geometry-and-trigonometry

[pure cape]

Sub this back in

What does sin^2(u)+cos^2(u)=?

[upper karma]

oh right

1

[pure cape]

Yes

So 1^2=1 is always true

[upper karma]

thanks

what about these ones

[pure cape]

Remember the identity of tan^2(x)+1?

[upper karma]

no

[pure cape]

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

Do you remember this?

[upper karma]

no, haven't ever seen it before

i'm pretty sure

[pure cape]

Ok lets prove it then

Firstly tho

Are you familiar with sec(x)?

[upper karma]

ye

[pure cape]

Oh alright

Ok so tanx=sinx/cosx

Correct?

[upper karma]

ye

[pure cape]

So can you simplify tan^2(x)+1 for me?

[upper karma]

sin^2 (x)/cos^2 (x) + 1

[pure cape]

Yes

Simplify it

Do common denominator and add them

[upper karma]

with what

[pure cape]

add tan^2(x) with 1

Or mathematically written tan^2(x)+1

[upper karma]

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

[pure cape]

Yes

And sin^2(x) + cos^2(x)=?

[upper karma]

1

[pure cape]

So we have 1/cos^2(x)

[upper karma]

so sec

[pure cape]

Yeah sec^2(x)

So now we have proven tan^2(x)+1=sec^2(x)

Lets use that

Posting this so i dont have to scroll up

So the denominator of tan^2(x)/[1+tan^2(x)] equals sec^2(x)

Which now becomes tan^2(x)/sec^2(x)

That can also be written as tan^2(x) * 1/sec^2(x)

Correct?

[upper karma]

ye

[pure cape]

And whats 1/sec^2(x)?

[upper karma]

tan^2(x)+1=sec^2(x)

[pure cape]

Uh no

[upper karma]

1/ (tan^2(x)+1)

[pure cape]

We used that already

Using it again is not gonna help

Remember how sec(x)=1/cos(x)

[upper karma]

ye

[pure cape]

So cos(x)=?

[upper karma]

1/sec^2(x) = cos(x)

^2

[pure cape]

Cos^2(x)

Yes

and tan^2(x)=sin^2(x)/cos^2(x)

Then whats tan^2(x)cos^2(x)?

[upper karma]

sin^2(x)

[pure cape]

Yes

So thats proven

Done for that one?

So we can move on to the last one

[upper karma]

|cos u| = 1/(sqrt(1+tan^2 (u))
|cos u| = 1/(sqrt(sec^2 (u))
cos^2 u = 1/(sec^2 (u)
cos^2 u = cos^2 (u)

[upper karma]

ye

[pure cape]

Yeah sure

[upper karma]

cos u = pi/2 - sin u = (2pi - sqrt(7) -1)/4 and don't know what to do now

[pure cape]

Try to use pythag identity

And cosu does not equal pi/2 - sinu

cosu=sin(pi/2-u)

Since u is in the first quadrant, cosu will be positive

So sqrt(cos^2(u))=cosu

Keep that in mind when using pythag identity

[upper karma]

cos^2 u + sin^2 u = 1
cos^2 u = 1 -sin^2 u
cos^2 u = 1 - ((sqrt(7)-1)/4)^2
dunno what to do now

@pure cape

[pure cape]

You didnt squared the value of sinu

It should be 1 - ((sqrt(7)-1)/4)^2

Now you have cos^2(u)

So how do you get cosu?

[upper karma]

square it

[pure cape]

Are you sure?

[upper karma]

cosu = (4- (sqrt(7)-1))/4

which is false

[pure cape]

You should be taking the square root on both sides

[upper karma]

oh wiat

[pure cape]

Square root and square cancels each other out

Rememember?

And no

Lets just do everything slowly

Simplify 1-((sqrt(7)+1)/4)^2 first

Whats that?

[upper karma]

((4 -sqrt(7)-1)/4)^2

[pure cape]

No...

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

[pure cape]

You can write this as 1-((sqrt(7)+1)^2/16

[upper karma]

so i can't include the 1 in there

[pure cape]

No

Can you expand (sqrt(7)+1)^2 first?

[upper karma]

(16 -7-2sqrt(7) - 1)/16

[pure cape]

Yes

Simplify?

Dont factor 2 out tho

Leave it there

[upper karma]

7+2sqrt(7) + 1 = 16 -7-2sqrt(7) - 1

[pure cape]

Yes

Which is (sqrt(7)-1)^2

Correct?

Oh wait

You made a mistake

Its 7-2sqrt(7)+1

[upper karma]

why

[pure cape]

7-2sqrt(7)+1=16-7-2sqrt-1

Why did the -2sqrt(7) suddenly changed into a +

[pure cape]

And notice how the left hand side and right hand side are not equal

[upper karma]

ye

ok thanks

[pure cape]

This is the correct one

Got it from here?

[upper karma]

the last two

[upper karma]

ye i understand everything

[pure cape]

Do you know the sum of angles formula?

[upper karma]

no

[pure cape]

Hmm

[upper karma]

i don't know what do you mean

[pure cape]

cos(a+b)=cos(a)cos(b)-sin(a)sin(b)

[upper karma]

no

[pure cape]

Hmmm

You might have to draw out the unit circle for this

[upper karma]

ok let's skip it

[pure cape]

Oh

Ok

[upper karma]

ig it's the same for the last q?

[pure cape]

No you have to prove it with similar triangles

Quite lengthy but not complicated

But i have to go

Maybe someone else has a better way

[upper karma]

ok thanks

[pure cape]

You can find these proofs online i think

[upper karma]

also, why is this the case

[peak flower]

$$\sin{(-x)}=-\sin{(x)}$$

[somber coyote]

𝔙eryhappyperson

[upper karma]

nvm, thanks

[peak flower]

https://math.stackexchange.com/questions/835460/prove-sin-x-sinx

[upper karma]

no need

just had a case of brainfog

[fast fern]

bro

[peak flower]

Either do it in your head, or split it up and do it in your head xD

[fast fern]

what about the log6

[peak flower]

$$\log{\frac{a}{b}}=\log{a}-\log{b}$$

[somber coyote]

𝔙eryhappyperson

[fast fern]

so what am i meant to do

[peak flower]

Split it up into 2 logs, with the formula I sent, then it should be quite obvious what the result is.

[fast fern]

wdym split

up 2

[fast fern]

i dont get it

[peak flower]

0.0

$$\log_6{\frac{1}{36}}=\log_6{1}-\log_6{36}$$

[somber coyote]

𝔙eryhappyperson

[peak flower]

Can you solve it now?

[fast fern]

so the answer is 6?/

[peak flower]

wut

no

why?

[fast fern]

2?

[peak flower]

-2

[fast fern]

how tho?

[peak flower]

because $6^{-2}=\frac{1}{36}$

[somber coyote]

𝔙eryhappyperson

[fast fern]

bro i h8 log

but

explain i wanna learn it

[peak flower]

Remember what a logarithm is actually asking.

"x to the power of 3/2 = 27"

This you can do in your head.

[fast fern]

do what?

[peak flower]

$$x^{\frac32}=27$$

[somber coyote]

𝔙eryhappyperson

[fast fern]

so the answer is 3

?

[peak flower]

,w x^{\frac32}=27

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by veryhappyperson#1152

[fast fern]

how?

[peak flower]

take the third root of both sides.

then square it.

then you have x

[fast fern]

yeah i get the idea

could u explain that for me?

bro

@peak flower

[peak flower]

Do you know what radians are?

[fast fern]

no

just tell me the equation

[peak flower]

-.- Of course, because you couldn't have just googled that

[fast fern]

what is it?

[peak flower]

$$Radians =\frac{Degrees \times \pi}{180}$$

[somber coyote]

𝔙eryhappyperson

[dark sparrow]

if you can convert between feet and inches you can convert between degrees and radians

it's really despicable how often i see radians vs degrees get presented as something fundamentally different from any other unit conversion

[upper karma]

idk

[shut venture]

@heavy storm what do you have to assume?

[upper karma]

where does pi - 6pi/7 come from?

6pi/7 is a bit less than pi

so wouldn't it make sense that the angle representing it be 6pi/7 - pi?

[upper karma]

what's up with that?

isn't it cos(cos^-1 (1/4)) + cos (-pi) = 1/4 + 1 = 5/4 ?

where does the pi come from? doesn't make much sense to me

9 = 2 + 7cos(4x) means that
cos(4x) = 1
cos(4 (pi/8)) = 1
cos(pi/2) = 1
which is false
so?

[pure cape]

yes, that is false because the solution to equation 9=2+7cos(4x) isnt x=pi/8

you just plugged in pi/8 for no reason because you think that since pi/8 is the upperbound which should output the largest value of the function

which is false

[upper karma]

right

[upper karma]

what about this one

[pure cape]

they use the supplementary identity

[upper karma]

idk what is it

[pure cape]

sin(pi-x)=sin(x)

[upper karma]

i don't see why

[pure cape]

there is a few proofs online

you could use sum of angles formula, but since you dont know that, the only way is to use the unit circle

since your exercise has mentioned this, i supposed you have went over this identity

[upper karma]

doubt it

axler only provides you with some knowledge to derive everything else

[pure cape]

then thats a terrible textbook

[upper karma]

nah, it's fun

in a chapter with polynomials he asked you to prove IRT in the exercises

without even mentioning it in the chapter itself

[pure cape]

welp thats what terrible textbooks are

[hollow plank]

prove that b+c=d also ⛛ ABC is equilateral using Ptolemy's theorem

without using Ptolemy's theorem

[dark sparrow]

is that even true?

also what exactly do we know about this picture? is anything known about triangle ABC?

@hollow plank

[hollow plank]

oh i forgot

the triangle is equilateral

[dark sparrow]

are you allowed to use complex numbers and/or trig

[hollow plank]

i guess

[dark sparrow]

okay

call the red point M

let α and β be the measures of arcs AM and MB respectively, and note that the measure of arc BC is 120°

[hollow plank]

how is arc BC 120*

[dark sparrow]

you said ABC is equilateral

arc BC is 1/3 of the whole circle

[hollow plank]

like this right /

[dark sparrow]

no

i didn't say those angles were alpha and beta respectively

[hollow plank]

then where is M ?

[dark sparrow]

you got M right

[hollow plank]

ok

[dark sparrow]

you did not get the angles right

those angles are not alpha and beta

[hollow plank]

oh

[dark sparrow]

the intersection of AB and MC is not the center of the circle

[hollow plank]

ok got it

[dark sparrow]

anyway, now we can use the formula for chord length to get the following:
b = 2r sin(α/2)
c = 2r sin(β/2)
d = 2r sin(60° + α/2)

what remains now is to prove sin(α/2) + sin(β/2) = sin(60° + α/2), knowing α+β=120°

[hollow plank]

ok

thanks

ok Here's another challenge
The Pentagon is regular.
||FIND b/a||

[dark sparrow]

what does the problem ask for this time

find the ratio between a and b?

...

why would you hide the goal behind a spoiler???

[hollow plank]

so that one could try to guess the question

[dark sparrow]

...

[hollow plank]

for dramatic effects

[dark sparrow]

😒

(b/a is just the golden ratio...)

[wise pawn]

lol

[hollow plank]

correct

how did u get

[dark sparrow]

it's probably possible to mess around with triangle similarity and angle chasing in some way

though to be honest i knew this particular one beforehand

[wise pawn]

one way is use complex numbers

$$\frac{|e^{i 2 \pi *2/5}-1|}{|e^{i 2 \pi /5}-1|} = |e^{i 2 \pi /5}+1| = 2|\frac{e^{i \pi/5}+e^{-i \pi/5}}{2}| = 2\cos(\pi/5) = \phi$$

[somber coyote]

Meρρa

[tiny snow]

@floral domeYou can use the fact that the triangle on the right is similar to the one on the left. Introduce y, such that x = 64 + y, then use ratios of the sides to determine that y = 36.

[winter helm]

https://gyazo.com/8505308b59f9d58b1beb65117422e480

[robust elm]

Hey! can anyone help me with question 35 and 36? ^T^

[dusky surge]

, rotate

[dusky surge]

Do you recall what is simltaenous equations?

[robust elm]

erm aint it like two equations that have the same values with the pronumerals or smth

[dusky surge]

Something like that, like the same unknowns d and h in both equations.

Well, you have Q33 and Q34

Those answers are equations

[robust elm]

oH

ok thank you 🍟 : )

[visual mantle]

How?

[dark sparrow]

is this a test?

[visual mantle]

warm up

Not rlly

[dark sparrow]

the 1 point in the corner caught me off guard

[visual mantle]

We do

We do worksheets in google forms

[dark sparrow]

mkay

so have you made any progress so far?

[visual mantle]

Ive done the triangle one

This one

[foggy talon]

Does this channel work @solid juniper

Can you explain a bit more if you don’t mind

[solid juniper]

@foggy talon You have a quadrilateral there, right?

[foggy talon]

Yes I do

[solid juniper]

OK, it's convex, right?

[foggy talon]

Yes it is

[somber coyote]

Chai T. Rex

[solid juniper]

So, we have 4 sides.

So the sum of angles is 360 degrees.

Now, your external lines are tangent to the circle, right?

[foggy talon]

I’ll plug it in

Yes

We don’t know n yet right

[solid juniper]

You mean n on the convex polygon angle sum formula?

It's the number of sides, so n = 4.

[foggy talon]

okay so 180(4-2)

Yes I meant that

[solid juniper]

OK, so the sum of angles is 360 degrees.

[foggy talon]

Yes 360

Do we use 88

To solve the problem

[solid juniper]

That's one of the four angles, yes.

What are two others?

[foggy talon]

I’m not sure how do we find S and U

[solid juniper]

Well, your problem says they're tangents, right?

ST and TU are tangent to the circle, right?

[foggy talon]

Correct

[solid juniper]

And the lines coming to meet them at S and U are radiuses, right?

[foggy talon]

Yes that’s right

[solid juniper]

Tangent lines are always at a right angle to the radiuses that meet them.

Does that make sense?

[foggy talon]

so S and U are 90 degrees

[solid juniper]

Yes.

[foggy talon]

So those three angles equal 268

[somber coyote]

Chai T. Rex

[solid juniper]

Right.

[foggy talon]

360-268

[solid juniper]

Yes.

[foggy talon]

92

[solid juniper]

Right.

[foggy talon]

So 92 is the answer

[solid juniper]

Yes, 92 degrees.

[foggy talon]

thank you @solid juniper

[solid juniper]

You're welcome.

[shut venture]

on a quadrilateral the supplementary right?

[echo hamlet]

Can i get some help

[shut venture]

i cant see a damn thing 😄

[echo hamlet]

@shut venture click on it then you can see

[shut venture]

im going blind

[echo hamlet]

Lol

[shut venture]

Its true

[echo hamlet]

Lemme get a better photo

[shut venture]

i can try and help

[echo hamlet]

Dont mind the middle 2 things

@shut venture

[shut venture]

let me study it

[echo hamlet]

Ok

[boreal narwhal]

does anyone know how to solve this?

[tardy dirge]

@boreal narwhal Is it 1/2 or -2

[boreal narwhal]

wym

[tardy dirge]

value of tan (theta)/2 is 1/2 or -2

You can easily solve for tan(theta) . For tan(theta)/2 use the identity

[echo hamlet]

@shut venture did you figure it out

[shut venture]

@echo hamlet still checking i need to remember some stuff

[echo hamlet]

Ok

[boreal narwhal]

@tardy dirge 1/2

[tardy dirge]

@boreal narwhal understood??

[boreal narwhal]

what do you mean

[tardy dirge]

You want me to explain or you got it yourself

[boreal narwhal]

Yeah please do, if you can

[tardy dirge]

Okk wait

@boreal narwhal I think this might help and neglecting -2 might be because we have taken sin theta to be in second quadrant

[boreal narwhal]

Yeah it helped a lot, thanks

[echo hamlet]

Its been 1 hour and 30 min

[coarse shoal]

could someone help me out with this?

[fast swift]

3 times square root of 2

[tiny monolith]

would anyone be able to help me with these

[silent plank]

applications of angle sums and equivalence

[worldly belfry]

can u help please

[main lintel]

@worldly belfry what have you tried? what is unclear?

[worldly belfry]

i dont know where to start

like i have no clue how to di it

[humble pulsar]

I already told you the general idea

in the other channel you asked in

[plucky willow]

can anyone help me w this i’m trying so many ways and i’m usually ending up w 48.1 but that’s not any of the answers

[humble pulsar]

just do it without rounding

$x^2+8^2=12^2$

[somber coyote]

Mosh

[plucky willow]

thanks sm I figured it out

[plucky willow]

can anyone help me w these last two questions

[echo hamlet]

@plucky willow let me see

[plucky willow]

Im good w the 2nd one now I just need help w problem 14

[echo hamlet]

Ok

We have tan^-1(x) = 50/20
Plug it in and you have your answer@plucky willow

[plucky willow]

Oh so option D. Thank u sm

[echo hamlet]

@humble pulsar are you good with circles

<@&286206848099549185> anyone that can help me with #1

[plucky willow]

nvm this is the last problem I need help with

[haughty spire]

Use length of arc formula

[plucky willow]

Thank you I figured it out

[haughty spire]

Cj did u understand

I am big smoke

[plucky willow]

lol

[haughty spire]

xD

Damn train

[long garnet]

1. plug those choices into the calculator, should come out as 48.1
2. if there's no choices, just put your answer in simplest radical form

[long garnet]

use the laws

[long garnet]

length of arc formula

[restive zodiac]

Could someone message me, I need help with vectors and the dot product. Thanks

[lusty mango]

any clue on how to solve this?

[dark sparrow]

well, a good first step might be to use the formula they literally give you here

and identify which line segment F is the midpoint of

[tight peak]

heyyo evewwyone

can you pwease send a few questions to hewp me

[sturdy yacht]

can someone help me with these 2

[supple rose]

Anyone know this?

[floral dome]

the last option

[supple rose]

Wrong picture

[supple rose]

Can you help me

[floral dome]

sorry im in a final right now

and im in geomotry too rn and i suck

i've been using calculator.com man

[supple rose]

I’m in finals too

I can’t fault

Fail

[vocal lodge]

55=(x-50)/2

solve for x

can someone help me with this?

[supple rose]

So@its D?

I got D

[vocal lodge]

yeah

its d

[supple rose]

Another question

To make sure

I got C

c correct or incorrect?

[weak lagoon]

Guys

I need help with the pythagorean theorem

[ornate iron]

go on

[weak lagoon]

And I need to learn how to get an answer

I'm in 8th and dont understand algebra

[raven gulch]

ok

so, now lets see we are given 2 sides of the triangle

[weak lagoon]

Ok

[raven gulch]

the base, and the height

how do u apply the theorem to this

[vocal lodge]

anyone?

[weak lagoon]

@raven gulch I dont know how

[raven gulch]

well, the theorem states, that

$Hypotenuse^2$=$Base^2$+$Height^2$

[somber coyote]

Shashwat

[raven gulch]

this is the pythagorean theorem

[supple rose]

Anyone know this?

[raven gulch]

stated differently

[supple rose]

Can you help me please?

You know this?

[raven gulch]

wait, channel is occupied

[supple rose]

Ok

[raven gulch]

i dont, sorry

[weak lagoon]

Hmm

[raven gulch]

So, @weak lagoon if i give u the length of the base, and the height, can u find the hypotenuse?

[weak lagoon]

Ok let's see

[raven gulch]

so

lets say the base is

6

[floral dome]

the length of those two are the same

[raven gulch]

and the height is 8

[weak lagoon]

Ok

[floral dome]

so set up an equation that sets them equal to eachother

[raven gulch]

try to plug these in the equation i just told u

what can u get

[supple rose]

So it’s A

[weak lagoon]

Okay lemme see

[supple rose]

?

[floral dome]

6+ 7x + 1 = 4x +1 + 10

[weak lagoon]

@raven gulch I don't know what to do 😔

[supple rose]

A?

[raven gulch]

dont worry

now, plugging those in the equation

we get

[floral dome]

i cant do the math right now im in a typing test for my buisness class

[supple rose]

Ok

[raven gulch]

$6^2+8^2=Hypotenuse^2$

[somber coyote]

Shashwat

[raven gulch]

now, @weak lagoon calculate 6^2+8^2 for me

[weak lagoon]

28?

[raven gulch]

we are adding the square of the numbers

whats 6^2

[weak lagoon]

12?

[raven gulch]

thats 6*2

6^2=6*6

[weak lagoon]

😔

[raven gulch]

this is a notation thing

dont mind it

but whenever i say

that anything^2

i mean that i am taking that anything, and multiplying it with itself

so in this case

6^2=6*6

=36

can u do 8^2 now?

[weak lagoon]

48?

wait no

64

[raven gulch]

yes

correct

now what is 6^2+8^2

[weak lagoon]

100

[raven gulch]

very good

[weak lagoon]

😅

[raven gulch]

now since, we know that 6^2+8^2=Hypotenuse^2

we can take the square root on both sides

to get the value of the hypotenuse

take the square root of 100

what is it?

[weak lagoon]

crap I forgot square roots

😅

[raven gulch]

ok now think one thing

since 8^2

is 64

[weak lagoon]

Ok

[raven gulch]

what would be 9^2

[weak lagoon]

84

Wait

81

[raven gulch]

yes

now do the same for 10

what is 10^2

[weak lagoon]

100

[raven gulch]

what do u think the answer to our question will be?

[weak lagoon]

10?

[raven gulch]

yes!

thats it

[weak lagoon]

woowowowowowooooo

[raven gulch]

now lets say we were given the hypotenuse

and one other side

we would do the same thing

set up the same pythagorean theorem equation

but we would solve it differently

[weak lagoon]

one question

[raven gulch]

yes?

[weak lagoon]

How do I know when I have to multiply the number by itself

Or by 2

[raven gulch]

when he number is written like this

$number^2$

[somber coyote]

Shashwat

[raven gulch]

when the 2 , or any number is written on the top right of any other number

we mean that the the main number, which we call the base number, is to be multiplied with itself

as many times as that number in the power wants us to

[weak lagoon]

Hmm

[raven gulch]

this is a notation thing

but in most cases

if any number is written in the top right

and is a little smaller in size

its telling u how many times to multiply that number thats bigger in size

by itself

[weak lagoon]

okay

@raven gulch now what

[raven gulch]

oh yeah

lets say we are given a side

like lets say the hypotenuse is 5

and one of the sides is 3

how do u think we can go about finding the other side

[weak lagoon]

Okay

Multiply?

[raven gulch]

well, we plug it into the original equation

this one

now we know hypotenuse

and one of the sides

so it will become

5^2=4^2+(x)^2

now how do u think we can move further

[weak lagoon]

Ok one sec

[raven gulch]

yea sure

[weak lagoon]

Hmm

I dont know

[raven gulch]

no worries

[raven gulch]

now, we subtract 4^2 from both sides

to get 5^2-4^2=x^2

we calculate the right side

and take the square root again

and thats it

@weak lagoon

this is all there is to the theorem, u can use what i just told u in a variety of different ways

[upper karma]

how?

[dark sparrow]

well you could do some boring unenlightening algebra, or a very nice geometric diagram

but knowing you, you're probably going to express your aversion to diagrams

let θ = arctan(t), then you know tan(θ) = t and -π/2 < θ < π/2; find cos(θ).

[upper karma]

i don't see how that helps

cosu = sqrt(1-sin^2 (u))

[dark sparrow]

consider not turning yourself into a glorified rewrite-rule machine

perhaps you could recall a trigonometric identity such as sin^2(θ) + cos^2(θ) = 1

and perhaps you could rewrite this into a form where only tan(θ) and cos(θ) appear

[upper karma]

this is how i derived it?

i had said that i don't know how to?

[dark sparrow]

try dividing both sides of $\sin^2(\theta) + \cos^2(\theta) = 1$ by $\cos^2(\theta)$

[somber coyote]

Ann

[dark sparrow]

tell me what you get

[upper karma]

tanu = sinu/cosu
tanu * cosu = sinu
(tanu * cosu) = sin^2 (u)
cosu = sqrt(1-(tanu * cosu))

or yours:
tan^2 (u) = 1/(cos^2 (u) - 1
tanu = sqrt(1/(cos^2 (u) - 1)
??????????????????????????????????
cosu = sqrt(1-(sqrt(1/(cos^2 (u) - 1)* cosu))
cosu = sqrt(1-(sqrt(1/(cosu - cosu)

I DON'T SEE HOW THAT HELPS

[dark sparrow]

you've gone astray almost immediately

[upper karma]

how am i supposed to know what to do

[dark sparrow]

tan^2 (u) = 1/cos^2 (u) - 1

you want to express cos(u) in terms of tan(u), so why not do your algebra accordingly?

why not isolate $\cos^2(u)$ in $\frac{1}{\cos^2(u)} - 1 = \tan^2(u)$?

[somber coyote]

Ann

[dark sparrow]

[also, you've got a lot of unbalanced parentheses]

[dark sparrow]

this is a difficult question to answer in general

it comes with experience, which your "cram cram cram nosleep allnighter cram cram cram" learning style seems almost counterproductive to building

[upper karma]

ok i admit it
idk how to isolate it

[dark sparrow]

okay, maybe all the trigonometry is making the algebra harder to see.

[upper karma]

1/cos^2 (u) - 1 = tan^2 (u)
(1 - cos^2 (u))/tan^2 (u) = cos^2
??????????????????????????
alternatively
cosu = sqrt(1-(tanu * cosu))
cos^2 u = 1 - tanu * cosu

[dark sparrow]

are you willing to set this problem aside for a moment

[upper karma]

?

[dark sparrow]

i was going to suggest a simpler, purely algebraic problem for you to solve so you could see what i was getting at.

[upper karma]

ok

[dark sparrow]

solve for $a$ in $\frac{1}{a} - 1 = b$

[somber coyote]

Ann

[upper karma]

and how is that different

it's not

[dark sparrow]

it is not different

[upper karma]

so it's not going to help me because i'm going to run into the same issue

[dark sparrow]

i'm just abstracting away all the trigonometry so you don't get sidetracked

you're running into the issue of applying too many trig identities all at once and making your own life harder

[upper karma]

(1 - a)/b = a

(1 - cos^2 (u))/tan^2 (u) = cos^2

[dark sparrow]

you have not isolated a in this equation

forget the trig

[upper karma]

ye

[dark sparrow]

forget the trig for now

all you have for now is the equation $-1 + \frac{1}{a} = b$

[somber coyote]

Ann

[upper karma]

.

[dark sparrow]

okay

add 1 to both sides in $-1 + \frac{1}{a} = b$. what do you get?

[somber coyote]

Ann

[dark sparrow]

(do ONLY this. do not progress further.)

[upper karma]

b + 1 = 1/a

[dark sparrow]

okay

do you see now how to isolate a from here?

it's one step away.

[upper karma]

^-1

[dark sparrow]

you can take the reciprocal of both sides, yes.

so you have a = 1/(b+1).

is this clear to you now?

[upper karma]

what else can you do

[dark sparrow]

you can do a lot of things

you can add 97 to both sides for no reason other than really wanting to

[upper karma]

?

[dark sparrow]

take the reciprocal of both sides,

[upper karma]

one step away

can take the reciprocal

[dark sparrow]

yes.
as in, "yes, that's exactly the one step i was going to suggest if you had said something else"

are you unhappy with my use of 'can'?

we seem to be getting sidetracked here again

this was intended to be no more than a tangent to remind you of basic algebra.

[upper karma]

i finished it

[dark sparrow]

oh, you finished the original problem?

[upper karma]

how?

[dark sparrow]

is that "how am i going to express aversion?" or "how do i do this with a diagram?"

[upper karma]

diagram

[dark sparrow]

can you please maybe be a little less terse

anyway, here

technically this only works for positive t, but once you have the result for that, handling t = 0 and t < 0 is a minor technicality by comparison

[upper karma]

not the opposite?

[dark sparrow]

...

[upper karma]

it's cos of tan?

[dark sparrow]

aaa

you ask for a diagram, i give you the diagram, you refuse to acknowledge the diagram and instead throw a question at me that you could've and should've asked way earlier

[upper karma]

now i review the question

[dark sparrow]

and this means that my answers to your questions are thrown into the void and given zero acknowledgment?

[upper karma]

what do you want

what am i supposed to say

[dark sparrow]

if the diagram makes the diagrammatic solution perfectly clear to you, then you should say "thank you" or "oh, that makes sense now" or something along those lines.
if there is something in the diagram that you would like me to explain in more detail, then ask me to explain whatever it is that needs explaining.

if you just say nothing, from my perspective it's as if you had not looked at it at all.

[upper karma]

i wouldn't have asked if i didn't look at it

[dark sparrow]

you asked a question which had nothing to do with the diagram.

[upper karma]

ok sry

[dark sparrow]

consider that other people are not telepaths and are not psychic

[upper karma]

you want to express cos(u) in terms of tan(u)
cos(u) in terms of tan(u)
Why is it
you want to express cos(u) in terms of tan(u)
if it's cos of tan

[dark sparrow]

it's not "cos of tan" it's "cos of inverse tan" if anything

[dark sparrow]

going back to this:

i suggested replacing the problem

show that cos(arctan(t)) = 1/sqrt(t^2 + 1)

with

let u = arctan(t). knowing that tan(u) = t and -pi/2 < u < pi/2, find cos(u).
[with the implication that you should end up with 1/sqrt(t^2 + 1)]

do you understand this transition? Y/N

[upper karma]

y

[dark sparrow]

this new problem requires you to find cos(u) [in terms of t] while knowing that tan(u) = t.

do you understand why this means finding cos(u) in terms of tan(u)? Y/N

[upper karma]

y

[dark sparrow]

have i answered your question in a manner which satisfies you? Y/N

[upper karma]

y

[dark sparrow]

do you have any further questions you would like to ask me pertaining to this problem? Y/N

[upper karma]

y

[dark sparrow]

okay, then ask them.

[upper karma]

how do you not get rid of the sin while still having tan

[dark sparrow]

it may be just the tiniest bit trickier

one purely algebraic way i see is this: from $\cos^2(u) = \frac{1}{\tan^2(u) + 1}$, multiply both sides by $\tan^2(u)$ to get $\sin^2(u) = \frac{\tan^2(u)}{\tan^2(u) + 1}$

[somber coyote]

Ann

[dark sparrow]

[of course, the diagram would have given clear answers to both the cos(arctan(t)) problem AND the sin(arctan(t)) problem, but guess who decided to throw that in the trash?]

[upper karma]

these diagrams helps with solving the problems and nothing else

[dark sparrow]

?

is 'solving the problems' not your goal here

[upper karma]

no

[dark sparrow]

also you're being overly dismissive of diagrams

but maybe i will not be able to convince you otherwise

[rapid tartan]

anyone have an idea

[dark sparrow]

@rapid tartan do you still need help with this?

[rapid tartan]

yes

[dark sparrow]

do you know what equations of circles look like, in general?

[rapid tartan]

not off the top of my head

but if you send it i’ll probably remember

[dark sparrow]

there are some regional differences in exact notation [i.e. the choice of exactly what letters appear as parameters], but the equation of a circle looks like this: $$(x-h)^2 + (y-k)^2 = r^2$$ where $(h,k)$ are the coordinates of the center, and $r$ is the radius

[somber coyote]

Ann

[dark sparrow]

does this ring any bells?

[rapid tartan]

nah not rlly

[dark sparrow]

okay, let's back up...

do you know the geometric definition of a circle?

@rapid tartan ?

[dark sparrow]

👻

[upper karma]

why is it the case ?

[dark sparrow]

the horizontal leg of your triangle has length 1, and the ratio of the vertical to the horizontal is tan(θ)

[upper karma]

right, thanks

[foggy talon]

Can someone explain this

[upper karma]

tangent is perpendicular to radius

which means angle qst = 90

triangle QSU is isoceles

sum of all angle in triangle = 180

those should be enough to solve the problem

@foggy talon

[foggy talon]

So 58+58+x=180

?

@upper karma

116+x=180

64?

[upper karma]

nope

angle tsu = 90-58 = 32

angle tus = 180-58=122

i'll leave it there for now

[foggy talon]

So should I subtract 122-32?

[upper karma]

tsu + tus + angle T = 180

32 + 122 + angle T = 180

[foggy talon]

So T is 26

@kitchen1112#3264

[rapid tartan]

my bad man i had a scrimmage

i can send question again if needed

@dark sparrow

[harsh panther]

Why do pyramidals

have base * height * 1/3rd

I think it comes pretty easily from just running some calculus

which works for arbitrary shapes

[acoustic epoch]

is angle T equal to 22.086 and 157.914?

becuase i think this is an ambigous triangle

[normal finch]

Could anyone help me wit

[vast fractal]

I understand the unit circle and circular motion, but all of these ratio thingies I don't get

[quasi badge]

Does anyone know how to set the equation up to find X

[silent plank]

sec-tan theorerm

[quasi badge]

So if i set it up using sec-tan theorem would the equation be 4*2=x(x+6)?

[silent plank]

no

[silent plank]

that is the theorem that could be applied here,
you aren't using that properly

wheres 4*2 coming from?

[quasi badge]

segment cd?

[dark sparrow]

where is the multiplication by 2 coming from?

[quasi badge]

oh do i not square the external segment?

[dark sparrow]

you do, but you did not actually write that

* represents multiplication, NOT exponentiation

[quasi badge]

ohh mb

so is my equation set up correctly? 4^2=x(x+6)

[dark sparrow]

yes, now it is.

[quasi badge]

ok thank you

[upper karma]

hey

henlo

[dull thorn]

by this definition, isnt' spherical distance just the angle between two points?

[dark sparrow]

it is

think of it like great-circle distance, if you're allowed to move only along the surface of the sphere.

[oak elbow]

had this question on my quiz review, how may i go about solving this?

[dark sparrow]

how familiar are you with the basics of trigonometry?

[oak elbow]

I'm not sure how to answer that 🤷‍♂️, we just started this chapter in Geometry

[dark sparrow]

does the SOH-CAH-TOA mnemonic ring any bells to you?

[oak elbow]

Yes

[dark sparrow]

okay

that's the sort of thing i meant when i talked about the basics of trigonometry

so, look at your diagram

you have here an isosceles triangle being split in two equal halves (which are themselves right triangles) by the dotted line

here's one of them, drawn separately from the other

do you understand why the bottom side of this triangle is 7.5 ft?

[dark sparrow]

@oak elbow ?

[upper karma]

anyone help me wtih this homework problem

[silent plank]

apply the appropriate formula for volumne

[inland dock]

I wanna ask, is most graphs of functions composed by «e» looks odd? Or there are more cleaned up graphs compare to ,e^x/cos(x) for example?

[wise pawn]

dunno, what do you mean by 'looks odd'?

[inland dock]

Like e^x/cos(x), it’s reciprocal, with sines

Especially then you divide trigonometric functions by exponent or vice versa

[upper karma]

How to I find how large this is? (Left to right)

[inland dock]

Go to help channel

[upper karma]

Idk where that is

[inland dock]

After the introduction chapter, there is “math help”

Find free channel

And ask question

Do not guarantee that it will be answered

[inland dock]

I mean strange

Compare to the simple curve exponent or trigonometric functions stretch

[wise pawn]

lol

[wise pawn]

what's strange about any of this

[inland dock]

That’s definitely not like others

[wise pawn]

e^x is smaller on the left and larger on the right

whenever you graph f(x)*g(x) you take the values of the function at every point and multiply their values there

[inland dock]

That doesn’t explain how from sine - periodic function, i get something like this

[wise pawn]

yes it does

e^x is not periodic

[inland dock]

But if you combine periodic and not-periodic, you can get periodic

[wise pawn]

give me one example

[inland dock]

Wait

I can multiply y=x and y=1/x which is hyperbola and not periodic (I guess) and get periodic line

[wise pawn]

you said periodic and not periodic, but both of those are not periodic

[inland dock]

Why y=x not periodic?

[wise pawn]

what's it's period?

f(x)=f(x+T) for all x and some constant T is what you need it to satisfy

[inland dock]

I think every point

[wise pawn]

what do you mean

what's T=?

T can't be 0 or infinite

[inland dock]

Can it be any number?

[wise pawn]

yeah

for sine, T=2pi

just as an example

[inland dock]

Yeah, for tg it is pi

And for a line, you basically can take any interval, and on each it will be the same

So wonder that line is a periodic

We can prove it by slope

So every line has unified slope on all its existing

How do you image that line can changing along his path in Euclid geometry?

[wise pawn]

false

not unless it's y=constant

like your y=x earlier won't work

f(x)=x if you try to plug it in and solve for T, f(x)=f(x+T) you get x=x+T which only allows T=0, which is not a period

[inland dock]

You can take plane, and rotate it so it will be look like constant

[wise pawn]

lol

[inland dock]

I think that the T not equal to zero, it’s rule that formalized it meaning

But line is not like the others periodic functions

[wise pawn]

if T=0 you'd just be writing f(x)=f(x+0) which is true for every function

[inland dock]

That means that T is bad understanding of period

[wise pawn]

no that means you have a bad understanding of period

[inland dock]

So there are functions , that don’t match periodic and not periodic?

[wise pawn]

a function is either periodic or not periodic

there's no in between

[inland dock]

Then what is line?

[wise pawn]

not periodic

[inland dock]

Because it has 0=period?

[wise pawn]

yep

no to your edited thing

there's no such thing as 0 period

period is a number > 0

[inland dock]

What if its period approaching -> 0

?

[wise pawn]

it can have a period but not a least period

like y=1

[inland dock]

It is a function, right?

[wise pawn]

you tell me, what is a function?

[inland dock]

No, I can’t understand why x=y is not periodic and y=0 is periodic, but just rotated

[wise pawn]

y=x is constantly increasing, it never touches the same height more than once

periodic functions touch the same height infinitely often at regular intervals

[inland dock]

What if i add new coordinate which measures by the distance from y=x?

It will be the same geometry

And it will hit that coordinate height infinitely times

[wise pawn]

sure do that, but that's not periodic it's something else

call it 'super periodic' or 'rotated periodic' or something if you want

I give you permission

[inland dock]

So you agree that line , at least in some meaning is periodic?

[wise pawn]

lol

sure

seems like you've lost sight of the point though

[inland dock]

I remember

[wise pawn]

you were wanting to understand what the graph of f(x)*g(x) looks like

trying to redefine 'periodic' isn't going to change that e^x * trig function isn't periodic

just accept the terms as they are, build on more stuff on top of it later if you want/need to

but there's no point in trying to redefine well established terms, they're just words that mean specific things

[inland dock]

But still we all can’t just accept what we got without reconsidering
The math follows the rules of nature, but it would be better to directly refer to nature itself

[wise pawn]

wrong

[inland dock]

Firstly, i dont know the rules about periodic function composition

[wise pawn]

if you have some concept different from 'periodic' that's like periodic, then you should be able to write it down clearly to describe it

then you can start to talk about what functions satisfy this condition

[dark sparrow]

wait what are we even talking about

[inland dock]

I know it formal formulation, but i prefer to just think of it as the same on certain intervals, greatly I do study it all myself, so dont have any restrictions with strictly posed tasks, and can just explore connections in world

[inland dock]

About periodic functions, and composed functions with exponent and trigonometric functions

[brave gust]

Can someone help me with this question please

[trim breach]

Find the interior angles of each polygon. Then, see if you can determine what the angle of that “gap” is using what you found.

Once you have that gap angle, you can see which regular polygon has that measure of interior angles.

[brave gust]

I found it as 40 degrees

9 sides

missing polynomial is equilateral triangle

Wait, interior angle 140 for nonagon.

[trim breach]

Interior angle for the nonagon should 140. Interior angle for the triangle should be 60.

[brave gust]

@trim breach Thanks.

[trim breach]

Were you able to determine what polygon to fit in there?

[dull blade]

I need help on this problem

[silent plank]

where are you stuck?

have you been introduced to the basics of right triangle trig?

sine, cosine, tan?

[dull blade]

8.60364654527

i got that as an answer

but im not sure if i am right

and i am supposed to solve for h

I used cosine

[upper karma]

correct

[dull blade]

yay

thanks for telling me

[upper karma]

no problem

[dull blade]

[plush creek]

Answer and explanation would be appreciated 😃

[trim breach]

Draw PT. P and T are parallel, so angles PTS and TPQ are supplementary. The shape is a pentagon, so there are 540 degrees internally.

Then you just solve for m.

@plush creek

[plush creek]

Ohhhhhhhh

Makes sense

Tysm for your help

[trim breach]

Glad to help.

[dull thorn]

does this just mean that a spherical line is an arc?

[dark sparrow]

a spherical line segment is a great-circle arc, yeah.

[hollow isle]

how?

[dark sparrow]

@hollow isle do you still need help w/ this?

[onyx granite]

If I have two plane with a given surface normal N1 and N2, how do you find a vector along the intersection of them?

can I cross product it?

<@&286206848099549185>

Wait, it looks so.

the result of N1 x N2 has to be perpendicular to both N1 and N2. As long as N1 and N2 are from the same origin, the resulting line is correct

[somber coyote]

Liria ^(;,;)^

[umbral sand]

Can anyone help

[bright gate]

you should try to search things on internet.

[fickle tundra]

do u know the answer?

[dark sparrow]

are you looking for just the answer?

cause we don't give out answers here.

[fickle tundra]

yes please

[dark sparrow]

the answer is ||go find somewhere else, because as i just stated, we do not give out answers here.||

#❓how-to-get-help says as much, in case you've chosen not to read it.

[weary drift]

👢

[near imp]

someone help me with these two questions please

[rotund linden]

@near imp refer this

[dark sparrow]

is this a test?

this looks like a test

[near imp]

yeah i got them figured out and solved! but no it’s not i was overthinking the questions once again

[eager rock]

hi

I have a bunch of math that I have to get dont in less than a hour

anyone able to just help?

A ship travels due west for 87 miles. It then travels in a northern direction for 78 miles and ends up 155 miles from its original position. How many degrees did it turn when it changed direction (inside the triangle)? Round your answer to the nearest tenth.

[wise pawn]

what have you tried

[tranquil shuttle]

just construct a triangle and use vector addition

[light lynx]

Solve right ∆ ABC, given that b = 5, c = 8, and ∠ A = 70°. C is the right angle. whos tryna help. solving for all three angles and sides

[trim breach]

That is way more information than your really need to solve that triangle. Solve a by using Pythagorean Theorem. Solve angle B by using the total interior degrees in a triangle.

[haughty eagle]

does anyone know how to solve for the A angle and the B angle

[trim breach]

Utilize inverse trigonometric functions.

[haughty eagle]

what

[trim breach]

For example, to solve for B, we know that cos(B) = 6/14.

[haughty eagle]

ok

[trim breach]

To “cancel” the argument of cosine, we use inverse cosine, usually written as cos^(-1) on calculators.

So the equation becomes cos^(-1)(6/14) = B.

[haughty eagle]

so 64.62