#silly little mess up

151 messages · Page 1 of 1 (latest)

jovial kelp
#

i mightve messed up a lil with this one

7pfbsK1.png
ruby knotBOT
#
  1. Ask your question and show the work you've done so far. If you've posted a screenshot of a question, specify which part you need help with.
  2. Wait patiently for a helper to come along.
  3. Once someone helps you, say thank you and close the thread with:
    +close
  4. Feel free to nominate the person for helper of the week in #helper-nominations
  5. Do not ping the mods, unless someone is breaking the rules.
  6. If you're happy with the help you got here, and the server overall, you can contribute financially as well:
jovial kelp
#

im curious as to how much i fucked up

sharp root
#

Let's start with a mire general case.

jovial kelp
#

i was gonna actually follow that up

sharp root
#

What are the solutions of cos(x) = a?

jovial kelp
#

im missing the + n * k stuff

sharp root
#

Yes. After all, sine and cosine are 2π-periodic (and π-antiperiodic).

jovial kelp
#

yeah because like

#

a simple way i think of it is like saying when did something happen

#

and then just saying like, at 3 oclock

#

it's like yeah but what day? month?

#

since it keeps on rotating yk

sharp root
#

Well, somewhat, yeah.

jovial kelp
#

but the problem is

#

with my case

#

idk how to determine the multiplicative for n

#

ive seen +n*2pi or x360

#

but i suspect thats wrong here

#

since im dividing

#

so i suspect it's...

#

for 6.a) it's 15deg + n*180

sharp root
#

Don't use degrees.

jovial kelp
#

for 6.b) it's +n*120 (or whatever the radian translation is)

#

ok wait

#

dont spoil it

sharp root
#

Of course.

#

Let's recall the general case first.

#

What are the solutions of cos(x) = a?

jovial kelp
#

for 6.a) it'd be + n * pi

#

for 6.b) it'd be + n * 2pi/3

sharp root
#

Yes, but that's not all.

jovial kelp
#

that's what we're taught to do?

sharp root
jovial kelp
#

is there more? 0_0

jovial kelp
sharp root
#

I want to gauge where you're missing something.

jovial kelp
#

i'd immediately be like, that cos is ugly and i'd hit it with a cos^-1 or arccos

#

which means that a would also be arccos'd

#

so x = cos^-1(a)

#

but also + n*2pi

sharp root
#

That's only half.

#

cos(x) = a has two series of solutions:
x = arccos(a) + 2πn, n ∈ ℤ
x = -arccos(a) + 2πk, k ∈ ℤ

#

That is due to the fact that cos(x) is even, so if x is a solution of cos(x) = a, then so is -x.

jovial kelp
#

oh RIGHT

#

because it can spin backwards as well

sharp root
#

Spin?

jovial kelp
#

unit circle 😂

#

i think of it that way

sharp root
#

Oh, ok.

#

So, that's cosine.

#

What about sin(x) = a?

jovial kelp
#

uhm isnt it the same?

sharp root
#

No.

#

sin(x) isn't even.

#

However, sin(x) is just cos(x), but with a phase shift.

#

Note that sin(x) is symmetric around the line x = π/2 (for that exact reason), so if x is a solution of sin(x) = a, then so is π - x.

#

Thus, the solutions of sin(x) = a are:
x = arcsin(a) + 2πn, n ∈ ℤ
x = π - arcsin(a) + 2πk, k ∈ ℤ

jovial kelp
#

we were never taught that?

#

the phase shift part

sharp root
#

Well, clearly, sin(x) = cos(x - π/2).

jovial kelp
#

ok well

#

real quick just so i understand

#

for 6.a) it'd be pi/12 +- n*pi

sharp root
#

No.

#

We've just discussed what the solutions of cos(x) = a look like.

jovial kelp
#

right wait wait

#

6.a)
x = π/12 + nπ, n ∈ ℤ
x = -π/12 + kπ, k ∈ ℤ

#

more like it? @sharp root

sharp root
#

And what about (b)?

jovial kelp
#

well thats the thing

#

i talked to my classmate and he reminded me

#

we account for phase shifting in our curriculum in a different matter

#

and i was gonna ask about that as well

#

so she showed me her answer and uh

#

for 6.a)

#

instead of the -

#

she had

#

x = 11π/12 + πn​ , n​ ∈ Z
x = π/12 + πn​ , n​ ∈ Z

#

and the same but pi/12

sharp root
#

After all, -1/12 + 1 = 11/12.

#

I just prefer when the "fundamental" solutions are in (-π, π], not [0, 2π).

#

Doesn't matter, of course. You can follow whatever convention you prefer.

jovial kelp
#

ah right

#

and the same for 6.b)

#

but that one mindfucked me

#

that's the one that has real phase shifting

#

i'll just send the pic she sent me

qdusCQ8.png
#

i just dnt get the whole

#

5pi/12

#

pi/4 is my org answer not accounting for all the answers which u taught me about

#

but the phase shifting accounted with both of u and my friend feeding me alternative solutions got my head spinning 😭

#

so can u pls elaborate on how she got 5pi/12 😭

sharp root
#

Let's try solving it as usual.

#

Oh, actually.

#

Here it's easier if we notice an identity.

#

What is sin(x - π/2)?

jovial kelp
#

well uh

#

it's a sine wave that's moved on the x axis

#

by ...

#

90 deg?

#

backwards...

#

or forwards?

sharp root
#

Wait, you haven't studied the basic identities yet?

jovial kelp
#

well im studying it as we speak lol

sharp root
#

Hm.

#

No, I mean, those are usually studied before trigonometric equations.

jovial kelp
#

our professor is known for giving us questions out of our league so we have to research shit on our own

sharp root
#

Oh...

jovial kelp
#

then she goes through those stuff later once we have a slight understanding

#

so instead of her hitting us with new stuff shes just kind of repairing what we know lol

#

and like

#

let's take what we got which is

#

where
A = amplitude
d = pushing on y-axis
v = pushing on x-axis
k = how many periods fit in 2pi

sharp root
#

Anyway, these are the basic identities related to symmetries of trigonometric functions.
So, sin(x - π/2) = -cos(x).

image.png
sharp root
# jovial kelp

Hm, quite an odd form... I'm more used to y = B + A cos(ωx + φ0).

#

Anyway, using the identity sin(x - π/2) = -cos(x), the equation becomes:
-cos(3x) = 1/√(2)
So, cos(3x) = -1/√(2).

#

The rest is easy.

jovial kelp
#

ima be real

#

im still confused on how she got 5pi/12

#

😭

#

because i looked it up and ik shes right but HOWWW

sharp root
#

So, you can also write the solution like this (this is how I would write it):
x = π/4 + 2πn/3, n ∈ ℤ
x = -π/4 + 2πk/3, k ∈ ℤ

jovial kelp
#

but thats wrong as u said

sharp root
#

No, why?

jovial kelp
#

we're working with sine here so ur not accounting for a phase shift

sharp root
#

Of course, you don't have to use that identity right away.

#

You can solve it as sin(x) = a first.

#

Doesn't matter, the result will be the same.

jovial kelp
#

so 6.b) can be written as...
x = pi/4 +n * 2pi/3, n -> Z
x = - pi/4 + k * 2pi/3, k -> Z

sharp root
#

Yes.

jovial kelp
#

right...

sharp root
#

I just usually prefer to write the solutions so that their symmetry can be seen more clearly.

#

That's mostly why I prefer the (-π, π] convention instead of [0, 2π).

#

If you want, we can try to solve (b) without using identities - just using the solutions for sin(x) = a.

jovial kelp
#

nah were good

sharp root
#

Alright, nice!

#

Oh, and as for tangent and cotangent, their solutions are even simpler.

jovial kelp
#

thanks for all the help man <3

#

cya

sharp root
#

tan(x) = a has solutions x = arctan(a) + πn, n ∈ ℤ, while cot(x) = a has solutions x = arccot(a) + πn, n ∈ ℤ.

#

The reason why they are simpler is because tan(x) is monotone on each period (where it's continuous), so there's only one series of solutions instead of two.

#

You're welcome!

oblique cypress
#

@jovial kelp er du norsk?