Question 461701
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Use the Half-Angle Formula for Sine which follows directly from the Double Angle Formula for Cosine which follows directly from the Sum Formula for Cosine.  See <a href="http://oakroadsystems.com/twt/double.htm
">Double Angle and Half Angle Formulas</a> for a complete discussion.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 22.5^\circ\ =\ \frac{45^\circ}{2}]


And


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(45^\circ)\ =\ \frac{\sqrt{2}}{2}]


From the unit circle:


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


Since the half-angle formula for sine is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin\left(\frac{\varphi}{2}\right)\ =\ \pm\sqrt{\frac{1\ -\ \cos(\varphi)}{2}}]


where you select the sign based on the quadrant in which *[tex \Large \sin\left(\frac{\varphi}{2}\right)] exists.


We can write, since *[tex \Large 22.5^\circ] is in the first quadrant:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin\left(22.5^\circ\right)\ =\ \sqrt{\frac{1\ -\ \frac{\sqrt{2}}{2}}{2}}\ =\ \frac{\sqrt{2\ -\ \sqrt{2}}}{2}]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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