geometry-and-trigonometry

[spark stag]

but those authors

are idiots

and best ignored

[fossil ledge]

yea i dont like the circle notation

[hard shard]

mathematical snowflakes

[spark stag]

darkrifts

i too love getting my function composition

mixed up with higher derivatives

[vagrant elk]

k e k

just use parenthesis smh

[hard shard]

just stack em up on each other

[vagrant elk]

But uh, if you take your little superscript to mean how many times you iterate the function, you then get f^-1 as the way to un-iterate it

not the same as the $x^{-1} = \frac 1 x$ you see in real numbers

[supple abyss]

{}

[somber coyote]

Darkrifts:

[spark stag]

darkrifts sure

but this notation

is stupid

[hard shard]

use 👏 arcsin 👏

[vagrant elk]

i agree, esp with trigonometric stuff

[spark stag]

at least while theres ambiguity

[fossil ledge]

arcsin^-1 x = sinx

[spark stag]

if it was more standardized

it'd be fine

[hard shard]

reee

[vagrant elk]

standardization would be nice for notation

[supple abyss]

$\sin^{-1}$ is abuse of notation because $\sin$ isn't injective, don't @ me

[fossil ledge]

Kill the heretics, easy clap

[somber coyote]

CaptainLightning:

[vagrant elk]

nah fam, it's just the preimage modulo an equivalence relation

Anyway, sin^-1 x is not the same as 1/(sin x)

but sin^2 x = (sin x)^2 so fuck why not make everything horrible

arcsin or bust

[devout wagon]

How was the sine and inverse sine function invented?

[spark stag]

what do you mean, "invented"?

[devout wagon]

e.g. looking at x = π/3, sin x = x - x^3/3! + 5π/5! - ...
But how do you get inverse sine?

[spark stag]

you mean how they're calculated?

[devout wagon]

Well I'm trying to learn where the trigonometric functions come from

yes

[spark stag]

well, literally we define inverse sine as just... the inverse of sine (over a certain domain), and that definition is enough. that doesnt really help us computationally, though

we can define it as:

$\sin^{-1} x = -i \ln \left(ix + \sqrt{1 - x^2} \right)$

[somber coyote]

Namington:

[spark stag]

(with branch cutting)

that said, this isnt really how its calculated in practice

the way computers calculate trig functions is universally using a lookup table for arctan values paired with an algorithm

that algorithm is CORDIC

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

this is really efficient, but it's also super impractical and hard to follow for humans

and its imprecise

the precise values are given by the monstrosity i gave above

(accounting for branch cuts)

but again, all these ways to calculate it sort of distract from its "purest" definition

the arcsin is, over a certain domain

the inverse of sine.

that's it.

its perfectly fine to define a relationship like that

although it may seem unsatisfying

oh and btw the i definition i gave above works buuut

for real x, you might find it more practical to use

$\sin^{-1} x = \int_{0}^{x} \frac{1}{\sqrt{1-t^2}} \dd{t}$

[somber coyote]

Namington:

[spark stag]

...i mean, neither of these expressions are ever gonna be practical

but that one's a bit nicer

you can evaluate this via binomial theorem to get an infinite series

but why would you

this does let you find the maclaurin series of arcsin, but again, why would you

[upper karma]

imagine asking why would you

[random spear]

How do you solve trig equations like sinθ = -0.3?
I know how to do it for unit circle values but I'm missing something for decimals.

[fringe dirge]

With a calculator

[random spear]

I have to show my work. Find all solutions, yknow?

I take the inverse function and then I'm not sure whether to subtract from π or 2π

Then add 2kπ

[hard shard]

Drawing a graph helps with inverse trigs

Draw a sine graph and a line for -0.3

It should help to know where the symmetry is and stuff

[worthy root]

I Agree that $sin\theta <1$ but some people same $ -1\leq sin\theta \leq 1$

[stuck torrent]

That’s also true though

[worthy root]

How can sin be negative bruh

Also how can it be equal to 1

[stuck torrent]

,w graph sin(x)

[somber coyote]

Display results online and refine query

Requested by Whoever#5822

What I have started will end:

[worthy root]

I sold know graph

Dont*

But side opposite to angle< hypotenuse

So sin is less than 1

Also side can never be negative so how 🤔

[dark sparrow]

the triangle definition of sin only really works when your angle is between 0 and 90°

afterwards you need the unit circle

[worthy root]

Ok

[worthy root]

I am trying to be done with right angle trig today .

[native night]

how I find intersection point of sphere tangent to plane and line between its "north pole" and point on plane

[stuck torrent]

Riemanns sphere?

Maybe try this

https://en.m.wikipedia.org/wiki/Stereographic_projection?wprov=sfti1

@native night

[native night]

that 1

[raw willow]

Hiii, I got a math problem which I found in my Unity programming code

I have to reflect p1 as p2

We know coords of: s1, e1, s2, e2 and p1

We have to calculate p2 (x2, y2)

as you can see s is a starting point and e is end point of segment

second segment is rotated and decreased length to visualize a problem

[supple abyss]

$p_1(3+0.8\cdot (5-3),5+0.8\cdot (9-5))\ p_2(4+0.8\cdot (2-4),14+0.8\cdot (16-14))$

[somber coyote]

CaptainLightning:

[lilac fox]

It’s the 2nd one. Halp pls, super easy question but I can’t applied math, thanks 😄

[fossil ledge]

draw a diagram

ladder is 9m long

thats the hypontenuse

its also 3m above the ground

you know dx/dt = 3m and you are solving for dy/dt

you also know x^2 + y^2 = 9^2

@green coral in 5 mins if you cant figure it out from there

[lilac fox]

Thank you

Ah ok so implicit differentiation, I think I got it now

[fossil ledge]

mhm

with respect to t

@lilac fox completed?

your dy/dt should be negative

[lilac fox]

umm ok apparently I got it wrong

[fossil ledge]

mind showing me your steps

the implicit differention mostly

[lilac fox]

ah wait I think I see my mistake

I didnt do implicit wrt t, I did with y

on the x^2 + y^2 = 81

[fossil ledge]

ah

dont do that, do it with t

[gentle fog]

Can somebody help me with this? I've been doing it for so long and I did the work and I get the answer 1. But apparently that's wrong and I'm honestly just confused

[spark stag]

are you given the identity $\tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}$?

[somber coyote]

Namington:

[gentle fog]

Yeah.

[spark stag]

so, can you find tan(theta)?

[gentle fog]

Yeah. According to the graph, it's 5/-sqrt(39) (Rationalized -5sqrt(39)/39)

[spark stag]

yep, and tan(pi/4)?

[gentle fog]

I dunno how to use the TeXit sorry lol

But that should be 1.

[spark stag]

yes

so we have $\frac{\frac{-5\sqrt{39}}{39} + 1}{1 - \frac{-5\sqrt{39}}{39}}$, yes?

[gentle fog]

Yeah

[spark stag]

can you simplify that?

[gentle fog]

Okay, give me a sec

[somber coyote]

Namington:

[spark stag]

just making it look nicer

i think you somehow flipped a negative to a positive, or vice versa, which is how you got 1

[gentle fog]

So simplified, it should be (-5sqrt(39)+39)/39+5sqrt(39))

[spark stag]

you should be able to get $\frac{-5\sqrt{39} + 39}{39 + 5\sqrt{39}}$

yeah

[somber coyote]

Namington:

[gentle fog]

Yeah

[spark stag]

now, we should probably rationalize that denominator

[gentle fog]

So multiply by the Reciprocal(I think it's called but 39-5sqrt(39))

[spark stag]

yes

[supple abyss]

conjugate

[spark stag]

^

the reciprocal is the fraction "flipped"

so like the reciprocal of 5/8 is 8/5

the reciprocal of 12 is 1/12

(more precisely, it's what you get when you raise something to -1)

here, conjugate is the correct term

[gentle fog]

Ohh the Conjugate. My bad lol

But I did the math

And it should be

(2496-390sqrt(39))/546)?

[spark stag]

yeah, but those are some big numbers

can you simplify them at all?

(divide by GCD)

...though ew, this GCD sucks

[gentle fog]

Yeah lol. It really does

[spark stag]

probably gonna have to use multiple steps

[gentle fog]

(32-5sqrt(39))/7

[spark stag]

yep, that seems good to me, unless i made a mistake somewhere

this is a bit of a convoluted derivation

,w tan(2pi - arcsin(5/8) + pi/4) = (32-5sqrt39)/7

[somber coyote]

Display results online and refine query

Requested by Namington#1983

[spark stag]

yep, seems good.

[gentle fog]

Alright, Thank you so much man lol

I needed the help

[worthy root]

I have to prove this

I have no idea

How to start

[zenith ember]

Cross multiply?

[worthy root]

Umm

I have to prove that R.H.S =L.H.S

[zenith ember]

Yes...

[worthy root]

Also I figured out how

[zenith ember]

Ok

[worthy root]

Eh I got stuck lol @zenith ember

I divided the numerator and denominator by costheta

And I am stuck

[zenith ember]

You're trying to go from the lhs to the rhs.

[worthy root]

Yes

[zenith ember]

And I am saying, instead of doing that you can just treat it like an equation.

And show that it reduces to the same thing.

[worthy root]

But the exercise is like . Prove the L.H.s=r.h s

[zenith ember]

So start by getting rid of the denominators with cross-multiplication.

I mean, if you need to, you can later back out the appropriate steps.

But honestly it's totally valid to do it the easy way.

[worthy root]

Yeah working backwards. Ok

@zenith ember ? Is there any use of proving these identities in physics ?

[zenith ember]

I wouldn't bother to memorize them

But they might be useful practice for manipulating trig functions.

[worthy root]

I mean

I know the Pythagorean identities

And I can use them

But proving two trigonometric identies defined for acute angle is a pain . Thinking of dropping the section

[final garnet]

if pie= 180 degree then what is 3.14 stand for?

so if i do i will get 360/pie and 1 pie is 180 so the answer will be 2

is no solution acceptable in this case

[honest bay]

can you be clearer

you were asked to convert each radians measure to degrees

which part do you have problems with?

[final garnet]

yes, im not sure how to converting a number without pie

[honest bay]

the same way with converting a number with pie

[final garnet]

so the answer will be 360/pie

[honest bay]

yes

[final garnet]

and degree can have pie in the value?

[honest bay]

pie is just a number

a constant

not every radians measure has pie

and not every degrees measure does not has pie

[final garnet]

i see, thanks you

[honest bay]

also, pie does not equal 180

pie radians = 180 degrees

so 360/pie degrees is not 360/180

@final garnet

it's like saying 1km = 1000 m

but 1 is not equal to 1000

[chilly ocean]

when making a triangle

can I have a the side lengths of 1 2 and 3?

or is it not possible because 1+2=3?

[zenith ember]

Not possible, for the reason you gave

[chilly ocean]

ok, thanks

[gritty flare]

@honest bay i would like to believe that pie is a dessert

[zenith ember]

A delicious dessert

[gritty flare]

Ssssssssss

[zenith ember]

And not a sandy, dry place

[gritty flare]

I to this day forget which one is which

[zenith ember]

Hehe

[final garnet]

so im evaluating the inverse tangent of 0
tan is y/x
but does inverse tangent effect it

the inverse trigonometric giving

do i need to apply this formula to the equation?

[supple abyss]

no...

[dark sparrow]

arctan(0) is the answer to the question "which angle between -pi/2 and +pi/2 has 0 as its tan?"

also, pi != pie

[plush moat]

Anyone online, I can't figure out this word problem.

[dark sparrow]

post it

[plush moat]

Is it okay if I write it or do I need to find a diagram

[dark sparrow]

there is always someone online on a server of thousands

[plush moat]

Nvm

[dark sparrow]

if your problem includes a diagram you'd do well to post that

[plush moat]

I've solved e, so I have 1 angle and 1 side length.

I don't really understand what to do with the ratio but I know it's important

[dark sparrow]

well

the ratio you're given is the ratio between the rise of the roof (here, d) and the corresponding horizontal run (here, e)

in other words, you're given $\frac{d}{e} = \frac{5}{12}$

[somber coyote]

Ann:

[plush moat]

I'm more confused, how can e be 12 if I solved it and got 5

[dark sparrow]

i'm not saying e = 12, though

[plush moat]

Oh okay, two different e's?

[dark sparrow]

i'm not saying there are two different e's either

just because two fractions are equal does not mean that their numerators and denominators are equal individually.

after all, $\frac{4}{6} = \frac{2}{3}$, yet $4 \neq 2$ and $6 \neq 3$

[somber coyote]

Ann:

[plush moat]

Ohh

Okay, makes sense.

If the run is 12, does that mean c is 12?

[dark sparrow]

no

your run is not 12

and the run is e anyway, not c

[plush moat]

So I have one length, so I need one more length to solve the rest of them?

Basically, I need to obtain an angle besides 90

[dark sparrow]

you can solve for $d$ using the equation $\frac{d}{e} = \frac{5}{12}$ and your value of $e$

[somber coyote]

Ann:

[dark sparrow]

after that, once you know all the sides, it's a matter of elementary trigonometry to get the angles

[plush moat]

I'm a little too dumb to solve that equation

Unless it's 5 divided by 12

[dark sparrow]

wdym by "it"

yes, the right hand side is 5/12. that's what i've written.

[plush moat]

5/12 = 0.41

[dark sparrow]

no

no decimals

leave the fraction as is

[plush moat]

So 5/12 is one of the sides?

[dark sparrow]

in the equation as it is now, 5/12 is the right-hand side. i don't see how explicitly stating that information would help you, but whatever floats your boat.

[plush moat]

I'm just pretty lost right now sorry

[dark sparrow]

ok let's start over

do you understand where the equation $\frac{d}{e} = \frac{5}{12}$ comes from in the first place?

[somber coyote]

Ann:

[plush moat]

d is the rise, the straight line, or the height of the shed, e is the run, the line on the bottom?

[dark sparrow]

yes

ok so like

do you understand where the equation comes from

[plush moat]

Not really

[dark sparrow]

the problem tells you that the roof pitch is 5:12

what this means is that the ratio of the rise to the run is

well

5:12

do you understand that

[plush moat]

Yes

[dark sparrow]

ok

so

$\frac{d}{e} = \frac{5}{12}$

[somber coyote]

Ann:

[dark sparrow]

is this clear now

i don't want to have to come back to this and explain it a second time. i want to make sure you're good on this before we move on, and i'm not going anywhere until you explicitly give me the OK to do so.

[plush moat]

the letters are the rise and run, since the ratio is 5:12, that means those two letters are 5/12 respectively (not as in the solution, just in this ratio)

[dark sparrow]

...ok, that's some weird wording if i ever saw any

ok but like

do you understand where this equation comes from, yes or no

[plush moat]

Yeah

[dark sparrow]

ok

good

so now

can we continue

[plush moat]

yeah

[dark sparrow]

ok

so i take it that earlier, you had obtained e = 5. is that correct?

[plush moat]

yeah

[dark sparrow]

so to cut through the bullshit, we might as well just insert that into our equation

to get $\frac{d}{5} = \frac{5}{12}$

[somber coyote]

Ann:

[dark sparrow]

do you understand what just happened

again i'm not continuing until i make sure you understand everything we're doing. the last thing i want is to lose you again.

[plush moat]

the five on the left side is the e

[dark sparrow]

i'm not asking you to explain what happened in your own words (as beneficial as that'd be). i'm asking you whether you understand what is happening.

and i expect a yes or no answer.

do you understand what just happened, yes or no

[plush moat]

no

[dark sparrow]

$\frac{d}{e} = \frac{5}{12}$

[somber coyote]

Ann:

[dark sparrow]

we started with this equation

then, we remembered that you had found $e$ earlier, and you had found $e = 5$

[somber coyote]

Ann:

[dark sparrow]

as such, we rewrote the equation to $\frac{d}{5} = \frac{5}{12}$, replacing $e$ with $5$.

[somber coyote]

Ann:

[plush moat]

Okay I understand

[dark sparrow]

brb

[plush moat]

sure

[dark sparrow]

ok, i'm back

@plush moat are you still here?

[plush moat]

yep

If this is 10 feet wide, wouldn't it not be 12

[dark sparrow]

what do you mean by "it"

[plush moat]

5/10

[dark sparrow]

what 5/10

where did you get 5/10 from?

[plush moat]

5:12 roof pitch, but the roof is only 10 feet

[dark sparrow]

no

look

we've got an equation

forget about the geometry for a moment. we are doing algebra right now.

if you try to bring the geometry into this, you risk both confusing yourself and going far astray.

[plush moat]

ok

[dark sparrow]

we now have the equation $\frac{d}{5} = \frac{5}{12}$

[somber coyote]

Ann:

[dark sparrow]

do you understand this

yes or no

i don't want you to launch into an explanation

i just want an answer

yes or no

[plush moat]

Not really

No

[dark sparrow]

:/

you said earlier that you did.

why did you say you understood it when you do not

[plush moat]

well I'm not sure if I do* understand it or if my understanding is wrong

[dark sparrow]

we started with the equation $\frac{d}{e} = \frac{5}{12}$, which you claim you understand. we then replaced $e$ with $5$, using the fact that you found $e = 5$, which you also said you understand. we obtained the equation $\frac{d}{5} = \frac{5}{12}$.

[somber coyote]

Ann:

[dark sparrow]

i was under the impression that this is spelled out as explicitly as possible. what is there to not understand

[plush moat]

D is 5 inches for every 12 inches, so 5 feet for every 12 feet, e is 10 feet so d is 4.1666... feet?

[dark sparrow]

no, e is not 10, e is 5!

...this is frustratingly difficult for me

[plush moat]

I'm sorry

We can stop I'll just try and figure it out

[dark sparrow]

do attempt to figure it out yourself

[plush moat]

I was doing fine with trig until they added a ratio and imperial units

[dark sparrow]

you may need to review ratios and proportions, and probably elementary algebra too

[plush moat]

Probably

I think I got d anyways

[worthy root]

Ratio is basically relative measurement of quantities. For example in a school the ratio of mathy to non mathy people is 1:3 . Note here we mean out of every 4 students 1 will by mathy and 3 will be non mathy . Ratio doesn't tells how many mathy ,non mathy or students were there . It's just says for every 3 non mathy student there is a mathy stud .

@plush moat

[upper karma]

is trigonometría used in triangles?

of course

[worthy root]

Yes

https://drive.google.com/file/d/11514wg5mX4GFTzFG7eC1n7LPhYYFQMKQ/view?usp=drivesdk @upper karma

[night karma]

Made a math problem for those who'd like one

[dusk gate]

quick question

i got vector u = (-6, 5)

i gotta find a perpendicular vector with norm 10

i found the vector v = ( 5, 6)

the norm is 7,81...

i hv no idea how to proceed

[devout shell]

trying to run through some combos in my head as well lol

-6a+5b=0
sqrt(a^2 + b^2)=10

you can always try solving that

[dusk gate]

:v ok

unfortunetly i got no brain power to solve that system

[devout shell]

both work

I graphed the system to get the answers because I didn't want to analytically solve (though I could have done that just fine)

[dusk gate]

ah thx

but where did this equation come from -6a + 5b = 0

[devout shell]

that is what happens when you have two vectors, one of which is normal to the other.

If the vectors are perpendicular to each other, then they have a zero dot product

[dusk gate]

OH

i see

I get negative root

[devout shell]

strange

what's going on in this line?

turned the + into a - there, but that is a mistake

[somber coyote]

⚡Amphy⚡:

[devout shell]

this would be the next correct step @dusk gate

[dusk gate]

oh Dx

i didnt even notice ty

[plush moat]

@worthy root I managed to figure it out by just reducing the formula to mutliplication 😃

[upper karma]

hello, can anyone tell me how i calculate this? https://gyazo.com/7af5878907c302640116a0ea2a5ba070

please ignore the norwegian text, and just read the numbers:L

[fossil ledge]

so from 270 to 360 degrees means we are in the 4th quadrant

keep that in mind

otherwise, notice that tan v = -45/28

tan v = opposite side/ adjacent side

so if you draw a triangle, with sides 45,28 and the hypotenuse

you should be able to derive sin and cos

@upper karma

[upper karma]

im completely stuck i would guess that sin or cos would become 60 and the other 45?

[worthy root]

What ?

@upper karma wdym?

[upper karma]

i am not sure myself, i have 30 minutes of experience xd

[worthy root]

Lol

https://drive.google.com/file/d/11514wg5mX4GFTzFG7eC1n7LPhYYFQMKQ/view?usp=drivesdk

Check this pdf and read it @upper karma

[upper karma]

need too request access

[worthy root]

Ok

@upper karma can you view it ?

[upper karma]

now i can

[worthy root]

If -SinA/CosA then can (-SinA/cosA)^2=tan^2A

[dark sparrow]

yes, tan^2(A) = (sin(A)/cos(A))^2 = (-sin(A)/cos(A))^2

[lost parcel]

Need help with this sum

[dark sparrow]

what is giving you trouble? @lost parcel

[lost parcel]

Idk what to do with that f(16).f(29) I know I have to but the values and multiply but im not getting anything solid enough.

[dark sparrow]

do you have any work to show

[lost parcel]

I tried to simplify it a bit by using the properties of sin2x=2sinxcosx and cos2x=cos²x-sin²x
..

Then I seperated the denominator for each of the terms

So by doing that I got
1/2cos2x- cosxsinx/cos2x + 1/2

I can try to solve it a bit more but it seems I won't be getting anything good

[dark sparrow]

$\frac{(1 - \sin(32^\circ) + \cos(32^\circ))(1 - \sin(58^\circ) + \cos(58^\circ))}{4 \cos(32^\circ)\cos(58^\circ)}$

[somber coyote]

Ann:

[dark sparrow]

simplify this

[lost parcel]

I'll try.

By further solving it I got
Tan58+1/4cos32 - tan32+1/4cos58
And im stuck ..

@dark sparrow senpai help

[dark sparrow]

that looks weird to me, can you show your work

[lost parcel]

I'll get a better picture wait a sec

[dark sparrow]

,rotate -90

[lost parcel]

Multiplied the brackets to get this idk I think im Wrong

[dark sparrow]

........ i'll have you know that you missed at least 4 pairs of parentheses when transcribing the thing into plaintext

[lost parcel]

Ummm..
I don't think I understand you.

[dark sparrow]

Tan58+1/4cos32 - tan32+1/4cos58

this reads like $\tan(58) + \frac14\cos(32) - \tan(32) + \frac14\cos(58)$

[somber coyote]

Ann:

[dark sparrow]

not what you intended it to be

also, your work is really unclear. but i think it's likely that you made some sign errors

[lost parcel]

$\frac{(1 - \sin(32^\circ) + \cos(32^\circ))(1 - \sin(58^\circ) + \cos(58^\circ))}{4 \cos(32^\circ)\cos(58^\circ)}$

[somber coyote]

Beast97:

[lost parcel]

How did I got this?

*u

[dark sparrow]

......i wrote out f(16°), then wrote out f(29°), then multiplied the two fractions together

[lost parcel]

Wait nvm

[dark sparrow]

was that not obvious

also, i strongly recommend rewriting cos(58°) and sin(58°) as sin(32°) and cos(32°) respectively.

[lost parcel]

Ok

[lost parcel]

@dark sparrow. Can I DM u?

[dark sparrow]

uh

sure?

[lost parcel]

Find the value of sin^-1(3/√73)+cos^-1(11/√146)+cot^-1(√3) is.?
A)5pi/12
B)17pi/12
C)7pi/12
D) none of these

So I have substituted the value of arcsin(3/√73) as arctan(3/8) and arccos(11/√146) as arctan (5/11)

Also substituted the value of arccot √3 as pi/6

[upper karma]

$arcsin(3/√73)+arcos(11/√146)+arcos(√3)$

[somber coyote]

fractally:

[upper karma]

do you mean that?

[lost parcel]

Arccot

[narrow sleet]

You just need to find the sum of arctan(3/8) and arctan(5/11) then

[upper karma]

$arcsin(3/√73)+arcos(11/√146)+arcot(√3)$

[somber coyote]

fractally:

[upper karma]

now?

[lost parcel]

Ye

How?

[dark sparrow]

$\arcsin \left( \frac{3}{\sqrt{73}} \right) + \arccos \left( \frac{11}{\sqrt{146}} \right) + \arccot (\sqrt{3})$

[somber coyote]

Ann:

[dark sparrow]

😒

[upper karma]

sorry Ann

[narrow sleet]

@lost parcel Do you know the formula for tan(a+b)?

[lost parcel]

Yeah

[upper karma]

I think it is A

[lost parcel]

Tan a+tan b/1-tana tanb

[upper karma]

you need to racionalize and calculate

[dark sparrow]

!!!!!!

[dark sparrow]

\texttt{a+b/c+d} will ALWAYS be read as $a + \frac{b}{c} + d$ and NEVER as $\frac{a+b}{c+d}$!!!!!!!! @lost parcel

[somber coyote]

Ann:

[narrow sleet]

Find tan y first, then find y

[lost parcel]

Ok

So (a+b)/(c+d) is the correct way?

[narrow sleet]

Yeap

[dark sparrow]

yes!

[lost parcel]

Ok I'll keep that in mind

[narrow sleet]

Remember that tan(arctan(x)) = x

[lost parcel]

Tysm

[narrow sleet]

👍

[royal pawn]

in this problem, where does the 2 come from https://gyazo.com/56657e806a6c301485754bfe517eb4f2

[narrow sleet]

Multiply both sides by 2

[worthy root]

Is sl loney a good book to learn trigonometry?

[shut crest]

Dumb question how many points are needed to find a sine wave?

[devout shell]

I would say 5

three zeros, min, and max point

[shut crest]

ok

[gentle fog]

Hey, can someone help me understand this? I'm completely lost. I understand the formula and how it works but we never worked on a problem with a square root in the Cosine so I'm not sure how to go about this

[gentle fog]

I tried and got that answer in the picture. It was wrong so

[plucky breach]

@upper karma

[plucky breach]

<@&286206848099549185>

[dark sparrow]

what is giving you trouble here

[plucky breach]

Pretty much everythin'

[dark sparrow]

ok but like where are you stuck exactly

do you have any work to show

[plucky breach]

Mostly trash..

Grouping is the deal..

[dark sparrow]

...ok i can't even make sense of what you were attempting to do there

[plucky breach]

Multiplied both of em' n added them

The trouble I'm having is that I can't approach the question properly

@dark sparrow

[dark sparrow]

are you familiar with vectors?

[plucky breach]

Ofc

It's prohibited thou

[dark sparrow]

what do you mean prohibited

[plucky breach]

We can't use vectors,Trigo only!

[dark sparrow]

what do you mean you can't use vectors

[plucky breach]

Our teachers have prohibited us to do so

[dark sparrow]

yuck

🤢

[plucky breach]

There's not much I can do there

[dark sparrow]

idk how they expect you to do this

[plucky breach]

Just by trigo

[dark sparrow]

they expect you to somehow wade through a forest of algebra

[plucky breach]

Or maybe quadratics

I feel you

But I'm confined

Tell me the vector approach,I'll try to translate into Algebra

Ping me when you're done.

[dark sparrow]

i have yet to fully flesh it out but my idea is to use the following vectors:

[cos(a), sin(a)], [cos(b), sin(b)], [cos(c), sin(c)], [cos(d), sin(d)]

naming them, idk, u_1 through u_4 i guess

your system of equations then becomes the vector equation u_1 + 7u_2 = 4u_3 + 8u_4

and i guess cos(a-d) and cos(b-c) can be recovered as u_1·u_4 and u_2·u_3

@plucky breach

[plucky breach]

Yea,I'm admiring em'

Go ahead,solve those please..

[dark sparrow]

...ok, so now that i've given it a little thought it actually becomes trivial with the vector approach

because now you can write $u_1 - 8u_4 = 4u_3 - 7u_2$ and then consider the squared length of both sides

[somber coyote]

Ann:

[dark sparrow]

using $|v|^2 = v \cdot v$

[somber coyote]

Ann:

[dark sparrow]

the purely algebraic equivalent i guess would be writing $\begin{cases} \cos(a) - 8 \cos(d) = 4\cos(c) - 7\cos(b) \ \sin(a) - 8\sin(d) = 4\sin(c) - 7\sin(b) \end{cases}$ and then squaring both sides of each equation and then adding

[somber coyote]

Ann:

[plucky breach]

How will it group to what I've to proof?

[dark sparrow]

why don't you go ahead and try it for yourself

[plucky breach]

I did,that's why I'm asking

[dark sparrow]

no like

try what i just suggested

[plucky breach]

Got it!

Thanks

[dark sparrow]

oh god. not this crap again.

$\sum_{n=1}^n$ is bad and ambiguous notation

[somber coyote]

Ann:

[upper karma]

lmao how idiotic

$\sum_{k=1}^n \frac{\tan \frac{x}{2^k}}{2^{k-1} \cos \frac{x}{2^{k-1}}}$

[subtle gate]

Just really quickly: How do you describe an angle by the two lines that make it up

Like if you have 3 points it's ∢ABC

But what if you describe it with the 2 intersecting lines?

[dark sparrow]

@upper karma yeah, but that requires "cHaNgInG iT tO k"

[upper karma]

lmao

didn't he argue about this before

saying it (sum from n=1 to n) was correct notation

[dark sparrow]

no, that was euleroid

[upper karma]

lol

[mighty onyx]

How to solve the ques

Any idea anyone

[trail minnow]

@subtle gate Normally you'd just label the point at which the two lines intersect "C" or something and then have points A and B on the first and second lines respectively

[subtle gate]

so i am just gonna say let B be any point on this part of the line?

[mighty onyx]

What does ques 4 want to say

[dark sparrow]

maybe there's a typo and they meant + instead of = in the last thing?

[mighty onyx]

Hmm

[wide carbon]

Is 270° = 3pi/2 radians?

[fossil ledge]

yes

[upper karma]

hi

i need help with geometry

its for summer school

i failed it

[quiet mason]

you try

and show your attempt

we will'help' not 'solve'

[dark sparrow]

@mighty onyx please do not post your question across multiple channels.

i've seen you do this on at least three occasions.

[mighty onyx]

Okk

[mighty onyx]

I've ticked the answers but don't know how they are correct anyone any idea?

[narrow sleet]

What the value in column II and III mean?

[mighty onyx]

That only we have to figure out

To match

Using options of the questions

[dark sparrow]

i don't think that's what wolf asked

what's the relationship between the function in column I and the number in column II supposed to be?

[worthy root]

@dark sparrowHow would you define Radian?

[dark sparrow]

wdym how would i define a radian

[upper karma]

just how anyone else would define it

[dark sparrow]

1 radian is the central angle subtended by an arc of length 1 in a circle of radius 1

there are many ways to essentially say the same thing

[worthy root]

My textbook says that

$\theta=\frac{l}{r} Radians $ Where "l" is a measure of arc

[somber coyote]

Krishna 🙎:

[upper karma]

this definition is equivalent

if the radius is 1, and the measure of the arc is 1, then that's 1 radian being subtended at the centre

[dark sparrow]

arc length is proportional to radius and angle

[worthy root]

Yes

[worthy root]

Why $sin(n\pi+\theta)=(-1)^n$

[somber coyote]

Krishna 🙎:

[worthy root]

@dark sparrow

Also, Why $cos(n\pi+\theta)=(-1)^n$

[somber coyote]

Krishna 🙎:

[worthy root]

These doesn't makes sense to me

[upper karma]

that's not true though

[dark sparrow]

$\cos(\theta + n \pi) = (-1)^n \cos(\theta)$

[somber coyote]

Ann:

[worthy root]

[narrow sleet]

Where n is an integer

[worthy root]

Algebra was so simple

So was geometry

So was right angle trigonometry

But this

[narrow sleet]

Walking was so simple
So was jumping
So was running and sprinting
But long jump

[worthy root]

😭

I do good long jumping

Btw

[upper karma]

riding a bicycle was so simple
so was driving a car
but flying a plane

[narrow sleet]

I don't

That's why I said that

[worthy root]

Lmao good one

Yes

I am having a hard time these days

Concepts don't come as obvious to me as were before

It takes time to make sense

[narrow sleet]

Yeap

[worthy root]

By the end of this month

In 30 days I have to be done with calc

Basics*

To do physics

Oxide @upper karma is it ok to skip inverse trig

I don't think they will be off any use

In physics

@narrow sleet is inverse trig of any use in mechanics

[narrow sleet]

To find the angle before forces?

[worthy root]

[upper karma]

it is not ok to skip inverse trig

[worthy root]

ok

by today I will do trig function

[narrow sleet]

Like this and the q tells u to find theta and F?

[narrow sleet]

It is not ok to skip anything

[worthy root]

Sorry

I meant function

[upper karma]

it's just funny

[buoyant oasis]

muscle memory 🤣

[worthy root]

🤓

[worthy root]

Trig functions are very easy

I finally understood it

[narrow sleet]

Lol

[quiet mason]

...

krishna which grade you in?

and what do you pursue to take in higher studies

[upper karma]

$\cos^2(x) - \cos(x)\sin^2(5x/4 - 5\pi/12) + 1/4 = 0$

[somber coyote]

Uncle_Bitch:

[upper karma]

can you see the path to solution right away

it's so obvious after you solve it 😔

[tranquil birch]

yeah x is pi/3

[upper karma]

x is 7pi/3 + 4pi*n

i don't think i'd solve it on an exam 😔

and it's in two weeks

[worthy root]

@quiet mason I will go in class 11 next month

[glass harness]

https://cdn.discordapp.com/attachments/331718580070645760/597288112930160646/unknown.png what is this formula for

[worthy root]

,w sin((2\pi)/3)

[somber coyote]

Display results online and refine query

Requested by Krishna 🙎#4971

[worthy root]

[worthy root]

Why can't the cosx attain the negative value ?

In this probkme

Problem*

$-1\leq cosx\leq 1$

[somber coyote]

My homies are good at math:

[narrow sleet]

If b^2-4ac < 0, then the quadratic equation has no real root

[worthy root]

Bro

But cosx can attain a negative value and become positive?

[upper karma]

the point is that cos**^2** cannot attain negative values

[narrow sleet]

^

[upper karma]

for real number inputs

[worthy root]

Woops

Sorry u are smarter then me homie @upper karma

[upper karma]

lol

[narrow sleet]

Lol

[worthy root]

cosx=0

For

x=n*pi/2 Where n=2k+1

Right?

[narrow sleet]

n = 2n + 1?

[upper karma]

lol

[worthy root]

Yes

Fixed

[narrow sleet]

Why can't you just say "where n is an odd number"?

[worthy root]

Yea

But

In my textbook cos x=0 For $x=2k\pi +\pi/2$

[narrow sleet]

n = 2n + 1, so you saying that n can only be -1 in order to satusfies the equation

[somber coyote]

My homies are good at math:

[worthy root]

I fixed that

Why?

Both are odd integers

Which is correct ?

I think the one I posted is right because it's an ncert textbook

[upper karma]

it's equivalent

[worthy root]

🤔

[upper karma]

or actually

[worthy root]

They. Have defined n to be an integer

[upper karma]

it should be kpi + pi/2

[worthy root]

Even I think so

It's typo I guess

[narrow sleet]

^

[quiet mason]

cos^2 theta is always positive and a negative no. multiplying it always gives the term negative except when x=0

[worthy root]

Yea I didn't noticed that the cosine was squared

[worthy root]

$ tan(\theta)=1$ for all $\theta = n\pi +\frac{\pi}{4}$ is true?

Where n is am integer

[somber coyote]

My homies are good at math:

[worthy root]

,w Is tan(\pi/4)=tan(3\pi+\pi/4)

[somber coyote]

Display results online and refine query

Requested by My homies are good at math#4971

[formal bolt]

tan(pi/4)=1
tan has a period of pi

[worthy root]

Yes

[formal bolt]

These two statements together answer the question

[worthy root]

I don't know what do you mean by period of pi

[formal bolt]

A function f has period P if
For all x, f(x+P)=f(x)

[worthy root]

Ok gotcha

[formal bolt]

Oh, my bad, I actually gave an improper definition
The period is the least positive real number such that what I wrote occurs

If f has period P, then for any integer multiple of P
f(x+kP) = f(x)

[spark stag]

intuitively, its the size of one repeating "piece" of the function

[formal bolt]

Yeah, the minimum size

[worthy root]

FINS THE values of x between 0 and $2\pi$ \
cosx=$-\frac{\sqrt{3}}{2}$

[somber coyote]

My homies are good at math:

[worthy root]

One value of x is $7\pi/6$

Can't figure out the other one

[somber coyote]

My homies are good at math:

[lost parcel]

5pi/6?

,w is cos(5pi/6)= -√3/2

[somber coyote]

Display results online and refine query

Requested by Beast97#9909

[lost parcel]

@worthy root

[timid imp]

anyone here?

I have a difficult math problem that I need to solve

not difficult for you at least hahaha, it is high school math

[stuck torrent]

I’m not anyone

I’m just whoever

@timid imp

[fringe dirge]

@timid imp just ask

[timid imp]

Already solved

Thanks to @tranquil birch

[worthy root]

Why $sin(\pi/2+\theta)=cos\theta$

[somber coyote]

My homies are good at math:

[worthy root]

@upper karma oxidwhats the reason behind this relation

[upper karma]

what do you mean by 'reason'? like a proof?

[worthy root]

Maybe

Because o don't know why this relation holds

[upper karma]

i mean there are a couple of proofs

[worthy root]

😭 why did I ask

Oxide are you also a night person ?

[upper karma]

yes

[worthy root]

At what time in morning do you wake up buddy ? I wake up like 10:00 am @upper karma

[upper karma]

idk when i wake up

[worthy root]

Also Is it ok to bunk college lectures

[upper karma]

i don't believe in clocks

[worthy root]

1°=60' 1'=60''

Come out of that cat world

[unborn grove]

Beginner at geometry here, is there a way to find angle c in the following diagram if angles a and b are known? If more measurements are needed tell me the method that uses those measurements and what they are

sorry for my terrible drawings

[grim pecan]

Im not sure but I don't think there can be multiple tangents coming from one point on a circle.

[idle bloom]

^

but given the tangents that form point C you can find the angle if you know the angle between the points of tangency on the circle

I'm trying to find the theorem but it can be done

Think this is what you need

[royal glade]

i got the radius of the biggest circle as tan(22.5), how do you solve the second radius?

[tawdry pivot]

that's not what I got hm

ye i just checked over my working, still getting different from tan(22.5)

@royal glade

[narrow sleet]

I get tan 22.5°/(1 + tan 22.5°) for largest radius

[royal glade]

oh

what do you do for the second radius?

what I did to solve for biggest radius is I split it in half then drew a line from the bottom left corner to the middle and solved for opposite. why cant i do that?

[tawdry pivot]

@royal glade this might be useful

[worthy root]

Need help with 3rd

Need help with 2.c cos\theta=-\sqrt{3}/2

[narrow sleet]

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

[worthy root]

I figured out one angle

[dark sparrow]

draw a unit circle

[worthy root]

🤔

Ok

[narrow sleet]

You mean 3 or 2c)?

[worthy root]

2c

[narrow sleet]

Oh ok

What answer have you gotten?

[worthy root]

One answer isn7pi/6

Other is in 2nd quadrant

But don't know what it is

@dark sparrow I did

[worthy root]

Should I minus the the reference angle from angle theta

I am confused

😭

[upper karma]

You should first find the Eigenvalues of the inverted möbius strip

[worthy root]

Reeee

Wot

[narrow sleet]

@upper karma Hey spoiler alert

[upper karma]

@upper karma spoiler alert

[worthy root]

[narrow sleet]

@worthy root First find the reference angle

[upper karma]

messing with the planck scale triggers the deutsch proposition

[worthy root]

It's \pi/6

[upper karma]

Oops I forgot @upper karma

Iron man dies by killing thanos in the endgame

[worthy root]

What next boi

[narrow sleet]

@worthy root Then you want to find the angle in Q2,so you can just pi - reference angle

[worthy root]

Thanks boi

[upper karma]

Don't assume genders please

[worthy root]

Oh

[narrow sleet]

He know that I'm a guy

[upper karma]

Stop being a sexist bigot

Don't assume people's gender

@worthy root

[worthy root]

🤔

I knew that he was a guy

What's your problem are you having a bad day ? If yes then I can understand your frustration.

[upper karma]

Yes

My dog died

[worthy root]

Sorry to hear that

[upper karma]

I killed him

[worthy root]

You hooman

[upper karma]

He was annoying af

Like actually

He always barked just because I locked him in the basement for only one week without food

stop trolling

[worthy root]

He is so cruel

[upper karma]

Ikr

[worthy root]

That's why I hate hooman

[upper karma]

Dogs are so cruel

They bark at you

They must die

[worthy root]

Ok you can stop trolling now

[upper karma]

<@&268886789983436800>

[worthy root]

For what values of $\theta$ lying between 0 and $2\pi$ satisfy $cosec\theta=\frac{2}{\sqrt{3}}$

[somber coyote]

My homies are good at math:

[worthy root]

One theta is \pi/3

Other is pi-pi/3=?

[gritty siren]

2π/3 is fine

[worthy root]

Yes

Prove that $sec^2\beta -sec^2\alpha=\tan^2\beta-\tan^2\alpha$

[somber coyote]

My homies are good at math:

[worthy root]

Should I take the trig ratios with conman angle on one side

🤔

Is this valid ?

@welpers is this valid

[gritty siren]

Well yeah

[worthy root]

Tank u

[supple abyss]

[gritty siren]

You're welcome

[worthy root]

4

,w simplify (tanA+secA-1)/(tanA-secA+1)

Is it hard to prove this

I found this expression equal to 1

<@&286206848099549185>

I need help with 4 and 5

Only 4*

Done with 5

[gritty siren]

for 4, prove $(\cos A) (\tan A+\sec A-1)=(1+\sin A)(\tan A-\sec A+1)$ instead

[somber coyote]

Tuong:

[worthy root]

ok

Bro

How to solve a quadratic with a teigononreic ratio

$sin\theta=\frac{1}{2}[x+\frac{1}{x}]$

[somber coyote]

My homies are good at math:

[worthy root]

x^2+1-2xsinx=0

x^2-2sinx+1=0

How do I solve this

[tawdry pivot]

for the latex, sin(theta) is a constant in terms of x

[worthy root]

How to solve x^2-2sin\thetax+1=0

Or wait

[marsh rune]

it looks like a quadratic doesnt it

[worthy root]

Cos^2\theta=0 for what values of x

[dark sparrow]

[worthy root]

I used the quadratic formula ann

[tawdry pivot]

your latex is different from your recent things

[worthy root]

Yea sorry about that

x^2-2sin\thetax+1=0

Ann for what values of$\theta$ $cos^2\theta =0$

[somber coyote]

My homies are good at math:

[gritty siren]

cos²θ = 0 iff cos θ = 0

[worthy root]

I mean I know that one of the value is pi/2

What are the other values

There are I finitely many

Infinitely*

What is form

[gritty siren]

(some odd integer)×π/2

[worthy root]

$2n\pi +\pi/2$

[somber coyote]

My homies are good at math:

[worthy root]

Where n is a integer

Is this correct aswell

[gritty siren]

yea, but you're skipping some

[worthy root]

Wot

[gritty siren]

-π/2 works as well but can't be written in the form you wrote

[worthy root]

Ok

,w x^2-2sin(\theta)*x+1=0

[somber coyote]

Display results online and refine query

Requested by My homies are good at math#4971

[worthy root]

Really?

[gritty siren]

Yea

[worthy root]

Mo

No

I got answer 1 if \theta =npi/2+pi/2 (n is a odd number)

Toung can you solve that for me please

I am really confused

[worthy root]

😭

Find the real values of x and $\theta$ satisfying the equation $x^2-2sin(\theta)x+1=0$

[somber coyote]

My homies are good at math:

[worthy root]

This is my original question @gritty siren

[gritty siren]

if you do the discriminant thing, you find that it has real solutions iff sin(θ)=±1

dig deeper and you find x=1 and θ=(odd integer)×π/2

[worthy root]

The roots are real iff sin(θ)=1 I.e θ=pi/2+2n\pi

Right?

[dark sparrow]

@gritty siren sin(θ) = ±1, maybe?

[gritty siren]

Oh yea

fixed, thanks

[worthy root]

Yws

Sin@=-1 if @= 2npi-pi/2

7

[gritty siren]

t = x/(v cos(θ))

[worthy root]

Ok

Got an idea

How would you isloate t in the second expression

[gritty siren]

No need, just replace t with x/(v cos(θ))

[worthy root]

Okko

[dark sparrow]

"sin θt" looks like sin(θt) >.>

[worthy root]

🤔

We just find y

Do y describes the trajectory bro?

[gritty siren]

an equation between y and x describes a trajectory

[worthy root]

I don't know lot physics but I believe u ol

Ok*

Yes got it sorry

Why I am so dumb

Welp me frens

<@&286206848099549185>

🤔

[somber coyote]

JY1853:

[worthy root]

Yes

Hmm

[dark sparrow]

\omega

[worthy root]

It's ok I understand

[tawdry pivot]

\sin

[worthy root]

I know that I tried all those