设f(a)=2sin(π+a)cos(π-a)-cos(π+a)/1+sin^2a+cos(3π/2+a)-cos^2(π/2+a),求f(-23π/6)的值.

设f(a)=2sin(π+a)cos(π-a)-cos(π+a)/1+sin^2a+cos(3π/2+a)-cos^2(π/2+a),求f(-23π/6)的值.
设f(a)=2sin(π+a)cos(π-a)-cos(π+a)/1+sin^2a+cos(3π/2+a)-cos^2(π/2+a),求f(-23π/6)的值.

设f(a)=2sin(π+a)cos(π-a)-cos(π+a)/1+sin^2a+cos(3π/2+a)-cos^2(π/2+a),求f(-23π/6)的值.
f(a)=[2sin( π+a)cos(π-a)-cos(π+a)]/[1+sin^2a+sin(π-a)-cos^2(π-a)]
=(2sinacosa+cosa)/(1+sin²a+sina-cos²a)
=cosa(2sina+1)/[sina(2sina+1)]
则f(-23π/6)=f(π/6)=√3


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