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Why is $\sin(30^\circ)$ exactly $0.5,$ when it could be 0.49999 or something else? There must be an easy geometric explanation?

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4 Answers

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Draw an equilateral triangle $ABC$ with $AB = BC = CA = 1$. Let $D$ be the midpoint of $BC$.

Now, $AD$ is the perpendicular bisector of $BC$, so $BD = CD = \tfrac{1}{2}$ and $\angle ADB = 90^{\circ}$.

Also, $AD$ is the angle bisector of $\angle BAC$, so $\angle BAD = \angle CAD = 30^{\circ}$.

Hence, triangle $ABD$ is a right triangle. The hypotenuse has length $AB = 1$ and the side opposite $\angle BAD$ has length $BD = \tfrac{1}{2}$. Hence $\sin 30^{\circ} = \sin \angle BAD = \dfrac{BD}{AB} = \dfrac{1}{2}$.

EDIT: Here I have implicitly used the fact that in an isosceles triangle, the altitude to the base is both the perpendicular bisector of the base and the angle bisector of the vertex opposite of the base. I overlooked this fact as being "trivial", but it does need to be stated. A proof of that fact can be found here. Thanks DonAntonio for pointing that out.

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Perhaps the easiest, most elementary explanation is from basic Euclidean Geometry, with the basic theorem:

"In a right-angled triangle $\;30^\circ-60^\circ-90^\circ\;$, the length of the leg opposite to the $\;30^\circ\;$ angle equals half the hypotenuse's length".

The above gives you at once that, with $\;x:=$ the hypotenuse's length:

$$\;\sin 30^\circ=\frac{\text{opposite leg}}{\text{hypotenuse}}=\frac{\frac x2}x=\frac12$$

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The way $\pi$ was originally defined was in geometric terms: the ratio of a circle's circumference to its radius. If you divide a circle into two equal halves and then divide one of those halves into 6 equal slices, then it stands to reason that the hypotenuse of the triangle is equal to the radius of the original circle, and therefore it follows that $\sin(\frac{\pi}{6}) = \frac{1}{2}$. You can do this without knowing the value of $\pi$ in decimal or sexagesimal or any numeral system whatsoever.

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Here's another point of view to consider.

I'm thinking that what you're really wondering is why $\sin(\pi/6)$ is such a "nice" number as 1/2.

Instead of looking at it that way, ask yourself, is there some number $x$ such that $\sin(x) = 1/2$? Now if you look at the graph for sin, which shows that it's a smooth function that has a minimum of -1 and a maximum of 1, then it's clear that some such $x$ must exist. Hence, you could say that $\pi/6$ is almost defined to be the number such that $\sin(\pi/6) = 1/2$. Of course, this isn't quite right, but the point is that there's an intricate relationship between the $\sin$ function and $\pi$ that makes the identity $\sin(\pi/6) = 1/2$ true.

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