已知cos(π/3+α)=-3/5 ,sin(2π/3-β)=5/13,且0<α<π/2<β<π,求cos(β-α).
问题描述:
已知cos(π/3+α)=-3/5 ,sin(2π/3-β)=5/13,且0<α<π/2<β<π,求cos(β-α).
答
0<α<π/2
0+π/3<α+π/3<π/2+π/3
π/3<α+π/3<5π/6
cos(π/3+α)=-3/5 ,
sin(π/3+α)=4/5 ,
π/2+π/3<β+π/3<π+π/3
5π/6<β+π/3<4π/3
cos(π/3+α)=-3/5 ,
sin(2π/3-β)=5/13
sin(π-π/3-β)=5/13
sin(π/3+β)=5/13
cos(π/3+β)=-12/13
cos(β-α)
=cos[π/3+β-(π/3+α)]
=cos(π/3+β)cos(π/3+α)+sin(π/3+β)sin(π/3+α)
=-12/13*(-3/5)+5/13*4/5
=36/65+20/65
=56/65