean here

"root of a expression"

why did it specifically choose 2/1

[fleet yew]

because it is a root of the expression

lol

[trim fable]

question

for this

would this be right to isolate for x

16^x+2y =256
2^4x+8y=2^8x-8y
16y=4x
4y=x

[viscid thistle]

@upbeat prairie you pretty much just have to guess, but of the possibilities that the rational root theorem gives, if there are rational roots, they will be of that form

and there are usually not very many to check

so you can try 1/1 or 13/1, you just might find that they arent roots

or -13/1 or -26/1

etc

[fleet yew]

@trim fable wrong

[trim fable]

oh what

why

what mistake did I make?

[fleet yew]

how did you get 256

[trim fable]

uh

i gotfot to write

256^x-y

i multiplied 32 by 8^x-y

[fleet yew]

$32*8^{x-y}\neq256^{x-y}$

[viscid thistle]

needs {}

[obsidian monolith]

AMD:

[fleet yew]

good enough

you can't do that because of the order of operations

[trim fable]

oh ok

[fleet yew]

there is a natural order to things that mustn't be violated

[trim fable]

how would we do this

ye i figured that one out i complicate things when u can easily simplify lol

i make my life harder

[fleet yew]

looks like they just want you to use a calculator

given the rounding

[trim fable]

oh

so

how would i put it in

like i dont

[fleet yew]

you still have to isolate x

[trim fable]

rearrange

oh ok

question

(10^3)^x

?

then

1000^x=415?

[fleet yew]

it's 10^3x

not 10^3^x

[trim fable]

ye but

cant u break apart 3x

coz powers get

multiplied

so do that?

but then u would multiply 10 too? coz its in the brackets so nvm?

[fleet yew]

just use the log lol

[trim fable]

oh..

question tho

how do u know

when to use log

i dont really get the concept of logs

[fleet yew]

no you're right mb you can do 1000^x=415

you can

but it's not the most convenient way to do it

[trim fable]

oh

[fleet yew]

take the base 10 logarithm of both sides

so 10^3x=415

3x=log(415)

[trim fable]

wait

what law is that?

or did u just do log

coz i get so confused

[fleet yew]

just take the logarithm of both sides

[trim fable]

oh

can i be honest

i still dont get why

we do that

we just add a log in

[fleet yew]

think about it this way

if i tell you that 3x=15, how do you solve for x

[trim fable]

divide by 3

to isolate for x

so x =5

ik that\logs are

the inverse of

exponentials

which kinda gets confusing coz like

why do u have to write the word log

what does log itself mean

like ik its an exponent

technically

but why cant u use a variable? lol

[fleet yew]

you can't use a variable because it's an operation

[trim fable]

hmm

so like

but with exponential u just

raise the base to a power but

like we dont write

exponential of something

so like the log itself makes it confusing

[fleet yew]

it's just mathematical convention

[trim fable]

ok

ill accept it

lol

[fleet yew]

it's not based on any divine law or anything

[trim fable]

oh

then

how did log become known as log?

[fleet yew]

it's short for logarithm

[trim fable]

yeah

ik

lol

[fleet yew]

i think it's from arabic or smth idk

[trim fable]

wow

[fleet yew]

no wait it's greek

[trim fable]

oh wow

so

how would u solve that

LOL

[fleet yew]

so 10^3x=415
3x=log(415)

[trim fable]

why LOL

anyone here

rip

[wise kelp]

What's the problem?

[trim fable]

heyy

uh

how did we get from

10^3x=415

to 3x=log(415) 😅

[wise kelp]

What does log mean?

[trim fable]

oh

logs..

logs are logs ;-;

logarithm

[wise kelp]

Do you know what a logarithm is

[languid crane]

@trim fable inverse of exponential

just like multiplication and division

or

in calc, differentiation and integration

[trim fable]

oh

[languid crane]

etc

[trim fable]

oh

ok

how do u know

how to write it out tho

for that question

[languid crane]

10^(3x) = 415?

Use properties of logs

[trim fable]

yeah

oh ok

[red tree]

For any invertible function $f$,[f(3x)=415\implies 3x=f^{-1}(415)]

[obsidian monolith]

Icy001:

[languid crane]

yea

[red tree]

(and vice versa)

[stark willow]

so $log_b(xy)=log_bx+log_by$
is there any identity for
$log_b(x+y)$
i couldn't find any on rapidtables.
much appreciated.

[obsidian monolith]

tryingtolearn:

[willow bear]

no there isn't

[stark willow]

thank you

[pseudo sonnet]

is this solvable without a calculator

if so, how

[torn swift]

would still need a trig table to tell you what arccos(-3/5) is

[pseudo sonnet]

well cos = xr

x/r **

then u find the Y value

so ud need a calculator?

[viscid thistle]

look at the title of the exercises

but yea you would need a calculator to get a very accurate answer

[viscid thistle]

https://gyazo.com/b507ab88e791d2fc8da3d3029b451144 how would i reduce this?

its not possible right?

i leave it as that right?

i don't think i can simplify that any more lolz

[undone pawn]

yeah you're good

[uncut mulch]

depends how they want it. you can rationalise the denom using sum of 2 cubes if required

[meager hemlock]

im not sure if this is calc or precalc but calc is being used anyway so

i'm totally lost on this question https://cdn.discordapp.com/attachments/559769393995841543/654593094603243532/20191212_015852.jpg

where should i start with it

[willow bear]

for each point, write out the sign of f at that point

then do the same to f'

then do the same to f''

[meager hemlock]

what..?

[undone pawn]

@meager hemlock

what Ann said^^

pretty easily you can see that f(x) is negative everywhere

and you know that f'(x) can be deduced from the slope of the line, which is tan(angle made with x axis) so using that you can see if f'(x) is negative or positive at each point

and I think you can find f''(x) sign by observing the concavity at the point

so use all that to answer the Q

[meager hemlock]

i can't really see that super easily? it raises up and accelerates up in a few places

[willow bear]

what part of what i said is unclear?

[meager hemlock]

because i'm really bad at math and don't understand anything lmao

idk the sign of f

[willow bear]

do you know how to tell the value of a function at a point from its graph

[meager hemlock]

I... don't think so?

[undone pawn]

hm

[willow bear]

wait what

aight let's forget this questoin for a moment

here's the graph of a function, g

[meager hemlock]

i think im misreading you

lol

[willow bear]

are you telling me you would be unable to say what g(1) is

from this graph

[meager hemlock]

its 1

[willow bear]

how do you know

[meager hemlock]

there are no numbers on the picture i gave you

[undone pawn]

they didn't ask for numbers though

only sign

[meager hemlock]

casue that's what the y value is at the same point

ohh like positive or negative sign?

[undone pawn]

yeah.. what else would they be asking

[meager hemlock]

well i dont know

doesnt mean i know what they are asking

[undone pawn]

well now you do

[meager hemlock]

cool

i still dont know what to do with that though

okay.. so if its negative its not negative anymore?

now what

[undone pawn]

okay.. so if its negative its not negative anymore?

what?

[meager hemlock]

when you say remove the sign i assume you mean if its a negative y value you just make it positive

all of the points are below the x axis

[undone pawn]

who said remove the sign?

what are you talking about

[meager hemlock]

what else does "write out" mean

[undone pawn]

she meant write down*

as in note the sign of f at each point

[meager hemlock]

if the value of the function is increasing from that point, it's positive right

[willow bear]

what's "it"

[meager hemlock]

the selected points

[undone pawn]

the point is positive??

how can a point be positive or negative

[meager hemlock]

right

god damnit i dunno sorry

[languid crane]

is this precalculus wth

[eternal flume]

if I need to find all x, for which this is valid: log0.5 x is more or equal to log0.5 2
can I say that x can be in this interval (0; 2>

[willow bear]

$\log_{1/2}(x) \geq \log_{1/2}(2)$?

[obsidian monolith]

Ann:

[willow bear]

is this your ineq?

[eternal flume]

yes, exactly, thanks 😄

so to make this valid, the x must be in an interval (0; 2>, right?

[willow bear]

are you including 2 itself or not

[eternal flume]

I'm including 2

[willow bear]

why are you not writing (0, 2] then

[eternal flume]

Well, I'm not sure

We're using > in math the way I posted it

[willow bear]

...

where are you even from

this is legit the first time i've seen that notation used like that

[eternal flume]

Middle Europe

Really, first time? 😂

Very common in my country 😂

Anyways, back to the question

is then (0,2] a correct answer?

[willow bear]

seems so

[eternal flume]

thanks 😄

[upbeat prairie]

How are the numbers between -8 and -7 going to be positive for the zeroes?

-7 for example in x - 7 would just equal -14

[valid violet]

What does "positive for the zeroes" mean @upbeat prairie

[nova dew]

$\lim_{x\to\infty} \frac{n^2 + n}{n^2 + n + 1}$

[obsidian monolith]

CoolShot:

[nova dew]

How would one evaluate this

The answer is (I'm guessing) supposed to be 1? But how

[rigid beacon]

You mean $\lim_{n\to\infty} \frac{n^2 + n}{n^2 + n + 1}$ ?

[obsidian monolith]

Spamakin🎷:

[nova dew]

yes sorry

[rigid beacon]

Ok

Try partial fractions

Oh wait

No no nvm

Look at the terms of largest power in the top and bottom of fraction

[willow bear]

lol imagine using partial fractions to evaluate a limit

[rigid beacon]

Ok so LaTeX failed but

I was thinking integrals for some reason @willow bear

But yea @nova dew do you remember any properties of limits to infinity

[willow bear]

[rigid beacon]

Of rational functions

[nova dew]

it's Q/Q

oh

ok nice

thx

I'm retarded

[rigid beacon]

You good

So the answer is?

[nova dew]

¹

[heady jewel]

no 1

[nova dew]

y no ¹

[whole fox]

How do I do part c I think I've done it wrong

And I dont understand what I've done

<@&286206848099549185>

[whole flint]

you done it right

[whole fox]

oh

but i dont know why haha

whats the difference between what i calculated

https://gyazo.com/b9bef17ccfb4c7b5a5288eb1581ec4ba

and the difference between the point and the green circle

[pure flame]

$Tan(a+b) = (Tan(a) + Tan(b))/(1+Tan(a)Tan(b))$

Hi guys! Could you demonstrate me this?

[lost pawn]

Write tan(a+b)=sin(a+b)/cos(a+b) then expand and simplify

[obsidian monolith]

FNDN:

[pure flame]

I've done that but I didn't get the right.

I'll try it again.

Now, I've realized that I can separete the fraction.

[vestal marlin]

What is the lim as x approaches 4 from the positive

[valid violet]

@vestal marlin the question doesn't make sense.

x can't approach 4 from the right

[vestal marlin]

It was on my hw so

No answer?

[valid violet]

so the answer is to explain to the reader why the question doesn't make sense

[vestal marlin]

Or dne?

[valid violet]

explain why the question doesn't have an answer

[trim fable]

question

how would u do this

anyone?

[normal sundial]

@trim fable let y = 3^x and see if you can solve for y

[trim fable]

nvm

i got it

😛

yeahhhh

thankssss!

[cerulean cape]

Heyo

I need to know how many coin flips it would take to be 50% and 90% likely to hit 7 total heads

ah, 13 for 50%

and 18-19 for 90% I think

[trim fable]

how would u do

4c

[wise kelp]

I think you can prove it using the change of base rule

[trim fable]

ye

ik

i was able to do the first two

but idk how to do c

@uncut mulch hiii can u help me 😛

[wise kelp]

@trim fable Use the change of the base rule and the rule that log(a^b) = b log(a)

[trim fable]

yeah

what i did so far was

logx(x)=1000x^2

now idk what to doooo

[wise kelp]

Uh

Are we looking at the same problem

[trim fable]

yes

[wise kelp]

Can I see how you got to that

[trim fable]

oh wait

oops

im talking about this ONE LOL

[wise kelp]

Ahh

Lemme see

[wise kelp]

Yeah, see if you can get anywhere with that

[viscid thistle]

Sin Ø = (root 3)/2, Ø = pi/3 and 2pi/3 right?

[stuck lark]

Yes if phi is restricted to between 0 & 2pi (btw don’t use empty set for phi)

[viscid thistle]

Thx

[slow roost]

how is sec29pi/4 equal to -root2
and not root 2

http://tutorial.math.lamar.edu/Solutions/CalcI/TrigFcns/TrigFcns_Files/image011.gif

this is the thing i got

so like

for the "bigger" trigs

if its 29pi/4

i would break it down to

28pi/4 + pi/4

so then pi/4 is where i look

and i have to look diagonal from that?

to find the right quadrant

bc i was under the impression that if it was pi/4 then the answer would be positive cause (+x,+y)

but in the pic, it shows you using pi/4 as a director to the quadrant of 29pi/4

[weak pagoda]

have u tried simplifying to stay in the quadrant?

instead of just breaking it straight down

[slow roost]

isnt 28pi/4 +pi/4 simplifying?

i only now how to break it down

[weak pagoda]

ye it is.

[slow roost]

ye ok

so then do i take the sign of whatever the coordinate pi/4 is

or do i look across

like diagonal

cause that seems to be working on every problem i try now

cause normally, if i got pi/4

i would be like ok so then is its cos, then it would be positive root2/2

but in this case, theyre showing to look across the pi/4 quadrant

so thats quadrant 3

[weak pagoda]

u want to stay in quad 3

[slow roost]

yes

ok

[weak pagoda]

with sec

[slow roost]

mhm

ok can i give u a sample problem and can u confirm if my thinking is right

[weak pagoda]

sure

[slow roost]

tan -121pi/4

would be quad 4

wait

or is it quad 2

[weak pagoda]

well shit now u got me trying to find it

because i see quad 2 but everything is split up for reasoning

quad 2

[fluid shore]

$\frac{121\pi}{4} = 30\pi + \frac{\pi}{4}$

[obsidian monolith]

Abhijeet Vats:

[fluid shore]

Lmao

[weak pagoda]

ye

but the negative

tan

[fluid shore]

Yea, just account for that

[weak pagoda]

$\frac{-121\pi}{4} = 30\pi + \frac{-pi}{4}$

oof wait

'

[slow roost]

yeah its quad 2 right

[fluid shore]

You have -pi/4 below the x-axis and you're rotating it clockwise 15 times in a full circle.

[slow roost]

quad 2

and u get like -1

cause its (-root2/2,root2/2)

and since its tan

you divide it

and it cancels and left w -1

y/x

but if u think of it in terms of quad 4

it results in the same answer

however

the first question i asked

is not

sec29pi/4

it would be quad 3

oops

yes

quad 3?

so then x is -

so its 1/root2/2

which is -root2

whereas if u said it was in quad 1, the answer would be root2

[fluid shore]

$\frac{29\pi}{4} = 6\pi + \frac{5\pi}{4}$

[obsidian monolith]

Abhijeet Vats:

[fluid shore]

Okay, it's very simple

^The idea is to make it something like the above

[slow roost]

uhuh

[fluid shore]

Notice that 6pi is a multiple of 2pi?

[slow roost]

yes

6pi * 4

gives u 24pi

[weak pagoda]

you want to simplify to keep it in the same spot

[slow roost]

so multiply by common denominator?

like the 6pi

cause 5pi/4 is in quad 3 if im not wrong

[fluid shore]

So, if you imagine a line joining the origin to a given point, then the 5pi/4 is the angle that the line should make with the x-axis. Then, any multiple of 2pi added to that will just be equivalent to a given number of rotations

Lmao, just calculate it

What's pi/4 in degrees?

[slow roost]

45

[fluid shore]

What's 5*45?

[slow roost]

225?

225 in

[fluid shore]

So which quadrant is that in?

[slow roost]

quad

3

[fluid shore]

Yea

[slow roost]

interesting

but how do u come up with the

5pi/4 tho

like

what im thinking is

you asked yourself what 29/4 was

closest is 6

so thats 6pi right

[fluid shore]

Ye

[slow roost]

leaving u with a remainder of 5/4

[fluid shore]

I was actively looking for a multiple of 2*pi

[slow roost]

ah wow! thats actually a nice way of thinking of it

but is my way incorrect?

thats what im trying to figure out

would it work in any situation

[fluid shore]

Uh i'm not sure. Describe it to me again?

[slow roost]

shit its hard to like

put in words

https://cdn.discordapp.com/attachments/363224154469826562/654894590066950146/unknown.png

ok

so that problem was sec29pi/4

the way i do it is

28pi/4 +pi/4

pi/4 would be in quad 1

[willow bear]

careful

[slow roost]

i would look-

[willow bear]

7π isn't a whole number of full turns

[fluid shore]

Yes

[willow bear]

so 29pi/4 won't be in the first quad

[fluid shore]

The idea is to look for a multiple of 2*pi

[slow roost]

yeah i get that

[fluid shore]

But it's a good start.

[slow roost]

ok

can i try w a diff problem

[fluid shore]

Sure

[slow roost]

er lets se

see

tan-23pi/6

would that be like

-3pi -5pi/6?

[willow bear]

it'd make more sense to write it as -4pi + pi/6

[fluid shore]

Is 3pi a multiple of 2pi?

[slow roost]

ah

ok

[willow bear]

but tan is pi-periodic

[slow roost]

mhm

ok so -4pi + pi/6

[weak pagoda]

you want full rotations to be taken out

[willow bear]

so you don't need to look for EVEN multiples of pi for tan or cot

for tan half-rotations are fine

[slow roost]

pi/6 is in quad 1

so is it quad 1?

or

[weak pagoda]

dont throw multiples and words out there. big angles = full rotations + leftovers

[slow roost]

got it

full rotations

full rotations are multiples of 2pi

[weak pagoda]

oh yea tan half rotations work but ya still learning this part.

yup

[slow roost]

so its in quad 1?

or nada

[fluid shore]

I mean, it's better to understand the general principles rather than learning specific things

But whatever lel

What do you think?

[slow roost]

er i think it is cause pi/6 would be in quad 1

so the answer would be root3?

[fluid shore]

Are you sure?

[slow roost]

wait

o shiz

is it root3/3

[fluid shore]

Idk, you tell me

[slow roost]

er

yes

i believe it is

[fluid shore]

You believe it is or are you sure it is?

[slow roost]

oml

im sure it is

[fluid shore]

Be a bit more confident with your answers. Confidence takes you quite far.

Good.

[slow roost]

ok

er can i try one more

o

cos 19pi/4

4pi + 3pi/4

3pi/4 is quad 2

cos will be negative

-root2/2

[fluid shore]

Are you sure?