21° (=7π60)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。
sin21°
sinの加法定理より
sin21°=sin(36°−15°)=sin36°cos15°−cos36°sin15°
sin36°=√10−2√54cos36°=√5+14sin15°=√6−√24cos15°=√6+√24
なので
sin21°=√10−2√54⋅√6+√24−√5+14⋅√6−√24=√2{(√3+1)√10−2√5−(√5+1)(√3−1)}16=√2(√3+1)2(10−2√5)−√30+√10−√6+√216=2√20+10√3−4√5−2√15−√30+√10−√6+√216−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−(a)
cos21°
cosの加法定理より
cos21°=cos(36°−15°)=cos36°cos15°+sin36°sin15°=√5+14⋅√6+√24+√10−2√54⋅√6−√24=√2{(√5+1)(√3+1)+(√3−1)√10−2√5}16=√2(√3−1)2(√10−2√5)+√30+√10+√6+√216=2√20−10√3−4√5+2√15+√30+√10+√6+√216−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−(b)
tan21°
三角関数の相互関係
tanθ=sinθcosθ
より、
(a),(b)を代入して
tan21°=sin21°cos21°=√2{(√3+1)√10−2√5−(√5+1)(√3−1)}16√2{(√5+1)(√3+1)+(√3−1)√10−2√5}16=(√3+1)√10−2√5−(√5+1)(√3−1)(√5+1)(√3+1)+(√3−1)√10−2√5×(√5+1)(√3+1)−(√3−1)√10−2√5(√5+1)(√3+1)−(√3−1)√10−2√5=(√5+1)√10−2√5−42(2√3+√5−1)×2√3−(√5−1)2√3−(√5−1)=(√15+√3−2)√10−2√5−2−4√3+2√52(3+√5)×3−√53−√5=(√15+√5−√3−3)√10−2√5−8−6√3+4√5+2√154=√(√15+√5−√3−3)2(10−2√5)−8−6√3+4√5+2√154=√110+60√3−46√5−28√15−4−3√3+2√5+√152−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
それぞれの近似値は以下のようになります。
sin21°=0.35837cos21°=0.93358tan21°=0.38386