How Do You Evaluate cos(arccos((32pi)/3))?

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Evaluate cos(arccos((32pi)/3))

From my understanding, I need to subtract (32pi)/3 by pi until the answer falls within the domain of [-1, 1]. Is this the correct way to solve this problem?
 
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Yes, I am sure. It IS possible, however, that the problem has no solution, if that's what you're getting at.
 
What is the domain of arccos? Also, this shouldn't be in the "Calculus and Beyond" section of the forum...
 
Ianfinity said:
Evaluate cos(arccos((32pi)/3))

From my understanding, I need to subtract (32pi)/3 by pi until the answer falls within the domain of [-1, 1]. Is this the correct way to solve this problem?
No. That's not the way to solve this.

See Ansatz7's post above.