$21°\ (=\dfrac{7\pi}{60})$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。
$\sin21°$
$\sin$の加法定理より
\begin{align*}\sin21°&=\sin(36°-15°)\\[0.5em]&=\sin36°\cos15°-\cos36°\sin15°\end{align*}
\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{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*}
$\cos21°$
$\cos$の加法定理より
\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
\[\tan\theta=\frac{\sin\theta}{\cos\theta}\]
より、$\text{(a),(b)}$を代入して
\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。
\begin{align*}\sin21°&=0.35837\\[1em]\cos21°&=0.93358\\[1em]\tan21°&=0.38386\end{align*}
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