How do I find what arccos(cos4) and arctan(tan4) is?
Is it 4 degrees or 4 radians. If radians:
$\displaystyle \text{arccos} (\cos4) \neq 4$ because the range of arcos is $\displaystyle 0 \leq \text{arcos} x \leq \pi $ ...... 4 radians is in the third quadrant - outside the range of arcos. So you need the 'equivalent' angle that's in the second quadrant - the second quadant is in the range of arcos.
So in fact, $\displaystyle \text{arccos} (\cos4) = \pi - (4 - \pi) = 2 \pi - 4$.
Do arctan(tan4) in a similar way.
My fault for not clarifying this. Since there was no pi term given for this question (although, i assume there originally was), i assumed 4 to mean 4 degrees. Yes, I doubt the 4 was originally given in terms of degrees, (it was very most likely given as 4pi radians), but I did what was given to me. Fantastic already solved the arccos problem its radian form, so I thought it would be good pedegogy nonetheless to do this excercize with 4 degrees. (4 degrees is out there, but it is a factor pair along with 45 of 180).
So, 4 degrees (which is 4*(pi/180) = pi/45 radians) is clearly in the first quad, thus we can do cos(4deg). The result is .9999 something, which is terribly close to the upper limit of the domain for arccos if we want an answer in the reals. (-1 to 1). Computing the arccos of .999something takes us back to 4 degrees. So, pi/45 (same as 4 deg) radians must satisfy the range requirements of arccos 0 to 180 deg (0 to pi radians). It does, and we get 4.
I apologize if I pre-empted any confusion.
The arccos of the cos of 4 is simply 4.
Always check your restrictions/limits. All Students Take Calculus for the first, second, third, and fourth quadrants, respectively.