geometry-and-trigonometry

[drowsy spoke]

oh wait, cot(pi/2-y)

[meager acorn]

@hard gale yes

wait

so no one can help me or

......

[hard gale]

we can help

[drowsy spoke]

turns out cot(pi/2-x)=tanx

[hard gale]

just mega confuzzeld at the beginning

[meager acorn]

for (pi/2-y)

do i have to solve for y?

[hard gale]

the goal is to show left hand side = right hand side for any values of y

[meager acorn]

yes

sec²(y) - cot²(pi/2 - y) = 1

i have to show the work left side

i am so lost at the moment

[hard gale]

it's not an equation here, it just involves rearranging LHS into RHS or vice versa (or more exotic strategies)

so yeah

[meager acorn]

i have to use identies of trig and analytic trig

[hard gale]

let's just rewrite the lhs with sin and cos for the moment

[meager acorn]

ok

[drowsy spoke]

it's like this: can you show that cosxtanx = sinx?

[meager acorn]

wait

sec re-write is = 1/cos theata

[drowsy spoke]

yes

[meager acorn]

cot = cos theta/sin theta

[hard gale]

yus so we have

$$\sec^2(y) - \cot^2(\frac{\pi}{2}-y) = \frac{1}{\cos^2(y)} - \frac{\cos^2(\frac{\pi}{2}-y)}{\sin^2(\frac{\pi}{2}-y)}\ \sec^2(y) - \cot^2(\frac{\pi}{2}-y) = \frac{1}{\cos^2(y)} - \frac{\sin^2(y)}{\cos^2(y)}$$

[meager acorn]

let me write this down

[drowsy spoke]

do you know how to expand sin(a + b)?

[meager acorn]

no sir

[drowsy spoke]

then I don't know any other method to simplify cot^2(pi/2-y)

[meager acorn]

wait

[hard gale]

(or just remembering cos(pi/2-y) = sin(y) and sin(pi/2 - y) = cos(y), but that's kinda cheating with addition formula)

[meager acorn]

can we change the cot^2 to tann^2

tan^2[pi/2-(pi/2-y)]

[hard gale]

that cot^2(pi/2 - y) = tan^2(y)

[meager acorn]

oh oops

😦

my bad

nononono

wait wait wait

cot theta = tan theta ?

if that happen

[drowsy spoke]

that's not going to simply things
cot theta = 1/tan theta

[hard gale]

it's illegible wtf

[meager acorn]

i am really out of my option

i need help T_T

[drowsy spoke]

are you Malaysian?

just curious

[meager acorn]

no

[drowsy spoke]

anywho

[meager acorn]

can i ask why cant you change that as tan

sec^2y-tan^2

[drowsy spoke]

so you've not studied sin(a + b) = sinacosb - sinbcosa ?

right?

[meager acorn]

i studied but i am not that strong

i need help

[drowsy spoke]

ohhh

yeah

that works

my bad

[meager acorn]

tan = cot

cot = tan

because cancels out

1

[drowsy spoke]

tan = 1/cot

[meager acorn]

yep

[drowsy spoke]

cot = 1/tan

[meager acorn]

so

sec^2y - tan ^2(pi/2-y)

[hard gale]

tan = cot isn't generally true, cot(pi/2 - x) = tan(x) is true

[drowsy spoke]

@hard gale i think i'm confusing things, i'll leave it to you

[meager acorn]

i have a question

[hard gale]

don't forget the arguments they're very important

[meager acorn]

whats pi/2 -pi/2 change that to?

1?

[hard gale]

err where pi/2-pi/2 ?

[meager acorn]

oops wrong type

pi/2-y

sec^2 -tan^2(pi/2-y)

[hard gale]

$$\cot(\frac{pi}{2}-y) = \frac{\cos{\frac{pi}{2}-y}}{\cos{\frac{pi}{2}-y}} = \frac{\sin(y)}{\cos(y)} = \tan(y)$$

[hard gale]

that's the thing : cot(y) isn't equal to tan(y)

[meager acorn]

so your saying my teacher is wrong

what she wrote in chuck board

[hard gale]

ah

1 sec i didn't actually read what was written

no that's true

[meager acorn]

ok

[hard gale]

doesn't mean cot = tan tho

[edgy tide]

Hey guys, this may or may not sound really dumb, but do sin, cos and tan have definite formulas, like for example f(x) = x + 1, or are they more complex? if so, how?

[meager acorn]

should i skip this question @hard gale

[hard gale]

(lemme show you the end of it if you want)

[meager acorn]

sure

please do

[hard gale]

https://cdn.discordapp.com/attachments/326138757474680852/510531663781036042/latex.png

so yeah now i just showed how cot(pi/2 - y) = tan(y) so cot^2(pi/2 - y) = tan^2(y) = sin^2(y)/cos^2(y)

so $$\sec^2(y) - \cot^2(\frac{\pi}{2}-y) = \frac{1}{\cos^2(y)} - \frac{\sin^2(y)}{\cos^2(y)}$$

[hard gale]

(our fractions have same denominator so we can add them)

$$\sec^2(y) - \cot^2(\frac{\pi}{2}-y) = \frac{1-\sin^2(y)}{\cos^2(y)}$$

[hard gale]

now don't you recognise something in the numerator?

[meager acorn]

i am bit confused on 2nd part

[hard gale]

confuzzled on what?

it's just fraction addition/substraction

[meager acorn]

so wait

OH OKOKOK

one sec

ok wait

i have a quesion

how did you end it up sin^2y/cos^2y

[hard gale]

cos(pi/2 - y) = sin(y) and sin(pi/2 - y) = cos(y) (those identities are seeable pretty easily on a trig circle)

[meager acorn]

so your saying..

shit

I SEE IT

i see the trig cofunction identies

@hard gale

is this where that come from correct?

[hard gale]

this is correct ye

[meager acorn]

wait

so than

so its going to be

1/cos^2 -sin^2/cos^2

[hard gale]

yus

[meager acorn]

-y

so

one sec

is this mean

you will get

since sin^2/cos^2 = tan^2y

1/cos^2y -tan^2y

@hard gale

which mean now we have to solve for 1/cos^2y

[hard gale]

it's useless to put it in this form tho

[meager acorn]

than how is it has to equal to 1

[hard gale]

$$\frac{1-sin^2(y)}{\cos^2(y)}$$

[meager acorn]

wait

can we not change sec^2

because

nvm

nvm

where did 1-sin come from

[hard gale]

substracting fractions :/

[meager acorn]

????

[hard gale]

$$\frac{1}{cos^2(y)} -\frac{sin^2(y)}{cos^2(y)} = \frac{1-sin^2(y)}{\cos^2(y)}$$

[hard gale]

(like basic arithmetic/algebra dope)

[meager acorn]

so i cant change it to tan yet

let me erase a step

[hard gale]

going with only sin and cos is usually easier that's the thing

[meager acorn]

now i see where i kinda got confused

one moment please

ok once its done

is there a another identies for 1-siny

or theta

is this mean i have to use power -reducing formulas

[hard gale]

cos^2(y) + sin^2(y) = 1 (that's one of the most basic identities :/)

[meager acorn]

i dont see that on my book

[hard gale]

cause they just think it's so basic they don't have to put it in their book i believe

[meager acorn]

let me write this on my note

so i dont for get

cos^2(y) + sin^2(y) = 1 (that's one of the most basic identities

one sec

but we have fraction here

doesnt matter if its fraction?

[hard gale]

(just rearrange that identity a little bit....)

[meager acorn]

1-sin^2y = 1>

[hard gale]

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

[meager acorn]

isnt it?

[hard gale]

:/

[meager acorn]

or sin

now i see

thank you

cos^2y/cos^2y = 1

:/

[hard gale]

://

[meager acorn]

thank you so much

[hard gale]

(screw trig)

[meager acorn]

this is my 1st time taking pre-calc

we got into trig

i got lost for bit

i have another question....

tan^2 theta/sec theta = sin theta tan theta

[hard gale]

put the left hand side in terms of sin and cos

(this one's defo easier)

[meager acorn]

ok thisis what i gotten so far

sin theta/cos theta / 1/cos theta

i gotten sin^2 theta /cos theta

i am stuck on next step

[hard gale]

it's $$\frac{\frac{\sin^2(\theta)}{\cos^2(\theta)}}{\frac{1}{\cos(\theta)}}$$

[hard gale]

tho

[meager acorn]

thats what i meant

so when you solve that

i gottn sin^2 theta/cos theta

1/cos flips

cancels out

just 1 cos

sin^2/cos

[hard gale]

yus taht's correct

[meager acorn]

since i have sin/cos

= tan theta

correcT?

one sec

[hard gale]

almost

[meager acorn]

sin theta x tan theta

?

OH SNAP

I TYPED without knowing it

O,O

because

[hard gale]

🍻

[meager acorn]

sin^2

its like

algebra

sin x sin = sin^2

sin theta tan theta

OH SNAP

💥

@hard gale how do you rep people in here?

[drowsy spoke]

t!rep @hard gale

[loud cedar]

🆙 | Kelsier, you can award more reputation in 3 hours, 51 minutes and 13 seconds.

[drowsy spoke]

Like that

[meager acorn]

t!rep @hard gale

[loud cedar]

🆙 | Yupki has given @hard gale a reputation point!

[meager acorn]

t!rep @drowsy spoke

oh

[drowsy spoke]

Hue hue

No need

[hard gale]

so many rep

& u can give rep only every 3h or so

so kek :/

t!rep @drowsy spoke

[loud cedar]

🆙 | emeric75 has given @drowsy spoke a reputation point!

[hard gale]

bcuz

[meager acorn]

can you help me with next problem too?

[hard gale]

go fur it

[meager acorn]

let me write the problem on sec please

csc(-x)/sec(-x) = - cot x

nso for csc and sec becomes

1/coes / 1/sine

[hard gale]

don't forget the (-x) :/

[meager acorn]

i got the -x

so i got

sin -x /cos -x

sin -x / cos -x = tan

wait

[hard gale]

tan(-x)

remember

[meager acorn]

so inverse of tan-x?

nono

i solved it wrong

suppose to be sin /cos

because csc /sec

i think i solved this one as sec/csc

[hard gale]

well let's just stay with sin and cos for the moment, you should have identities for sin(-x) and cos(-x)

[meager acorn]

so i would be getting cos -x /sin -x

which is cot -x

is this mean do you have to re-write the cot - x to - cot x?

because you cant have cot -

maybe i m wrong..?

o,o

[hard gale]

yes, that's why i prefer staying with sin and cos

[meager acorn]

oh

so either way i was doing ok.. than

[hard gale]

just easier to work with

[meager acorn]

i see

at least now i am getting this

i understand so much better

[hard gale]

"you should have identities for sin(-x) and cos(-x)" search in your head (or in your book)

[meager acorn]

because last test we had extra credit like this

i am pretty sure you cant have sin -4

we were doing graphing

finding amp period p.shift . v.shift etc

a+bx+c+d

now i understand thank you!

[hard gale]

🍻

[meager acorn]

going to take a break and solve some more

my head is starting to hurt

[sick veldt]

x = 10 correct??

Question 2:

Are these three selected answers correct??

[little osprey]

Nein

x≠10

@sick veldt

[sick veldt]

aww man

[little osprey]

But question 2 seems right.

[sick veldt]

alright

Hmm

[little osprey]

115=5x+15

What's the first step?

[sick veldt]

50

omg

my bad

[little osprey]

No.

[sick veldt]

nooo ik it lol

20

[little osprey]

Yeah.

[sick veldt]

ty I got 100

[little osprey]

np

[meager acorn]

whats
( 1 - cos^2 x) = ?

whats the trig identies?

sin?

or 1

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

how do you verify the identity on this question?

sin^4x + cos^4 x = 1-2cos^2x + 2 cos^4 x

[thorn valley]

the right side is all cosine

so maybe transform the sin on the left to cosine using

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

and that squared should give you what u want 😃

[upper karma]

Hey could I have some help?

I was homeschooled for 3 years and have just gone back to mainstream school (in sixth form) so am struggling to get used to this way of teaching (previously had 1 to 1 maths tutors)

I have some trig to do and am struggling

PM me if you can help me out please

[dire rampart]

post your questions

and sm1 will help if available

[upper karma]

Solve the following equations giving all solutions in the range 0<=x<=360
a) 3cosx+2=0
b) tan^2 x=1/2
c) sin(3x+20)=0.3

[dire rampart]

what have u tried

[upper karma]

I have literally no idea where to start

I think it's to do with trigonometric identities

[dire rampart]

no all of them are just rearranging

first one for instance

[upper karma]

cosx=-2/3

[dire rampart]

yes

now

inverse cos both sides

what do you get

[upper karma]

Am I meant to use a calculator for these questions or not

The teachernever specifies :/

[dire rampart]

yeah u do

to find the inverse cos of -2/3

[upper karma]

131.8

[dire rampart]

unless the question explicity states

dont use a calc

[upper karma]

so teh next one x = inverse tan (sqrt1/2)

is that right?

[dire rampart]

yup

[upper karma]

Okay I overcomplicated thta in my head so much

[dire rampart]

this one we gotta be a bit more careful tho

cuz the q asks for all solutions

[upper karma]

Yes

[dire rampart]

we gotta make sure we get all of them within that range

0 to 360

[upper karma]

tan(x)=tan(180+x)

[dire rampart]

yes

or

-180

both ways

make sure theyre in the range tho

do you know why that is?

why tanx=tan(180+x) i mean

[upper karma]

the graph

I don't know how to explain it, but the values on the graph are the same if you apply that

[dire rampart]

the tan function is a periodic function

meaning it repeats

every 180 degrees

oke what did u get for your inverse

[upper karma]

215.3 and 35.3

This last one looks quite weird

I have to separate the x

[dire rampart]

what do you think

[upper karma]

So I could do sinx(3+20/x)

[dire rampart]

we should do

why would you do that

[upper karma]

to separate the x?

Since it's the value I need to find

Or is that wrong

[dire rampart]

try takin the inverse

and see what happens

what do you get

[upper karma]

Ohhhh

Inverse of 0.3?

[dire rampart]

without altering anything in the brackets

yes

indeed

[upper karma]

17.5

So then 3x+20=17.5?

[dire rampart]

oke

yup

there ya go

but

remember

solutions

[upper karma]

Between 0 and 360

That will give me a negative solution so I need to apply a rule

[dire rampart]

note that sinx=sin(180-x)

[upper karma]

ja

ty :)

Now I have a REAL question

[dire rampart]

and also

[upper karma]

These 3 questions were worth 10 marks altogether

[dire rampart]

yea go on

[upper karma]

Meaning each one is worth 3.3333 marks???

Interesting

[dire rampart]

or one is worth 4

and the others 3 3

[upper karma]

Oh yeah I'm dumb

so I got x=-2.5/3

Meaning now I need to add that to 180 since double negative

[dire rampart]

?

for which q

[upper karma]

c

[dire rampart]

how did u get -2.5/3

[upper karma]

3x+20=17.5

3x=-2.5

[dire rampart]

oh yes

[upper karma]

x=-2.5/3

[dire rampart]

my bad

thats one of them

whats the other

[upper karma]

180.83

[dire rampart]

is that x

or 3x+20

your other solution should be 180-17.5

[upper karma]

uh

[dire rampart]

so 3x+20=180-17.5

[upper karma]

Ohhh

Okay

[hard gale]

The thing is you'll have more than 2sols here

[dire rampart]

would u

its between 0 and 360

inclusive

oooh

the range

must be changed

cuz its 3x

[hard gale]

But it's sin(3x+....)=...

Ye

[dire rampart]

lol i always forget that

my bad

you wanna take over octonion?

ion wanna forget somthin else

[hard gale]

That's why I prefer just solving generally then finding the sols that fit

I'm eating a burrito tho

[upper karma]

I like this it's like when I used to have online lessons :)

[dire rampart]

oke enjoy yo burrito

[upper karma]

If you're eating I should go eat as well, I haven't had any breakfast or lunch yet

I'll be right back

[dire rampart]

oke enjoy yo food

[upper karma]

Okay back

[dire rampart]

that was some quick eating

you should join an eatin competition

[upper karma]

I havent eaten silly

I brought food up to my ROOM

[hard gale]

And eating while discord

That's the obvs strat

[dire rampart]

ikr

[upper karma]

Exactly

[dire rampart]

oke

you see how we have 3x inside the sine instead of x

that means we gotta alter our range

to fit that transformation

since we're solving for x

[upper karma]

so it becomes 0<=3x<=540

?

[dire rampart]

360x3

and 0x3

[upper karma]

360 oops

I did 180

1080

[dire rampart]

yea

our new range is 0 to 1080 now

[upper karma]

So how would I write that

0<=3x<=180

I meant 1080 instead of 180 oops

[dire rampart]

yea like that

so when we inverse initally

we keep adding 360

until we we go over the limit

[upper karma]

Can we go back a step a minute

[dire rampart]

sure

[upper karma]

I got inv sin(0.3) = 17.5

Then 3x+20=17.5

[dire rampart]

yes

yup

[upper karma]

Then x=-2.5/3

[dire rampart]

now we gotta find other solutions tho

[upper karma]

But what do I do with the -2.5/3

Because that's lower than 0

Do I just ignore it

[dire rampart]

yeah u gotta a fair point

i think that means

we ignore 17.5

[hard gale]

Or there's the strategy of writing 3x+20=17.5+360k or 3x+20=162.5+360k, k in Z (using the periodicity of the sine function)

[dire rampart]

yeah u could do that

17.5 isnt a valid solution tho is it

within the range

[upper karma]

what do I do with the 20 though

That's what I'm confused about

[dire rampart]

you just take that away from both sides

when solving

like youre rearranging for x

[hard gale]

(That's why I solve generally the equation first, less confusion going on)

[dire rampart]

which is what we're doing

[upper karma]

I might just skip this question

[dire rampart]

why wheres the confusion?

[upper karma]

I'm just confused by the whole 3x+20 th ing

What is the next step?

I might understand it better

Do I inverse tan 17.5 or something

[dire rampart]

you simply rearrange for x

[upper karma]

I did that though

[dire rampart]

so if 3x+20=17.5

[upper karma]

And got -2.5/3

[dire rampart]

yea thats the right approach

that just means 17.5 isnt valid

[upper karma]

Okay

[dire rampart]

theres other solution

we can find

[upper karma]

OH and then I add 360 onto 17.5?

Or 180

[dire rampart]

using the fact that sinx=sin(180-x)

and sin(x)=sin(x+360)

so one of our other solutions would be 180-17.5

which is 162.5

now we keep adding 360

till we go over our range

so 162.5 is one solution

[upper karma]

Don't we have to minus the 20

[dire rampart]

yea after we find all of them

[upper karma]

So 360+17.5

[hard gale]

Adding 360, you forget to take care of the transformation here

[upper karma]

ds]jgsijG[jggh

[dire rampart]

wdym oct

[hard gale]

$$3x+20=17.5+360k\iff x=-2.5/3 +120k$$

[hard gale]

For example

That 120 dope....

[dire rampart]

yea but you could do all that after you add 360 surely

so after you find all solutions to 3x+20

either way works

[hard gale]

(Shit I forgot you changed your range 🙃 )

[dire rampart]

oh

is what u did

to account

for the range?

[hard gale]

Yus

[dire rampart]

oke then we just keep addin 360s

dan u still here?

[hard gale]

My head doesn't want to comply with your sol it seems (but it works NP)

[upper karma]

YEs

I am

Sorry I was taking my plate downstairs

I think he will go over it in the lesson so I'll just write what I understood and hopefully at least get a couple marks for it

[dire rampart]

erm sure

gl

[upper karma]

and for the next question (i'll show u)

i just use the cosine rule right

Oh

It's sideways

Sorry

[dire rampart]

😠

lemme see

yus

you can

use cosine for part a

thing is

i dont think your calc

will give an exact value

for cos15

which is what u need

[upper karma]

Any other ways I could do it?

Oh the sin rule maybe?

[dire rampart]

you can find the third angle

[upper karma]

OH OF COURSE

[dire rampart]

then use sine rule ya

[upper karma]

sinA/a = sinB/b = sinC/c right

[dire rampart]

there is a way to find exact value of cos15

if youre interested

[upper karma]

How :O

[dire rampart]

have u done angle sum formulas

so cos(A+B)

[upper karma]

Uh can I have an example

I think I touched on them briefly in year 10 with my tutor

[dire rampart]

cos(A+B)=cosAcosB-sinAsinB

yes?

no?

[upper karma]

no rip

[dire rampart]

rip

[upper karma]

One day

Can we PM

I feel like I'm clogging up chat

[dire rampart]

youre not lol dw

theres no one here

[hard gale]

This chat is made to be clogged

[dire rampart]

question channels are empty

[upper karma]

Okay good

[dire rampart]

theres multiple channels

dw

also oct

[hard gale]

Shit I took the subway the wrong way 🙃

[dire rampart]

why are the channel names not in order

its triggerin me

alpha should be at the top

[upper karma]

sin120 = sqrt3/2 right

[dire rampart]

yes

its the same as sin60

which is root3/2

[hard gale]

Ask June if it really bothers you

[dire rampart]

na im just jk

[upper karma]

I feel so dumb not being able to do these questions

[dire rampart]

just practice

it will come to you

[upper karma]

Would I do sin120 / 8x-3 to start with

[dire rampart]

lemme see

ya thats right

wot bout the third angle

wt did u get for that

[upper karma]

Yeah

45

[dire rampart]

oke

thats good

thats an exact value

now

u can use sine rule

[upper karma]

Yes

[dire rampart]

and rearrange for x

[upper karma]

Okay so

Can I write it down and take a pic

Also excuse my scruffy handwriting

[hard gale]

You can np

[dire rampart]

no

no scruffy

[dire rampart]

yes

now rearrange

for x

[upper karma]

Okay so rt3/2 / 8x-3

sqrt3(8x-3)/2

?

[dire rampart]

no

[upper karma]

fridsp[gjs[g

I'm sorry that I'm being so annoying

[dire rampart]

sqrt3/2(8x-3)

[upper karma]

I haven't done trig since before summer and I wasn't great at it then

[dire rampart]

its ok

better to make mistakes

thats how one learns

now

[upper karma]

isn45 = sqrt2/2

[dire rampart]

yes

[upper karma]

so sqrt2/2(4x-1)

[dire rampart]

it is

yup

cross multiply

[upper karma]

so you mean

sqrt3(8x-2) / sqrt2(16x-6)

?

[dire rampart]

wot

wt did u do

[upper karma]

idk DUDE

[dire rampart]

you have $$\frac{\sqrt{3}}{2(8x-3)}=\frac{\sqrt{2}}{2(4x-1)}$$

yes?

[upper karma]

Yes

I expanded them

[dire rampart]

first

lets get rid of that 2

at the bottom

[upper karma]

wait wait wait

you mean

[dire rampart]

of rach

[upper karma]

8x-3

[dire rampart]

multiply both sides by 2

ooops

ya

[dire rampart]

is this right?

[upper karma]

Yes

[dire rampart]

now multiply both sides by 2

[upper karma]

So then 2sqrt3 / 8x-3 = 2sqrt2 / 4x-1

[dire rampart]

2sqrt3?

just sqrt3

[upper karma]

o ye

my bad

oops

[dire rampart]

oke

now we have

$$\frac{\sqrt{3}}{(8x-3)}=\frac{\sqrt{2}}{(4x-1)}$$

[dire rampart]

ye?

cross multiply

[upper karma]

Yes

so sqrt3 (4x-1)

over sqrt2 (8x-3)

[dire rampart]

$$\sqrt{2}\cdot (8x-3)=\sqrt{3}\cdot (4x-1)$$

[upper karma]

Dot means multiply right

[dire rampart]

yup

can you expand and simplify that

[upper karma]

uhm

Probably but I've not an idea how

something -3sqrt2

[dire rampart]

just treat the square roots as any other number

and expand normally

[upper karma]

= something -sqrt3

8xsqrt2

[dire rampart]

$$8\sqrt{2}x-3\sqrt{2}=4\sqrt{3}x-\sqrt{3}$$

[upper karma]

yes

Wait

I wrote them at 8x sqrt2

and 4x sqrt3

Is that okay

[dire rampart]

yes

thats fine

multiplication is commutative

[upper karma]

yes

And now rearrange for x

[dire rampart]

now puttin x on one side

$$8\sqrt{2}x-4\sqrt{3}x=3\sqrt{2}-\sqrt{3}$$

[upper karma]

lol

[dire rampart]

agree?

[upper karma]

yes

Then separate x

[dire rampart]

yup

[upper karma]

so you get x(8sqrt2-4sqrt3)

[dire rampart]

then divide

[upper karma]

wait

[dire rampart]

and ratiolanze

[upper karma]

Can't I do it by 4x instead

Would that make it easier

[dire rampart]

how do you mean

oh

like take a factor of 4 out

[upper karma]

yes

[dire rampart]

yes actually

that would help

good thinkin

so we have $$4x=\frac{3\sqrt{2}-\sqrt{3}}{2\sqrt{2}-\sqrt{3}}$$

yea?

[upper karma]

Hang on

Writing stuff down

lol

[dire rampart]

there we go

[upper karma]

oh God

now to rationalise

brb

Gotta go do smth

[dire rampart]

ping me plz

when ur back

[upper karma]

Oki

@dire rampart

[dire rampart]

helo

ok so we gotta rationalise this

what we do is

multiply top and bottom by $$2\sqrt{2}+\sqrt{3}$$

[dire rampart]

try it

and lemme know

[upper karma]

Yes I know rationalising it's just a pain

:p

[dire rampart]

oke

you should get

after rationalizing

$$\frac{9-3\sqrt{6}}{5}$$

[dire rampart]

that

[upper karma]

uhh

That's not right is it

[dire rampart]

?

[upper karma]

9+sqrt6

[dire rampart]

is that what it asks for

[upper karma]

Yes and what you get when you rationalise

I just did it :p

[dire rampart]

i prolly messed up hang on

yup you are right

my bad

i messed up

[upper karma]

Now part b

lemme read it

find the area of ABC giving your answer to 2 decimal places

Is it 1/2 ab sinC

[dire rampart]

yup

ez pzy

[upper karma]

not ezpz

Because I have to substitute x

Which is 9+sqrt6/20

[dire rampart]

just use your calculator

be careful tho

use brackets

[upper karma]

I'll come back to it

And do a couple of easier questions first

Because I've been doing work all morning

[upper karma]

Okay I need some more help

Find the set of values of x for which -32 < 7x+3 < -11 and 2x^2 >= 18

Can you solve each one indiv and then intersect the sets?

Yes so

x^2 >= 9

so then x<=-3 and x>=3

But then how do I do the -32 < 7x+3 < -11

[dire rampart]

what ever you do to the middle

do to both sides

[upper karma]

I don't get what u mean

OH

First -3?

From both sides?

and then they perfectly divide by 7

excellent

It's like solving equalities, except that you can't multiply by a negative number

So I get -5 < x < -2

Do I just leave it

like that

And write x <= -3, x >= 3 and -5 < x < -2

Well, intersect those two intervals. Where on the number line are both conditions true?

x <= -3

Er... what about -6.. that's not in the second interval?

[upper karma]

wut

so x<-5

?

Well, ok, not quite. What numbers have both (x < -3 or x>=3) AND (-5 < x < -2)?

OH

Wait no idea

Sorry

I'm too tired now :(

Sorry about the delays, doing too much stuff at once :( OK, I'll give you this one, but won't do that generally: -5 < x <= -3

That's where both conditions are met

[upper karma]

OHH

Okay I get it

Thank you

[rain tulip]

how would i maximize 3x^2 + 6y^2 + 5z^2, given that x^4 + y^4 + z^4 = 3?

[upper karma]

lagrange multipliers

[keen aspen]

^

Take the partial derivative of x y and z ln both and set equal to each other with a scalar multiple on your constraint (lambda)

You'll get a system after doing the derivatives and you can find what x y and z are

[rain tulip]

this showed up on a precalculus test

so is there a solution without calc or above?

@keen aspen @upper karma

[keen aspen]

Not that I know of

[upper karma]

I'd probably guess (0,1,0) to maximize the y term with the largest coefficient or solve for x=y=z case, the extreme or balanced cases usually tend to win out

probably some theorem frm class you're expected to use

[rain tulip]

the answer was sqrt(210) but idk how they got that

this wasn't in class its precalc competition test

[upper karma]

Can someone help me with this

I want to know the radius of the small circle

A Radius is 4
B radius is 2

The drawing is not proportional

[rain tulip]

12 - 8sqrt2

@upper karma

[upper karma]

Solving?

@rain tulip

[rain tulip]

use the property 1/sqrt(a) + 1/sqrt(b) = 1/sqrt(c)

[eager pendant]

@rain tulip I’m getting the answer to be sqrt(70)

[upper karma]

What's that property

[rain tulip]

im pretty sure it was given as a japanese temple problem

here lemme look it up

@eager pendant

how?

[eager pendant]

are you absolutely sure the answer is sqrt(210)?

[upper karma]

Well

That's the same result as a friend

Are your sure it is correct?

[rain tulip]

jazza yes the answer is sqrt(210)

besides, if you let x, y, z = 1 then it's already 14

which is bigger than sqrt(70)

@upper karma im sure its correct

[upper karma]

But how to Apply the property

[rain tulip]

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

click on it

[upper karma]

Well yea

Thank you

I understand the result now

[rain tulip]

@upper karma

where did you get this problem from

[latent canopy]

My math teacher

[upper karma]

Yup

What he said

It's an exercise that @latent canopy Will use to get +0.5 in his final evaluation

[rain tulip]

oh

nice

what course are you taking right now?

[eager pendant]

@rain tulip what do you mean set everything to 1

[rain tulip]

because it equals 3

[eager pendant]

You wrote x^4+y^4+z^4=1

[rain tulip]

x^4 + y^4 + z^4 = 3

[latent canopy]

Science and technology 10°

[eager pendant]

ohhhh

[latent canopy]

Scientific course

[eager pendant]

then the answer is sqrt(210), maximum is sqrt(70) for x^4+y^4+z^4=1

[rain tulip]

how?

[eager pendant]

CS

[rain tulip]

ugh

cauchy

[eager pendant]

lol it’s not too bas

bad

[rain tulip]

i forgot it

i learned it like

a year ago

long time

[drifting temple]

can someone hel me with c

[coarse aurora]

yeah sure pal

[rain tulip]

just get common denominators

and simplify

[drifting temple]

dw

got it already

[rain tulip]

ok

[languid lodge]

It should lead to $$\sin^2 x+ \cos^2x = 1$$

[steady sleet]

@drifting temple Note don't post the same question in multiple channels

Read #❓how-to-get-help for rulez

[drifting temple]

ok

[upper karma]

Need help with trig <@&286206848099549185>

[steady sleet]

a) post what you actually need help with.
b) wait a minimum of 15 minutes before pinging helpers.
Read #❓how-to-get-help for rules.

[frigid wasp]

does anyone know how to do this (from another discord server) https://media.discordapp.net/attachments/411573884597436416/511248974779383828/unknown.png?width=1037&height=365

[upper karma]

I think you need to break the triangle into two right triangles and then chase numbers

[limpid basin]

replace sin c with sqrt(1 - cos^2 c) and use law of cosines

[frigid wasp]

whats law of cosines? a^2=b^2+c^2-2bc cosA? @limpid basin

[drifting temple]

@obsidian moon yeah

[thorny mantle]

https://gyazo.com/01008b4cab5ccc332988f995b97824e9

wouldnt this be 180

cause literally every triangle is 180

[wet fulcrum]

@thorny mantle no, the sum of angles is 180. this is asking for angle JKL

[thorny mantle]

then wait

how would i solve this then

100-180=80

for angle L

[wet fulcrum]

yes

[thorny mantle]

it all of them together still gives me 180

4x+16+80

[wet fulcrum]

yup, that equals 180, so you solve for x

then plug back in to get K

[thorny mantle]

how do i solve for x

[wet fulcrum]

4x+16+80 = 180

[upper karma]

oh okay

[wet fulcrum]

4x = 180 - 80 - 16

[thorny mantle]

x=21

[upper karma]

to get x is easy. you take all values given to you inside of the triangle and equate them to 180

[wet fulcrum]

yup

then, plug back into (2x-11) to get K

[thorny mantle]

thatd be 31

but it asking JK and L

i did that

plugged it in

gives me 180

then why would 180 be wrong

[upper karma]

180 is the sum of all the angles put together

[thorny mantle]

then what would be the right answer though

they asking for all angles basically

[upper karma]

180

that is the sum of all angles

[thorny mantle]

then why did he say it was wrong when i said the answer is 180

im confused

[upper karma]

alright try putting in J

[thorny mantle]

i did bro

[upper karma]

all of them?

[thorny mantle]

yeah

added it all

still gives me 180

i dont get it

[upper karma]

Did you put in the value of K?

[thorny mantle]

what you mean by that

2(21)-11

=31

[upper karma]

did you put 31 in it? the program is supposed to put a ^above the angle it wants

[wet fulcrum]

@thorny mantle what are you confused about?

the answer is 31

[thorny mantle]

oh

but why though

[wet fulcrum]

it's asking for that angle

[thorny mantle]

they asked all of angles J K AND L

[wet fulcrum]

No

That's a way you denote a single angle

[thorny mantle]

oh

so the middle angle they talking about inbetween that is K

[wet fulcrum]

angle ABC means the angle formed by the POINTS A, B, and C

[upper karma]

You are supposed to put a ^ above the angle you want found

[wet fulcrum]

yes

[upper karma]

program is dumb

[thorny mantle]

geometry is confusing for me

ugh

[upper karma]

what grade are you in?

[thorny mantle]

sophomore

planning to take calc during senior year

summer school the way 2 go

[frigid wasp]

ik im late but angle JKL means the angle at point K or between the lines JK and KL

(which are the same thing)

[winter ingot]

#20

[upper karma]

Anyone good with quadratic functions?

The Y intercepting X, X intercepting Y, End points and Turning points

Im stuck at X intercepting Y

need help

[astral hornet]

@winter ingot tip: label the sides with their lengths and slopes first