$3°$刻みで$93°～177°$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。
$0°<α<90°$を使って$90°<β<180°$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(180°-\alpha)\\[0.5em]&=\mathbf{\sin\alpha}\\[1em]\mathbf{\cos\beta}&=\cos(180°-\alpha)\\[0.5em]&=\mathbf{-\cos\alpha}\\[1em]\mathbf{\tan\beta}&=\tan(180°-\alpha)\\[0.5em]&=\mathbf{-\tan\alpha}\end{align*}
$93°\ (=\frac{31}{60}\pi)$ $(=180°-87°)$
\begin{align*}\sin93°&=\frac{2\sqrt{20+10\sqrt{3}+4\sqrt{5}+2\sqrt{15}}+\sqrt{2}+\sqrt{30}-\sqrt{6}-\sqrt{10}}{16}\\[1em]\cos93°&=-\frac{\sqrt{30}+\sqrt{10}-\sqrt{6}-\sqrt{2}-2\sqrt{20+4\sqrt{5}-10\sqrt{3}-2\sqrt{15}}}{16}\\[1em]\tan93°&=-\frac{\sqrt{110+60\sqrt{3}+46\sqrt{5}+28\sqrt{15}}+4+3\sqrt{3}+2\sqrt{5}+\sqrt{15}}{2}\end{align*}
$96°\ (=\frac{8}{15}\pi)$ $(=180°-84°)$
\begin{align*}\sin96°&=\frac{\sqrt{10-2\sqrt{5}}+\sqrt{3}+\sqrt{15}}{8}\\[1em]\cos96°&=-\frac{\sqrt{30-6\sqrt{5}}-\sqrt{5}-1}{8}\\[1em]\tan96°&=-\frac{\sqrt{50+22\sqrt{5}}+3\sqrt{3}+\sqrt{15}}{2}\end{align*}
$99°\ (=\frac{33}{60}\pi)$ $(=180°-81°)$
\begin{align*}\sin99°&=\frac{\sqrt{2}+\sqrt{10}+2\sqrt{5-\sqrt{5}}}{8}\\[1em]\cos99°&=-\frac{\sqrt{2}+\sqrt{10}-2\sqrt{5-\sqrt{5}}}{8}\\[1em]\tan99°&=-\sqrt{5}-1-\sqrt{5+2\sqrt{5}}\end{align*}
$102°\ (=\frac{17}{30}\pi)$ $(=180°-78°)$
\begin{align*}\sin102°&=\frac{\sqrt{30+6\sqrt{5}}+\sqrt{5}-1}{8}\\[1em]\cos102°&=-\frac{\sqrt{10+2\sqrt{5}}+\sqrt{3}-\sqrt{15}}{8}\\[1em]\tan102°&=-\frac{\sqrt{10+2\sqrt{5}}+\sqrt{3}+\sqrt{15}}{2}\end{align*}
$105°\ (=\frac{7}{12}\pi)$ $(=180°-75°)$
\begin{align*}\sin105°&=\frac{\sqrt{6}+\sqrt{2}}{4}\\[1em]\cos105°&=-\frac{\sqrt{6}-\sqrt{2}}{4}\\[1em]\tan105°&=-2-\sqrt{3}\end{align*}
$108°\ (=\frac{9}{15}\pi)$ $(=180°-72°)$
\begin{align*}\sin108°&=\frac{\sqrt{10+2\sqrt{5}}}{4}\\[1em]\cos108°&=-\frac{\sqrt{5}-1}{4}\\[1em]\tan108°&=-\sqrt{5+2\sqrt{5}}\end{align*}
$111°\ (=\frac{37}{60}\pi)$ $(=180°-69°)$
\begin{align*}\sin111°&=\frac{2\sqrt{20+2\sqrt{15}-10\sqrt{3}-4\sqrt{5}}+\sqrt{2}+\sqrt{6}+\sqrt{10}+\sqrt{30}}{16}\\[1em]\cos111°&=-\frac{2\sqrt{20+10\sqrt{3}-4\sqrt{5}-2\sqrt{15}}+\sqrt{2}+\sqrt{10}-\sqrt{6}-\sqrt{30}}{16}\\[1em]\tan111°&=-\frac{\sqrt{110+28\sqrt{15}-60\sqrt{3}-46\sqrt{5}}+3\sqrt{3}+2\sqrt{5}-4-\sqrt{15}}{2}\end{align*}
$114°\ (=\frac{19}{30}\pi)$ $(=180°-66°)$
\begin{align*}\sin114°&=\frac{\sqrt{5}+1+\sqrt{30-6\sqrt{5}}}{8}\\[1em]\cos114°&=-\frac{\sqrt{3}+\sqrt{15}-\sqrt{10-2\sqrt{5}}}{8}\\[1em]\tan114°&=-\frac{\sqrt{10-2\sqrt{5}}+\sqrt{15}-\sqrt{3}}{2}\end{align*}
$117°\ (=\frac{13}{20}\pi)$ $(=180°-63°)$
\begin{align*}\sin117°&=\frac{2\sqrt{5+\sqrt{5}}+\sqrt{10}-\sqrt{2}}{8}\\[1em]\cos117°&=-\frac{2\sqrt{5+\sqrt{5}}+\sqrt{2}-\sqrt{10}}{8}\\[1em]\tan117°&=1-\sqrt{5}-\sqrt{5-2\sqrt{5}}\end{align*}
$120°\ (=\frac{2}{3}\pi)$ $(=180°-60°)$
\begin{align*}\sin120°&=\frac{\sqrt{3}}{2}\\[1em]\cos120°&=-\frac{1}{2}\\[1em]\tan120°&=-\sqrt{3}\end{align*}
$123°\ (=\frac{41}{60}\pi)$ $(=180°-57°)$
\begin{align*}\sin123°&=\frac{2\sqrt{20+10\sqrt{3}+4\sqrt{5}+2\sqrt{15}}+\sqrt{6}+\sqrt{10}-\sqrt{2}-\sqrt{30}}{16}\\[1em]\cos123°&=-\frac{2\sqrt{20+4\sqrt{5}-10\sqrt{3}-2\sqrt{15}}+\sqrt{10}+\sqrt{30}-\sqrt{2}-\sqrt{6}}{16}\\[1em]\tan123°&=-\frac{\sqrt{110+60\sqrt{3}+46\sqrt{5}+28\sqrt{15}}-4-3\sqrt{3}-2\sqrt{5}-\sqrt{15}}{2}\end{align*}
$126°\ (=\frac{7}{10}\pi)$ $(180°-54°)$
\begin{align*}\sin126°&=\frac{\sqrt{5}+1}{4}\\[1em]\cos126°&=-\frac{\sqrt{10-2\sqrt{5}}}{4}\\[1em]\tan126°&=-\frac{\sqrt{25+10\sqrt{5}}}{5}\end{align*}
$129°\ (=\frac{43}{60}\pi)$ $(=180°-51°)$
\begin{align*}\sin129°&=\frac{2\sqrt{20+10\sqrt{3}-4\sqrt{5}-2\sqrt{15}}+\sqrt{6}-\sqrt{10}+\sqrt{30}-\sqrt{2}}{16}\\[1em]\cos129°&=-\frac{\sqrt{2}+\sqrt{6}+\sqrt{10}+\sqrt{30}-2\sqrt{20+2\sqrt{15}-10\sqrt{3}-4\sqrt{5}}}{16}\\[1em]\tan129°&=-\frac{\sqrt{110+60\sqrt{3}-46\sqrt{5}-28\sqrt{15}}+4+3\sqrt{3}-2\sqrt{5}-\sqrt{15}}{2}\end{align*}
$132°\ (=\frac{11}{15}\pi)$ $(180°-48°)$
\begin{align*}\sin132°&=\frac{\sqrt{10+2\sqrt{5}}+\sqrt{15}-\sqrt{3}}{8}\\[1em]\cos132°&=-\frac{\sqrt{30+6\sqrt{5}}+1-\sqrt{5}}{8}\\[1em]\tan132°&=-\frac{\sqrt{50-22\sqrt{5}}+3\sqrt{3}-\sqrt{15}}{2}\end{align*}
$135°\ (=\frac{3}{4}\pi)$ $(=180°-45°)$
\begin{align*}\sin135°&=\frac{\sqrt{2}}{2}\\[1em]\cos135°&=-\frac{\sqrt{2}}{2}\\[1em]\tan135°&=-1\end{align*}
$138°\ (=\frac{23}{30}\pi)$ $(=180°-42°)$
\begin{align*}\sin138°&=\frac{\sqrt{30+6\sqrt{5}}+1-\sqrt{5}}{8}\\[1em]\cos138°&=-\frac{\sqrt{10+2\sqrt{5}}+\sqrt{15}-\sqrt{3}}{8}\\[1em]\tan138°&=-\frac{\sqrt{3}+\sqrt{15}-\sqrt{10+2\sqrt{5}}}{2}\end{align*}
$141°\ (=\frac{47}{60}\pi)$ $(=180°-39°)$
\begin{align*}\sin141°&=\frac{\sqrt{2}+\sqrt{6}+\sqrt{10}+\sqrt{30}-2\sqrt{20+2\sqrt{15}-10\sqrt{3}-4\sqrt{5}}}{16}\\[1em]\cos141°&=-\frac{2\sqrt{20+10\sqrt{3}-4\sqrt{5}-2\sqrt{15}}+\sqrt{6}+\sqrt{30}-\sqrt{2}-\sqrt{10}}{16}\\[1em]\tan141°&=-\frac{\sqrt{110+28\sqrt{15}-60\sqrt{3}-46\sqrt{5}}+4+\sqrt{15}-3\sqrt{3}-2\sqrt{5}}{2}\end{align*}
$144°\ (=\frac{4}{5}\pi)$ $(=180°-36°)$
\begin{align*}\sin144°&=\frac{\sqrt{10-2\sqrt{5}}}{4}\\[1em]\cos144°&=-\frac{\sqrt{5}+1}{4}\\[1em]\tan144°&=-\sqrt{5-2\sqrt{5}}\end{align*}
$147°\ (=\frac{49}{60}\pi)$ $(=180°-33°)$
\begin{align*}\sin147°&=\frac{2\sqrt{20+4\sqrt{5}-10\sqrt{3}-2\sqrt{15}}+\sqrt{10}+\sqrt{30}-\sqrt{2}-\sqrt{6}}{16}\\[1em]\cos147°&=-\frac{2\sqrt{20+10\sqrt{3}+4\sqrt{5}+2\sqrt{15}}+\sqrt{6}+\sqrt{10}-\sqrt{2}-\sqrt{30}}{16}\\[1em]\tan147°&=-\frac{\sqrt{110+46\sqrt{5}-60\sqrt{3}-28\sqrt{15}}+3\sqrt{3}+\sqrt{15}-4-2\sqrt{5}}{2}\end{align*}
$150°\ (=\frac{5}{6}\pi)$ $(=180°-30°)$
\begin{align*}\sin150°&=\frac{1}{2}\\[1em]\cos150°&=-\frac{\sqrt{3}}{2}\\[1em]\tan150°&=-\frac{\sqrt{3}}{3}\end{align*}
$153°\ (=\frac{17}{20}\pi)$ $(=180°-27°)$
\begin{align*}\sin153°&=\frac{2\sqrt{5+\sqrt{5}}+\sqrt{2}-\sqrt{10}}{8}\\[1em]\cos153°&=-\frac{2\sqrt{5+\sqrt{5}}-\sqrt{2}+\sqrt{10}}{8}\\[1em]\tan153°&=\sqrt{5-2\sqrt{5}}+1-\sqrt{5}\end{align*}
$156°\ (=\frac{13}{15}\pi)$ $(=180°-24°)$
\begin{align*}\sin156°&=\frac{\sqrt{3}+\sqrt{15}-\sqrt{10-2\sqrt{5}}}{8}\\[1em]\cos156°&=-\frac{\sqrt{5}+1+\sqrt{30-6\sqrt{5}}}{8}\\[1em]\tan156°&=-\frac{\sqrt{50+22\sqrt{5}}-3\sqrt{3}-\sqrt{15}}{2}\end{align*}
$159°\ (=\frac{53}{60}\pi)$ $(=180°-21°)$
\begin{align*}\sin159°&=\frac{2\sqrt{20+10\sqrt{3}-4\sqrt{5}-2\sqrt{15}}+\sqrt{2}+\sqrt{10}-\sqrt{6}-\sqrt{30}}{16}\\[1em]\cos159°&=-\frac{2\sqrt{20+2\sqrt{15}-10\sqrt{3}-4\sqrt{5}}+\sqrt{2}+\sqrt{6}+\sqrt{10}+\sqrt{30}}{16}\\[1em]\tan159°&=-\frac{\sqrt{110+60\sqrt{3}-46\sqrt{5}-28\sqrt{15}}+2\sqrt{5}+\sqrt{15}-4-3\sqrt{3}}{2}\end{align*}
$162°\ (=\frac{9}{10}\pi)$ $(=180°-18°)$
\begin{align*}\sin162°&=\frac{\sqrt{5}-1}{4}\\[1em]\cos162°&=-\frac{\sqrt{10+2\sqrt{5}}}{4}\\[1em]\tan162°&=-\frac{\sqrt{25-10\sqrt{5}}}{5}\end{align*}
$165°\ (=\frac{11}{12}\pi)$ $(=180°-15°)$
\begin{align*}\sin165°&=\frac{\sqrt{6}-\sqrt{2}}{4}\\[1em]\cos165°&=-\frac{\sqrt{6}+\sqrt{2}}{4}\\[1em]\tan165°&=\sqrt{3}-2\end{align*}
$168°\ (=\frac{14}{15}\pi)$ $(=180°-12°)$
\begin{align*}\sin168°&=\frac{\sqrt{10+2\sqrt{5}}+\sqrt{3}-\sqrt{15}}{8}\\[1em]\cos168°&=-\frac{\sqrt{30+6\sqrt{5}}+\sqrt{5}-1}{8}\\[1em]\tan168°&=-\frac{3\sqrt{3}-\sqrt{15}-\sqrt{50-22\sqrt{5}}}{2}\end{align*}
$171°\ (=\frac{19}{20}\pi)$ $(=180°-9°)$
\begin{align*}\sin171°&=\frac{\sqrt{2}+\sqrt{10}-2\sqrt{5-\sqrt{5}}}{8}\\[1em]\cos171°&=-\frac{\sqrt{2}+\sqrt{10}+2\sqrt{5-\sqrt{5}}}{8}\\[1em]\tan171°&=\sqrt{5+2\sqrt{5}}-\sqrt{5}-1\end{align*}
$174°\ (=\frac{29}{30}\pi)$ $(=180°-6°)$
\begin{align*}\sin174°&=\frac{\sqrt{30-6\sqrt{5}}-\sqrt{5}-1}{8}\\[1em]\cos174°&=-\frac{\sqrt{10-2\sqrt{5}}+\sqrt{3}+\sqrt{15}}{8}\\[1em]\tan174°&=-\frac{\sqrt{10-2\sqrt{5}}+\sqrt{3}-\sqrt{15}}{2}\end{align*}
$177°\ (=\frac{59}{60}\pi)$ $(=180°-3°)$
\begin{align*}\sin177°&=\frac{\sqrt{30}+\sqrt{10}-\sqrt{6}-\sqrt{2}-2\sqrt{20+4\sqrt{5}-10\sqrt{3}-2\sqrt{15}}}{16}\\[1em]\cos177°&=-\frac{2\sqrt{20+10\sqrt{3}+4\sqrt{5}+2\sqrt{15}}+\sqrt{2}+\sqrt{30}-\sqrt{6}-\sqrt{10}}{16}\\[1em]\tan177°&=-\frac{\sqrt{110+46\sqrt{5}-60\sqrt{3}-28\sqrt{15}}+4+2\sqrt{5}-3\sqrt{3}-\sqrt{15}}{2}\end{align*}
(2022/8)tan156°に誤りがありましたので修正しました。
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