How I show that the function cos: [0,pi] is a bijection?I am curious about the various sugestions

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nathanshields | High School Teacher | (Level 1) Associate Educator

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It's been awhile since I've done these, so extra help would be appreciated.

To prove cos(x) from 0 to pi is a bijection, we need to prove:
(a) injection (1-to-1)
(b) surjection (onto)

(a) To show the function is 1-to-1, let's assume that it's not.  That is, assume f(u) = f(v) for some u and v in the domain.  Then we have

cos(u) = cos(v).

For this to be true, either u = v or `v=2pi*k+-u` .  The first case is trivial; the second case is impossible if u and v are both in the domain 0 to pi.

(b) To show the function is onto, we must show that for any element y in the range (-1 to 1), there exists an x between 0 and pi such that f(x) = y.

Let y be an number in [-1, 1].  Then y = cos(x), so x = arccos(y), whose default range is 0 to pi.

I've probably overlooked something, but it's a start.

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