sin21°、cos21°、tan21°はどんな数？ ~ 数学について考えてみる

2022年1月10日

sin21°、cos21°、tan21°はどんな数？

PLUMBAGO

　 $21°\ (=\dfrac{7\pi}{60})$ のときの三角関数がどんな式で表されるのかを調べてみました。

$\sin21°$

\sin

$\sin$ の加法定理より

\begin{aligned} \sin 21 ° & = \sin (36 ° - 15 °) \\ = \sin 36 ° \cos 15 ° - \cos 36 ° \sin 15 ° \end{aligned}

$\begin{align*}\sin21°&=\sin(36°-15°)\\[0.5em]&=\sin36°\cos15°-\cos36°\sin15°\end{align*}$

「

\sin 36 °

$\sin36°$ がどんな数になるか求めてみよう」、「

\sin 15 °, \cos 15 °, \tan 15 °

$\sin15°,\cos15°,\tan15°$ はどんな数？」より

\begin{aligned} \sin 36 ° & = \frac{\sqrt{10 - 2 \sqrt{5}}}{4} \\ \cos 36 ° & = \frac{\sqrt{5} + 1}{4} \\ \sin 15 ° & = \frac{\sqrt{6} - \sqrt{2}}{4} \\ \cos 15 ° & = \frac{\sqrt{6} + \sqrt{2}}{4} \end{aligned}

$\begin{align*}\sin36°&=\frac{\sqrt{10-2\sqrt{5}}}{4}\\[0.5em]\cos36°&=\frac{\sqrt{5}+1}{4}\\[1em]\sin15°&=\frac{\sqrt{6}-\sqrt{2}}{4}\\[0.5em]\cos15°&=\frac{\sqrt{6}+\sqrt{2}}{4}\end{align*}$

なので

\begin{aligned} \sin 21 ° & = \frac{\sqrt{10 - 2 \sqrt{5}}}{4} \cdot \frac{\sqrt{6} + \sqrt{2}}{4} - \frac{\sqrt{5} + 1}{4} \cdot \frac{\sqrt{6} - \sqrt{2}}{4} \\ (a) & = \frac{\sqrt{2} {(\sqrt{3} + 1) \sqrt{10 - 2 \sqrt{5}} - (\sqrt{5} + 1) (\sqrt{3} - 1)}}{16} \\ = \frac{\sqrt{2 (\sqrt{3} + 1)^{2} (10 - 2 \sqrt{5})} - \sqrt{30} + \sqrt{10} - \sqrt{6} + \sqrt{2}}{16} \\ = \underline{\frac{2 \sqrt{20 + 10 \sqrt{3} - 4 \sqrt{5} - 2 \sqrt{15}} - \sqrt{30} + \sqrt{10} - \sqrt{6} + \sqrt{2}}{16}} \end{aligned}

$\begin{align*}\sin21°&=\frac{\sqrt{10-2\sqrt{5}}}{4}\cdot\frac{\sqrt{6}+\sqrt{2}}{4}-\frac{\sqrt{5}+1}{4}\cdot\frac{\sqrt{6}-\sqrt{2}}{4}\\[0.5em]&=\frac{\sqrt{2}\left\{(\sqrt{3}+1)\sqrt{10-2\sqrt{5}}-(\sqrt{5}+1)(\sqrt{3}-1)\right\}}{16}\tag{a}\\[0.5em]&=\frac{\sqrt{2(\sqrt{3}+1)^2(10-2\sqrt{5})}-\sqrt{30}+\sqrt{10}-\sqrt{6}+\sqrt{2}}{16}\\[0.5em]&=\underline{\frac{2\sqrt{20+10\sqrt{3}-4\sqrt{5}-2\sqrt{15}}-\sqrt{30}+\sqrt{10}-\sqrt{6}+\sqrt{2}}{16}}\end{align*}$

$\cos21°$

\cos

$\cos$ の加法定理より

\begin{aligned} \cos 21 ° & = \cos (36 ° - 15 °) \\ = \cos 36 ° \cos 15 ° + \sin 36 ° \sin 15 ° \\ = \frac{\sqrt{5} + 1}{4} \cdot \frac{\sqrt{6} + \sqrt{2}}{4} + \frac{\sqrt{10 - 2 \sqrt{5}}}{4} \cdot \frac{\sqrt{6} - \sqrt{2}}{4} \\ (b) & = \frac{\sqrt{2} {(\sqrt{5} + 1) (\sqrt{3} + 1) + (\sqrt{3} - 1) \sqrt{10 - 2 \sqrt{5}}}}{16} \\ = \frac{\sqrt{2 (\sqrt{3} - 1)^{2} (\sqrt{10 - 2 \sqrt{5}})} + \sqrt{30} + \sqrt{10} + \sqrt{6} + \sqrt{2}}{16} \\ = \underline{\frac{2 \sqrt{20 - 10 \sqrt{3} - 4 \sqrt{5} + 2 \sqrt{15}} + \sqrt{30} + \sqrt{10} + \sqrt{6} + \sqrt{2}}{16}} \end{aligned}

$\begin{align*}\cos21°&=\cos(36°-15°)\\[0.5em]&=\cos36°\cos15°+\sin36°\sin15°\\ &=\frac{\sqrt{5}+1}{4}\cdot\frac{\sqrt{6}+\sqrt{2}}{4}+\frac{\sqrt{10-2\sqrt{5}}}{4}\cdot\frac{\sqrt{6}-\sqrt{2}}{4}\\[0.5em]&=\frac{\sqrt{2}\left\{(\sqrt{5}+1)(\sqrt{3}+1)+(\sqrt{3}-1)\sqrt{10-2\sqrt{5}}\right\}}{16}\tag{b}\\[0.5em]&=\frac{\sqrt{2(\sqrt{3}-1)^2(\sqrt{10-2\sqrt{5}})}+\sqrt{30}+\sqrt{10}+\sqrt{6}+\sqrt{2}}{16}\\[0.5em]&=\underline{\frac{2\sqrt{20-10\sqrt{3}-4\sqrt{5}+2\sqrt{15}}+\sqrt{30}+\sqrt{10}+\sqrt{6}+\sqrt{2}}{16}}\end{align*}$

$\tan21°$

　三角関数の相互関係

\tan θ = \frac{\sin θ}{\cos θ}

$\tan\theta=\frac{\sin\theta}{\cos\theta}$

より、

(a),(b)

$\text{(a),(b)}$ を代入して

\begin{aligned} \tan 21 ° & = \frac{\sin 21 °}{\cos 21 °} \\ = \frac{\frac{\sqrt{2} {(\sqrt{3} + 1) \sqrt{10 - 2 \sqrt{5}} - (\sqrt{5} + 1) (\sqrt{3} - 1)}}{16}}{\frac{\sqrt{2} {(\sqrt{5} + 1) (\sqrt{3} + 1) + (\sqrt{3} - 1) \sqrt{10 - 2 \sqrt{5}}}}{16}} \\ = \frac{(\sqrt{3} + 1) \sqrt{10 - 2 \sqrt{5}} - (\sqrt{5} + 1) (\sqrt{3} - 1)}{(\sqrt{5} + 1) (\sqrt{3} + 1) + (\sqrt{3} - 1) \sqrt{10 - 2 \sqrt{5}}} \\ \times \frac{(\sqrt{5} + 1) (\sqrt{3} + 1) - (\sqrt{3} - 1) \sqrt{10 - 2 \sqrt{5}}}{(\sqrt{5} + 1) (\sqrt{3} + 1) - (\sqrt{3} - 1) \sqrt{10 - 2 \sqrt{5}}} \\ = \frac{(\sqrt{5} + 1) \sqrt{10 - 2 \sqrt{5}} - 4}{2 (2 \sqrt{3} + \sqrt{5} - 1)} \times \frac{2 \sqrt{3} - (\sqrt{5} - 1)}{2 \sqrt{3} - (\sqrt{5} - 1)} \\ = \frac{(\sqrt{15} + \sqrt{3} - 2) \sqrt{10 - 2 \sqrt{5}} - 2 - 4 \sqrt{3} + 2 \sqrt{5}}{2 (3 + \sqrt{5})} \times \frac{3 - \sqrt{5}}{3 - \sqrt{5}} \\ = \frac{(\sqrt{15} + \sqrt{5} - \sqrt{3} - 3) \sqrt{10 - 2 \sqrt{5}} - 8 - 6 \sqrt{3} + 4 \sqrt{5} + 2 \sqrt{15}}{4} \\ = \frac{\sqrt{(\sqrt{15} + \sqrt{5} - \sqrt{3} - 3)^{2} (10 - 2 \sqrt{5})} - 8 - 6 \sqrt{3} + 4 \sqrt{5} + 2 \sqrt{15}}{4} \\ = \underline{\frac{\sqrt{110 + 60 \sqrt{3} - 46 \sqrt{5} - 28 \sqrt{15}} - 4 - 3 \sqrt{3} + 2 \sqrt{5} + \sqrt{15}}{2}} \end{aligned}

$\begin{align*}\tan21°&=\frac{\sin21°}{\cos21°}\\[0.5em]&=\frac{\cfrac{\sqrt{2}\left\{(\sqrt{3}+1)\sqrt{10-2\sqrt{5}}-(\sqrt{5}+1)(\sqrt{3}-1)\right\}}{16}}{\cfrac{\sqrt{2}\left\{(\sqrt{5}+1)(\sqrt{3}+1)+(\sqrt{3}-1)\sqrt{10-2\sqrt{5}}\right\}}{16}}\\[0.5em]&=\frac{(\sqrt{3}+1)\sqrt{10-2\sqrt{5}}-(\sqrt{5}+1)(\sqrt{3}-1)}{(\sqrt{5}+1)(\sqrt{3}+1)+(\sqrt{3}-1)\sqrt{10-2\sqrt{5}}}\\ &\qquad×\frac{(\sqrt{5}+1)(\sqrt{3}+1)-(\sqrt{3}-1)\sqrt{10-2\sqrt{5}}}{(\sqrt{5}+1)(\sqrt{3}+1)-(\sqrt{3}-1)\sqrt{10-2\sqrt{5}}}\\[0.5em]&=\frac{(\sqrt{5}+1)\sqrt{10-2\sqrt{5}}-4}{2(2\sqrt{3}+\sqrt{5}-1)}×\frac{2\sqrt{3}-(\sqrt{5}-1)}{2\sqrt{3}-(\sqrt{5}-1)}\\[0.5em]&=\frac{(\sqrt{15}+\sqrt{3}-2)\sqrt{10-2\sqrt{5}}-2-4\sqrt{3}+2\sqrt{5}}{2(3+\sqrt{5})}×\frac{3-\sqrt{5}}{3-\sqrt{5}}\\[0.5em]&=\frac{(\sqrt{15}+\sqrt{5}-\sqrt{3}-3)\sqrt{10-2\sqrt{5}}-8-6\sqrt{3}+4\sqrt{5}+2\sqrt{15}}{4}\\[0.5em]&=\frac{\sqrt{(\sqrt{15}+\sqrt{5}-\sqrt{3}-3)^2(10-2\sqrt{5})}-8-6\sqrt{3}+4\sqrt{5}+2\sqrt{15}}{4}\\[0.5em]&=\underline{\frac{\sqrt{110+60\sqrt{3}-46\sqrt{5}-28\sqrt{15}}-4-3\sqrt{3}+2\sqrt{5}+\sqrt{15}}{2}}\end{align*}$

　それぞれの近似値は以下のようになります。