geometry-and-trigonometry

[upper karma]

which should be u = arccos(1) and v = arcsin(-1) right?

[vivid ridge]

well there may be no solution to that since it has to match up at the same time

this is pretty difficult :(

[upper karma]

Yeah doing that gives pi and -pi/2

which uhh, can't both be x lol

wolfram is able to find roots, but it's using a heuristic to approximate the values

it offers an absolutely nutty method of finding the roots through
x = 2 (tan^(-1)(root of x^6 + 3 x^5 - x^4 - 10 x^3 - x^2 + 3 x + 1) + π n) {n is an integer}
and has no explanation for how it gets here

And this method doesn't help in the general case because of the abel-ruffini theorem

[vivid ridge]

you may be digging yourself into a deeper and deeper hole

[upper karma]

Yes lol

[vivid ridge]

I salute you on your quest

[upper karma]

T_T

what are you trying to do?

like factor trig functions as polynomials?

Find exact roots for
sin(3x) + cos(2x) + 1

is this for school or your own fun

Own fun

It may not even be possible

ok good answer

it might be nicer to rewrite these in terms of e^{ix}

1 + 1/2 e^(-2 i x) + 1/2 e^(2 i x) + 1/2 i e^(-3 i x) - 1/2 i e^(3 i x)

then you can do substitutions like a=e^{ix} and b=e^{-ix} or something to get polynomials in two variables

I don't know if that'd be any more helpful, but I think it gives you a more of a path towards trying to prove it can't be solved for exact roots maybe

idk you have seemed to have thought about it much more than me

Yeah haha at least for the better half of this afternoon

I haven't looked too hard into the e^ method yet

hmm so generically what cases are you wanting to solve?

[rain tulip]

use sum to product

wait

nvm

im stupid

sry

[upper karma]

relax

Blind I'm just trying to solve
sin(ax) + cos(bx) + c where a > b > 1 and a ∈ N, b ∈ N

also where a != b (i guess that's already implied l0l)

ok cool

wait no constraint on c?

Ehh not really

Although |c| greater than the amplitude of sin(x) + cos(x) is trivial as there will be no roots

yeah

equal to the amplitude is also trivial I think, since the roots will just be some function of the periodicity

a sort of silly idea that might work

I'm down for some silliness

rearrange it to solve for x

in one of them, so specifically,

sin(3x) + cos(2x) = -1

=tex x= \frac{1}{a} \sin^{-1}(-c-\cos(bx))

[upper karma]

then plug it into itself

lol wut

=tex x= \frac{1}{a} \sin^{-1}(-c-\cos(b \frac{1}{a} \sin^{-1}(-c-\cos(b \frac{1}{a} \sin^{-1}(-c-\cos(b\cdots))))))

[upper karma]

see if this converges

Wait I'm a little confused

Is this recursive?

yeah

it's like trying to solve x=sinx by looking at x=sin(sin(sin(...)))

if it's an attracting fixed point

well there can (and will be) multiple points in this case

I mean yeah that should work, but it wouldn't find exact roots would it?

depends on what makes you happy

what is "exact"

a power series or something with roots or elementary functions only etc

I'd be okay with any representation that doesn't use an infinite sum or product. roots, transcendental functions, all that's cool

I don't know how to converge recursive functions

How how I would figure out how to represent that convergent point

ehhh I don't think it will be like how you want it to be even if we did think about that type of way more

I don't know how I'd go about that myself either

There's already heuristics to find the roots. Newtons method can approximate them pretty easily

and I don't want to try to think about it to figure it out, I'd rather just try to find a different approach

I guess to my mind, square roots are usually just another infinite process lurking in the background anyways

That's true

so what's so great about writing a sqrt symbol instead of saying you can just use newton's method?

Really any transcendent number is infinite

although in practice using newton's method might be quite hard here

[mossy vine]

the wolfram method looks like it's using e^ix

[upper karma]

I think this can have arbitrarily many roots

Well yeah, but it has a maximum number of roots per pediod

[mossy vine]

because you'll get (e^(3ix) - e^(-3ix))/2i + (e^(2ix) + e^(-2ix))/2 + 1 = 0

[upper karma]

A modified version of newtons method should be able to approximate all the roots fairly quickly since you have a bounded region and pretty tame derivative movement

[mossy vine]

u = e^ix

(u^3 - u^(-3))/2i + (u^2 + u^-2)/2 + 1 = 0

multiply through by u^3

[upper karma]

Ahh that's where they get the polynomial from then

[mossy vine]

yeah

[upper karma]

and then the rest is just uhh "desubstituting"

[mossy vine]

I guess

[upper karma]

good word

[mossy vine]

the tan^-1 is probably from some 1/2 ln((1-x)/(1+x)) or something, idk

[upper karma]

Yeah

So would that mean that for the general case a > b > 1, a ∈ N, b ∈ N there's no solution in radicals as shown by the abel-ruffini theorem

[mossy vine]

in general I'd expect there not to be

but there could very well be some with solutions in radicals

or radicals * pi or something

[upper karma]

does abel-ruffini apply here for complex coefficients

pi is a freebie here

Yes

thought it was only rational coefficients

wikipedia says arbitrary, good enough for me

Oh I thought you meant complex solutions durr reading is hard

I think it does apply to coefficients too though

by rearranging the polynomial I get is,

=tex -iu^{2a} +u^{a+b} + 2cu^a +u^{a-b} + i = 0

[mossy vine]

in fact something stronger than abel-ruffini is true if you allow non-rational coefficients

for example, x - pi = 0 has no solutions in radicals

[upper karma]

lmfao

depends on how you mean it, since I'm thinking rationals adjoined with i only, since that is the smallest field extension I can think of so we don't need reals

im notta galois theorist ok lol

[mossy vine]

me neither :^)

[wary stag]

What is x?

[rain tulip]

1/8

[upper karma]

^

[meager acorn]

is this mean i have to solve it as like this?
-3.3 x 180/pi ?

[meager acorn]

<@&286206848099549185>

[upper karma]

yeah

[clear falcon]

is it necessary to draw the angle

like this or no?

[clear falcon]

<@&286206848099549185>

but do i need to draw the angle

is it necessary?

so it's not necessary?

oh ok thx

[meager acorn]

I dont know where to start on this problem..

[clear falcon]

t!rep salty boi

[loud cedar]

🆙 | Comet0605 has given @upper karma a reputation point!

[meager acorn]

salt

can you help me with my question...

[clear falcon]

👌

[meager acorn]

so do you put that on ti-84 = answer?

i am so bad at word problem..

where did 1/aqrt2 come from

pi/4

samething right?

thank you

how do you rep

[loud cedar]

🆙 | Salty Boi, you can award more reputation in 17 hours, 7 minutes and 31 seconds.

[meager acorn]

t!rep @upper karma

[loud cedar]

🆙 | WhyTempFate has given @upper karma a reputation point!

[meager acorn]

thank you so much

if i have a more question may i ask

:/

just got into trig from pre-calc

something i just dont get it... i mean SOH CAH TOA helps

a lot

^

helps so much

comp angle = 90 degrere
supp angle = 180 degree

correct?

o,o

awesome 👍

😄

[clear falcon]

is it x= tan (12/10)

[meager acorn]

This helps so much

[clear falcon]

OOOOOO ye i forgot

thx

[meager acorn]

@upper karma the answer says 4.2N but how..

F = .6 x 9.6 x 1/sqrt 2

i got -40.729

how o,o

so.. like this? (.6 x 9.6) 1/sqrt 2

i did but i'vegotten 4.07

[charred spear]

Runtime error in iterm_1
On line 1 at position 16

0.6 * 9.8 * (1/sqrt2)
               ^
Failed to access variable sqrt2

Wolfram|Alpha didn't send a result back.
Maybe your query was malformed?

[meager acorn]

yikes

sorry

my bad

4.15 =4.2.

[distant sonnet]

ice bear!

@meager acorn

brother

[uneven ruin]

I need to prove that N is 90 degrees.

[left folio]

I don't see where M is

lol yea

[uneven ruin]

Sorry

M is under N

It’s not aesthetic cuz I’m bad at drawing but I can try to draw it better if it’s hard to understand my lines

[vapid mica]

i think i can help if you draw it correctly i dont think it will be that hard

and btw what grade are you in just to know what skill set you use

[uneven ruin]

Basically everything is allowed as long it is correct mathematically. This is like 10/11th grade math but I finished high school. I will draw a better sketch soon.

[eager pendant]

@uneven ruin M is defined as hte midpoint of DE?

and you're trying to prove that AM is perpendicular to BE

[uneven ruin]

It’s the midpoint.
However, DE isn’t perpendicular to AM.

And I’m trying to prove that BE is perpendicular to AN

That way, N will be 90 degrees @eager pendant

[silent sparrow]

Can someone help me with a problem

I forgot how to do it

Number 24

[small onyx]

can someone run me through how to attempt this, A vertical tower, OA, casts a shadow OB on the level ground when
the sun is elevated above the horizon at an angle of θ = 15o
. Later in the
day when θ = 6o
, the shadow has moved to OC, where OC is longer than
OB by 122m. You may assume that OBC is a straight line. Find the height
of the tower and the lengths of OB and OC.

[hard gale]

have you tried drawing the situation?

[zenith pulsar]

🤔 this is used for proofs, too?

I’m pretty horrible at that.

[hard gale]

a drawing is not a proof

it helps to get ideas in this case tho

[zenith pulsar]

No. I’m trying to get a problem right, loading it right now

There it is

[hard gale]

ah those 2-column things :/

[zenith pulsar]

Yes, I hate them

[hard gale]

(i just know about them cause i'm the server so lel)

[zenith pulsar]

Oh lol

Well, I’ll wait for someone who can help with them, that is all I need

[small onyx]

i have yez

[limpid basin]

use AA

ASA*

[small onyx]

whats that?

[steady hull]

similarity

i hope

[meager acorn]

how do you solve this?

[umbral snow]

Gotta know your unit circle

[meager acorn]

i got it

sqrt3/2 / 1

its ref of 60 degree

[umbral snow]

Very nice, that makes sense to me @meager acorn

[silent sparrow]

How do i do number 24?

[uneven ruin]

Hi, can someone help me out on this one ? I need to prove that N is 90 degrees/that BE is perpendicular to AN which is basically the same, details given in left side. Thanks in advance.

[meager acorn]

@umbral snow thank you i just need to make sure !

[left folio]

Uhm how to "Find the derivative of sin(cos(6x)) "?

[left folio]

=text sin(cos(6x))

[silent sparrow]

How do i start this problem

24?

[left folio]

@silent sparrow which

[silent sparrow]

24

The temperature

[left folio]

I've got no idea honestly

[silent sparrow]

Dang

[meager acorn]

you neeed
y = sin
amp = 40
phase shift
vertical shift
period

since its sin you need 2pi/b to find your pierod

[silent sparrow]

Im not graphing it

I have to find the temperature

[meager acorn]

i knw

[silent sparrow]

The correct answer is B but i forgot how to solve this

[left folio]

I'm going to say something stupid probably but doesn't he need a limit there?

[meager acorn]

no

[left folio]

Ok

[silent sparrow]

Ya this one is hard

[meager acorn]

the answer is telling me Quadrant 1
Can someone explain to me why?

[zenith pulsar]

Thanks, plum that helped

@small onyx that is Angle Side angle so when you’re trying to do a proof of a triangle you’re trying to find the angles and sides

If you haven’t been thru proofs yet it’s gonna be hell

[limpid basin]

Uh? Yw I guess

[keen aspen]

@meager acorn which quadrant is both sin and cos positive

[meager acorn]

(+,+) Q1

i cant think of any...

[keen aspen]

Yep Q1

[meager acorn]

because its greater than >

[keen aspen]

Yes

[meager acorn]

if it was less than it would be in?

Q4?

[keen aspen]

No

Cos is positive in Q4

Think of the quadrant thats (-,-)

[meager acorn]

Q1

but i have a question than

[keen aspen]

No

[meager acorn]

hm...

[keen aspen]

Think of a negative coordinates

(-1, -1)

Where would you plot that

[meager acorn]

Q4

no

[keen aspen]

Do you know your quadrants?

[meager acorn]

lksfjaslkfjasdlf

yes i do know one sec

something like that?

[keen aspen]

Yes

Now look to see where cos and sin are negative

[meager acorn]

q1?

this is confusing bit

-_-

[keen aspen]

Bruh

Look at each of the quadrants

And tell me which one you dont get BOTH cos and sin

[meager acorn]

sin -1,1
cos 1,-1

since sin teata > 0

1,1

so its Q1

[keen aspen]

Yes for both positive

[meager acorn]

cos teata >0

[keen aspen]

Now what about for both negative

[meager acorn]

1.1

sin <0 Q2

cos q4

[keen aspen]

No

Which quadrant do you plot (-1, -1)

[meager acorn]

Q3

[keen aspen]

Exactly

So sin and cos are both negative in Q3

[meager acorn]

sin and cos only deals with x-axis?

[keen aspen]

No

[rain tulip]

Sin is y axis

Cos is x axis

[meager acorn]

😮

[meager acorn]

can someone explain or how to do 1st step..

[keen aspen]

Make a reference triangle

What do you know about sin^(-1)

[meager acorn]

thats the inverse right?

so arc sin (-1)

theta = sin^(-1)

[keen aspen]

Yep

[meager acorn]

whats the next step..

[keen aspen]

So you are finding the angle

[meager acorn]

yes

[keen aspen]

And you have (3/5)

So make a reference triangle

Sin=opp/hyp

[meager acorn]

OH WAIT

so you solve inside 1st

and solve for cot

?

[keen aspen]

You make a triangle and then you solve for cot(theta)

[meager acorn]

let me solve it and post in pic one sec

[keen aspen]

You use the pythagorean theorem to find the 3rd

[meager acorn]

one sec please

@keen aspen you there

so somthing like this?

[keen aspen]

Yes

Now what's cot?

[meager acorn]

cot adj/opp

4/3

[keen aspen]

Yes

That's the answer

[meager acorn]

awesome thank you 👍

[keen aspen]

Jp

Np

[meager acorn]

is this mean i can set my x as adj and y as opp and r as my theata ? to find SOH CAH TOA ?

or...

[keen aspen]

r is the hypotenuse

[meager acorn]

oh

i dont have to use pyth

just need to rewrite the soh cah toa form

sec csc cot

[keen aspen]

Yes

If cos is adj/hyp, sec is hyp/adj

[meager acorn]

yep they are opposite

as long as you know soh cah toa

[keen aspen]

Yes

[meager acorn]

is this mean

i have to put where Q1 Q2 Q3 Q4 is in graph?

[keen aspen]

Yes and which trig functions are positive

[meager acorn]

so something like this?

[keen aspen]

Ya

[meager acorn]

i am understanding better now thank you

since its radian to degree

-3.3 x pi/180?

nvm

its -3.3 degree x 180/pi

[keen aspen]

Yes

[meager acorn]

so i have a question

if your solving

Rad to degree you use 180/pi
degree to rad you use 180/pi
or its differernt?

[keen aspen]

No that's right

Radians involve pi

So to get rid of pi you divide by it

Hence why when you convert from radians to degrees, pi is in the denominator

[meager acorn]

so let say you have - pi/3
you should solve it as

• pi/3 x 180/pi

you will get -60 degree

[keen aspen]

Yes

[meager acorn]

sweeet

thank you !!!

if you have sin 11pi/2

do you solve 11/2 = 5.5 pi

sin 5.5pi ?

put in calc ?

how do you solve this one

[keen aspen]

11pi/2-2pi

7pi/2-2pi

3pi/2

You know what 3pi/2 is

[meager acorn]

one sec please

where did you get 2pi from

2pi is the 180 degree ?

pi = 90?

[dire rampart]

@meager acorn no 2pi is 360 degrees meaning pi is 180

[meager acorn]

so wait

[dire rampart]

the reason he can just subtract 2pi like that is because sine is a periodic function

[meager acorn]

11 pi - 2pi

[dire rampart]

with a period of 2pi

[meager acorn]

to solve

i am bit confused

[dire rampart]

sine has a period of 2pi

meaning

it repeats every 2pi degrees

[meager acorn]

11pi/2 -2pi ?

[dire rampart]

so $$\sin(x)=\sin(x+2\pi)$$

[dire rampart]

11pi/2 minus 2pi is 7pi/2

[meager acorn]

i got 7pi/2

thats it or

i dont have to do anything with sin?

OH

okok

you converted sin to 2pi

11pi/2 -2pi x 2
11pi/2 -4pi/2
= 7pi/2

?

[keen aspen]

Yes

And then do it again

-4pi/2

[meager acorn]

i kinda seeit now

i need more practice

thank you!

you guys explain so much better than college tutors

dang

at least college that i go to..

i have a question for this problem since its P= (x,y)
i can draw a triangle and put my x = adj as 2/9 ,and y = hyp as - sqrt 77/9
its asking for trig function of t and find cos t

or.. how do i solve this?

[meager acorn]

nvm i got it

can someone explain to me how to solve this?

[upper karma]

=tex cos(\frac{5 \pi}{6})

[meager acorn]

the answer says 3/2 but how

[upper karma]

=wolf (5 times 180)/6

[charred spear]

Do more with Wolfram|Alpha pro

Query made by

@sage sonnet

[upper karma]

Ah 150 this'll be easy now

[meager acorn]

um.. i dont get it

[upper karma]

Angle starting from left side of x axis on unit circle is 30

So set of the right triangle

It follows that:

$$ cos(30) = \frac{\sqrt{3}}{2}

$$ cos(30) = \frac{\sqrt{3}}{2} $$

[upper karma]

But since we are on the left side of the x axis that turns negative

So

$$ cos(\frac{5 \pi}{6}) = -\frac{\sqrt{3}}{2} $$

[meager acorn]

but how do you know cos is 30 degree

[upper karma]

Because half way around the unit circle is 180°

180° - 150° = 30°

[meager acorn]

wait

cos is only deals with x- axis?

cos is like supp angle?

also where is this 150 come from

[upper karma]

A ray intersects the unit circle given an angle x
cos(x) = x coordinate of the point where the ray intersects unit circle
sin(x) = y coordinate of the point where the ray intersects unit circle

150° = 5pi/6 radians

[meager acorn]

ohohohohoho nwait

wait

so

cos (x) x coorinate

0 to 180 degree

but 150 where did it come from

nvm i see it

5 pi /6

cos (5pi/6)

cos becomes 180

5x 180 /6 =150 ?

180-150 = 30 degree?

cos of 30 = sqrt 3/2 ?

[rain tulip]

No

Negative

[upper karma]

This video might be helpful: https://youtu.be/1m9p9iubMLU

[rain tulip]

Since second quadrant

[meager acorn]

• sqrt 3 /2

?

[rain tulip]

Yes

[meager acorn]

because -x

is cos

trig is so complicated...

i have another question

is this the same problem as last problem?

[upper karma]

"If people don't realize that mathematics is simple, it's because they don't realize how complex life is." - John Von Neumann

=wolf arctan(-7)

[charred spear]

Do more with Wolfram|Alpha pro

Query made by

@sage sonnet

[upper karma]

=wolf sin(-7)

[charred spear]

Do more with Wolfram|Alpha pro

Query made by

@sage sonnet

[upper karma]

=wolf tan(-7)

[charred spear]

Do more with Wolfram|Alpha pro

Query made by

@sage sonnet

[meager acorn]

to solve this

pi/2 x 4/1 ?

[umbral snow]

The only thing that affects the period is the 1/4

Uhh, yeah, actually that's it. You got it

[meager acorn]

but

the answer says 8 pi

how

[umbral snow]

Wait no, the period is 2π on a normal sine curve, and this one has a horizontal stretch by 4

So think 2π × 4/1

[meager acorn]

2π × 4/1 = 8pi

wait

wait

wait

how did you get 2pi

[umbral snow]

2π is the regular period on the sine curve

[meager acorn]

did you solve for pi?

for pi/2

fuck

i dont get it

i am sorry

[umbral snow]

And this sine curve has a period 4× the regular period

[meager acorn]

sin = 2pi/b

[umbral snow]

I'm not solving for anything here. For example:
sin(x) has a period 2π
sin(2x) has a period π
sin(x/2) has a period 4π

This one is similar to sin(x/4)

[meager acorn]

pi/2 is my ps

for period

i am still not understanding

[umbral snow]

What's a ps?

[meager acorn]

phase shift

pi/2

[umbral snow]

Your phase shift doesn't affect the period

[meager acorn]

oh....

since its sin1/4

2pi/b

[umbral snow]

Period is the length necessary for the sin curve to "restart"

[meager acorn]

plug 1/4 in b

[umbral snow]

This length doesn't change if the phase changes

[meager acorn]

2pi/1/4

2pi x 4

8pi..

oh

p.s doesnt not effect the bx

which is period

so for this one

2pi/b = 2pi/1 = 2 pi?

[umbral snow]

2π/4, since this is a horizontal compression by 4

[meager acorn]

1/2 pi

this is my 1st time learning trig in college never took trig in highschool

[umbral snow]

= pup graph 3sin(4(x + π/2))

[meager acorn]

i am sorry if i sound stupid

[meager acorn]

i am want to understand it

[umbral snow]

You can see the period on this is about 1.6

[meager acorn]

yes

[umbral snow]

== π/2

[charred spear]

π/2 = 1.5707963267949

[umbral snow]

So the graph agrees with our value!

[meager acorn]

i see..

i am sorry about this question i just want to pass my exam tomorrow

[umbral snow]

Feel free to ask as much as you need, that's what we're here for

[meager acorn]

thank you

i really want to take calc 1 next semaster

i have 1 more test after this and if i can pass that exam i can pass my class as B

sin = csc
cos = sec
tan = cot
@umbral snow is this correct?

[umbral snow]

You mean
1/sin(x) = csc(x)
ect?

[meager acorn]

ok so

[umbral snow]

1/cos(x) = sec(x)
1/tan(x) = cot(x)

[meager acorn]

SOH = csc
CAH = sec
TOA = cot

okok i see

Did i solve this correctly?
Its asking about 6 trig function

i have a question about this

if your trying to find exact value
so the answer is just pi/2

Is this correct?

[umbral snow]

I can't read your paper, but I agree the answer is just π/2

Note that cos¯¹ and cos cancel, leaving the π/2

[meager acorn]

i think i need break i studied straight 7hrs

thank you for help kaynex

[silent sparrow]

Hey guys im having a problem putting a problem in my ti 83 plus

Idk how to put it in correctly

[limpid basin]

~ rotate

~rotate

[silent sparrow]

Sorry will take another pic

[meager acorn]

there you go @limpid basin

[silent sparrow]

Oh ty very much

[meager acorn]

photoshop is nice sometimes

[silent sparrow]

Im sorry temp im not really good at trig i know you messaged me

Im sorry

[meager acorn]

i am not good at trig too 😦

[silent sparrow]

Yea i wana matain my B lol

I keep getting this

[meager acorn]

Did you change radian to degree

[silent sparrow]

Yes its in degrees

I feel like it has to do something with parentheses

@limpid basin

[limpid basin]

have an essay soz

[silent sparrow]

Its ok

Im sorry

[meager acorn]

how do you solve this?

[keen aspen]

So the maximum sin reaches is 1

And that's when the argument is pi/2

+2pi

2t/5=pi/2

5pi/4=t

[meager acorn]

where did you get pi/2 from

[keen aspen]

Because sin(x)=1 when x is pi/2

[rain tulip]

Because that’s when sin is maximized

[meager acorn]

i m lost

how is this possible..

[rain tulip]

When sin is 1

1050 + 120(1) = 1170

@meager acorn

[meager acorn]

so... if the time t is = 2

it would be
1050+120(2)

?

[rain tulip]

Noooo

It would be 1050 + 120(sin(4/5))

[meager acorn]

why

[rain tulip]

Because

[meager acorn]

how did you get sin4/5

[rain tulip]

U plug in 2

Dude

You plug in 2

Into the function

[meager acorn]

ok um

since time t = 1

you plug that into function

[rain tulip]

?

[meager acorn]

right?

[rain tulip]

What

Who said that time was 1

[meager acorn]

where did sine x =1 come from

[rain tulip]

You want to maximize

The

Function

The range of sin is [-1,1]

The max value sin can take on is 1

[meager acorn]

oh....

wait so for this question i can plug -1

one sec

[rain tulip]

No

[meager acorn]

this ^

they say largest and smallest elk

[rain tulip]

Ok

[meager acorn]

so i can plug in -1?

[rain tulip]

Range and domain are different

The max output is 1

To get that max output, you need to have sin(pi/2)

Since sin(pi/2) = 1

[meager acorn]

so for this one.. than

[rain tulip]

Just use min and max outputs for sin

[meager acorn]

can you type it in equation form

[upper karma]

Sin(4t/5) = 1 is when 4t/5 =pi/2 solve for t
sin(4t/5)=-1 is when 4t/5 = 3pi/2 solve for t

@meager acorn

[meager acorn]

can i ask how did you get pi/2

3pi/2

[upper karma]

Sin is max at every P/2 with 2pi period so it's pi/2 5pi/2 9pi/2 etc

Sin is min at every -pi/2 with 2pi period so it's -pi/2 3pi/2 7pi/2 etc

Alternatively U can draw sin function

Added the projections of angle

[meager acorn]

i got it 😄

next problem that i am working on

[upper karma]

I name axises like this , they actually y and X btw

[meager acorn]

i am not sure where to start

[upper karma]

Link

[upper karma]

Or screenshot

[meager acorn]

do i have to find hyp on right side?

[meager acorn]

is this mean i have to find both side of hyp

[upper karma]

U can use the tangents

So no need for the hyp

[meager acorn]

tang = opp/adj

[upper karma]

Yerp

It is actually arctan

[meager acorn]

sin/cos

[upper karma]

Wait a sec

[meager acorn]

arct means

[upper karma]

Overseas of tan

It's like what angle gives you the following tan value

[meager acorn]

so if its asking entire race

than i would add the,?

[upper karma]

Alpha and beta

Sum

[meager acorn]

ok one sec

[upper karma]

162.6 degrees

[meager acorn]

how did you get that

[upper karma]

Arctan(Alpha) + arctan(beta)

[meager acorn]

wait

so arctan (alpha) is 440/90 ?

can you tell me which alpha is as equation

[upper karma]

Nono arctan(440/90) is some angle

[meager acorn]

so arctan alpha + arctan beta

what is alpha as?

[upper karma]

NO

Arctan(440/90) + arctan(880/90)

Tan(Alpha) =440/90
Tan(beta) = 880/90

[upper karma]

It's arctan

Arctan is same as tan^(-1)

[meager acorn]

its inverse

[upper karma]

Just notations EU/NA

Yerp

But not the 1/tan

Americans use tan^(-1) (number)

Europeans use arctan(number)

[meager acorn]

i see..

[upper karma]

Power will be at the end like Tan(number)^(-1)

If U want 1/tan

[meager acorn]

wait how did you get

162.6 degrees

tan(440/90) + tan(880/90) ?

@upper karma

[upper karma]

Arctan

Not tan

[meager acorn]

so tan^-1

[upper karma]

Ye

But it's not the 1/tan

[meager acorn]

nono

thats different story if it was 1/tan

[upper karma]

Like this

Arctan of 1 is 45 degrees

Tan(45degrees)=1

[signal hemlock]

Can someone help me with the angle of the dangle

[upper karma]

TF is this

@signal hemlock post Ur question

[keen aspen]

Lmao

[uneven ruin]

Hi, can someone help me out on this one ? I asked about it yesterday, I need to prove that N is 90 degrees/that BE is perpendicular to AN which is basically the same, details given in left side. Thanks in advance.

[upper karma]

The angle N is not clearly seen.

[uneven ruin]

Look closely, it’s above the letter M

[left folio]

The triangle is isosceles and DA is perpendicular to BC. DE is also perpendicular to AC and makes a 90° angle. Now we know that the known part of the angle CED is 90°, and therefore the rest of E is 90° in total because the sum of all E should make 180°. We can't know BED possibily but we still know that BEA must be >90°. I don't have enough patience to do the rest sorry

[uneven ruin]

you mean BEA must be higher than or less than 90 degrees ?

I don’t know the difficulty level on this one, but should I post it in advanced geometry?

[left folio]

BEA is lower than 90

[uneven ruin]

Yeah that makes sense.

[left folio]

It's not advanced

[uneven ruin]

Ok thanks 🙏 I will be waiting then.

[left folio]

You should be able to prove EBA°+(90-BEA°)=BEA°

Sounds more complicate than it is

[round isle]

sorry, if this seems like a trivial question but does anybody here remember how to rotate a coordinate by a given angle about another coordinate in the same reference plane? I am not talking about the origin of my reference plane (I can simply use the rotation matrix to do that) but about another coordinate that is in my reference plane. I seem to recall there was some exact formula to do this but I can't remember exactly what it was.
Or will rotating about the origin of my reference plane using the rotation matrix and then simply adding the value of the coordinate, about which I wanted rotation in the first place, give me the exact same answer?

[jovial axle]

I don't see why the second paragraph wouldnt work

treat it as a vector where one part is one rotation and the other is translation

[viscid carbon]

Does anyone know about rotations around a point that is not at the origin

[dim vale]

Can someone help ?

[steady sleet]

Don't post the same question in multiple channels. Don't ping individual users.
#❓how-to-get-help @dim vale

[upper karma]

help I didn't study

not sure this pic even corresponds with this channel

[dusty coral]

@upper karma Which question do you need help in?

[upper karma]

both above

[dusty coral]

You can see that for problem a, both angles are vertical angles.

Therefore they are equivalent

2x + 4 = x + 12

[upper karma]

I got x = 8

[dusty coral]

correct

For the second equation, both angles are supplementary, therefore they add up to 180.

Therefore (5x) + (10x-20) = 180

[upper karma]

mm

I got that wrong

[dusty coral]

what did you get

and what was your process

[upper karma]

5x = 10x - 20

they aren't equivalent

[dusty coral]

?

They're not suppsoed to be

[upper karma]

yes ik

[dusty coral]

they add up to 180

[upper karma]

can you tell me the value of x then

im struggling

nvm

@dusty coral thanks

[upper karma]

help

[pseudo ferry]

They are supplimentary angles @upper karma

Since those lines are parallel

"Parallel Lines Theorem
If two lines are intersected by a transversal, then alternate interior angles, alternate exterior angles, and corresponding angles are congruent. The converse of the theorem is true as well. If two corresponding angles are congruent, then the two lines cut by the transversal must be parallel."

[left folio]

"Let C be a circle of radius r. Let A be an arc on C subtending a central angle θ. Let B be the chord of C whose endpoints are the endpoints of A. (Hence, B also subtends θ.) Let s be the length of A and let d be the length of B. Sketch a diagram of the situation and compute the following limit:"

[hard gale]

it's =tex

[left folio]

Oh thanks

=tex \lim_{θ\to 0}

[hard gale]

so lim of s/d ? (sorry i was busy for a sec) @left folio

[left folio]

Yup!

[hard gale]

so did you determine s and d (in function of theta)?

[left folio]

I haven't started it, i would've liked more to let someone else do this and take it as an eaxmple for futute ones since i don't really know how to start

[hard gale]

ah k lel

so that's our situation basically

[left folio]

lol neat

[hard gale]

(incredible paint skillz hehe)

so our arc A along with the chord B

and the goal is to find the lengths of A and B in function of theta as i kinda suggested

[left folio]

lol ok

[hard gale]

so let's start with A (cause it's the easiest)

it's just a fraction of the total perimeter of the circle

(and let's just use radians as our angle unit cause the expressions are much nicer than with degrees)

or we can start with degrees if you want?

[left folio]

Radians i prefer if you don't mind

I'm used to those, and sorry i was gone for a bit

[hard gale]

okok

so our A has length $$\theta r$$

[hard gale]

(it's like our usual 2 * pi * r generalized to any fraction of the circle)

and now there's that B.....

do you know the cosine rule ?

[left folio]

Ye i think

Cos(x)=2pi/3 or something similar? lol

[hard gale]

(it's a relationship between angles and lengths in a triangle)

[left folio]

Well then i think i don't have it

[hard gale]

cause i don't really see any other way to find B

[left folio]

Tell me please

I don't remember them atm, it's like 10:30pm and i have to go in a bit

[hard gale]

in any triangle

we find this relationship (ie a generalisation of pythagoras)

(and we can choose any side for our c, as long as the other sides/angle are selected accordingly)

here our c could be the length d of B we're searching for

https://cdn.discordapp.com/attachments/326138757474680852/509474154530078727/unknown.png

so we'd have $$d^2 = 2r^2 - 2r^2\cos{\theta}$$

[uneven ruin]

Bump, can someone help me out on this one ? I asked about it a few days ago and after countless tries you guys are my last hope, I need to prove that N is 90 degrees/that BE is perpendicular to AN which is basically the same, details given in left side. Thanks in advance.

[dire rampart]

there is no N in your diagram?

[upbeat hound]

I'm trying to find "k" in a problem where it's Find the value of k such that the line containing point (2, k) is perpendicular to the line y= 2x-3 at point (4, 5)

Any ideas on how to solve this? I'm stumped

Yeah I'm bad at math lmao

Find y and solve for x?

Ok lol

Oh ok, I think so

Yeah

Ok thanks man

t!rep saltyb boi

[loud cedar]

🆙 | Pantasaurus has given @upper karma a reputation point!

[upbeat hound]

np

[left folio]

@hard gale I'm very thenkful for the help, but i don't think that is the way it's supposed to be intended that way of solution

[hard gale]

well that's the rigourous way of proving it, you could also just say that the more you reduce the angle the more the section of arc identifies to just a straight line, hence lim(t->0) s/d = 1 @left folio

[left folio]

@hard gale I need to ask you one important thing i didn't get. What angle is it referring to in the whole limit? Te chord makes like 4 angles doesn't it?

[hard gale]

(look at my figure again, i put it in there)

[left folio]

Mmmh i see

I mean what do i do by proving the angle can become a line if limit tends to 1 as a solution?

But if the limit is calculated to be 1 i guess it would be a line?

[hard gale]

that's what i was doing yesterday : actually computing the two lengths then calc the limit of s/d

idk how formal your thing needs to be so :/

[left folio]

Well i don't know much about trigonometry, so not the "advanced" formula you showed me yesterday; i can't really tell you all i've done though

You can try compute it the easiest way you can, and i'll do it too. No problem for the method at this point @hard gale

[hard gale]

(actually when i have the two lengths idk how to calc the limit so kek : i just know it's 1 from wolfram alpha when we consider only positive values for the angle)

[left folio]

lol yikes

I guess i'll ask my teacher and let you know in 2 days

[hard gale]

ye lel ^^

[sick veldt]

Are my answers right

Oops

I don't think A and B can be right

Cause the angle numbers above an un-parallel (also known as C) can't be congruent right?

[sick veldt]

<@&286206848099549185>

[upper karma]

I am not a helper but your answers that you checked are correct

angles 9, 10, 11, and 12 are unknown and there is no way of knowing them or their relations with the line a and b with the given information except that none of the angles are congruent with 1,2,3,4,5,6,7, or 8

[sick veldt]

ok ty

[meager acorn]

Did i simplify the answer wrong?
Can someone help me with next step?

[rare scarab]

Guys is the normal reaction force of a rod hinged and resting on a smooth support 90 to the rod or the support ?

Like to the target of touching which is the rod

Or the surface at which the support is fixed

[meager acorn]

Tell me if i did this ok

[rare scarab]

Seems right

[quick grove]

Hi guys, what does the symbol "Phi" mean in this question? I don't understand these equations at all. The book seems to just throw them at you without explanation.

I'm really trying to understand all the math behind this graphics stuff, but there's so many curve balls haha

[umbral snow]

@quick grove
φ is taken as the angle from the top of the sphere, down to the point in question

[quick grove]

Ah! Thank you so much, is there anything I can research in particular to make this make more sense?

I am looking at "unit spheres" at the moment now

[umbral snow]

If you understand polar coordinates well, this is a pretty natural extension. If you don't, I recommend you start there.

If you have any questions in particular, feel free to post them

[quick grove]

Excellent, I will start there thank you again.

[winged arch]

Can someone explain how to do number 1 ?

I have been Trying to figure this out but I do not understand it

[winged arch]

like my teacher got 24= (x-24) /2

I do not understand how he got this equation or which of the angles go where

the answer is apparently x=72

[dry cedar]

Hi guys

It's a bit physics ish but I think it's ok to ask here too

https://cdn.discordapp.com/attachments/338590819151904768/510037445593661440/unknown.png

I'm looking to count the number of particles with speed outside of this cone, we assume a maxwellian distribution and align the velocity coordinate system $$(v,\theta,\phi)$$ such that $$\theta$$ is counted the same way as shown on the figure

[dry cedar]

We further assume that the distribution is independant of $$\phi$$

[dry cedar]

This is what they have in the PDF I found

I don't understand the boundaries of $$\int_{0}^{\theta{tr}}$$ or the factor of 2 in front of the integral

[dry cedar]

If I sum all the particles with speed between 0 and theta_tr, I get only the upper cone, I have a feeling that to account for the lower cone they just doubled that hence the factor of 2, but how do I know there are the same number of particles in the lower cone?

[dry cedar]

In case anyone is interested after reading this I got it

first of all, the distribution $$f$$ is assumed independant of $$\theta$$ and $$\phi$$

[dry cedar]

to get the two cones, we're integrating in this domain: $$[0, \theta_{tr}] \cup [\pi - \theta_{tr}, \pi]$$

[dry cedar]

but with the proper substitution we have $$\int_{\pi - \theta_{r}}^{\pi} \sin(\theta) = - \int_{\theta_{tr}}^{0} \sin(\pi - u) du = \int_{0}^{\theta_{r}} \sin(u) du$$

[upper karma]

Why is the answer sin3a

If sin(a+b) = sinBcosA + cosBsinA

Wouldn't it make sense for sin (2a+a) cos (2a + a)

So sin(3a)cos(3a) as the answer

[keen aspen]

No it would be sin(3a)

[upper karma]

Care to explain why

Wait nvm I figured it out

[marble perch]

Can someone help me

I have my exam today in geometry and am having trouble on this one type of problem

Number 6

the one the arrow is pointing to

[jaunty plume]

so you basically go from left to right

[marble perch]

I know that the angle vertical of the 45 is 45 degrees

[jaunty plume]

yeah, red is 45

[marble perch]

And then you use 180 for the green to get 76

[jaunty plume]

then you use the fact that angles in a triangle sum to 180 to work out green

[marble perch]

*5

[jaunty plume]

yup

[marble perch]

how do you do blue?

[jaunty plume]

angles on a straight line add up to 180 too

[marble perch]

where does the 68 come into play

[jaunty plume]

green + 68 + blue = 180

[marble perch]

so 37

is blue

and then 43 yellow

[jaunty plume]

yes

[marble perch]

can you explain purple

[jaunty plume]

it's the same as green and blue

[marble perch]

my teacher said it was 137 but i didn’t get how

[jaunty plume]

angles on a straight line add up to 180

except here there's only 2 angles

[marble perch]

would it be 37

not 137?

[jaunty plume]

no

yellow is 43

[marble perch]

yes

[jaunty plume]

and purple + yellow = 180

[marble perch]

oh

last question, I get it now but I was confused because I thought the 100 degree angle came into play with that

[jaunty plume]

well it sort of does

100 + blue = purple

that's only because they chose nice numbers though, and it typically doesn't work like that

[marble perch]

kk

can you explain the other arrow?

[jaunty plume]

yeah

the little square in the corner indicates that it's a right angle

so it's 90

[marble perch]

yes

[jaunty plume]

so step 1 would be to work out either purple or green

using angles on a straight line = 180

[marble perch]

yeah

[jaunty plume]

purple + 120 = 180

so purple is 60

[marble perch]

k

which makes both greens 60

[jaunty plume]

yeah

[marble perch]

cause of vertical < ‘s

1= 30

[jaunty plume]

ya

[marble perch]

and 2 is?

[jaunty plume]

green + blue = 180

[marble perch]

ok thx a lot 😊

[jaunty plume]

[marble perch]

that emoji is cute xD

@jaunty plume sorry 😂 I just want to know what they mean by classify

does it mean like ASA, SSS...?

actually that still doesn’t make sense lol

my teacher never told me if that was the right triangle to classify

i just assumed

[jaunty plume]

equilateral, scalene, right, isosceles, acute, obtuse

[marble perch]

oh alright 😂

I don’t think that triangle can be classified so hopefully if it’s on the test, it’ll be fixed

alright thx sorry for the trouble

[jaunty plume]

it can be classified

the three angles are 83 17 and 80 so it's scalene

[graceful oxide]

the max and min points for sine are the same as cosecant right?

So the derivative of one shd be equal to the derivative of the other

[meager acorn]

does anyone here good with Analytic Trignometry

sec^2y - cot^2(pi/2-y)
where do i start this problem ?

it says i need verify the identity

<@&286206848099549185>

[drowsy spoke]

what is it equivalent to?

[meager acorn]

i dont know where to start...

[hard gale]

where's the identity?

[drowsy spoke]

^^

[meager acorn]

sec ^2 = 1/cos^2 y

[drowsy spoke]

not that

[hard gale]

no i mean the whole thing

[meager acorn]

cot^2 =cos ^2y /sin ^2 y

[hard gale]

you want to show sec²(y) - cot²(pi/2 - y) = to what?

[meager acorn]

1

= 1

[drowsy spoke]

phew

[hard gale]

dem

[meager acorn]

?

[drowsy spoke]

they're just fractions

add them up