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Computation of cos(pi/5)

From jscl-meditor

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\mathrm{cos}\left(a\right)+i\mathrm{sin}\left(a\right) = eia
{\left({e}^{i\frac{2\pi }{5}}\right)}^{5} = 1
{\left(\mathrm{cos}\left(\frac{2\pi }{5}\right)+i\mathrm{sin}\left(\frac{2\pi }{5}\right)\right)}^{5} = 1
 − 1 + z5 = 0
-\left(1-z\right)\left(1+z+{z}^{2}+{z}^{3}+{z}^{4}\right) = 0
1 + z + z2 + z3 + z4 = 0
1+z+{z}^{2}+\frac{1}{z}+\frac{1}{{z}^{2}} = 0
-1+\left(z+\frac{1}{z}\right)+{\left(z+\frac{1}{z}\right)}^{2} = 1+z+{z}^{2}+\frac{1}{z}+{\left(\frac{1}{z}\right)}^{2}
\mathrm{solve}\left(-1+x+{x}^{2},x\right)
\left(\frac{1-\sqrt{5}}{2}+\left(z+\frac{1}{z}\right)\right)\left(\frac{1+\sqrt{5}}{2}+\left(z+\frac{1}{z}\right)\right) = 0
2+2{z}^{2}+z\left(1-\sqrt{5}\right) = 0
a) \mathrm{solve}\left(2+2{z}^{2}+z\left(1-\sqrt{5}\right),z\right)
z = -\frac{1-\sqrt{5}-\sqrt{2}\sqrt{-5-\sqrt{5}}}{4}
z = -\frac{1-\sqrt{5}+\sqrt{2}\sqrt{-5-\sqrt{5}}}{4}
2+2{z}^{2}+z\left(1+\sqrt{5}\right) = 0
b) \mathrm{solve}\left(2+2{z}^{2}+z\left(1+\sqrt{5}\right),z\right)
z = -\frac{1+\sqrt{5}-\sqrt{2}\sqrt{-5+\sqrt{5}}}{4}
z = -\frac{1+\sqrt{5}+\sqrt{2}\sqrt{-5+\sqrt{5}}}{4}
\mathrm{cos}\left(\frac{2\pi }{5}\right)+i\mathrm{sin}\left(\frac{2\pi }{5}\right)=z=-\frac{1-\sqrt{5}-\sqrt{2}\sqrt{-5-\sqrt{5}}}{4}=-\frac{1-\sqrt{5}-i\sqrt{2}\sqrt{5+\sqrt{5}}}{4}
\mathrm{cos}\left(\frac{2\pi }{5}\right) = -\frac{1-\sqrt{5}}{4}
\mathrm{cos}\left(2a\right) = -1+2{\mathrm{cos}}^{2}\left(a\right)
\mathrm{subst}\left(-1+2{\mathrm{cos}}^{2}\left(a\right),a,\frac{\pi }{5}\right)
-1+2{\mathrm{cos}}^{2}\left(\frac{\pi }{5}\right) = -\frac{1-\sqrt{5}}{4}
\mathrm{solve}\left(-1+\frac{1-\sqrt{5}}{4}+2{\mathrm{cos}}^{2}\left(\frac{\pi }{5}\right),\mathrm{cos}\left(\frac{\pi }{5}\right)\right)
\mathrm{cos}\left(\frac{\pi }{5}\right) = \frac{\sqrt{2}\sqrt{3+\sqrt{5}}}{4}
0.8090169943749475 = 0.8090169943749475