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

[Ianfinity]

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?

 

[D H]

Are you sure you don't mean \arccos(\cos(\frac{32\pi}3)) instead of \cos(\arccos(\frac{32\pi}3)) ?

 

[Ianfinity]

Yes, I am sure. It IS possible, however, that the problem has no solution, if that's what you're getting at.

 

[Ansatz7]

What is the domain of arccos? Also, this shouldn't be in the "Calculus and Beyond" section of the forum...

 

[SammyS]

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.