precalculus

[harsh cipher]

Then used double angle identity tangent

[uncut mulch]

what's your tan(x)

[harsh cipher]

-sqrt(21)/2

[uncut mulch]

what's the first thing you got after applying the dbl angle identity

[harsh cipher]

(-2sqrt(21)/2) / (1)+(21/2)

[willow bear]

parentheses fuckup again

[harsh cipher]

llollllll

where?

[uncut mulch]

around the denominator which is also wrong (even if there were parentheses)

[willow bear]

this reads as $\frac{-\frac{2\sqrt{21}}{2}}{1} + \frac{21}{2}$

[obsidian monolith]

Ann:

[harsh cipher]

oh yeah that's wrong

[willow bear]

i'm tempted to make a problem set which has formulas written in plaintext and asks students to determine their correct translation to proper mathematical text

[harsh cipher]

using what

[willow bear]

LaTeX

[harsh cipher]

ok so then...why is my denom wrong

ohhh

denom is

1-(21/4) ?

[uncut mulch]

that's better

[harsh cipher]

oh wait

denom has to be 1-(-21/4)?

other wise I get -4sqrt(21)/17 as my final answer

[uncut mulch]

when i said "that's better", i implied that you had it right

[harsh cipher]

hmm okay

[uncut mulch]

how are you getting -4sqrt(21)/17?

[harsh cipher]

(-2sqrt(21)/2)/(-17/4)

oh no

-8sqrt(21)/(-34)

is 4sqrt(21)/17

[uncut mulch]

a little inefficient but you got there 🙂

[harsh cipher]

haha

yes!

thank you for the help

the parentheses thing actually helps

hahahaha

[uncut mulch]

something like -2sqrt(21)/2 can be simplified to -sqrt(21) a lot earlier

[harsh cipher]

right, but I'm not at that stage yet. I will get used to doing that in the future.

Thanks!!!

@willow bear Thanks!

[viscid thistle]

Tan(3a)^2(1+cos(6a)) what should i do first? Open the brackets by multiplying?
Or use tan(3a)^2 as sin(3a)^2/cos(3a)^2 ?

[uncut mulch]

consider simplifying (1+cos(6a)) first

[eternal flume]

Hello, how can I solve this equation?
sin2x = tgx
I was able to calculate this
sinx (2cos^2(x) - 1) = 0
, but I do not know how to continue.

[viscid thistle]

Simplified it should be 1+cos(3a)^2-sin(3a)^2.
Rearrange: 1-sin(3a)^2+cos(3a)^2=cos(3a)^2+cos(3a)^2=2cos(3a)^2
From tan(3a)^2=》sin(3a)^2/cos(3a)^2 × 2cos(3a)^2=sin(3a)^2×cos(3a)^2=2sin(3a/2)^2×2cos(3a/2)^2=(1-cos(3a))(1+cos(3a)=1+cos(3a)-cos(3a)-cos(3a)^2=1-cos(3a)^2

Thats the answer I got. Correct answer should be 1-cos(6a). Can somebody tell me where i went wrong

[warped dagger]

Why is x (va) = 1, x = 2, & x = 4, i just dont understand how he got those values?

[jagged glade]

@warped dagger do you understand what does log with base 3 means?

[warped dagger]

Yes, i do

[willow bear]

@warped dagger the other two are just easy-to-plot points on a set of axes

[warped dagger]

my bad, brainfart

[jagged glade]

No problem😀

[harsh cipher]

Hi

Question

-(csc^2x)= (2)/(cos2x-1)

left side is equal to (-1)/(sin^2x)

the right side

(2)/((1-2sin^2x)-1))

,help

[obsidian monolith]

A brief description and guide on how to use me was sent to your DMs! Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[serene heath]

@harsh cipher yes

[harsh cipher]

@serene heath ty

[harsh cipher]

Hi, question again

cos(2x) = 1-(tan^2x)/(1+tan^2x)

then I have cos^2(x)-sin^2(x)/ cos^2(x)+sin^2(x)

which is Cos^2(x)-sin^2(x) /1

which is equal to Cos2(x)

left side = right side

[tacit bane]

Can anyone explain how to find the intervals of increase in this graph?

Of 1/h(x)

[fleet yew]

@tacit bane where is the function increasing

[tacit bane]

in 1/h(x) i see it increasing at the end behaviour at the left and between asymptote +1 and +3

reason i asked is because if i were to write it for h(x), lets say at point -2; (+infinity,-2)

but i dont know how to write it for 1/h(x)

[fleet yew]

@tacit bane i want you to put dots on all the bumps in the graph

Lets just look at h(x) for now

Because i dont think you understand what increasing means

[tacit bane]

Sorry I messed up I assumed anything above the x axis is increasing

I have drawn arrows at where I think it is increasing

[fleet yew]

Yes thats right

Yeah dont confuse positive with increasing lol

I used to do that a lot

[tacit bane]

i think i figured it out for 1/h(x)...so between asymptote -2 and 1 there is a bump. so (-2+1)/2 is the point at which it will increase then decrease?

[patent beacon]

If h is increasing, what is 1/h doing?

[tacit bane]

i think it also increases?

[fleet yew]

Draw arrows

Along the path of the function

You can think of increasing as "going up"

[patent beacon]

When h gets large, 1/h gets smaller. It's actually the opposite. 1/h is increasing when h is decreasing

[tacit bane]

Oh so @patent beacon is my graph for h(x) wrong?

@fleet yew i drew the arrows in orange

[fleet yew]

Wait this is a different function

One problem at a time lol

[tacit bane]

sorry i drew it for 1/h(x)

[patent beacon]

Oh mb. I see, you drew 1/h for the graph h to the right there

[tacit bane]

1/h(x) graph i mean @patent beacon

is it right?

[patent beacon]

So when h is negative, 1/h will be too. Check your signs

You have the right idea to set the asymptotes to 0

[tacit bane]

@fleet yew so did i understand increasing function yet?

oh so can you please pin point where i was wrong?

@patent beacon

[patent beacon]

You've correctly identified the increasing intervals on the graph you drew.

Your 1/h should always have the same sign as h

[tacit bane]

i assumed that the bumps would flip except for the end behaviour

[patent beacon]

At x = 0 for example. The two graphs have different signs, which is wrong

[tacit bane]

ohhh

okay i understand

[patent beacon]

But the idea is right, otherwise

[tacit bane]

Now?

[patent beacon]

Wait, which one are you supposed to set? Lol

[tacit bane]

the graph for h(x)

[patent beacon]

But yes that looks much better

[tacit bane]

oh, is there something that i must fix? when you asked which one i was supposed to set?

[patent beacon]

I assumed you were given h, and had to graph 1/h

Yeah, nvm I'm getting loopy

[tacit bane]

ah yes, i was given h(x) and i had to draw 1/h(x)

thanks so much! @fleet yew and @patent beacon

[fleet yew]

@tacit bane no problem

Just remember for the love of God

[tacit bane]

yes increasing is not the same as above the x axis

[fleet yew]

Do not confuse "positive" with "increasing"

Or "negative" with "decreasing"

[tacit bane]

i had a feeling you would say that

[fleet yew]

Yeah im just projecting i guess

Because i made that mistake like a million times myself back in alg 1

[tacit bane]

thanks so much! i learnt something very important today

[fleet yew]

👍

[harsh cipher]

Hi

I have a question.

[fleet yew]

I have an answer

[harsh cipher]

lol

okay let me write the question

[fleet yew]

Provided that your question is within the scope of my knowledge

Which is fairly likely considering that this is the precalc channel

[harsh cipher]

(2cosx)/(sec^2x) = (1+cos2x)/(secx)

[vernal moon]

ya

whats the question

[harsh cipher]

can i write either side of the equation

in fraction form?

[vernal moon]

uh what

[harsh cipher]

I can only write the answer in identity form

[fleet yew]

$\frac{2cosx}{sec^2x}= \frac{1+cos2x}{secx}$

[harsh cipher]

(cosx)(1+cos2x)

[obsidian monolith]

AMD:

[vernal moon]

identity form?

[fleet yew]

Is this what you mean

Just being clear

Or do you mean cos^2x

[vernal moon]

i think cos(2x)

[stuck lark]

$\frac{2\cos(x)}{\sec^2(x)}= \frac{1+\cos(2x)}{\sec(x)}$

[obsidian monolith]

RokettoJanpu:

[fleet yew]

How do you make it black

[stuck lark]

,tex --color

[obsidian monolith]

Valid colour schemes are: transparent, trans_black, trans_white, default, gray, darkgrey, light, grey, black, white, dark

[vernal moon]

also why does he have an aleph in his name

[fleet yew]

Bro thats flipping my text input

It is weird

[vernal moon]

lmao

it makes sense

hebrew is right to left

[fleet yew]

But i dont like it

[harsh cipher]

okay so

[fleet yew]

What

[harsh cipher]

is there another proof

[fleet yew]

$\frac{2\cos(x)}{\sec^2(x)}= \frac{1+\cos(2x)}{\sec(x)}$

[obsidian monolith]

AMD:

[fleet yew]

Is this the equation

[harsh cipher]

yes

[fleet yew]

What are you trying to do

Prove the identity?

[harsh cipher]

yes

this is a question from double angle identities

[fleet yew]

@harsh cipher

Multtiply both sides by secx

[harsh cipher]

ok

okay I couldn't get the answer

1 more question

2/((-cos(x)/(sinx)) + ((sin(x)/cos(x))

$2\frac{-cosx/sinx}+{sin(x)/cos(x)}$

[obsidian monolith]

אewb64:
Compile Error! Click the reaction for details. (You may edit your message)

[harsh cipher]

oh no

lollllll

$\frac{2/(-cosx/sinx}+\frac{sinx/cosx}$

[obsidian monolith]

אewb64:
Compile Error! Click the reaction for details. (You may edit your message)

[harsh cipher]

hmmm

$\frac{2

$\frac{2}/frac{-cosx/sinx)}$

[obsidian monolith]

אewb64:
Compile Error! Click the reaction for details. (You may edit your message)

[willow bear]

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

[obsidian monolith]

Ann:

[willow bear]

is this what you were attempting to write @harsh cipher

[harsh cipher]

yes

not complete, but yea

I've been trying to solve this for the last 2 hours

$\frac{2}{\frac{-\cos(x)}{\sin(x)}}+{\frac{sinx}{\cos(x)}}$

[obsidian monolith]

אewb64:

[harsh cipher]

oh no

lol

hmmm

[willow bear]

uhh

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

[obsidian monolith]

Ann:

[willow bear]

i take it you rewrote the RHS as this?

[harsh cipher]

yes

but can you explain why you changed -cot(x) to positive

[willow bear]

what

[harsh cipher]

oh no you didn't nm

[willow bear]

anyway, did i understand correctly that in those two hours it did not occur to you that this is a nested fraction and as such it would help to multiply the outer num and denom by sin(x)cos(x) to get rid of the inner denominators?

to make it not a nested fraction anymore?

[harsh cipher]

it did but I could not simplify properly

[willow bear]

but what did you get upon doing that?

[harsh cipher]

something crazy

[willow bear]

can you show me what the "something crazy" is

[harsh cipher]

-cos(x)cos(x)/sin(x)sin(x) + sin(x)sin(x)/sin(x)cos(x)

[willow bear]

... what.

can you write that down on paper or something

[vernal moon]

Or latex

[alpine narwhal]

i think he meant to group the last two expressions

-cos^2(x)

[willow bear]

you should have gotten $\frac{2\sin(x)\cos(x)}{(\frac{\sin(x)}{\cos(x)}-\frac{\cos(x)}{\sin(x)})\sin(x)\cos(x)}$ prior to further simplification

[alpine narwhal]

divided by (sin^2(x)-cos^2(x))/sinxcosx

[obsidian monolith]

Ann:

[willow bear]

and the denominator SHOULD have rather obviously transformed itself into sin^2(x) - cos^2(x)

but something prevented you from doing what at least in retrospect ought to seem like a natural sequence of algebraic moves to get to this point

[harsh cipher]

okay

hmm

i got that

but what I did was since there was 2 in the numerator

[willow bear]

you wrote the 2 as 2(cos^2(x)+sin^2(x))?

[harsh cipher]

lolllllll

ahahahha

okay

then we have 2/sin(x)cos(x)?

[willow bear]

no?

[harsh cipher]

why?

[willow bear]

why would we.

[harsh cipher]

denom cancels sin(x) in the numerator?

[willow bear]

what.

are you trying to cancel the $\sin(x)\cos(x)$ in $\frac{2\sin(x)\cos(x)}{(\frac{\sin(x)}{\cos(x)}-\frac{\cos(x)}{\sin(x)})\sin(x)\cos(x)}$

[obsidian monolith]

Ann:

[harsh cipher]

yes

[alpine narwhal]

i’ll stop you there

[harsh cipher]

okay

[willow bear]

...

congratulations

[harsh cipher]

........

[willow bear]

you've tried to UNDO MY LAST STEP.

[harsh cipher]

LOLLLLLLLL

[alpine narwhal]

thats what i was tryna say

lmao

you know why you multiply the numerator and denominator by sinxcosx?

[harsh cipher]

no

[alpine narwhal]

well how are you going to condense the denominator?

[harsh cipher]

by multiplying ?

[alpine narwhal]

by?

[harsh cipher]

sinxcosx

[alpine narwhal]

yes

but whatever you do to the denom

you must do to the numerator

[harsh cipher]

did not know that

[alpine narwhal]

keep it equal

[harsh cipher]

I think there was an explanation about adding fractions with binomial denom

[willow bear]

i was not doing any fraction addition

[alpine narwhal]

whoops

[willow bear]

also

but whatever you do to the denom
you must do to the numerator
this is vague and misleading

[alpine narwhal]

sorry bout that
i meant whatever you multiply the denom by you must multiply the numerator by

[willow bear]

yes, and THAT'S WHAT I DID, and i tried to make it explicit

but failed to do so i guess

[alpine narwhal]

man im sorry im bad at this

[harsh cipher]

so.....

[willow bear]

aleph-ewb64, do you understand how i went from $\frac{2}{\frac{\sin(x)}{\cos(x)}-\frac{\cos(x)}{\sin(x)}}$ to $\frac{2\sin(x)\cos(x)}{(\frac{\sin(x)}{\cos(x)}-\frac{\cos(x)}{\sin(x)})\sin(x)\cos(x)}$?

[obsidian monolith]

Ann:

[harsh cipher]

yes you mutlipled, cos(x)sin(x)

to the numerator

[undone pawn]

$a = a\cdot 1 = a \cdot \frac{b} {b} = \frac{ab} {b} $

[obsidian monolith]

FlynnXD:

[undone pawn]

for non zero b

[willow bear]

no, i multiplied THE NUMERATOR AND THE DENOMINATOR by cos(x)sin(x)

[harsh cipher]

and then... you multiple sin(x) (cos)x to each fraction with different denom

[undone pawn]

what

[harsh cipher]

in the bottom

[willow bear]

and then i SIMPLIFIED THE DENOMINATOR

[harsh cipher]

yes

[willow bear]

by EXPANDING and CANCELING

[harsh cipher]

yes

[willow bear]

do you understand how $\paren{\frac{\sin(x)}{\cos(x)} - \frac{\cos(x)}{\sin(x)}}\sin(x)\cos(x)$ simplifies to $\sin^2(x)-\cos^2(x)$?

[obsidian monolith]

Ann:

[willow bear]

yes or no

[harsh cipher]

yes

I can do that

[willow bear]

okay great

so now

we have

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

do you see how to proceed from here

[harsh cipher]

No

[willow bear]

look at the numerator

what can it be rewritten as

[harsh cipher]

hmmmm

[alpine narwhal]

take a look at your trig identities

[harsh cipher]

is there an identity there?

[willow bear]

you either know it or you don't

consider what your goal is.

[harsh cipher]

I don't know

[willow bear]

-tan(2x)

bolded on purpose

[harsh cipher]

I still don't know

[alpine narwhal]

should tell you what identity to look for

[harsh cipher]

oh

[willow bear]

does this not scream "double angle identities"

[alpine narwhal]

try looking there

[harsh cipher]

sin2(x)/cos2(x)?

ummm

[willow bear]

[alpine narwhal]

boiling hot

[willow bear]

never write "sin2(x)"

[harsh cipher]

cos2(x)

ok

[willow bear]

or "cos2(x)"

the numerator is just sin(2x).

[vernal moon]

oml

[alpine narwhal]

wait

nvm

dont mind that

[willow bear]

the denominator is not cos(2x)

[alpine narwhal]

im an idiot

did your simplifying wrong

[harsh cipher]

okay

I'm not sure what -tan(2x)

I don't know what -tan(2x) is

[alpine narwhal]

well

[willow bear]

-tan(2x) is -tan(2x)

[alpine narwhal]

you were there

just think

[willow bear]

@harsh cipher tell me, what double angle identities do you know of

urgh

the fucking aleph you insist on having in your name

[undone pawn]

$\sin(2x) = 2\cdot\sin x\cdot \cos x$

[obsidian monolith]

FlynnXD:

[undone pawn]

there's no way to proceed unless you know this

[harsh cipher]

well I know the sin identity and the other cos identities

[alpine narwhal]

well

[undone pawn]

you dont know this one?

that I wrote?

[alpine narwhal]

you did your simplifying wrong

[undone pawn]

well you know sin(x+y) right

just sub in x=y in that

[harsh cipher]

i know that one

[alpine narwhal]

lets take a look at the denominator

[willow bear]

cos(2x) = cos^2(x) - sin^2(x)
do you not know this one

[harsh cipher]

i know that one too

[undone pawn]

similarly for cos(x+y)

[alpine narwhal]

well

[undone pawn]

oh

[alpine narwhal]

take a look at your denominator

[willow bear]

and if you do then how can you simplify sin^2(x) - cos^2(x)

[harsh cipher]

sin(2x)

[willow bear]

WHAT THE FUCK

[harsh cipher]

i mean

cos(2x)

lol

[willow bear]

NO

[undone pawn]

lmao

no

[alpine narwhal]

i got this

[willow bear]

NO

NO NO NO NO NO NO NO

[undone pawn]

chilll

[alpine narwhal]

here here

we got this

take a breather

[undone pawn]

yeah

[harsh cipher]

LMAO

[willow bear]

yknow that part in bohemian rhapsody

"no, no, no, NO, NO, NO, NO!!!"

that's me rn

[vernal moon]

mama mia mama mia

[willow bear]

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

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

[alpine narwhal]

rewrite the expression

[harsh cipher]

cos(2x)?

[vernal moon]

oml

[willow bear]

NO!!!!

[harsh cipher]

why

[willow bear]

IT'S NOT cos(2x)!!!

[harsh cipher]

hahahahahaha

[willow bear]

SUBTRACTION ISN'T COMMUTATIVE!!!! WHY WOULD YOU THINK IT IS!!!

[alpine narwhal]

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

[willow bear]

X-Y IS NOT THE SAME AS Y-X!!!!!!!

[alpine narwhal]

what do you notice about y=x-1 and y=1-x?

[harsh cipher]

they are not the same

[vernal moon]

..

[alpine narwhal]

well

[harsh cipher]

hold on let me rewrite Ann's question first

[undone pawn]

$8-3\neq 3-8$

[obsidian monolith]

FlynnXD:

[alpine narwhal]

5 is not -5

[undone pawn]

indeed it is not

[alpine narwhal]

now

[vernal moon]

but do you see a similarity between em..

[harsh cipher]

no

[vernal moon]

5 and -5

[harsh cipher]

5 does not equal -5

[vernal moon]

yes but

[alpine narwhal]

what do you notice about their relationship

[vernal moon]

ok listen if a-b=k what does b-a equal

[alpine narwhal]

here

[willow bear]

k

[alpine narwhal]

tryn so hard not to laugh

okay

gamemode back on

[undone pawn]

xD

cmon lmao shouldn't make fun

[alpine narwhal]

i cant help it

im on the ground

[harsh cipher]

same

hahahahahahha

[alpine narwhal]

okay now

you can rewrite 1-x as -x+1, right?

[harsh cipher]

yea

[vernal moon]

ok

thats progress

[willow bear]

-cos(2x). it's fucking,,,
MINUS cos(2x).
seriously, i am surprisappointed that this is even an issue.

[alpine narwhal]

me too

but

[vernal moon]

surprisappointed

thats an unexpected portmanteu

[alpine narwhal]

anyways

now that’s done

[harsh cipher]

okay....its -cos(2x)

[alpine narwhal]

you see how to get to -tan(2x)?

you have your numerator

sin(2x)

and your denominator

-cos(2x)

one more step

[harsh cipher]

okay

[willow bear]

still like

bruh

[alpine narwhal]

me too

[willow bear]

[alpine narwhal]

i am assuming you got your answer

[harsh cipher]

no way

[alpine narwhal]

is that a joke or

an exclamation

[vernal moon]

...

[alpine narwhal]

please tell me its an exclamation

[willow bear]

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

[obsidian monolith]

Ann:

[willow bear]

come on. one last step.

you can do it.

[alpine narwhal]

what else can you write -tan(2x) as?

[harsh cipher]

-sin(2x)/cos(2x)

[willow bear]

this is basic algebra man

[harsh cipher]

no?NO!?

plz don't say oml.....

[willow bear]

yes -tan(2x) = -sin(2x)/cos(2x)

but like

bruh. algebra 101

[harsh cipher]

thank you all

[alpine narwhal]

no problem

[harsh cipher]

really appreciate the help.

[alpine narwhal]

thank @willow bear

[willow bear]

well thank you for that incredibly sarcastic remark

:|

[alpine narwhal]

i mean literally

i shouldve said shoutout

but

[harsh cipher]

did you really fall on the ground laughing though

[willow bear]

i was talking about aleph-ewb's remark

[alpine narwhal]

oh

and yes

[harsh cipher]

hahaha

[alpine narwhal]

if a-b=k

[harsh cipher]

laughing is the best medicine

[alpine narwhal]

then b-a is

[harsh cipher]

-k

[vernal moon]

yay

[alpine narwhal]

okay at least you learned something today

congrats

you get a medal

[harsh cipher]

lmao but Flynn's explanation nailed it for me

somewhere up there

[undone pawn]

[alpine narwhal]

the a1=a(b/b)?

[harsh cipher]

what?

[alpine narwhal]

nvm

[undone pawn]

8-3 I guess

[harsh cipher]

lmao

[alpine narwhal]

LOL

[harsh cipher]

ahhahahaaha

[alpine narwhal]

i asked my dad that question

what is 8-3

5

what is 3-8

5

[undone pawn]

maybe he thought |3-8|

[alpine narwhal]

i lost it there and i was laughing so hard

[undone pawn]

some people have absolute values turned on by default

xP

[willow bear]

[alpine narwhal]

odd

but still

i explained it to him once more and then he said -5

this was fun

i wanna do it again

[harsh cipher]

okay I have a lot more, i'll be here for the next 5 years

[undone pawn]

lmao

[harsh cipher]

I mean eventually I'll get better but it could be just as exciting as the previous one 😛

[alpine narwhal]

im looking forward to that

[viscid thistle]

i feel like i missed some fun stuff

[alpine narwhal]

yes you did

bro you should’ve seen everything that went down

[viscid thistle]

Can i write 2cot(a/2) as 2cot(a/2)/2sin(a/2)?

Lasg one is supposed to be cos over sin

[uncut mulch]

$2\cot(\frac a2) \umwhat \frac{2 \cos(\frac a2)}{2 \sin(\frac a2)}$

[obsidian monolith]

ramonov:

[uncut mulch]

$2\cot(\frac a2) \umwhat \frac{2 \cos(\frac a2)}{2}\cdot \sin(\frac a2)$

[obsidian monolith]

ramonov:

[uncut mulch]

no

[viscid thistle]

Can i write it as something else?

[uncut mulch]

you could write it in terms of sin and cos,
but for you had and extra 2

2/2 = 1

[viscid thistle]

Sorry i dont understand .
My problem is $\frac{2 \cos(\frac a2)sin (\frac 2a 2)}{cos a}

$\frac{2 \cos(\frac a2)sin (\frac 2a 2)}{cos a}

[uncut mulch]

needs $ on both sides

[viscid thistle]

$\frac{2 \cos(\frac a2)sin (\frac 2a 2)}{cos a}$

[obsidian monolith]

Yethhound:

[viscid thistle]

Its sin(2a/2)

Fucked up

[uncut mulch]

do you have a pic of the question?

$\frac{2 \cos(\frac a2)\sin (\frac{2a}{2})}{\cos a} ?$

[obsidian monolith]

ramonov:

[viscid thistle]

Im supposed to simplify

[uncut mulch]

the 2a/2 looks very suspicious

and how is this related to your question about cot

[viscid thistle]

If i can make sin (2a/2) into sin(a)
And 2cos(a/2) into cos(a). Then i gave the correct answer. But i dont know if i can manipulate it like that

Damn

[uncut mulch]

2cos(a/2) into cos(a)

no

$a \cdot f(x) \not\equiv f(ax)$

[viscid thistle]

$\frac{2 \cot(\frac a2)sin (\frac (2a)(2))}{cos a}$

[obsidian monolith]

ramonov:

Yethhound:

[uncut mulch]

$\frac{2 \cot(\frac a2) \cdot \sin (\frac{2a}{2})}{\cos a} ?$

is that the ORIGINAL expression?

[viscid thistle]

Cos is cot

And sin(2a/2)

Upper one is cot

Lower is cos.

[obsidian monolith]

ramonov:

[viscid thistle]

Yes

J think i have to manipulate cot somehow

[uncut mulch]

and are you 100% certain that the question actually says "2a/2" inside the sin?

[viscid thistle]

100% staring at me in the book

Photomath says its equal to sin(a)

[uncut mulch]

well yeh. 2a/2 = a
and sin(2a/2) = sin(a)

which gets you: $\frac{2 \cot(\frac a2) \cdot \sin (a)}{\cos(a)} $

[obsidian monolith]

ramonov:

[uncut mulch]

and you can simplify the sin(a)/cos(a) to tan(a)

to get: $2\cot(\frac a2)\tan(a)$

[obsidian monolith]

ramonov:

[uncut mulch]

what's the final answer supposed to be?

and or take a picture of both

[viscid thistle]

Tan(a)

[uncut mulch]

yeh, please take a picture of the question

[uncut mulch]

from what's being given, i highly suspect there's some sort of major misprint

[viscid thistle]

The third one

So far I'm running into problems with all of them.

[uncut mulch]

questions fked

[viscid thistle]

Hmm, ok

[uncut mulch]

as i suspected, if they actually gave you something like
sin(2a/2) something is very wrong

[viscid thistle]

Can i manipulate the cot somehow?

[uncut mulch]

you could write it as

$\cot(\frac a2) = \frac{\cos( \frac a2)}{\sin( \frac a2)} \ \
2\cot(\frac a2) = \frac{2\cos( \frac a2)}{\sin( \frac a2)}$

but that doesn't really help

[viscid thistle]

Isnt that just cot(a/2). What happens to the 2?

[uncut mulch]

you only asked about the cot

[viscid thistle]

Ok, i see

[obsidian monolith]

ramonov:

[viscid thistle]

Ok. I think i solved it as the book gives it.

[uncut mulch]

what does the book actually say?

[viscid thistle]

That simplified it should be tan(a)

Assuming i can make sin(a) into 2sin(a/2)

[uncut mulch]

no\

[viscid thistle]

Then fuck :D

[uncut mulch]

$a \cdot f(x) \not\equiv f(ax)$

[obsidian monolith]

ramonov:

[uncut mulch]

as i mentioned earlier

the question if fked

[willow bear]

he tried to replace sin(2a) with 2sin(a) the other day

[uncut mulch]

sin(2a/2)/cos(a) already simplifies to tan(a)

which gets you: $2\cot(\frac a2)\tan(a)$

[obsidian monolith]

ramonov:

[viscid thistle]

Yeah, im having a hard time with that particularity

[uncut mulch]

and IF you somehow got the books answer from the initial expression

then it means you're essentially implying that 2cot(a/2) = 1

the point is the question is completely fked

the typo is so major that i can't even tell what they actually intended

[willow bear]

probably bc you don't have a good understanding of what a function is

you wouldn't think that f(2x) = 2f(x) was automatically true for all functions f

[viscid thistle]

Ok. Thx for your help. I'll move on to other problems

[odd helm]

I’m stuck on this problem

[viral trail]

Consider f(x) = 4x/(3x + 4)

I found the inverse is f-1(y)= -4y/(3y-4)

But how can I find range of f(x)?

@odd helm
tan(x-2) = ± √3
Principal solutions ±π/3 then form general solutions

x-2 = ±π/3

[serene heath]

@viral trail what is the domain of f?

[copper vigil]

What is the maximum number of intersections that f(x) and g(x) can have if:

f(x) is a function of the form a|x-h|+k

and g(x) is a function of the form a(x-h)^2+k

i don't even know how i am supposed to begin trying to solve this

<@&286206848099549185>

[remote veldt]

@copper vigil note that both have the same "vertex". You know how |x| comes to a point, and x^2 has the vertex? because h and k are the same, that point on the absolute value function will be the same as the vertex on the quadratic. That's intersection #1, and it also gives you a very powerful tool: if you note that both are symmetric about the same point, you only need to look for intersections on one side of h. However many there are, there'll be just as many as the other side (make sure you don't double count the vertex!). See what you can do with that

[warped dagger]

This is probably a pretty dumb question, and its probably something that i missed out on so i'm not getting the whole picture. 2*log25-log125-log5=0 I have no clue how to solve this, im pretty confused, if anyone can just tell me how to, or what to do.

[fleet yew]

@warped dagger what are you trying to do with that equation

there are no variables, so there is nothing to solve for

are you just trying to prove the identity?

[warped dagger]

Yeah, it's proving identity, thank you for the help, that was a pretty retarded question, gonna go take a look on youtube

[fleet yew]

@warped dagger what base?

[warped dagger]

Theres no base given, so 10 i assume

[fleet yew]

i don't think it really matters

[warped dagger]

used base 5 and it was 0

[odd helm]

I’m lost on this question

I got this polynomial

But when I graph the polynomial I get a zero at x=5 but none of my possible zeroes give me 5

[alpine narwhal]

i believe you got the wrong polynomial

check your distribution

[harsh cipher]

Hello

I'm back

I guessed and got it wrong

oh no

new picture

Does the answer part make sense?

to me it does not make sense

[uncut mulch]

no it does not

missing parentheses

= cos ( A + B )

[harsh cipher]

loolllllll

where

[uncut mulch]

first line

compound angle identity for cosine

[harsh cipher]

what about the 4th line

[uncut mulch]

NFI WTF IS GOING ON THERE

[harsh cipher]

hahahahah

[undone pawn]

lmao

[uncut mulch]

from cos ( A + B )

its just a simple substitution

[undone pawn]

randomly B came out of the trig function

oh ok I get the problem

they don't like parenthesis

[harsh cipher]

this person that wrote this page for the school doesn't like to use parentheses or new line. I've this type of mistake from previous chapters

seen

[undone pawn]

lmao

cos(A+B) = cos(pi/2 - B + B) = cos(pi/2)=0

that's what they intended to write

they hate = too

[uncut mulch]

fk lol i see that now

geez

[harsh cipher]

yes!!

[uncut mulch]

gotta be psychic

[undone pawn]

yas

[undone pawn]

^

[harsh cipher]

hahaha

so I couldn't see where the parentheses or the equal sign goes

-_-

I flagged the question. never used it before though 🙂

[uncut mulch]

parentheses are important

they're functions. you wouldn't ever consider writing
f x instead of f(x)

[harsh cipher]

hell yeah they are

[hollow cobalt]

hello

got a question for yall

h(x)=f(x/2)-3

means that every x in H

would grow twice as bigger

and the y would lose 3

right?

[lost pawn]

If you're comparing to f(x), then yes. You make a horizontal stretch by a factor of 2 and a vertical shift of -3 units

[hollow cobalt]

factor?

i am sorry, i study maths in a foreign language

[lost pawn]

It means that every x gets multiplied by 2

In the graph

[hollow cobalt]

alright

ty

so i got another question

f(x) has asemptots in 1 and 4

if i said it right

x=1 x=4

and the extreme points are 2 and 7

so 2 becomes 4 in h(x)

but its impossible

f(x)=(e^x/2)/(x^2-5x+4)

[lost pawn]

I don't get the question, sorry

Can you rephrase?

What is your native langauge btw?

[hollow cobalt]

hebrew

ill explain

they asked me to sketch h(x)

after that h(x)=f(x/2)-3

and f'(x)=0 then x=2 x=7

now they become x=4 and x=14

but x=4 can't be because of x^2-5x+4

so what is happenin

[lost pawn]

Well

h(4) is actually fine

Because h(4)=f(2)-3

and f(2) is fine to compute

But h(8) on the other hand will run into that problem

[hollow cobalt]

nvm i got it

if you do h(x)=f(x/2)

mean when x=1 asymptote becomes 2

and 4 becomes 8

so 2 ain't no problem

[lost pawn]

yeah

[hollow cobalt]

why i thought it wouldn't change

so lesson learned

when ever there is a change in f(x), inside the brackets

it means every thing moves

on the x axis

[lost pawn]

yes, you can put it like that

[lilac scaffold]

ok there is chances n3 to get one for each playlist

[hollow cobalt]

c and d are false for sure

if i am not wrong

[lilac scaffold]

now someone tell me whats that c

[hollow cobalt]

c is n over 3

[lilac scaffold]

(n 3) is that combination or C ?

[hollow cobalt]

wym

same thing

[lilac scaffold]

like nC3

[hollow cobalt]

its the chances of picking n out of 3

yes

[lilac scaffold]

i used them a lot

[hollow cobalt]

it is

[lilac scaffold]

and (n 3) both are same right ?

[hollow cobalt]

i don't understand your answer

questions**%^4

n 3 and nC3

arre the same

[lilac scaffold]

thanks for clearing it

and the answer is B

i dont know why they divide it by 3^3

[hollow cobalt]

you have 3 playlists

each one contains n/3

songs

and your way of picking is n/3 * n/3 * n/3

the chances of picking a song from each playlist are the same

means the max you get is n^3/27

and it has equals sign to it

unlike A

so you pick B

[lilac scaffold]

now i got it

i read the question wrong

[hollow cobalt]

damn i didn't do combinatorics for long time

is it for sat?

[lilac scaffold]

i thought, every playlist has n no of songs

so nnn

thanks btw @hollow cobalt

[hollow cobalt]

👍

[robust island]

Hello.

Any answer for this ? ^

[willow bear]

could,,, you take a screenshot

[viral trail]

Consider f(x) = 4x/(3x + 4)

I found the inverse is f-1(y)= -4y/(3y-4)

But how can I find range of f(x)?
@viral trail

Domain of f is R

[narrow peak]

@robust island what do you not know

[robust island]

I'm confuse of how to substitute and when it cancel out

[narrow peak]

which

[robust island]

The first picture 1,2,3,4,5,6

[narrow peak]

aight lemme see

foe 2nd one

for

u do know

[robust island]

I know their value but i don't know when to cancel out

[narrow peak]

tanθ=sinθ/cosθ

@robust island is it ok now

for 2nd

[proud gate]

A vector with angle of -27.57 if different than the same vector magnitude and angle of 333.43? The angles, while 360 degrees apart still point the vector in the same direction.
But this got me wrong in a quiz. But is my reasoning right?

[proud gate]

Oh and also, the vector component form given is (4, -2)

[obtuse mulch]

Hey, I’m Max.

I’m having trouble with rationalising 1 over the square root of 6 could someone help?

I know the answer is the square root of 6 / 6 but I’m not quite sure why

[remote veldt]

@obtuse mulch $$\frac{1}{\sqrt{6}} = \frac{1}{\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}}$$

[obsidian monolith]

Nicholas:

@Max_The_Masterton $$\frac{1}{\sqrt{6}} = \frac{1}{\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}}$$
```Compile error! Output:

! Missing $ inserted.
<inserted text>
$
l.54 @Max_
The_Masterton $$\frac{1}{\sqrt{6}} = \frac{1}{\sqrt{6}} \cdot \fra...
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.

LaTeX Font Info: Calculating math sizes for size <14> on input line 54.
LaTeX Font Info: Try loading font information for U+msa on input line 54.
(/usr/local/texlive/2018/texmf-dist/tex/latex/amsfonts/umsa.fd

[proud gate]

@obtuse mulch multiply by sqrt(6) over sqrt(6)

This is the same as 1/1 but the sqrt(6) on the bottom cancels the sqrt to just 6.
With a sqrt(6) on the top.

[peak hill]

Question about this problem

[sharp marsh]

Help

[fleet yew]

@sharp marsh i want you to do something

[sharp marsh]

ya?

[fleet yew]

do you see that left hand side

[sharp marsh]

mhmm

[fleet yew]

make it have a base of e

[sharp marsh]

so e^1.618?

[fleet yew]

no

[sharp marsh]

err

[fleet yew]

you have 4^(x+3)

make it have a base of e

instead of a base of 4

[sharp marsh]

o

[fleet yew]

do you know how to do that

[sharp marsh]

wait I was putting it into the wrong problem It hink

nvm

Says it's wrong still

[fleet yew]

dude you have to work with me

do you know how to make that expression have a base of e

[sharp marsh]

So which part

Uhhh

Ln*?

[fleet yew]

wait which problem do you need help with

[sharp marsh]

First one

I insert into wrong section

I mean second

This new one

Ya

[fleet yew]

ok

just take ln of both sides

[sharp marsh]

I did

[fleet yew]

x^2-3=x-2

[sharp marsh]

(1+sqrt(5))/2

(1-sqrt(5))/2

Right?

Since x^2-x-1

One number was negative though do I put in that too?

Nvm figured it out

[fleet yew]

got it?

[sharp marsh]

Yep

Didn’t include -.618 bc I was thinking of log

With negative number

[sharp marsh]

How do I do log(x-5) = 13/7

[fleet yew]

@sharp marsh what base

[sharp marsh]

Sorry the original problem was

7log(x-5)=13

@fleet yew

[fleet yew]

what base logarithm?

[sharp marsh]

er 10

[fleet yew]

ok

let's look at this expression:

log_10(x-5)

what can you do with this

[sharp marsh]

um

Can I multiply by 10?

[fleet yew]

there is subtraction inside the logarithm

[sharp marsh]

oh

log(x-5)/log(10)

[fleet yew]

that's change of base

how do you simplify subtraction

[sharp marsh]

um

wym

[stuck lark]

@fleet yew no

log(a-b) is not equivalent to log(a)/log(b)

[fleet yew]

yeah my bad lol typed the wrong thing

[stuck lark]

log(a/b)=log(a)-log(b)

[sharp marsh]

uh

so I can't do that?

[stuck lark]

log(a-b) is not equivalent to log(a)/log(b)

[sharp marsh]

What can I do with it then?

[stuck lark]

7log(x-5)=13
log(x-5)=13/7 sounds fine

then you should know that exponentiation and logarithms are inverse functions

Specifically $x\mapsto10^x$ and $x\mapsto\log_{10}(x)$ for $x>0$ are inverse functions

[obsidian monolith]

RokettoJanpu:

[stuck lark]

this gives us that for all $x>0$,
$$\log_{10}(10^x)=10^{\log_{10}(x)}=x$$

[obsidian monolith]

RokettoJanpu:

[sharp marsh]

erm

So do I

10^13/7

?

So that I can get rid of log

[stuck lark]

the exact step you’re thinking of is setting both sides as the exponent of 10

[sharp marsh]

ya

[wheat condor]

Hey guys, I have 3 question from my homework that I don’t understand, can anyone help? It’s simple precalc

[remote veldt]

[wheat condor]

i have no idea what im doing on all 3 but ik for the 3rd question, quadrant one is both positives

[remote veldt]

for the first one, write csc(x) as 1/sin, and write cot(x) as cos/sin in the numerator, and write sec(x) as 1/cos(x) in the denominator. take each of the numerator and the denominator, and simplify/combine them individually, then see what cancels

[fleet yew]

[stuck lark]

@sharp marsh that leaves the left side as x-5, the rest is easy

[wheat condor]

ok

can i get help on the other 2 please?

<@&286206848099549185>

nvm i got them

thanks

[vernal spindle]

I'm confused what this question is asking for exactly

[remote veldt]

write [tan(x) + cot(x)]/[csc(x)] in terms of cos(x)

[fluid shore]

The question's just asking you to give sin(x) as the answer by stating that the given expression is axiomatically equal to it

[vernal spindle]

Is the first step ((sinx/cosx)+(cosx/sinx))/(1/sinx)

[vernal spindle]

?

[pale bison]

yes

wops nvm the message was several hours late

[fluid shore]

No just assume that it's sin(x)

[pale bison]

[vernal spindle]

What do you mean assume its sin(x) @fluid shore

[fluid shore]

lmao i was just messing around, ignore me

[vernal spindle]

I'm still lost on the question though

[willow bear]

rewriting your expression as (sin(x)/cos(x) + cos(x)/sin(x))/(1/sin(x)) will certainly be a sensible first step

as long as you know how to deal with nested fractions

[vernal spindle]

is

the first step

[willow bear]

there's no such thing as "the" first step because there are many ways to do this problem

each having a different first step

@vernal spindle

[vernal spindle]

hmm

[willow bear]

...

did you spend all this time waiting for me to give you explicit approval along the lines of "yes this is the first step, keep going"?

[vernal spindle]

Not exactly, I'm trying to figure out if im on the right path though

[willow bear]

well you've got your thing in terms of sines and cosines only

[vernal spindle]

I think its
((2sinx/cosxsinx)+(2cosx/cosxsinx))/(1/sinx)

[willow bear]

you what

[vernal spindle]

^as a step not answer

[willow bear]

$\frac{\sin(x)}{\cos(x)} \neq \frac{2\sin(x)}{\sin(x)\cos(x)}$

[obsidian monolith]

Ann:

[willow bear]

first off you're overthinking it already

like why not AT LEAST rewrite the division by (1/sin(x)) as a multiplication by sin(x)

[vernal spindle]

isnt 1/sinx = cscx

[sturdy haven]

Help with 1 2 and 3 please

[willow bear]

@vernal spindle 1/sin(x) = csc(x), but if you're interpreting that as "every time you see 1/sin(x) in any context whatsoever, the next step has to be to replace it with csc(x) no matter what" then you're wrong

plus you had just rewritten csc(x) as 1/sin(x), so what's the point in undoing that

what i'm saying is

$\frac{1}{\frac{1}{\sin(x)}} = \sin(x)$

[obsidian monolith]

Ann:

[vernal spindle]

So now I just have to get the sinx into terms of cos?

[willow bear]

you have to SIMPLIFY the expression, first and foremost.

get it into a form where it's not as much of a MESS as $\frac{\frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)}}{1/\sin(x)}$, and only THEN worry about how to write it in terms of $\cos(x)$ only.

[obsidian monolith]

Ann:

[willow bear]

wouldn't that be the natural course of action which almost suggests itself?

[vernal spindle]

Yeah

so the expression simplifies down to sinx since the top is 1 then 1 over a faction cancels leaving just sin

[willow bear]

no the expression doesn't simplify down to sin(x)

[fleet yew]

use the quotient identities

[willow bear]

the top isn't 1

[vernal spindle]

I'm struggling too much

[fleet yew]

look at the numerator

[vernal spindle]

tanx+cotx

[fleet yew]

very good

so you have $\frac{\tan+\cot}{1/\sin}$

[vernal spindle]

yeah

[obsidian monolith]

AMD:

[fleet yew]

so that denominator looks kind of tricky

we have division in the denominator

let's get rid of it

how?

[vernal spindle]

rewrite it as csc?

[willow bear]

you literally just rewrote csc as 1/sin. why would you undo that

[fleet yew]

you know what sure you can do that

[vernal spindle]

ohh i see that now

can i multiply sin to both top and bottom to rid it?

[fleet yew]

yes

[vernal spindle]

so (tanx+cotx)sinx

[fleet yew]

yes

[vernal spindle]

what would i do next

[fleet yew]

distributive property

[vernal spindle]

tanxsinx+cotxsinx

[fleet yew]

now what

[vernal spindle]

erm

[fleet yew]

what is the definition of tan

[vernal spindle]

sin/cos

[fleet yew]

so let's just use the definitions of tan and cot

(sin/cos) * sin + (cos/sin) * sin

[vernal spindle]

(2sin/cos)+cos

[fleet yew]

it's not 2sin

[vernal spindle]

sin^2

[fleet yew]

yes

[vernal spindle]

what can i do next with the sin^2

[fleet yew]

bring it all to a common denominator

cos = cos/1

so you have

sin^2/cos + cos/1

[vernal spindle]

ohh

sin^2+cos^2

[fleet yew]

that's the numerator

what's the denominator

[vernal spindle]

cos

(sin^2+cos^2)/cos

[fleet yew]

ok

now you should know something about that numerator

[vernal spindle]

its 1