anonymous
  • anonymous
H E L P : Cos (13pi/12 ) Use Half Angle ID to simplify ~
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
I got : -root 3 - root 6 / 4 Is that what you got..?
Venomblast
  • Venomblast
radian or degree?
anonymous
  • anonymous
What do you mean? You just use that ID to simplify?

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Venomblast
  • Venomblast
but i dont what ID.
jim_thompson5910
  • jim_thompson5910
\[\Large \cos\left(\frac{13\pi}{12}\right) = \cos\left(\frac{1}{2}\cdot\frac{13\pi}{6}\right)\] \[\Large \cos\left(\frac{13\pi}{12}\right) = -\sqrt{\frac{1+\cos\left(\cdot\frac{13\pi}{6}\right)}{2}}\] \[\Large \cos\left(\frac{13\pi}{12}\right) = -\sqrt{\frac{1+\frac{\sqrt{3}}{2}}{2}}\] \[\Large \cos\left(\frac{13\pi}{12}\right) = -\sqrt{\frac{1}{2}+\frac{\sqrt{3}}{4}}\] \[\Large \cos\left(\frac{13\pi}{12}\right) = -\sqrt{\frac{2}{4}+\frac{\sqrt{3}}{4}}\] \[\Large \cos\left(\frac{13\pi}{12}\right) = -\sqrt{\frac{2+\sqrt{3}}{4}}\] \[\Large \cos\left(\frac{13\pi}{12}\right) = -\frac{\sqrt{2+\sqrt{3}}}{\sqrt{4}}\] \[\Large \cos\left(\frac{13\pi}{12}\right) = -\frac{\sqrt{2+\sqrt{3}}}{2}\]
anonymous
  • anonymous
cool thnx! (:
jim_thompson5910
  • jim_thompson5910
yw

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