/sci/ - Science & Math

when two cosines are equated (cos(x)=cos(y)), their respective arguments are such as
x=y + 2*k*Pi
of x= -y + 2*k*Pi
here, you have y=1, so try to find a value of k such as x belongs to [pi,2pi]!

arccos(1)

2pi-1

No need for a book.
All you need to know is that cosine is periodic and symmetric.

>>4188273
The answer is 5.28.
arccos(1) = 0.

And theta must be <span class="math">\pi \leq \theta \leq 2\pi[/spoiler]

>>4188269
Theta equals -1. If theta is between pi and two pi, then we use the unit circle and focus on quadrants III and IV. we know that cos of theta is equal to cos of negative theta, so we select the appropriate cosine within the quadrants we are restricted to.

>>4188284
>implying 5.28=2pi-1

>>4188269

When you see this:

2x=2(5)

you simply drop the "2 operator" such that

x=5

The same case applies here.

cos(x)=cos(1)

Drop the cosine operator, but remember that on a circle values are identical if they differ by a number of full circulations around the circle. (multiples of 2pi)

If you stand in one spot on a circle and go around once (add 2 pi) you are in the same place.

Therefore, the answer is this:

cos(x)=cos(1)
ergo
x=1
But, keeping in mind the periodic nature of the function, the solution is completely:
x=1 + k(2pi) for k is an integer

>>4188285
>implying -1 is greater than pi and less than 2pi

>>4188284
spaz alert, sorry I goofed. Add 2 pi to the -1, so that it's in the right qualifications.

cos 1=0
x=0

>>4188298
0/10

>>4188290
It's what the book says.
And yes, it is.

>>4188277
I understand what you are saying, but what is k supposed to represent?

>>4188298
cos1=0
divide by 1
cos=0

>>4188304
>5.28=2pi-1
>OP can't into irrational numbers

confirmed for underage. reported. enjoy your ban.

>>4188304
k is an integer, positive or negative (or 0).

Alright, I see what's going on.

I got two answers, 4.28 and 5.28

4.28 = 1 + 2k<span class="math">\pi[/spoiler]
k = 1
4.28 = -1 + 2k<span class="math">\pi[/spoiler]
k = 2

But the book only gives the choice for 5.28, so how would I know which answer is correct?

I'm pretty sure I'm doing this wrong.

>>4188323
the second was supposed to be 5.28

>>4188323
It's not multiple choice, by the way.

OP can't into imagining swinging radius and tracing the sines and cosines as it swings along.

>>4188269

>>4188336
fuck i'm dumb

good thing I only need a 55 on this exam.

>>4188345

Hey, are you sure it's cos \theta = cos 1? Cos it would really make sense if it was just cos \theta = 1

>>4188350
yup

>>4188361

you can use a calc right?

>>4188365
Yes, for this section.

troll thread

> variable is bigger than pi
> therefore it's greater than 1
> cos theta = cos 1
> theta = 1

sure is typo error in there, son.

It's cosine and it's positive so first and fourth quadrant... the acute angle is 1, and so the actual angles are [/math]0 + 1 [/math] and <span class="math">2 \pi[/spoiler] radians for the interval <span class="math">[0,\ 2 \pi[/spoiler]. For <span class="math">0,\ 4 \pi[/spoiler], you'd just add <span class="math">2 \pi[/spoiler] to all the values you calculated for <span class="math">[0,\ 2 \pi[/spoiler].

OP, this is so easy.

All you have to do is think of the values of cosine all along the unit circle.
Do it and you will see what I mean, you should get an answer immediately

You will notice a symmetry, for example, sine has a vertical symmetry, values on the left and right of <span class="math">\theta=\pi[/spoiler] have the same values.
Can you figure out what it is for cosine?

It's cosine and it's positive so first and fourth quadrant... the acute angle is 1, and so the actual angles are /math]0 + 1 [/math] and <span class="math">2 \pi[/spoiler] radians for the interval <span class="math">[0,\ 2 \pi][/spoiler]. For <span class="math">[0,\ 4 \pi][/spoiler], you'd just add <span class="math">2 \pi[/spoiler] to all the values you calculated for <span class="math">[0,\ 2 \pi][/spoiler].

It's cosine and it's positive so first and fourth quadrant... the acute angle is 1, and so the actual angles are <span class="math">0 + 1 [/spoiler] and <span class="math">2 \pi[/spoiler] radians for the interval <span class="math">[0,\ 2 \pi][/spoiler]. For <span class="math">[0,\ 4 \pi][/spoiler], you'd just add <span class="math">2 \pi[/spoiler] to all the values you calculated for <span class="math">[0,\ 2 \pi][/spoiler].

It's cosine and it's positive so first and fourth quadrant... the acute angle is 1, and so the actual angles are <span class="math">0 + 1 [/spoiler] and <span class="math">2 \pi + 1[/spoiler] radians for the interval <span class="math">[0,\ 2 \pi][/spoiler]. For <span class="math">[0,\ 4 \pi][/spoiler], you'd just add <span class="math">2 \pi[/spoiler] to all the values you calculated for <span class="math">[0,\ 2 \pi][/spoiler].

Thanks for the help.