# In the unit circle, how are sine and cosine values derived for points on it?

As one can see here, in this unit circle (like this one: http://www.coolmath.com/sites/cmat/files/images/28-trigonometry-03.gif), and in the table that I linked to on Wikipedia below, there are several points on the unit circle at which one can use a radical divided by a value (for sine and cosine, what I ask about, that value is 2) to obtain an exact value for the sine or cosine of an angle. An example is for $\frac{\pi}{3}$ radians, where is sine of that angle is $\frac{\sqrt 3}{2}$.

What I do not understand is where the idea to use the square root of 3 came from, and equally so, where the idea to divide the value by 2 came from. I do understand that the unit circle has a radius of 1 and sides of triangles made within it must pertain to the pythagorean theorem (hence these values with radicals, for accuracy), but that is all I understand.

How would one know to put exactly $\frac{\sqrt 3}{2}$ for the sine of $\frac{\pi}{3}$ radians? This is unclear to me.

• You do not understand why the unit circle has a radius of 1? Do you understand why a triangle has 3 sides? What is your real question here, exactly? – Daniel W. Farlow Nov 23 '16 at 4:02
• I wrote this up recently for some precalculus notes...not sure if that will help you, but maybe it is a start. – Daniel W. Farlow Nov 23 '16 at 4:07
• I get a little bit confused when you ask "where the idea to use the square root of 3 came from, and equally so, where the idea to divide the value by 2 came from". Nobody came up with these arbritrarily, it's just the value that has been proven to be equal to $\sin \dfrac {\pi}{3}$, using the standard definition of sine. Are you asking for a proof that $\sin \dfrac {\pi}{3} = \dfrac {\sqrt 3}{2}$? – Ovi Nov 23 '16 at 4:13
• The standard geometric proof for $\sin\frac{\pi}{3}$ as well as $\sin\frac{\pi}{6}$ and cosines of those involves looking first at the equilateral triangle with all sides equal to one and then bisecting one of the angles, forming a $30^\circ$-$60^\circ$-$90^\circ$ triangle with one side length $\frac{1}{2}$ and the hypotenuse length $1$. The remaining side's length can be found via pythagorean theorem to be precisely $\frac{\sqrt{3}}{2}$. For more general inputs however, it is easiest done using the formal definition of sine in terms of power series. – JMoravitz Nov 23 '16 at 5:08

1) $\angle C = ?$; 2) $BC = ?$; and 3) $AC = ?$. 1) $\angle Q = ?$; 2) $\angle QPR = ?$; $QR = ?$ and 4) $PR = ?$.