sina(a/b+2k丌).cos(11/3丌)-tan(5/4丌+2k丌)化简并计算,具体怎么解?

sina(a/b+2k丌).cos(11/3丌)-tan(5/4丌+2k丌)化简并计算,具体怎么解?
sina(a/b+2k丌).cos(11/3丌)-tan(5/4丌+2k丌)
化简并计算,具体怎么解?

sina(a/b+2k丌).cos(11/3丌)-tan(5/4丌+2k丌)化简并计算,具体怎么解?
sin和cos周期是2π,tan是π
所以原式=sin(a/b)cos(4π-π/3)-tan(5π/4)
=sin(a/b)cos(-π/3)-tan(π+π/4)
=sin(a/b)cos(π/3)-tan(π/4)
=1/2*sin(a/b)-1