7.5°\ (=\dfrac{\pi}{24})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。
\sin7.5°
\sinの加法定理より
\begin{align*}\sin7.5°&=\sin(22.5°-15°)\\[0.5em]&=\sin22.5°\cos15°-\cos22.5°\sin15°\end{align*}
「\sin15°,\cos15°,\tan15°はどんな数？」、「\sin22.5°,\cos22.5°,\tan22.5°はどんな数？」より
\begin{align*}\sin15°&=\frac{\sqrt{6}-\sqrt{2}}{4}\\[0.5em]\cos15°&=\frac{\sqrt{6}+\sqrt{2}}{4}\\[1em]\sin22.5°&=\frac{\sqrt{2-\sqrt{2}}}{2}\\[0.5em]\cos22.5°&=\frac{\sqrt{2+\sqrt{2}}}{2}\end{align*}
なので
\begin{align*}\sin7.5°&=\frac{\sqrt{2-\sqrt{2}}}{2}\cdot\frac{\sqrt{6}+\sqrt{2}}{4}-\frac{\sqrt{2+\sqrt{2}}}{2}\cdot\frac{\sqrt{6}-\sqrt{2}}{4}\\[0.5em]&=\frac{\sqrt{2}\left\{(\sqrt{3}+1)\sqrt{2-\sqrt{2}}-(\sqrt{3}-1)\sqrt{2+\sqrt{2}}\right\}}{8}\tag{a}\end{align*}
ここで
\tan22.5°=\frac{\sqrt{2-\sqrt{2}}}{\sqrt{2+\sqrt{2}}}=\sqrt{2}-1
より
\sqrt{2-\sqrt{2}}=(\sqrt{2}-1)\sqrt{2+\sqrt{2}}
なので
\begin{align*}\sin7.5°&=\frac{\sqrt{2}\left\{(\sqrt{3}+1)(\sqrt{2}-1)\sqrt{2+\sqrt{2}}-(\sqrt{3}-1)\sqrt{2+\sqrt{2}}\right\}}{8}\\[0.5em]&=\frac{(1+\sqrt{3}-\sqrt{6})\sqrt{2+\sqrt{2}}}{4}\\[0.5em]&=\frac{\sqrt{(1+\sqrt{3}-\sqrt{6})^2(2+\sqrt{2})}}{4}\\[0.5em]&=\underline{\frac{\sqrt{8-2\sqrt{2}-2\sqrt{6}}}{4}}\end{align*}
\cos7.5°
\cosの加法定理より
\begin{align*}\cos7.5°&=\cos(22.5°-15°)\\[0.5em]&=\cos22.5°\cos15°+\sin22.5°\sin15°\\[0.5em]&=\frac{\sqrt{2+\sqrt{2}}}{2}\cdot\frac{\sqrt{6}+\sqrt{2}}{4}+\frac{2-\sqrt{2}}{2}\cdot\frac{\sqrt{6}-\sqrt{2}}{4}\\[0.5em]&=\frac{\sqrt{2}\left\{(\sqrt{3}+1)\sqrt{2+\sqrt{2}}+(\sqrt{3}-1)\sqrt{2-\sqrt{2}}\right\}}{8}\tag{b}\\[0.5em]&=\frac{\sqrt{2}\left\{(\sqrt{3}+1)\sqrt{2+\sqrt{2}}+(\sqrt{3}-1)(\sqrt{2}-1)\sqrt{2+\sqrt{2}}\right\}}{8}\\[0.5em]&=\frac{(\sqrt{3}+\sqrt{2}-1)\sqrt{2+\sqrt{2}}}{4}\\[0.5em]&=\frac{\sqrt{(\sqrt{3}+\sqrt{2}-1)^2(2+\sqrt{2})}}{4}\\[0.5em]&=\underline{\frac{\sqrt{8+2\sqrt{2}+2\sqrt{6}}}{4}}\end{align*}
\tan7.5°
三角関数の相互関係
\tan\theta=\frac{\sin\theta}{\cos\theta}
に\text{(a),(b)}を代入して
\begin{align*}\tan7.5°&=\frac{\sin7.5°}{\cos7.5°}\\[0.5em]&=\frac{\cfrac{\sqrt{2}\left\{(\sqrt{3}+1)\sqrt{2-\sqrt{2}}-(\sqrt{3}-1)\sqrt{2+\sqrt{2}}\right\}}{8}}{\cfrac{\sqrt{2}\left\{(\sqrt{3}+1)\sqrt{2+\sqrt{2}}+(\sqrt{3}-1)\sqrt{2-\sqrt{2}}\right\}}{8}}\\[0.5em]&=\frac{(\sqrt{3}+1)\sqrt{2-\sqrt{2}}-(\sqrt{3}-1)\sqrt{2+\sqrt{2}}}{(\sqrt{3}+1)\sqrt{2+\sqrt{2}}+(\sqrt{3}-1)\sqrt{2-\sqrt{2}}}\\
&\qquad\cdot\frac{(\sqrt{3}+1)\sqrt{2+\sqrt{2}}-(\sqrt{3}-1)\sqrt{2-\sqrt{2}}}{(\sqrt{3}+1)\sqrt{2+\sqrt{2}}-(\sqrt{3}-1)\sqrt{2-\sqrt{2}}}\\[0.5em]&=\frac{\sqrt{2}-1}{\sqrt{3}+\sqrt{2}}×\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}\\[0.5em]&=\underline{\sqrt{6}+\sqrt{2}-\sqrt{3}-2}\end{align*}
それぞれの近似値は以下のようになります。
\begin{align*}\sin7.5°&=0.13053\\[1em]\cos7.5°&=0.99144\\[1em]\tan7.5°&=0.13165\end{align*}
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