Understanding Arctan: Solving for 0 and 1/0

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Arctan(0) equals 0, as confirmed by calculators and the definition of the function, which states that arctan(x) returns the angle whose tangent is x. The confusion arises from the fact that tan(pi) also equals 0, but arctan is defined to return values in the range of -π/2 to π/2, making arctan(0) = 0 the correct answer. For arctan(1/0), the operation is undefined due to division by zero; however, the limit as x approaches 0 from the positive side results in arctan(1/x) approaching π/2. Understanding these concepts requires recognizing the specific properties and definitions of the arctan function.
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What is arctan(0)?!

I tired using my calculator (its a ti-83 plus) I hit [sec] button then [tan] then [ 0 ] and [ ) ] and it gives me 0 as answer It is in rad mode too (deg mode is same results)

I tried using the google calculator and it gives me 0 also!
http://www.google.com/search?hl=en&q=arctan(0)

BUT my math book says its pie!


Then it tried tan(pi) and indeed that does = 0!

WHY!??!? doenst arctan(0) = pi?



Another question
What is arctan(1/0) ? doens that = impossible?? (if i use my calc it says can't divide by 0)

However the answer is pi/2?!??! WHYYY


Thanks for the help
 
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Arctan(x) = Pi means x = tan(pi)

tan(\pi) = \frac{0up}{1left} = 0

Arctan(x) = 0 means x = tan(0)

tan(0) = \frac{0up}{1right} = 0

Get it?

2. The answer is in limits.

arctan(1/0) = undefined

However the limit is defined:

\lim x\rightarrow 0 \mbox{ of } arctan(1/x) = \frac{\pi}{2}

Try graphing the function arctan(x) and use the trace tool to see the values at those points.
 
Try translating tan(x) = \frac{sinx}{cosx}, you might see it more clearly.
 
tan (2pi) = 0 too. So why doesn't arctan(0) = 2pi?

Because there's an important part of the definition of arctan that you're forgetting -- go look at it again.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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