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

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Homework Help Overview

The problem involves evaluating cos(arccos((32pi)/3)), which raises questions about the domain of the arccos function and the validity of the input value.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need to adjust the input value to fit within the domain of the arccos function, with one participant questioning whether the problem may have no solution. Another participant suggests a potential misunderstanding regarding the expression being evaluated.

Discussion Status

The discussion is exploring different interpretations of the problem, particularly regarding the input value for the arccos function and its implications. Some participants are questioning the appropriateness of the problem's placement in the forum.

Contextual Notes

There is a focus on the domain of the arccos function, with participants noting that the input value (32pi/3) may not be valid within the expected range.

Ianfinity
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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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Are you sure you don't mean [itex]\arccos(\cos(\frac{32\pi}3))[/itex] instead of [itex]\cos(\arccos(\frac{32\pi}3))[/itex] ?
 
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.
 

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