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The sine function is defined as the opposite side of the angle in question over the hypotenuse of the $90^\circ$ triangle.

$$\sin(â) = \frac{\text{opposite side}}{\text{hypotenuse}} \tag{$0^\circ<â<90^\circ$}$$

Therefore, I cannot see why this is valid for angles equal or greater than $90^\circ$. I know that in the trigonometric cirle, we can just look for the simetric angle in the other side and then the sine value will be the same, since the radius of the circle is $1$ (then the sine is just the height of the point, or the $y$ value). But I cannot understand how they became with this definition. For me, there's no sense to admit that the sine of an angle greater than $90^\circ$ will be symmetric to a triangle on the other side.

I know that the Taylor expansion for sine can converge for any angle (in radians) but this expansion is derived from the derivative of the sine function, which should be only defined for $0^\circ<x<90^\circ$.

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We defined $\sin x := \sum_{i=0}^\infty {(-1)^n\cdot\frac{x^{2n+1}}{(2n+1)!}}\ (x\in\mathbb{R})$ – moose Sep 24 '13 at 5:41
More or less everything is a convention. Definitions are arbitrary, although they are subject to "natural selection". Useful ones carry on, others are forgotten. – Dan Shved Sep 24 '13 at 6:30

1 Answer 1

When $\sin$ is defined geometrically, it is typically as the vertical coordinate on the unit circle. Certainly this is the only sensible extension from $[0^\circ, 90^\circ]$ if you want to retain any of the nice properties like analyticity, angle addition formulas, Euler's formula, etc.

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