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Can somebody show me why $$-\arctan\left(\frac{2\pi}{1-\cos(2\pi)}\right)$$ equals to $-90^\circ$ degrees? Thanks.

Dennis Gulko
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Joe Allendale
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    $\cos (2\pi) = 1$, so you have an infinity as the argument of $\tan^{-1}$. – Daniel Fischer Nov 10 '13 at 16:01
  • Plus, the $\arctan$ of anything lies strictly between $-90$ and $+90$ degrees. It looks very much like this was originally a statement about limits that got "simplified", so to say. Is that so? – Dan Shved Nov 10 '13 at 16:04

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Since $\;1-\cos 2\pi=0\;$ the argument isn't defined, but I guess the intention here is

$$\lim_{x\to 2\pi}-\arctan\frac{2\pi}{1-\cos x}=-\frac\pi2$$

since we know $\;\tan x\xrightarrow[x\to\frac\pi2\,^-]{}\infty\;$

DonAntonio
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  • I see it this way @GitGud: $$\frac{2\pi}{1-\cos x}\xrightarrow[x\to 2\pi]{}+\infty\implies,-\arctan\left(\frac{2\pi}{1-\cos x}\right)\xrightarrow[x\to 2\pi]{}-\frac\pi2 $$ precisely because of what you say... – DonAntonio Nov 10 '13 at 16:14
  • Close to, and to the left of $;\pi/2;$, both $;\sin,,,\cos;$ are positive, so tangent must be non-negative there...! – DonAntonio Nov 10 '13 at 16:17
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    Sorry for bothering you with these wrong trivialities, I visualized it completly wrong.You're right, of course. – Git Gud Nov 10 '13 at 16:19
  • Welcome to the " Rats, I read something yet I understood something else!" club, @GitGud: I'm honorary president. – DonAntonio Nov 10 '13 at 16:20
  • $\ddot \smile{}$ – Git Gud Nov 10 '13 at 16:20