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