已知cosa=1/3,则cos(a+2b)+2sin(a+b)sinb=

问题描述:

已知cosa=1/3,则cos(a+2b)+2sin(a+b)sinb=

cos(a+2b)+2sin(a+b)sinb
=cos[(a+b)+b]+2sin(a+b)sinb
=[cos(a+b)cosb-sin(a+b)sinb]+2sin(a+b)sinb
=cos(a+b)cosb+sin(a+b)sinb
=cos[(a+b)-b]
=cosa
=1/3