What is the range of y=cos^-1x?

Question

...and why?...

Answer 1

[0, pi]

Answer 2

The answer is (0, pi)

If asked to evaluate arccos 0, same thing as cos^-1 0, you might answer that cosine is equal to 0 when x =pi/2 or 3pi/2. It is true that cos pi over 2 and 3pi over 2 are both equal zero. HOwever, this means that the function y=cos^-1 0 has two output when x=0, so y=cos^-1 x is not a function. This is remedied by restricting the ranges, and the answer for y=cos^-1 X is (0, pi)

Answer 3

The function cos x if you consider the whole real line has no inverse since it is periodic. However if you just consider values in [0, pi] this is a decreasing function and so has an inverse that you call cos^(-1).
So cos: [0, pi] ->R
and cos^(-1): R->[0, pi]

In this way:
cos(cos^(-1)(y))=y for any y in R
and
cos^(-1)(cos(x))=x for any x in [0, pi]

Answer 4

The range of y = cos^(-1) x is
[0, π].
For inverse of x = cos y to exist, the values of y should be so restricted that the many-one function becomes one-one without which the inverse cannot be written. Inverse of many-one will be one-many and one -many cannot be a function.

Answer 5

-infinity to + infinity, because is is not asking for the principal value Cos^-1. Therefore, it is not restricting the domain to be a function.

Answer 6

- inf to inf.

because cos is a periodic fn and can have a domain of all real numbers

Answer 7

cos(y) = x , i.e.,
x is in the range plus/minus one

Answer 8

sorry i dont know

Answer 9

do your own homework, my friend :)

all the best ;)