已知cos(π+a)=-1/2,计算(1).sin(2π-a) (2)sin(a+(2n+1)π)+a-(2n+1)π计算(1).sin(2π-a) (2){sin[a+(2n+1)π]+[sina-(2n+1)π]}/sin(a+2nπ)×cos(a-2nπ)

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已知cos(π+a)=-1/2,计算(1).sin(2π-a) (2)sin(a+(2n+1)π)+a-(2n+1)π计算(1).sin(2π-a) (2){sin[a+(2n+1)π]+[sina-(2n+1)π]}/sin(a+2nπ)×cos(a-2nπ)

已知cos(π+a)=-1/2,计算(1).sin(2π-a) (2)sin(a+(2n+1)π)+a-(2n+1)π计算(1).sin(2π-a) (2){sin[a+(2n+1)π]+[sina-(2n+1)π]}/sin(a+2nπ)×cos(a-2nπ)
已知cos(π+a)=-1/2,计算(1).sin(2π-a) (2)sin(a+(2n+1)π)+a-(2n+1)π
计算(1).sin(2π-a) (2){sin[a+(2n+1)π]+[sina-(2n+1)π]}/sin(a+2nπ)×cos(a-2nπ)

已知cos(π+a)=-1/2,计算(1).sin(2π-a) (2)sin(a+(2n+1)π)+a-(2n+1)π计算(1).sin(2π-a) (2){sin[a+(2n+1)π]+[sina-(2n+1)π]}/sin(a+2nπ)×cos(a-2nπ)
cos(π+a)=-cosa=-1/2
cosa=1/2
sin²a+cos²a=1
sina=±√3/2
所以.sin(2π-a)
=-sina
=±√3/2
{sin[a+(2n+1)π]+[sina-(2n+1)π]}/sin(a+2nπ)×cos(a-2nπ)
=(-sina-sina)/(sinacosa)
=-2tana
=±2√3

cos(π+a)=cos(π)cos(a)-sin(π)sin(a) = -cosa = -1/2.
所以,cos(a)=1/2;
sin(2π-a)=sin(-a)=-sin(a)=±√3/2;
{sin[a+(2n+1)π]+[sina-(2n+1)π]}/sin(a+2nπ)×cos(a-2nπ)
=[sin(a+π)+sin(a-π)]/sin(a)×cos(a)
=[-sin(a)-sin(a)]/sin(a)×cos(a)
=-2×cos(a)=-2