1.化简f(a)={[sin(π-α)cos(2π-α)cot(3π+α)]/[cot(-π-α)sin(-π-α)]}得：f(a)得值是多少?注意括号）2.已知f(x)=αsin2x+βtanx+1,且f(-2)=-2007,那么f(π-2)的值是多少?3.求cos(π/5)+cos(2π/5)+cos(3π/5)+cos(4π/5)的值.
问题描述：
1.化简f(a)={[sin(π-α)cos(2π-α)cot(3π+α)]/[cot(-π-α)sin(-π-α)]}得：f(a)得值是多少?注意括号）
2.已知f(x)=αsin2x+βtanx+1,且f(-2)=-2007,那么f(π-2)的值是多少?
3.求cos(π/5)+cos(2π/5)+cos(3π/5)+cos(4π/5)的值.
答
1.化简f(a)={[sin(π-α)cos(2π-α)cot(3π+α)]/[cot(-π-α)sin(-π-α)]}得：f(a)得值是多少?（过程要详细 注意括号） 解答如下：f(a)={[sin(π-α)cos(2π-α)cot(3π+α)]/[cot(-π-α)sin(-π-α)]}={[sinαc...