sin[α-(π⼀3)]+cos[α+(13π⼀6)]化简

sin[α-(π/3)]+cos[α+(13π/6)]
=sin(α-π/3)+cos(α+π/6)
=sinαcosπ/3-cosαsinπ/3+cosαcosπ/6-sinαsinπ/6
=(1/2)sinα-(√3/2)cosα+(√3/2)cosα-(1/2)sinα
=0
sin[α-(5π/4)]+cos[α+(π/2)]
=sinαcos5π/4-cosαsin5π/4+cosαcosπ/2-sinαsinπ/2
=-√2/2sinα+√2/2cosα+0-sinα
=(-1-√2/2)sinα+√2/2cosα

解：（1）因为 cos[a+(13派/6)]=cos[a+(派/6)]
=cos[(派/2)+(a--派/3)]
=--sin[a--(派/3)]
所以 sin[a--(派/3)]+cos[a+(13派/6)]
=sin[a--(派/3)]--sin[a--(派/3)]
=0.
(2) 因为 sin[a--(5派/4)]=sinacos(5派/4)--cosasin(5派/4)
=--sinacos(派/4)--cosasin(派/4)
=[--(根号2)/2](sina+cosa),
cos[a+(派/2)]=--sina
所以 sin[a--(5派/4)]+cos[a+(派/2)]
=[--1--(根号2)/2]sina--[(根号2)/2]cosa.