$\sin3°$
$\sin$の加法定理より
\begin{align*}\sin3°&=\sin(18°-15°)\\[0.5em]&=\sin18°\cos15°-\cos18°\sin15°\end{align*}
ここで、「$\sin18°,\cos18°,\tan18°$はどんな数？」、「$\sin15°,\cos15°,\tan15°$はどんな数？」より
\begin{align*}\sin15°&=\frac{\sqrt{6}-\sqrt{2}}{4}\\[1em]
\cos15°&=\frac{\sqrt{6}+\sqrt{2}}{4}\\[1.5em]
\sin18°&=\frac{\sqrt{5}-1}{4}\\[1em]
\cos18°&=\frac{\sqrt{10+2\sqrt{5}}}{4}\end{align*}
なので、
\begin{align*}\sin3°&=\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{5}-1)(\sqrt{6}+\sqrt{2})-(\sqrt{6}-\sqrt{2})\sqrt{10+2\sqrt{5}}}{16}\\[0.5em]&=\frac{\sqrt{2}(\sqrt{5}-1)(\sqrt{3}+1)-\sqrt{2}(\sqrt{3}-1)\sqrt{10+2\sqrt{5}}}{16}\tag{a}\\[0.5em]&=\frac{\sqrt{30}+\sqrt{10}-\sqrt{6}-\sqrt{2}-\sqrt{(\sqrt{6}-\sqrt{2})^2(10+2\sqrt{5})}}{16}\\[0.5em]&=\underline{\frac{\sqrt{30}+\sqrt{10}-\sqrt{6}-\sqrt{2}-2\sqrt{20-10\sqrt{3}+4\sqrt{5}-2\sqrt{15}}}{16}}\end{align*}
$\cos3°$
$\cos$の加法定理より
\begin{align*}\cos3°&=\cos(18°-15°)\\[0.5em]
&=\cos18°\cos15°+\sin18°\sin15°\\[0.5em]&=\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{6}+\sqrt{2})\sqrt{10+2\sqrt{5}}+(\sqrt{5}-1)(\sqrt{6}-\sqrt{2})}{16}\\[0.5em]&=\frac{\sqrt{2}(\sqrt{3}+1)\sqrt{10+2\sqrt{5}}+\sqrt{2}(\sqrt{5}-1)(\sqrt{3}-1)}{16}\tag{b}\\[0.5em]&=\frac{\sqrt{(\sqrt{6}+\sqrt{2})^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*}
$\tan3°$
三角関数の相互関係
\[\tan\theta=\frac{\sin\theta}{\cos\theta}\]
なので
\[\tan3°=\frac{\sin3°}{\cos3°}\]
これに$\text{(a),(b)}$を代入して
\begin{align*}\tan3°&=\frac{\frac{\sqrt{2}(\sqrt{5}-1)(\sqrt{3}+1)-\sqrt{2}(\sqrt{3}-1)\sqrt{10+2\sqrt{5}}}{16}}{\frac{\sqrt{2}(\sqrt{3}+1)\sqrt{10+2\sqrt{5}}+\sqrt{2}(\sqrt{5}-1)(\sqrt{3}-1)}{16}}\\[0.5em]&=\frac{\sqrt{2}(\sqrt{5}-1)(\sqrt{3}+1)-\sqrt{2}(\sqrt{3}-1)\sqrt{10+2\sqrt{5}}}{\sqrt{2}(\sqrt{3}+1)\sqrt{10+2\sqrt{5}}+\sqrt{2}(\sqrt{5}-1)(\sqrt{3}-1)}\\[0.5em]&=\frac{(\sqrt{5}-1)(\sqrt{3}+1)-(\sqrt{3}-1)\sqrt{10+2\sqrt{5}}}{(\sqrt{3}+1)\sqrt{10+2\sqrt{5}}+(\sqrt{5}-1)(\sqrt{3}-1)}\\
&\qquad\cdot\frac{(\sqrt{3}+1)\sqrt{10+2\sqrt{5}}-(\sqrt{5}-1)(\sqrt{3}-1)}{(\sqrt{3}+1)\sqrt{10+2\sqrt{5}}-(\sqrt{5}-1)(\sqrt{3}-1)}\\[0.5em]&=\frac{(\sqrt{5}-1)\sqrt{10+2\sqrt{5}}-4}{2\left\{(1+2\sqrt{3})+\sqrt{5}\right\}}\cdot\frac{\left\{(1+2\sqrt{3})-\sqrt{5}\right\}}{\left\{(1+2\sqrt{3})-\sqrt{5}\right\}}\\[0.5em]&=\frac{(1+2\sqrt{3}-\sqrt{5})(\sqrt{5}-1)\sqrt{10+2\sqrt{5}}-4(1+2\sqrt{3}-\sqrt{5})}{8(2+\sqrt{3})}\\
&\qquad\cdot\frac{2-\sqrt{3}}{2-\sqrt{3}}\\[0.5em]&=\frac{(8-\sqrt{3}+3\sqrt{5}-2\sqrt{15})\sqrt{10+2\sqrt{5}}+4-3\sqrt{3}+2\sqrt{5}-\sqrt{15}}{2}\\[0.5em]&=\frac{\sqrt{(8-\sqrt{3}+3\sqrt{5}-2\sqrt{15})^2(10+2\sqrt{5})}+4-3\sqrt{3}+2\sqrt{5}-\sqrt{15}}{2}\\[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*}\sin3°&=0.05234\\[1em]\cos3°&=0.99863\\[1em]\tan3°&=0.05241\end{align*}
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