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

數(shù)學人氣：668 ℃時間：2020-05-21 08:38:37

優(yōu)質解答

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展開得
則cosacos(π/6)+sinasin(π/6)+sina=4√3/5
即(√3/2)cosa+(3/2)sina=4√3/5
兩邊同除以√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