π已知f(x)=sin(nπ-x)cos(nπ+ x)/cos[(n+1)π-x]*tan(x-nπ)(n属于Z)求f(6/7π)
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π已知f(x)=sin(nπ-x)cos(nπ+ x)/cos[(n+1)π-x]*tan(x-nπ)(n属于Z)求f(6/7π)
π已知f(x)=sin(nπ-x)cos(nπ+ x)/cos[(n+1)π-x]*tan(x-nπ)(n属于Z)求f(6/7π)
π已知f(x)=sin(nπ-x)cos(nπ+ x)/cos[(n+1)π-x]*tan(x-nπ)(n属于Z)求f(6/7π)
f(x)=sin(nπ-x)cos(nπ+ x)/cos[(n+1)π-x]*tan(x-nπ)
=sin(nπ-x)cos(nπ+ x)/[-cos(nπ-x)]*[-tan(nπ-x)]
=cos(nπ+ x)
f(7π/6)=cos(nπ+7π/6)
=±√3/2