Need help with this

[harsh jungle]

Not sure where to start off, if someone could provide me some hints that'd be great

[dense pivot]

Well, you need to solve the equation cos(πx/2) = 1/2, as well as the inequality cos(πx/2) > 1/2 for x ∈ [0, 4]. Note that you don't actually need to solve the last inequality: just invert the result from cos(πx/2) > 1/2.

[harsh jungle]

Alright, thank you. After that, would I need to use Unit Circle to figure out what values are greater than 1/2?

[dense pivot]

Yeah, when solving inequalities, I usually imagine the unit circle.

[harsh jungle]

For these answers, would I need to put in one answer or two?

ie 1pi/2 and 2pi/3

etc

[dense pivot]

There shouldn't be any π-s left when you find x.
And you write however many values you get.

[harsh jungle]

Awesome, thanks for the help

[dense pivot]

You're welcome!

[harsh jungle]

It's been a while since I touched this, but I managed to figure out that g(x) = 1/2 when x is 10/3 and 2/3 on a domain between 0 and 4 on a line

So after graphing it for a visual in my head, I basically made this sketch

lemme take a pic rq

[dense pivot]

Yes, that is correct.

[harsh jungle]

So right now I'm trying to figure out which intervals between [0, 4] is g(x) > 1/2

And which interval is g(x) ≤ 1/2 as well

What's a good place to start from here? I know that there's a finite amount of exact answers between these lines because angles can get extremely precise on the Unit Circle.

[dense pivot]

Well, you can use the unit circle to solve elementary trigonometric inequalities.

[harsh jungle]

Gonna place this here as a placeholder

[dense pivot]

For example, you can infer these by looking at a unit circle:
1.
cos(x) > a (-1 ≤ a < 1)
x ∈ (-arccos(a) + 2πn, arccos(a) + 2πn)
2.
cos(x) ≤ a (-1 ≤ a ≤ 1)
x ∈ [arccos(a) + 2πn, 2π - arccos(a) + 2πn]

(or by looking at a graph of cos(x))

So, can you apply that to our case?

[harsh jungle]

uh

is 2πn in this case a full rotation * how many times it goes around

[dense pivot]

Yes.

[harsh jungle]

i have no clue why i'd apply arccos

[dense pivot]

Forgot to mention: n ∈ ℤ, as usual.

[dense pivot]

Well, for the same reason as when you solve cos(x) = a.

[harsh jungle]

oh alright

so n is an element of the integer

[harsh jungle]

im assuming thats what that means

[dense pivot]

Yes. After all, cos(x) is 2π-periodic.

[harsh jungle]

oke

in my case of looking for which integer is > 1/2, where would I apply a into this

just like any radian within the unit circle or

[dense pivot]

What do you mean?
arccos(1/2) has one value.

[harsh jungle]

woops

which is pi/3

[dense pivot]

Yes.

[harsh jungle]

so because arccos(1/2) is pi/3, in that case would i do (pi/3 + 2πn, 2π - pi/3 + 2πn) to account for arccos(a)

and just solve for x afterwards

er

5pi/3 + 2πn i mean

pi/3 + 2πn, 5pi/3 + 2πn

[dense pivot]

Ah, sorry! I forgot to put square brackets.After all, the inequality is non-strict.

[harsh jungle]

sorry if i seem stupid this has been a while since i brushed up on calc

[dense pivot]

Corrected.

[harsh jungle]

oh now i see ok

[π/3 + 2πn, 5π/3 + 2πn]

[dense pivot]

So, we have:
cos(πx/2) ≤ 1/2
πx/2 ∈ [π/3 + 2πn, 5π/3 + 2πn]

Then what will be the set for x?

[harsh jungle]

set for x?

what u mean

[dense pivot]

Well, we've obtained the set of solutions, but only for πx/2.

But we need just x.

[harsh jungle]

would i have to multiply both π/3 + 2πn and 5π/3 + 2πn by 2/π

to isolate x

[dense pivot]

Yes.

[harsh jungle]

ok and then from there i'd get 2/3 + 4n and 10/3 + 4n

but those are like the exact values for x = 1/2

i also need to find out what's greater and what's less/equal than those values

[dense pivot]

Yes.
x ∈ [2/3 + 4n, 10/3 + 4n]
However, remember that we're restricted to the interval x ∈ [0, 4].

So, we neet to cut down that solution set, leaving only the part in our domain.

[harsh jungle]

ok

i just need to figure out from that point whether or not [0, 2/3], [2/3, 10/3] or [10/3, 4] is > than 1/2

how do i go about finding that

would it just be like

[dense pivot]

Well, let's try some values of n.
n = 0: [2/3, 10/3]. This interval is completely contained inside [0, 4]. Also, considering that cos(πx/2) is 4-periodic, all other intervals will lie completely outside of [0, 4].
Thus, the solution of cos(πx/2) ≤ 1/2 for x ∈ [0, 4] is x ∈ [2/3, 10/3].

[harsh jungle]

alright so everything between 2/3 and 10/3 will be less/equal to x=1/2

[dense pivot]

Yes.

[harsh jungle]

how would i know that all other intervals lie outside of [0, 4]

[dense pivot]

Because all the other intervals are shifted by multiples of 4.

Which is the length of our interval.

Thus, if we have an interval completely inside it and shift it by at least 4 in any direction, it will be completely outside it.

[harsh jungle]

oh alright so anything higher than 10/3 would be 4x too great than 4 and anything lower than 2/3 would be 4x lesser than 0

that's what im getting rn

[dense pivot]

Well, not 4 times greater. It will be either at least 4 more or 4 less.

[harsh jungle]

oh alright

[dense pivot]

Alright. Now, for cos(πx/2) > 1/2.

And, as I said, there's no need to actually solve it.

For any number a and function f(x), we have either f(x) > a or f(x) ≤ a. There are no other options, right?

So, what we can do for cos(πx/2) > 1/2 is take our solution for cos(πx/2) ≤ 1/2 and subtract it from the whole interval [0, 4].

By subtract, I mean set subtraction, so you just remove the points, not subtract their coordinates.

[harsh jungle]

ok

let me try to catch up real quick

[dense pivot]

Just to recall: our total domain is [0, 4] and the solution of cos(πx/2) ≤ 1/2 is [2/3, 10/3].

So, the solution of cos(πx/2) > 1/2 must be everything from [0, 4] except [2/3, 10/3].

[harsh jungle]

so [0, 2/3] and [10/3, 4]

because anything in between 2/3 and 10/3 is ≤ 1/2

[dense pivot]

Yes.

Oh, not exactly, though.

Note that the points x = 2/3 and x = 10/3 are included in the solution of cos(πx/2) ≤ 1/2.

So, we need to exclude them too.

[harsh jungle]

er

[0, 2/3) and (10/3, 4]?

[dense pivot]

Yes!

[harsh jungle]

oh thank you

i understand now

[dense pivot]

Great! You're welcome.

[harsh jungle]

there's a part of the problem that allows me to graph g(x) and mark where y > 1/2 as well

[dense pivot]

Well, do you know what a sine wave looks like?

[harsh jungle]

so im guessing in that case i'd just have to draw a line and show what intervals is > 1/2, etc

[harsh jungle]

yeah

middle up middle down

repeat over

[dense pivot]

We know the points where it is equal to zero (x = 2/3, x = 10/3) and we know where it's positive and negative.

So, that allows you to draw it.

[harsh jungle]

so from there i'd just have to draw a sine wave that goes as high as 10/3 and as low as 2/3

[dense pivot]

Oh, no-no.

Its zeroes are at x = 2/3 and x = 10/3.

We have cos(πx/2), so its amplitude is still 1. So, it still goes from 1 to -1, etc.

[harsh jungle]

oh right

i just have to draw the zeroed points at 2/3 and 10/3 on that x-axis

[dense pivot]

Yeah. Also, I recommend putting + and - signs where the function should be positive or negative.

Moreover, recall that the peaks and valleys of a sine wave must occur exactly between its zeroes, so that gives you some other points to mark.

[harsh jungle]

hmm

not sure if im doing this right

one sec ill take a pic

would it look somewhat like that

i feel like that sine wave is too big

[dense pivot]

Mostly yes, but a couple of comments:

1. Considering what I said about peaks and valleys, you can also mark points (0, 1), (2, -1) and (4, 1).
2. We must be confined to [0, 4], so don't draw anything outside of that.

[harsh jungle]

gotcha

[dense pivot]

Oh, also! I think I've said that x = 2/3 and y = 10/3 are zeroes, which is wrong. Those are the points at which g(x) is equal to 1/2.

Sorry about that.

[harsh jungle]

all good

[dense pivot]

The zeroes should be located at x = 1 and x = 3, which you can also find by solving cos(πx/2) = 0 in our interval.

So, the points you can mark on the graph are:
(0, 1)
(2/3, 1/2)
(1, 0)
(2, -1)
(3, 0)
(10/3, 1/2)
(4, 1)

With that many points it should be pretty easy to draw a sine wave.

[harsh jungle]

alright, thank you so much for the help!

if i have any more questions, i'll be sure to let you know

you've done a great job today 🙂

[dense pivot]

You too!

[harsh jungle]

Sorry to bother, but this also stumped me a little bit

On E where it says that I need to use my sine wave to graph that piecewise function, what's it asking for exactly?

[dense pivot]

First, mark the regions where g(x) > 1/2 and g(x) ≤ 1/2. For example, you can dash them in different directions.
Then, for the region where g(x) > 1/2, your f(x) should be 1, and it should be 0 everywhere else on [0, 4].

[harsh jungle]

Good afternoon! I was looking at this earlier and I wondered if the sine wave I drew was actually a cos wave

I'm used to cos(x) usually sloping down when it starts at 0, rather than it going up to 1

so I was thinking that I might've written a cos wave on accident

[dense pivot]

Sine wave and cosine wave are the same thing, just shifted. I prefer not to differentiate these terms.

[harsh jungle]

Alright, so the answer is still acceptable then?

[dense pivot]

Of course.

[harsh jungle]

Alright, thanks. I'll try to work on the next wave

[dense pivot]

Good luck!

[harsh jungle]

I have two subsequent assignments at the moment, so I'm trying to deal with the other sheet that's due on Wednesday