Question

化简：sin(a-B)cosa-1/2[sin(2a+B)-sinB]=？ 已知cos(a-π/6)+sina=4√3/5则sin（ a+π/6)的值？

Answer 1

sin(a-B)cosa-1/2[sin(2a+B)-sinB]
=[sin(a-B+a)+sin(a-B-a)]/2-sin(2a+B)/2+sinB/2
=sin(2a-B)/2+sin(-B)/2-sin(2a+B)/2+sinB/2
=sin(2a-B)/2-sinB/2-sin(2a+B)/2+sinB/2
=sin(2a-B)/2-sin(2a+B)/2
=1/2*[sin(2a-B)-sin(2a+B)]
=1/2*2*cos[(2a-B+2a+B)/2]sin[(2a-B-2a-B)/2]
=cos[(4a/2]sin[(-2B)/2]
=cos2asin(-B)
=-cos2asinB

cos(a-π/6)+sina=4√3/5
cos(-π/6+a)+sina=4√3/5
cos(π/6-a)+sina=4√3/5
cos(π/2-π/3-a)+sina=4√3/5
sin(π/3+a)+sina=4√3/5
2sin[(π/3+a+a)/2]cos[(π/3+a-a)/2]=4√3/5
sin(π/6+a)cos(π/6)=2√3/5
sin(π/6+a)*√3/2=2√3/5
sin(π/6+a)=4/5