Question

高中数学小题两则（需详细过程）.

Answer 1

1）[sin5π/18(1+√3tanπ/18)-cosπ/9]/[cos4π/9√(1-cosπ/9)]
=[sin5π/18(1+√3tanπ/18)-cosπ/9]/cos4π/9√2(sinπ/18)^2]
=[2sin5π/18(1/2cosπ/18+√3/2sinπ/18)-cosπ/9cosπ/18]/√2cosπ/18cos4π/9sinπ/18]
=2[2sin5π/18sin(π/18+π/6)-cosπ/9cosπ/18]/2*√2cosπ/18cos4π/9sinπ/18]
=2[2sin5π/18sin2π/9-cosπ/9cosπ/18]/√2cos4π/9sinπ/9]
=[4sin5π/18sin2π/9-2cosπ/9cosπ/18]/√2cos4π/9sinπ/9
=[2cos(5π/18-2π/9)-2cos(5π/18+2π/9)-cos(π/9+π/18)-cos(π/9-π/18)]/√2cos4π/9sinπ/9
=[2cosπ/18-2cosπ/2-cosπ/6-cosπ/18]/[√2/2*sin(4π/9+π/9)-sin(4π/9-π/9)]
=[cosπ/18-cosπ/6]/[√2/2*(sin5π/9-sinπ/3]
=√2[cos(π/2-4π/9)-cos(π/2-π/3)]/*(sin5π/9-sinπ/3)
=√2(sin4π/9-sinπ/3)/(sin5π/9-sinπ/3)
=√2[(sin(π-5π/9)-sinπ/3]/(sin5π/9-sinπ/3)
=√2(sin5π/9)-sinπ/3]/(sin5π/9-sinπ/3)
=√2
2)cosw=-√3/3 ,sinw=-√6/3
sin2w=2sinwcosw=2*(-√6/3 )*(-√3/3)=2√2/3
cos2w=2(cosw)^2-1=2*(-√3/3)^2-1=-1/3
tan2w=sin2w/cos2w=2√2/3/-1/3=-2√2