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Suppose I were to define the notion of angle using the unit circle. By elementary geometry, I realise I could use the unit circle's area, which is an intrinsic part of the circle, as my "new" measure of angle.

Thus, "a whole turn" would correspond to $\pi$ (the area of my unit circle), half a turn would be $\frac{\pi}{2}$, and so on.

Here comes the problem: according to my angle definition, $\sin(x+\pi)=\sin(x)$. But the infinite series of $\sin$, where $\sin(x) = x - \frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!} +\ ...$, doesn't agree with my angle definition. Does this mean that the infinite series of the trigonometric functions were defined on the basis of measuring angles using arclengths of the unit circle?

Maxis Jaisi
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  • Yes, the radian is defined in terms of the arc length, and all the usual trig relations assume radians. If you wish to use a different unit, substitute $x = \lambda u$ and derive equivalent relations in $u$. – dxiv Dec 22 '16 at 07:47
  • @dxiv If we define the radian in terms of area of sector, we wouldn't need the factor of $2$ in the period of the trigonometric functions (which is in some sense simpler?). Why do mathematicians agree to use arc length as the default measure, then? – Maxis Jaisi Dec 22 '16 at 08:05
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    @MaxisJaisi Some things are simpler, others are not. You no longer have $\frac{d}{dx} \sin = \cos$, for instace. – user7530 Dec 22 '16 at 08:09
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    @user7530 You are right. On the side of advantages, taking the definition of the radian as the sector area would unify the two trigonometries (circular and hyperbolic). Indeed, considering reference hyperbola $(H) x^2-y^2=1$, the area of the curvilinear triangle determined by $M_t(\cosh(t), \sinh(t)), O(0,0)$, and $A(1,0)$ is precisely $t$ (linear segments $M_tO$ and $OA$, curve arc along (H) from $A$ to $M_t$) – Jean Marie Dec 22 '16 at 08:30
  • @MaxisJaisi You may enjoy reading The Tau Manifesto and, of course, The Pi Manifesto. – dxiv Dec 22 '16 at 16:50

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