作單位圓O，∠AOB＝ß，∠BOC＝ạ，半徑OA＝OB＝OC＝R，

AD⊥OC於D，交OB於E，AF⊥OB於F，∠EAF=∠BOC=ạ

求證：sin(ạ+ß)=sinạcosß+sinßcosạ,

sin(ạ+ß)=(AE+ED)/R

∵AF/AE=cosạ, AF=Rsinß, ∴AE=AF/cosạ=Rsinß/cosạ,

∵ED=OEsinạ, OE=OF-EF, OF=Rcosß, EF=AEsinạ=Rsinạsinß/cosạ,

∴OE=OF-EF=Rcosß-Rsinạsinß/cosạ,

ED=OEsinạ=(Rcosß-Rsinạsinß/cosạ)*sinạ=Rsinạcosß-Rsinạ²ạsinß/cosạ,

∴sin(ạ+ß)=(AE+ED)/R

=(Rsinß/cosạ+Rsinạcosß-Rsinạ²ạsinß/cosạ)/R

= sinß/cosạ+sinạcosß-sin²ạ sinß/cosạ

=sinạcosß+sinß/cosạ(1-sin²ạ)

=sinạcosß+cosạ²sinß/cosạ

=sinạcosß+sinßcosạ.