sin1°,cos1°を求めてみた！

1.$\cos 10°$の値を求める。
2.$\cos 9°$の値を求める。
　・36°を求める。
　・9°を求める。
3.加法定理で$\cos 1°$を求める。

$$\begin{array}{rl}\cos \frac{\pi }{18}& =\frac{e^{i\pi /18}+e^{-i\pi /18}}{2}\\ & =\frac{1}{2}(\sqrt[3]{e^{i\pi /6}}+\sqrt[3]{e^{-i\pi /6}})\\ & =\frac{1}{2}(\sqrt[3]{\frac{\sqrt{3}+i}{2}}+\sqrt[3]{\frac{\sqrt{3}-i}{2}})\end{array}$$
$$\begin{array}{rl}\sin \frac{\pi }{18}& =\frac{e^{i\pi /18}-e^{-i\pi /18}}{2i}\\ & =\frac{1}{2i}(\sqrt[3]{e^{i\pi /6}}-\sqrt[3]{e^{-i\pi /6}})\\ & =\frac{1}{2i}(\sqrt[3]{\frac{\sqrt{3}+i}{2}}-\sqrt[3]{\frac{\sqrt{3}-i}{2}})\end{array}$$

${\displaystyle x=\frac{\pi }{5}}$であるとすると、$2x=\pi -3x$であるから、
$$\begin{array}{rl}\sin 2x& =\sin (\pi -3x)\\ & =\sin 3x\\ 2\sin x\cos x& =3\sin x-4{\sin }^{3}x (二倍角と三倍角の公式)\\ 2\cos x& =3-4{\sin }^{2}x (\sin x\ne 0より)\\ 2\cos x& =3-4(1-{\cos }^{2}x)\\ 2\cos x& =-1+4{\cos }^{2}x\\ \therefore 4{\cos }^{2}x-& 2\cos x-1=0\end{array}$$
この二次方程式を解いて、
$$\cos \frac{\pi }{5}=\frac{1+\sqrt{5}}{4}$$
また、${\displaystyle {\sin }^{2}x=1-{\cos }^{2}x}$より、
$$\begin{array}{rl}\sin \frac{\pi }{5}& =\sqrt{1-\frac{(1+\sqrt{5}{)}^2}{16}}\\ & =\sqrt{1-\frac{6+2\sqrt{5}}{16}}\\ & =\frac{\sqrt{10-2\sqrt{5}}}{4}\end{array}$$

$$\begin{array}{rl}\cos \frac{\pi }{20}& =\cos (\frac{\pi }{4}-\frac{\pi }{5})\\ & =\cos \frac{\pi }{4}\cos \frac{\pi }{5}+\sin \frac{\pi }{4}\sin \frac{\pi }{5}\\ & =\frac{1}{\sqrt{2}}(\cos \frac{\pi }{5}+\sin \frac{\pi }{5})\\ & =\frac{1}{\sqrt{2}}(\frac{1+\sqrt{5}}{4}+\frac{\sqrt{10-2\sqrt{5}}}{4})\\ & =\frac{1+\sqrt{5}}{4\sqrt{2}}+\frac{\sqrt{5-\sqrt{5}}}{4}\end{array}$$

$$\begin{array}{rl}\sin \frac{\pi }{20}& =\sin (\frac{\pi }{4}-\frac{\pi }{5})\\ & =\sin \frac{\pi }{4}\cos \frac{\pi }{5}-\cos \frac{\pi }{4}\sin \frac{\pi }{5}\\ & =\frac{1}{\sqrt{2}}(\cos \frac{\pi }{5}-\sin \frac{\pi }{5})\\ & =\frac{1}{\sqrt{2}}(\frac{1+\sqrt{5}}{4}-\frac{\sqrt{10-2\sqrt{5}}}{4})\\ & =\frac{1+\sqrt{5}}{4\sqrt{2}}-\frac{\sqrt{5-\sqrt{5}}}{4}\end{array}$$

$$\begin{array}{rl}\sin 1°& =\sin (10°-9°)\\ & =\sin 10°\cos 9°-\cos 10°\sin 9°\\ & =\frac{1}{2i}(\sqrt[3]{\frac{\sqrt{3}+i}{2}}-\sqrt[3]{\frac{\sqrt{3}-i}{2}})(\frac{1+\sqrt{5}}{4\sqrt{2}}+\frac{\sqrt{5-\sqrt{5}}}{4})\\ & -\frac{1}{2}(\sqrt[3]{\frac{\sqrt{3}+i}{2}}+\sqrt[3]{\frac{\sqrt{3}-i}{2}})(\frac{1+\sqrt{5}}{4\sqrt{2}}-\frac{\sqrt{5-\sqrt{5}}}{4})\end{array}$$
$$\begin{array}{rl}\cos 1°& =\cos (10°-9°)\\ & =\cos 10°\cos 9°+\sin 10°\sin 9°\\ & =\frac{1}{2}(\sqrt[3]{\frac{\sqrt{3}+i}{2}}+\sqrt[3]{\frac{\sqrt{3}-i}{2}})(\frac{1+\sqrt{5}}{4\sqrt{2}}+\frac{\sqrt{5-\sqrt{5}}}{4})\\ & +\frac{1}{2i}(\sqrt[3]{\frac{\sqrt{3}+i}{2}}-\sqrt[3]{\frac{\sqrt{3}-i}{2}})(\frac{1+\sqrt{5}}{4\sqrt{2}}-\frac{\sqrt{5-\sqrt{5}}}{4})\end{array}$$