Explain why tan pi=0 does not imply that arctan0=pi?

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Explain why tan pi=0 does not imply that arctan0=pi?

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Answer 1 · 2015-09-10T21:08:40

$π$ $pi$ is not in the interval used for $arctan$ $arctan$

Explanation:

We want $arctan$ $arctan$ to be a function. We want it to give one, never two, values for a single input.

By definition:

$y = arctan x$ $y = arctan x$ if and only if ( $tan y = x$ $tan y = x$ and $- \frac{π}{2} < y < \frac{π}{2}$ $-pi/2 < y < pi/2$ )

Since $π$ $pi$ is not in $(- \frac{π}{2}, \frac{π}{2})$ $(-pi/2, pi/2)$ there is no $x$ $x$ for which $arctan x = π$ $arctanx = pi$

(The situation is similar to: Explain why ${(- 3)}^{2} = 9$ $(-3)^2 = 9$ does not imply that $\sqrt{9} = - 3$ $sqrt9 = -3$ . It ( $- 3$ $-3$ ) is the wrong kind of number to be a principal square root.)

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Explain why tan pi=0 does not imply that arctan0=pi?

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