$3°$刻みで$183°～267°$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$180°<β<270°$の三角関数を表すと以下のようになります。
\begin{align*}\sin\beta&=\sin(270°-\alpha)\\[0.5em]&=-\sin\alpha\\[1em]\cos\beta&=\cos(270°-\alpha)\\[0.5em]&=-\cos\alpha\\[1em]\tan\beta&=\tan(270°-\alpha)\\[0.5em]&=\tan\alpha\end{align*}
$183°\ (=\frac{61}{60}\pi)$ $(=270°-87°)$
\begin{align*}\sin183°&=-\frac{2\sqrt{20+10\sqrt{3}+4\sqrt{5}+2\sqrt{15}}+\sqrt{2}+\sqrt{30}-\sqrt{6}-\sqrt{10}}{16}\\[1em]\cos183°&=-\frac{\sqrt{30}+\sqrt{10}-\sqrt{6}-\sqrt{2}-2\sqrt{20+4\sqrt{5}-10\sqrt{3}-2\sqrt{15}}}{16}\\[1em]\tan183°&=\frac{\sqrt{110+60\sqrt{3}+46\sqrt{5}+28\sqrt{15}}+4+3\sqrt{3}+2\sqrt{5}+\sqrt{15}}{2}\end{align*}
$186°\ (=\frac{16}{15}\pi)$ $(=270°-84°)$
\begin{align*}\sin186°&=-\frac{\sqrt{10-2\sqrt{5}}+\sqrt{3}+\sqrt{15}}{8}\\[1em]\cos186°&=-\frac{\sqrt{30-6\sqrt{5}}-\sqrt{5}-1}{8}\\[1em]\tan186°&=\frac{\sqrt{50+22\sqrt{5}}+3\sqrt{3}+\sqrt{15}}{2}\end{align*}
$189°\ (=\frac{21}{20}\pi)$ $(=270°-81°)$
\begin{align*}\sin189°&=-\frac{\sqrt{2}+\sqrt{10}+2\sqrt{5-\sqrt{5}}}{8}\\[1em]\cos189°&=-\frac{\sqrt{2}+\sqrt{10}-2\sqrt{5-\sqrt{5}}}{8}\\[1em]\tan189°&=\sqrt{5}+1+\sqrt{5+2\sqrt{5}}\end{align*}
$192°\ (=\frac{31}{30}\pi)$ $(=270°-78°)$
\begin{align*}\sin192°&=-\frac{\sqrt{30+6\sqrt{5}}+\sqrt{5}-1}{8}\\[1em]\cos192°&=-\frac{\sqrt{10+2\sqrt{5}}+\sqrt{3}-\sqrt{15}}{8}\\[1em]\tan192°&=\frac{\sqrt{10+2\sqrt{5}}+\sqrt{3}+\sqrt{15}}{2}\end{align*}
$195°\ (=\frac{13}{12}\pi)$ $(=270°-75°)$
\begin{align*}\sin195°&=-\frac{\sqrt{6}+\sqrt{2}}{4}\\[1em]\cos195°&=-\frac{\sqrt{6}-\sqrt{2}}{4}\\[1em]\tan195°&=2+\sqrt{3}\end{align*}
$198°\ (=\frac{6}{5}\pi)$ $(=270°-72°)$
\begin{align*}\sin198°&=-\frac{\sqrt{10+2\sqrt{5}}}{4}\\[1em]\cos198°&=-\frac{\sqrt{5}-1}{4}\\[1em]\tan198°&=\sqrt{5+2\sqrt{5}}\end{align*}
$201°\ (=\frac{67}{60}\pi)$ $(=270°-69°)$
\begin{align*}\sin201°&=-\frac{2\sqrt{20+2\sqrt{15}-10\sqrt{3}-4\sqrt{5}}+\sqrt{2}+\sqrt{6}+\sqrt{10}+\sqrt{30}}{16}\\[1em]\cos201°&=-\frac{2\sqrt{20+10\sqrt{3}-4\sqrt{5}-2\sqrt{15}}+\sqrt{2}+\sqrt{10}-\sqrt{6}-\sqrt{30}}{16}\\[1em]\tan201°&=\frac{\sqrt{110+28\sqrt{15}-60\sqrt{3}-46\sqrt{5}}+3\sqrt{3}+2\sqrt{5}-4-\sqrt{15}}{2}\end{align*}
$204°\ (=\frac{19}{15}\pi)$ $(=270°-66°)$
\begin{align*}\sin204°&=-\frac{\sqrt{5}+1+\sqrt{30-6\sqrt{5}}}{8}\\[1em]\cos204°&=-\frac{\sqrt{3}+\sqrt{15}-\sqrt{10-2\sqrt{5}}}{8}\\[1em]\tan204°&=\frac{\sqrt{10-2\sqrt{5}}+\sqrt{15}-\sqrt{3}}{2}\end{align*}
$207°\ (=\frac{23}{20}\pi)$ $(=270°-63°)$
\begin{align*}\sin207°&=-\frac{2\sqrt{5+\sqrt{5}}+\sqrt{10}-\sqrt{2}}{8}\\[1em]\cos207°&=-\frac{2\sqrt{5+\sqrt{5}}+\sqrt{2}-\sqrt{10}}{8}\\[1em]\tan207°&=\sqrt{5-2\sqrt{5}+\sqrt{5}-1}\end{align*}
$210°\ (=\frac{7}{6}\pi)$ $(=270°-60°)$
\begin{align*}\sin210°&=-\frac{\sqrt{3}}{2}\\[1em]\cos210°&=-\frac{1}{2}\\[1em]\tan210°&=\sqrt{3}\end{align*}
$213°\ (=\frac{71}{60}\pi)$ $(=270°-57°)$
\begin{align*}\sin213°&=-\frac{2\sqrt{20+10\sqrt{3}+4\sqrt{5}+2\sqrt{15}}+\sqrt{6}+\sqrt{10}-\sqrt{2}-\sqrt{30}}{16}\\[1em]\cos213°&=-\frac{2\sqrt{20+4\sqrt{5}-10\sqrt{3}-2\sqrt{15}}+\sqrt{10}+\sqrt{30}-\sqrt{2}-\sqrt{6}}{16}\\[1em]\tan213°&=\frac{\sqrt{110+60\sqrt{3}+46\sqrt{5}+28\sqrt{15}}-4-3\sqrt{3}-2\sqrt{5}-\sqrt{15}}{2}\end{align*}
$216°\ (=\frac{6}{5}\pi)$ $(=270°-54°)$
\begin{align*}\sin216°&=-\frac{\sqrt{5}+1}{4}\\[1em]\cos216°&=-\frac{\sqrt{10-2\sqrt{5}}}{4}\\[1em]\tan216°&=\frac{\sqrt{25+10\sqrt{5}}}{5}\end{align*}
$219°\ (=\frac{73}{60}\pi)$ $(=270°-51°)$
\begin{align*}\sin219°&=-\frac{2\sqrt{20+10\sqrt{3}-4\sqrt{5}-2\sqrt{15}}+\sqrt{6}+\sqrt{30}-\sqrt{2}-\sqrt{10}}{16}\\[1em]\cos219°&=-\frac{\sqrt{2}+\sqrt{6}+\sqrt{10}+\sqrt{30}-2\sqrt{20+2\sqrt{15}-10\sqrt{3}-4\sqrt{5}}}{16}\\[1em]\tan219°&=\frac{\sqrt{110+60\sqrt{3}-46\sqrt{5}-28\sqrt{15}}+4+3\sqrt{3}-2\sqrt{5}-\sqrt{15}}{2}\end{align*}
$222°\ (=\frac{37}{30}\pi)$ $(=270°-48°)$
\begin{align*}\sin222°&=-\frac{\sqrt{10+2\sqrt{5}}+\sqrt{15}-\sqrt{3}}{8}\\[1em]\cos222°&=-\frac{\sqrt{30+6\sqrt{5}}+1-\sqrt{5}}{8}\\[1em]\tan222°&=\frac{\sqrt{50-22\sqrt{5}}+3\sqrt{3}-\sqrt{15}}{2}\end{align*}
$225°\ (=\frac{5}{4}\pi)$ $(=270°-45°)$
\begin{align*}\sin225°&=-\frac{\sqrt{2}}{2}\\[1em]\cos225°&=-\frac{\sqrt{2}}{2}\\[1em]\tan225°&=1\end{align*}
$228°\ (=\frac{19}{15}\pi)$ $(=270°-42°)$
\begin{align*}\sin228°&=-\frac{\sqrt{30+6\sqrt{5}}+1-\sqrt{5}}{8}\\[1em]\cos228°&=-\frac{\sqrt{10+2\sqrt{5}}+\sqrt{15}-\sqrt{3}}{8}\\[1em]\tan228°&=\frac{\sqrt{3}+\sqrt{15}-\sqrt{10+2\sqrt{5}}}{2}\end{align*}
$231°\ (=\frac{77}{60}\pi)$ $(=270°-39°)$
\begin{align*}\sin231°&=-\frac{\sqrt{2}+\sqrt{6}+\sqrt{10}+\sqrt{30}-2\sqrt{20+2\sqrt{15}-10\sqrt{3}-4\sqrt{5}}}{16}\\[1em]\cos231°&=-\frac{2\sqrt{20+10\sqrt{3}-4\sqrt{5}-2\sqrt{15}}+\sqrt{6}+\sqrt{30}-\sqrt{2}-\sqrt{10}}{16}\\[1em]\tan231°&=\frac{\sqrt{110+28\sqrt{15}-60\sqrt{3}-46\sqrt{5}}+4+\sqrt{15}-3\sqrt{3}-2\sqrt{5}}{2}\end{align*}
$234°\ (=\frac{13}{10}\pi)$ $(=270°-36°)$
\begin{align*}\sin234°&=-\frac{\sqrt{10-2\sqrt{5}}}{4}\\[1em]\cos234°&=-\frac{\sqrt{5}+1}{4}\\[1em]\tan234°&=\sqrt{5-2\sqrt{5}}\end{align*}
$237°\ (=\frac{79}{60}\pi)$ $(=270°-33°)$
\begin{align*}\sin237°&=-\frac{2\sqrt{20+4\sqrt{5}-10\sqrt{3}-2\sqrt{15}}+\sqrt{10}+\sqrt{30}-\sqrt{2}-\sqrt{6}}{16}\\[1em]\cos237°&=-\frac{2\sqrt{20+10\sqrt{3}+4\sqrt{5}+2\sqrt{15}}+\sqrt{6}+\sqrt{10}-\sqrt{2}-\sqrt{30}}{16}\\[1em]\tan237°&=\frac{\sqrt{110+46\sqrt{5}-60\sqrt{3}-28\sqrt{15}}+3\sqrt{3}+\sqrt{15}-4-2\sqrt{5}}{2}\end{align*}
$240°\ (=\frac{4}{3}\pi)$ $(=270°-30°)$
\begin{align*}\sin240°&=-\frac{1}{2}\\[1em]\cos240°&=-\frac{\sqrt{3}}{2}\\[1em]\tan240°&=\frac{\sqrt{3}}{3}\end{align*}
$243°\ (=\frac{27}{20}\pi)$ $(=270°-27°)$
\begin{align*}\sin243°&=-\frac{2\sqrt{5+\sqrt{5}}+\sqrt{2}-\sqrt{10}}{8}\\[1em]\cos243°&=-\frac{2\sqrt{5+\sqrt{5}}+\sqrt{10}-\sqrt{2}}{8}\\[1em]\tan243°&=\sqrt{5}-1-\sqrt{5-2\sqrt{5}}\end{align*}
$246°\ (=\frac{41}{30}\pi)$ $(=270°-24°)$
\begin{align*}\sin246°&=-\frac{\sqrt{3}+\sqrt{15}-\sqrt{10-2\sqrt{5}}}{8}\\[1em]\cos246°&=-\frac{\sqrt{5}+1+\sqrt{30-6\sqrt{5}}}{8}\\[1em]\tan246°&=\frac{\sqrt{50+22\sqrt{5}}-3\sqrt{3}-\sqrt{15}}{2}\end{align*}
$249°\ (=\frac{83}{60}\pi)$ $(=270°-21°)$
\begin{align*}\sin249°&=-\frac{2\sqrt{20+10\sqrt{3}-4\sqrt{5}-2\sqrt{15}}+\sqrt{2}+\sqrt{10}-\sqrt{6}-\sqrt{30}}{16}\\[1em]\cos249°&=-\frac{2\sqrt{20+2\sqrt{15}-10\sqrt{3}-4\sqrt{5}}+\sqrt{2}+\sqrt{6}+\sqrt{10}+\sqrt{30}}{16}\\[1em]\tan249°&=\frac{\sqrt{110+60\sqrt{3}-46\sqrt{5}-28\sqrt{15}}+2\sqrt{5}+\sqrt{15}-4-3\sqrt{3}}{2}\end{align*}
$252°\ (=\frac{7}{5}\pi)$ $(=270°-18°)$
\begin{align*}\sin252°&=-\frac{\sqrt{5}-1}{4}\\[1em]\cos252°&=-\frac{\sqrt{10+2\sqrt{5}}}{4}\\[1em]\tan252°&=\frac{\sqrt{25-10\sqrt{5}}}{5}\end{align*}
$255°\ (=\frac{17}{12}\pi)$ $(=270°-15°)$
\begin{align*}\sin255°&=-\frac{\sqrt{6}-\sqrt{2}}{4}\\[1em]\cos255°&=-\frac{\sqrt{6}+\sqrt{2}}{4}\\[1em]\tan255°&=2-\sqrt{3}\end{align*}
$258°\ (=\frac{43}{30}\pi)$ $(=270°-12°)$
\begin{align*}\sin258°&=-\frac{\sqrt{10+2\sqrt{5}}+\sqrt{3}-\sqrt{15}}{8}\\[1em]\cos258°&=-\frac{\sqrt{30+6\sqrt{5}}+\sqrt{5}-1}{8}\\[1em]\tan258°&=\frac{3\sqrt{3}-\sqrt{15}-\sqrt{50-22\sqrt{5}}}{2}\end{align*}
$261°\ (=\frac{87}{60}\pi)$ $(=270°-9°)$
\begin{align*}\sin261°&=-\frac{\sqrt{2}+\sqrt{10}-2\sqrt{5-\sqrt{5}}}{8}\\[1em]\cos261°&=-\frac{\sqrt{2}+\sqrt{10}+2\sqrt{5-\sqrt{5}}}{8}\\[1em]\tan261°&=\sqrt{5}+1-\sqrt{5+2\sqrt{5}}\end{align*}
$264°\ (=\frac{22}{15}\pi)$ $(=270°-6°)$
\begin{align*}\sin264°&=-\frac{\sqrt{30-6\sqrt{5}}-\sqrt{5}-1}{8}\\[1em]\cos264°&=-\frac{\sqrt{10-2\sqrt{5}}+\sqrt{3}+\sqrt{15}}{8}\\[1em]\tan264°&=\frac{\sqrt{10-2\sqrt{5}}+\sqrt{3}-\sqrt{15}}{2}\end{align*}
$267°\ (=\frac{89}{60}\pi)$ $(=270°-3°)$
\begin{align*}\sin267°&=-\frac{\sqrt{30}+\sqrt{10}-\sqrt{6}-\sqrt{2}-2\sqrt{20+4\sqrt{5}-10\sqrt{3}-2\sqrt{15}}}{16}\\[1em]\cos267°&=-\frac{2\sqrt{20+10\sqrt{3}+4\sqrt{5}+2\sqrt{15}}+\sqrt{2}+\sqrt{30}-\sqrt{6}-\sqrt{10}}{16}\\[1em]\tan267°&=\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)tan246°に誤りがありましたので修正しました。
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