若f(a)=[1+sin^2a+cos((3/2)π+a)-sin^2((π/2)+a)]/[2sin(π+a)cos(π-a)-cos(π-a)],且1+2sina≠a

问题描述:

若f(a)=[1+sin^2a+cos((3/2)π+a)-sin^2((π/2)+a)]/[2sin(π+a)cos(π-a)-cos(π-a)],且1+2sina≠a
①化简f(a)
②求f(π/3)的值
③若a∈(-π/2,0)且f(a)=-根号3,求cos(7π/2-a)的值

①f(a)=tana
②f(π/3)=3^(1/2)
③a=-π/3,cos(7π/2-a)=-3^(1/2)/2不会化简的过程 麻烦讲一下1+sin^2a+cos((3/2)π+a)=sin^2a+sina+1sin^2((π/2)+a)=cos^2a2sin(π+a)cos(π-a)=2(-sina)(-cosa)=2sinacosa-cos(π-a)=cosasin^2a+sina+1-cos^2a=2sin^2a+sina=sina(2sina+1)2sinacosa+cosa=cosa(2sina+1)(sina(2sina+1))/(cosa(2sina+1))=tana