Question 366650
<font face="Garamond" size="+2">


Half-angle formula for sin:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2\left(\frac{a}{2}\right)\ =\ \frac{1}{2}\left\[1\ -\ \cos(a)\right\]]


Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2\left(\frac{5\pi}{12}\right)\ =\ \frac{1}{2}\left\[1\ -\ \cos\left(\frac{5\pi}{6}\right)\right\]]


Looking at the unit circle:


<img src="http://www.math.ucsd.edu/~jarmel/math4c/Unit_Circle_Angles.png">


we see that *[tex \Large \cos\left(\frac{5\pi}{6}\right)\ =\ -\frac{\sqrt{3}}{2}]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2\left(\frac{5\pi}{12}\right)\ =\ \frac{1}{2}\left\[1\ +\ \frac{\sqrt{3}}{2}\right\]]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2\left(\frac{5\pi}{12}\right)\ =\ \frac{2\ +\ \sqrt{3}}{4}]


Since sin is positive on *[tex \Large (0,\pi)] and *[tex \Large \frac{5\pi}{12}\ <\ \pi], take the positive root:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin\left(\frac{5\pi}{12}\right)\ =\ \frac{\sqrt{2\ +\ \sqrt{3}}}{2}]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
<div style="text-align:center"><a href="http://outcampaign.org/" target="_blank"><img src="http://cdn.cloudfiles.mosso.com/c116811/scarlet_A.png" border="0" alt="The Out Campaign: Scarlet Letter of Atheism" width="143" height="122" /></a></div>
</font>