Question

高一数学 三角函数正余弦急问！！！谢谢各位了！！！！拜托！！！！！

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）

2.已知f(x)=αsin2x+βtanx+1，且f(-2)=-2007，那么f(π-2)的值是多少？（详细过程）

3.求cos(π/5)+cos(2π/5)+cos(3π/5)+cos(4π/5)的值.（过程越详细越好）

Answer 1

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)]}
={[sinαcosacot(2π+π+α)]/[cot(2π-π-α)sin(2π-π-α)]}
={[sinαcosacot(π+α)]/[cot(π-α)sin(π-α)]}
={-[sinαcosacot(α)]/-[cot(α)sinα]}
={[sinαcosacot(α)]/[cot(α)sinα]}
=cosa

2.已知f(x)=αsin2x+βtanx+1，且f(-2)=-2007，那么f(π-2)的值是多少？（详细过程）
解答如下：
f(-2)=αsin2(-2)+βtan(-2)+1
=-αsin4-βtan(2)+1
f(π-2)=αsin2(π-2)+βtan(π-2)+1
=αsin(2π-4)+βtan(π-2)+1
=-αsin4-βtan(2)+1=-2007;

3.求cos(π/5)+cos(2π/5)+cos(3π/5)+cos(4π/5)的值.（过程越详细越好）
解答如下：
原式=cos(π/5)+cos(3π/5)+cos(2π/5)+cos(4π/5)
=2cos(2π/5)sin(-π/5)+2cos(3π/5)sin(-π/5)
=-2sin(π/5)*[cos(2π/5)+cos(3π/5)]
=-2sin(π/5)*2cos(π/2)sin(-π/10)
=0

Answer 2

f(a)={[sin(π-α)cos(2π-α)cot(3π+α)]/[cot(-π-α)sin(-π-α)]}

sin(π-α）=sinα cos(2π-α)=cosα cot(3π+α)]=cotα

cot(-π-α)=cot(-α) sin(-π-α)]=sinα
所以f(a)=（sinαcosαcotα）/(cot(-α)sinα )
=-cosα

αsin2(-2）+βtan(-2）+1=αsin2（π-2）+βtan（π-2）+1
f(π-2)=f(-2)=-2007