.若f(x)={sin(n派-x)cos(n派+x)/cos[(n+1)派-x]}*tan(x-n派)cot[(n派/2
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.若f(x)={sin(n派-x)cos(n派+x)/cos[(n+1)派-x]}*tan(x-n派)cot[(n派/2)+x],n属于z,求f(7派/6)的值
f(x)={sin(n派-x)cos(n派+x)/cos[(n+1)派-x]}*tan(x-n派)cot[(n派/2)+x]
={sin(-x)cosx/cos[(n+1)派-x]}*tanx*cot[(n派/2)+x]
若n奇数,
f(x)={sin(-x)cos(x)/cos[(n+1)派-x]}*tanxcot[(n派/2)+x]
={sin(-x)cosx/cosx}*tanx*cot[(派/2)+x]
=-sinx*tanx*(-tanx)
=sinx*(tanx)^2
f(7派/6)=-1/6
若n偶数,
f(x)={sin(-x)cos(x)/cos[(n+1)派-x]}*tanxcot[(n派/2)+x]
={sin(-x)cosx/(-cosx)}*tanx*cotx
=sinx
f(7派/6)=-1/2
={sin(-x)cosx/cos[(n+1)派-x]}*tanx*cot[(n派/2)+x]
若n奇数,
f(x)={sin(-x)cos(x)/cos[(n+1)派-x]}*tanxcot[(n派/2)+x]
={sin(-x)cosx/cosx}*tanx*cot[(派/2)+x]
=-sinx*tanx*(-tanx)
=sinx*(tanx)^2
f(7派/6)=-1/6
若n偶数,
f(x)={sin(-x)cos(x)/cos[(n+1)派-x]}*tanxcot[(n派/2)+x]
={sin(-x)cosx/(-cosx)}*tanx*cotx
=sinx
f(7派/6)=-1/2
.若f(x)={sin(n派-x)cos(n派+x)/cos[(n+1)派-x]}*tan(x-n派)cot[(n派/2
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