Question

化简:√3 sin[(π/6)-α)]-[cos(π/6)-α]

Answer 1

原式=2[sin(π/6-a)*√3/2-cos(π/6-a)*1/2]
=2[sin(π/6-a)cosπ/6-cos(π/6-a)sinπ/6]

=2sin(π/6-a-π/6)
=2sin(-a)
=-2sina

Answer 2

:√3 sin[(π/6)-α)]-[cos(π/6)-α]
=√3(sin(π/6)cosa-cos(π/6)sina)-[cos(π/6)cosα+sin(π/6)sinα]
=√3(1/2cosa-√3/2sina)-[√3/2cosα+1/2sinα]
=√3/2cosα-3/2sina)-√3/2cosα-1/2sinα
=-2 sinα