Can someone give me an example of how cos and arccos are inverses?

>>8588431
your first two inverses are inverses under multiplication
the function 1/x is certainly not the inverse of the function x. see, apply them both in any order and you just get 1/x.
inverses are like x^2 and sqrt(x). like so: (sqrt(x))^2 = x
so just like that, acos(cos(x)) = x

>>8588431
There are different kinds of inverse, brainlet.

>>8588447
please explain cause im dumb

>>8588446
I dont get what you mean here
>the function 1/x is certainly not the inverse of the function x. see, apply them both in any order and you just get 1/x.

1/x * x = 1

>>8588431
>arccos(cos(x)) does not equal x?
this is literally how arccos is defined

[math] cos(x) [/math]

Dom: [math] \mathbb{R} [/math]

Range: [math] [-1,1] [/math]

now if [math] y = acos(x) [/math] then [math] cos(y) = x [/math] with [math] x \in [-1,1], y \in [0,\pi] [/math]

[math] cos(x) [/math] is not injective so it doesn't have a true inverse, but we can restrict the domain to produce [math] acos(x) [/math]

>>8588456
he means nesting the functions
f(x) = x ; g(x) = 1/x

f(g(x)) = f(1/x) = 1/x
g(f(x)) = g(x) = 1/x

>>8588456
do you know what a function is? read closely, all the information you need is in that post

>>8588491
no please explain you seem to understand

>>8588469
ok i see, so not multiplcation but nesting of the functions.

when you say inverse, we mean function nesting?

>>8588504
you cannot just say "inverse," you have to be more specific.
the additive inverse of 5 is -5 since 5 + (-5) =0
the multiplicative inverse of cos(x) is 1/cos(x) since cos(x) * 1/cos(x) = 1
the funtion inverse of f(x) is g(x) where f(g(x)) = x
arccos(x) happens to be the function inverse of cos(x). you could talk about "nesting" the functions, but the formal term for the operation is "function composition"

>>8588753
Okay this makes sense thank you.