What is Cos(Cos^-1(4/5) and why?

Question

What is Cos(Cos^-1(4/5) and why?

Answer 1

Let Ө = cos^(-1) (4/5)
cos Ө = (cos ) [ cos^(-1)(4/5) ]
cos Ө = 4/5
Ө = 36.9°

Answer 2

its 4/5 , as cos(cos^-1 (x)) = x

Answer 3

from the definition of inverse cos , if cos(y) = x , then cos^-1(x) = y

similarly if cos-1(4/5) = x, then

cosx = 4/5

therefore cos(cos^-1(4/5)) = cosx = 4/5

Answer 4

The cos and cos^-1 functions undo each other, so
the answer is 4/5.

Answer 5

To elaborate on what nicholasm40 said, because cosine is not an injective function, it does not have an inverse function in the traditional sense. What is called the inverse cosine is the inverse of the restriction of the cosine function to some convenient interval on which the cosine function is injective -- in the case of calculators, this is always the interval [0, π]. Thus, arccos x is the unique angle in the interval [0, π] whose cosine is x. But if θ is not actually in the interval [0, π], then arccos (cos θ) will not equal θ.

Note that none of this applies to the case of cos (arccos x) -- the arccos function is 1-1, and the restriction of cos to its range is indeed injective, so for all x∈[-1, 1] (which is the domain of the arccos function), cos (arccos x) = x. Period. And the person who gave thumbs down to the first two (correct) answers needs to be thwacked upside the head.

Answer 6

A function of an inverse function is equal to 1.

f( f^-1(x) ) = x

cos(cos^-1 (4/5)) = 4/5

Answer 7

Cos^-1(4/5) = 36.67º
cos 36.67º = 0.80 = 4/5

Answer 8

This is true, but to the person who asked, be careful when calculating

cos^-1 ( cos x),

which is not always x because of how the inverse cosine is defined.

Answer 9

cos^-1 or arccos is defined on the interval [-1,1]
Since 4/5 is on [-1,1], then the number is well defined
Now, cos^-1: [-1,1] ---> [0, pi] is the inverse of the cosine function restricted to the interval [0,pi]:
cos : [0,pi] ---> [-1,1]

Therefore, cos^-1( cos(4/5)) =4/5 and
cos(cos^-1(4/5)) = 4/5,
by the definition of inverse function.