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Take half a square with side length $1$. The resulting right-angled triangle ABC has two angles of $45^\circ$. By Pythagoras’ theorem, the hypotenuse AC has length $\sqrt{2}$. Applying the definitions on the previous page gives the values in the table below. that $\sin 30^\circ= \frac{1}{2}$

Sorry I cannot provide diagram, but from my understanding $\sin =$ opposite / hypotenuse. How is the value $0.5$ derived then? No possible combination. What point of reference should I be looking from?

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You have 45 degrees? Not 30.. $sin(45) = \frac{1}{\sqrt{2}}$ – Noxet Jul 29 '14 at 13:30

Take an equilateral triangle of side $1$.

Bisect it through a vertex and the midpoint of the opposite side.

You now have two right-angled triangles with angles $30^\circ, 60^\circ, 90^\circ$ with the edge opposite the $30^\circ$ of length $\frac12$ and hypotenuse $1$. So $$\sin (30^\circ)=\frac{\frac12}{1}=\frac12.$$

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The trigonometric function values for all the special angles are derived from the first three regular polygons (equilateral triangle, square, and regular pentagon). The trigonometric function values for the first two shapes make use of the Pythagorean theorem. The values for the pentagon make use of similar triangles.

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Nice catch. I'll make the change shortly. :) – John Joy May 16 at 13:23

The right angled triangle you describe has angles of $45$ degrees, not $30$ degrees. What you proved is $\sin 45=\frac{1}{\sqrt2}$, which is correct.

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you mean incorrect i believe? – MonK Jul 29 '14 at 13:33
@Sid Typo. Fixed. – 5xum Jul 29 '14 at 13:39

First draw a equilateral triangle and find the sine or cosine of the angle of $60^\circ$. Now for the angle of $30$ degrees apply the identity $$\cos x=\sin (90^\circ -x)$$

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