Question

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)cosa-1/2[2cos(a+b)sina]=sin(a-b)cosa-cos(a+b)sina=sinacosbcosa-cosasinbcosa-cosacosbsina+sinasinbsina=sinb(sinasina-sinasina)=-sinbcos2a
已知cos(a-π/6)+sina=4√3/5展开得
则cosacos(π/6)+sinasin(π/6)+sina=4√3/5
即(√3/2)cosa+(3/2)sina=4√3/5
两边同除以√3得
(√3/2)sina+(1/2)cosa=4/5，由于cos(π/6)=
(√3/2),sin(π/6)=1/2.最后再合并得sin(a+π/6)=4/5

Answer 2

sin(a-B)cosa-1/2[sin(2a+B)-sinB]
=(1/2)[sin(2a-B)+sin(-B)]-1/2[sin(2a+B)-sinB]
=(1/2)[sin(2a-B)-sin(2a+B)]
=(1/2)*2cos(2a)sin(-B)
=-sinBcos(2a)
已知cos(a-π/6)+sina=4√3/5
则cosacos(π/6)+sinasin(π/6)+sina=4√3/5
(√3/2)cosa+(3/2)sina=4√3/5
两边同乘以√3
(√3/2)sina+(1/2)cosa=4/5
所以sin(a+π/6)=4/5