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)](/uploads/image/z/6072371-35-1.jpg?t=%E5%B7%B2%E7%9F%A5cos%EF%BC%88%CF%80%2Ba%EF%BC%89%3D-1%2F2%2C%E8%AE%A1%E7%AE%97%EF%BC%881%EF%BC%89.sin%EF%BC%882%CF%80-a%EF%BC%89+%EF%BC%882%EF%BC%89sin%EF%BC%88a%2B%282n%2B1%EF%BC%89%CF%80%EF%BC%89%2Ba-%282n%2B1%EF%BC%89%CF%80%E8%AE%A1%E7%AE%97%EF%BC%881%EF%BC%89.sin%EF%BC%882%CF%80-a%EF%BC%89+%EF%BC%882%EF%BC%89%EF%BD%9Bsin%5Ba%2B%282n%2B1%EF%BC%89%CF%80%5D%2B%5Bsina-%282n%2B1%EF%BC%89%CF%80%5D%EF%BD%9D%2Fsin%EF%BC%88a%2B2n%CF%80%EF%BC%89%C3%97cos%EF%BC%88a-2n%CF%80%EF%BC%89)
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