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Without producing the maclaurin series for $\arctan x$, how would determine whether it was odd or even?

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It is the inverse function of $tan(x)$ on $(-\pi/2,\pi/2)$, which is odd. – TTY May 22 '13 at 19:37
Alternatively, you also know that $$ \arctan(x) + \arctan(y) = \arctan\left(\frac{x+y}{1-xy}\right).$$ This implies that $\arctan(x)+\arctan(-x) = 0$. – Tyler Holden May 22 '13 at 19:41
Thanks guys!! I didn't realise that the inverse of an odd function is also odd but just proved it to myself haha ... @Tyler Is that an identity I should know? I have no idea how I'd derive it – James Stott May 22 '13 at 19:49
If you are defining $\arctan x$ analytically by $\arctan x=\int_0^x\frac{dt}{1+t^2}$, can use the fact that an integral from $0$ to $x$ (posibly negative) of an even function is odd. – André Nicolas May 22 '13 at 19:56
@James You can show that $\tan(x+y) = \frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}$ fairly quickly by using the angle sum identities. In that case, the identity above is the dual statement about the inverses. – Tyler Holden May 22 '13 at 19:58

From a geometrical definition of $\tan$ and $\arctan$, their oddness follows by reflecting the right triangle $ABC$ with $AB=1$ and $BC=\tan \angle A$ along the line $AB$.

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You could also use some differential calculus. Applying the Chain Rule:

$$\left(\arctan(-x)\right)'=-\frac1{1+x^2}=-\left(\arctan x\right)'\implies\arctan(-x)=-\arctan x+C$$

Now just evaluate at $\;x=0\;$ the last equality and get $\;C=0\implies\arctan(-x)=-\arctan x\;$ .

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