# Half-Angle formulas

• May 2nd 2010, 09:16 AM
GUNSLINGAZ
Half-Angle formulas
find the exact value of sin(5pi/12)

here's what I have, can you tell me if I'm doing it correctly and if I've simplified. I'm more than a little confused.

[sin(5pi/6)]/2 = +/- √[(1-cosx)/2]

= +/- √{[1-cos(5pi/6)]/2}

= +/- √ ({1-[-√(3)/2]}/2)

= +/- √ {[1+√(3)/2]/2}

And that's as far as I got. I'm not sure if I'm doing it correctly or if i am, i have no idea how to simplify further. :/

Thanks ahead of time, all help is very much appreciated.
• May 2nd 2010, 09:57 AM
Can't quite tell exactly what you're doing.

Here's a way...

$\sin(5\pi/12)=\sin(\pi/4 + \pi/6)$

= $\sin(\pi/4)\cos(\pi/6)+\cos(\pi/4)\sin(\pi/6)$

= $\frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}}\frac{1}{2}$

= $\frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}}$

= $\frac{(1 + \sqrt{3})}{2\sqrt{2}}$

= $\frac{\sqrt{2}(1 + \sqrt{3})}{4}$