$3°$刻みで$3°～87°$のときの三角関数がどんな式になるのかをまとめてみました。
$3°\ (=\frac{\pi}{60})$
\begin{align*}\sin3°&=\frac{\sqrt{30}+\sqrt{10}-\sqrt{6}-\sqrt{2}-2\sqrt{20+4\sqrt{5}-10\sqrt{3}-2\sqrt{15}}}{16}\\[1em]\cos3°&=\frac{2\sqrt{20+10\sqrt{3}+4\sqrt{5}+2\sqrt{15}}+\sqrt{2}+\sqrt{30}-\sqrt{6}-\sqrt{10}}{16}\\[1em]\tan3°&=\frac{\sqrt{110+46\sqrt{5}-60\sqrt{3}-28\sqrt{15}}+4+2\sqrt{5}-3\sqrt{3}-\sqrt{15}}{2}\end{align*}
$6°\ (=\frac{\pi}{30})$
\begin{align*}\sin6°&=\frac{\sqrt{30-6\sqrt{5}}-\sqrt{5}-1}{8}\\[1em]\cos6°&=\frac{\sqrt{10-2\sqrt{5}}+\sqrt{3}+\sqrt{15}}{8}\\[1em]\tan6°&=\frac{\sqrt{10-2\sqrt{5}}+\sqrt{3}-\sqrt{15}}{2}\end{align*}
$9°\ (=\frac{\pi}{20})$
\begin{align*}\sin9°&=\frac{\sqrt{2}+\sqrt{10}-2\sqrt{5-\sqrt{5}}}{8}\\[1em]\cos9°&=\frac{\sqrt{2}+\sqrt{10}+2\sqrt{5-\sqrt{5}}}{8}\\[1em]\tan9°&=\sqrt{5}+1-\sqrt{5+2\sqrt{5}}\end{align*}
$12°\ (=\frac{\pi}{15})$
\begin{align*}\sin12°&=\frac{\sqrt{10+2\sqrt{5}}+\sqrt{3}-\sqrt{15}}{8}\\[1em]\cos12°&=\frac{\sqrt{30+6\sqrt{5}}+\sqrt{5}-1}{8}\\[1em]\tan12°&=\frac{3\sqrt{3}-\sqrt{15}-\sqrt{50-22\sqrt{5}}}{2}\end{align*}
$15°\ (=\frac{\pi}{12})$
\begin{align*}\sin15°&=\frac{\sqrt{6}-\sqrt{2}}{4}\\[1em]\cos15°&=\frac{\sqrt{6}+\sqrt{2}}{4}\\[1em]\tan15°&=2-\sqrt{3}\end{align*}
$18°\ (=\frac{\pi}{10})$
\begin{align*}\sin18°&=\frac{\sqrt{5}-1}{4}\\[1em]\cos18°&=\frac{\sqrt{10+2\sqrt{5}}}{4}\\[1em]\tan18°&=\frac{\sqrt{25-10\sqrt{5}}}{5}\end{align*}
$21°\ (=\frac{7}{60}\pi)$
\begin{align*}\sin21°&=\frac{2\sqrt{20+10\sqrt{3}-4\sqrt{5}-2\sqrt{15}}+\sqrt{2}+\sqrt{10}-\sqrt{6}-\sqrt{30}}{16}\\[1em]\cos21°&=\frac{2\sqrt{20+2\sqrt{15}-10\sqrt{3}-4\sqrt{5}}+\sqrt{2}+\sqrt{6}+\sqrt{10}+\sqrt{30}}{16}\\[1em]\tan21°&=\frac{\sqrt{110+60\sqrt{3}-46\sqrt{5}-28\sqrt{15}}+2\sqrt{5}+\sqrt{15}-4-3\sqrt{3}}{2}\end{align*}
$24°\ (=\frac{2}{15}\pi)$
\begin{align*}\sin24°&=\frac{\sqrt{3}+\sqrt{15}-\sqrt{10-2\sqrt{5}}}{8}\\[1em]\cos24°&=\frac{\sqrt{5}+1+\sqrt{30-6\sqrt{5}}}{8}\\[1em]\tan24°&=\frac{\sqrt{50+22\sqrt{5}}-3\sqrt{3}-\sqrt{15}}{2}\end{align*}
$27°\ (=\frac{3}{20}\pi)$
\begin{align*}\sin27°&=\frac{2\sqrt{5+\sqrt{5}}+\sqrt{2}-\sqrt{10}}{8}\\[1em]\cos27°&=\frac{2\sqrt{5+\sqrt{5}}+\sqrt{10}-\sqrt{2}}{8}\\[1em]\tan27°&=\sqrt{5}-1-\sqrt{5-2\sqrt{5}}\end{align*}
$30°\ (=\frac{\pi}{6})$
\begin{align*}\sin30°&=\frac{1}{2}\\[1em]\cos30°&=\frac{\sqrt{3}}{2}\\[1em]\tan30°&=\frac{\sqrt{3}}{3}\end{align*}
$33°\ (=\frac{11}{60}\pi)$
\begin{align*}\sin33°&=\frac{2\sqrt{20+4\sqrt{5}-10\sqrt{3}-2\sqrt{15}}+\sqrt{10}+\sqrt{30}-\sqrt{2}-\sqrt{6}}{16}\\[1em]\cos33°&=\frac{2\sqrt{20+10\sqrt{3}+4\sqrt{5}+2\sqrt{15}}+\sqrt{6}+\sqrt{10}-\sqrt{2}-\sqrt{30}}{16}\\[1em]\tan33°&=\frac{\sqrt{110+46\sqrt{5}-60\sqrt{3}-28\sqrt{15}}+3\sqrt{3}+\sqrt{15}-4-2\sqrt{5}}{2}\end{align*}
$36°\ (=\frac{\pi}{5})$
\begin{align*}\sin36°&=\frac{\sqrt{10-2\sqrt{5}}}{4}\\[1em]\cos36°&=\frac{\sqrt{5}+1}{4}\\[1em]\tan36°&=\sqrt{5-2\sqrt{5}}\end{align*}
$39°\ (=\frac{13}{60}\pi)$
\begin{align*}\sin39°&=\frac{\sqrt{2}+\sqrt{6}+\sqrt{10}+\sqrt{30}-2\sqrt{20+2\sqrt{15}-10\sqrt{3}-4\sqrt{5}}}{16}\\[1em]\cos39°&=\frac{2\sqrt{20+10\sqrt{3}-4\sqrt{5}-2\sqrt{15}}+\sqrt{6}+\sqrt{30}-\sqrt{2}-\sqrt{10}}{16}\\[1em]\tan39°&=\frac{\sqrt{110+28\sqrt{15}-60\sqrt{3}-46\sqrt{5}}+4+\sqrt{15}-3\sqrt{3}-2\sqrt{5}}{2}\end{align*}
$42°\ (=\frac{7}{30}\pi)$
\begin{align*}\sin42°&=\frac{\sqrt{30+6\sqrt{5}}+1-\sqrt{5}}{8}\\[1em]\cos42°&=\frac{\sqrt{10+2\sqrt{5}}+\sqrt{15}-\sqrt{3}}{8}\\[1em]\tan42°&=\frac{\sqrt{3}+\sqrt{15}-\sqrt{10+2\sqrt{5}}}{2}\end{align*}
$45°\ (=\frac{\pi}{4})$
\begin{align*}\sin45°&=\frac{\sqrt{2}}{2}\\[1em]\cos45°&=\frac{\sqrt{2}}{2}\\[1em]\tan45°&=1\end{align*}
$0°＜α＜45°$の三角関数を利用して$45°＜β＜90°$の三角関数を表すと以下のようになります。
\begin{align*}\mathbf{\sin\beta}&=\sin(90°-\alpha)\\[0.5em]&=\mathbf{\cos\alpha}\\[1em]\mathbf{\cos\beta}&=\cos(90°-\alpha)\\[0.5em]&=\mathbf{\sin\alpha}\\[1em]\mathbf{\tan\beta}&=\tan(90°-\alpha)\\[0.5em]&=\mathbf{\frac{1}{\tan\alpha}}\end{align*}
$48°\ (=\frac{4}{15}\pi)$ $(=90°-42°)$
\begin{align*}\sin48°&=\frac{\sqrt{10+2\sqrt{5}}+\sqrt{15}-\sqrt{3}}{8}\\[1em]\cos48°&=\frac{\sqrt{30+6\sqrt{5}}+1-\sqrt{5}}{8}\\[1em]\tan48°&=\frac{\sqrt{50-22\sqrt{5}}+3\sqrt{3}-\sqrt{15}}{2}\end{align*}
$51°\ (=\frac{17}{60}\pi)$ $(=90°-39°)$
\begin{align*}\sin51°&=\frac{2\sqrt{20+10\sqrt{3}-4\sqrt{5}-2\sqrt{15}}+\sqrt{6}+\sqrt{30}-\sqrt{2}-\sqrt{10}}{16}\\[1em]\cos51°&=\frac{\sqrt{2}+\sqrt{6}+\sqrt{10}+\sqrt{30}-2\sqrt{20+2\sqrt{15}-10\sqrt{3}-4\sqrt{5}}}{16}\\[1em]\tan51°&=\frac{\sqrt{110+60\sqrt{3}-46\sqrt{5}-28\sqrt{15}}+4+3\sqrt{3}-2\sqrt{5}-\sqrt{15}}{2}\end{align*}
$54°\ (=\frac{3}{10}\pi)$ $(=90°-36°)$
\begin{align*}\sin54°&=\frac{\sqrt{5}+1}{4}\\[1em]\cos54°&=\frac{\sqrt{10-2\sqrt{5}}}{4}\\[1em]\tan54°&=\frac{\sqrt{25+10\sqrt{5}}}{5}\end{align*}
$57°\ (=\frac{19}{60}\pi)$ $(=90°-33°)$
\begin{align*}\sin57°&=\frac{2\sqrt{20+10\sqrt{3}+4\sqrt{5}+2\sqrt{15}}+\sqrt{6}+\sqrt{10}-\sqrt{2}-\sqrt{30}}{16}\\[1em]\cos57°&=\frac{2\sqrt{20+4\sqrt{5}-10\sqrt{3}-2\sqrt{15}}+\sqrt{10}+\sqrt{30}-\sqrt{2}-\sqrt{6}}{16}\\[1em]\tan57°&=\frac{\sqrt{110+60\sqrt{3}+46\sqrt{5}+28\sqrt{15}}-4-3\sqrt{3}-2\sqrt{5}-\sqrt{15}}{2}\end{align*}
$60°\ (=\frac{\pi}{3})$ $(=90°-30°)$
\begin{align*}\sin60°&=\frac{\sqrt{3}}{2}\\[1em]\cos60°&=\frac{1}{2}\\[1em]\tan60°&=\sqrt{3}\end{align*}
$63°\ (=\frac{7}{20}\pi)$ $(=90°-27°)$
\begin{align*}\sin63°&=\frac{2\sqrt{5+\sqrt{5}}+\sqrt{10}-\sqrt{2}}{8}\\[1em]\cos63°&=\frac{2\sqrt{5+\sqrt{5}}+\sqrt{2}-\sqrt{10}}{8}\\[1em]\tan63°&=\sqrt{5}+\sqrt{5-2\sqrt{5}}-1\end{align*}
$66°\ (=\frac{11}{30}\pi)$ $(=90°-24°)$
\begin{align*}\sin66°&=\frac{\sqrt{5}+1+\sqrt{30-6\sqrt{5}}}{8}\\[1em]\cos66°&=\frac{\sqrt{3}+\sqrt{15}-\sqrt{10-2\sqrt{5}}}{8}\\[1em]\tan66°&=\frac{\sqrt{10-2\sqrt{5}}+\sqrt{15}-\sqrt{3}}{2}\end{align*}
$69°\ (=\frac{23}{60}\pi)$ $(=90°-21°)$
\begin{align*}\sin69°&=\frac{2\sqrt{20+2\sqrt{15}-10\sqrt{3}-4\sqrt{5}}+\sqrt{2}+\sqrt{6}+\sqrt{10}+\sqrt{30}}{16}\\[1em]\cos69°&=\frac{2\sqrt{20+10\sqrt{3}-4\sqrt{5}-2\sqrt{15}}+\sqrt{2}+\sqrt{10}-\sqrt{6}-\sqrt{30}}{16}\\[1em]\tan69°&=\frac{\sqrt{110+28\sqrt{15}-60\sqrt{3}-46\sqrt{5}}+3\sqrt{3}+2\sqrt{5}-4-\sqrt{15}}{2}\end{align*}
$72°\ (=\frac{2}{5}\pi)$ $(=90°-18°)$
\begin{align*}\sin72°&=\frac{\sqrt{10+2\sqrt{5}}}{4}\\[1em]\cos72°&=\frac{\sqrt{5}-1}{4}\\[1em]\tan72°&=\sqrt{5+2\sqrt{5}}\end{align*}
$75°\ (=\frac{5}{12}\pi)$ $(=90°-15°)$
\begin{align*}\sin75°&=\frac{\sqrt{6}+\sqrt{2}}{4}\\[1em]\cos75°&=\frac{\sqrt{6}-\sqrt{2}}{4}\\[1em]\tan75°&=2+\sqrt{3}\end{align*}
$78°\ (=\frac{13}{30}\pi)$ $(=90°-12°)$
\begin{align*}\sin78°&=\frac{\sqrt{30+6\sqrt{5}}+\sqrt{5}-1}{8}\\[1em]\cos78°&=\frac{\sqrt{10+2\sqrt{5}}+\sqrt{3}-\sqrt{15}}{8}\\[1em]\tan78°&=\frac{\sqrt{10+2\sqrt{5}}+\sqrt{3}+\sqrt{15}}{2}\end{align*}
$81°\ (=\frac{9}{20}\pi)$ $(=90°-9°)$
\begin{align*}\sin81°&=\frac{\sqrt{2}+\sqrt{10}+2\sqrt{5-\sqrt{5}}}{8}\\[1em]\cos81°&=\frac{\sqrt{2}+\sqrt{10}-2\sqrt{5-\sqrt{5}}}{8}\\[1em]\tan81°&=\sqrt{5}+1+\sqrt{5+2\sqrt{5}}\end{align*}
$84°\ (=\frac{7}{15}\pi)$ $(=90°-6°)$
\begin{align*}\sin84°&=\frac{\sqrt{10-2\sqrt{5}}+\sqrt{3}+\sqrt{15}}{8}\\[1em]\cos84°&=\frac{\sqrt{30-6\sqrt{5}}-\sqrt{5}-1}{8}\\[1em]\tan84°&=\frac{\sqrt{50+22\sqrt{5}}+3\sqrt{3}+\sqrt{15}}{2}\end{align*}
$87°\ (=\frac{29}{60}\pi)$ $(=90°-3°)$
\begin{align*}\sin87°&=\frac{2\sqrt{20+10\sqrt{3}+4\sqrt{5}+2\sqrt{15}}+\sqrt{2}+\sqrt{30}-\sqrt{6}-\sqrt{10}}{16}\\[1em]\cos87°&=\frac{\sqrt{30}+\sqrt{10}-\sqrt{6}-\sqrt{2}-2\sqrt{20+4\sqrt{5}-10\sqrt{3}-2\sqrt{15}}}{16}\\[1em]\tan87°&=\frac{\sqrt{110+60\sqrt{3}+46\sqrt{5}+28\sqrt{15}}+4+3\sqrt{3}+2\sqrt{5}+\sqrt{15}}{2}\end{align*}
(2022/4)tan3°の式に誤りがありましたので修正しました。
(2022/8)tan24°に誤りがありましたので修正しました。
Share:
