杭州交通信息网互换资源吧:三角函数问题2
来源：百度文库 编辑：杭州交通信息网 时间：2024/04/29 05:30:48
已知a+b=2pai/3,求y=[1-cos(pai-2a)]/[cot(a/2)-tan(a/2)] - cos^(pai/4-b) 的值
全部利用倍角公式
∵1-cos(pai-2a) = 1+cos(2a)
= 2cos^(a)
cot(a/2)-tan(a/2) = cos(a/2)/sin(a/2)-sin(a/2)/cos(a/2)
= (cos^(a/2)-sin^(a/2))/(sin(a/2)cos(a/2))
= cos(a)/[sin(a)/2]
= 2cos(a)/sin(a)
cos^(pai/4-b) = [cos(π/2-2b)+1]/2
= [sin(2b)+1)/2
= sin(b)cos(b)+1/2
∴ y = [1-cos(pai-2a)]/[cot(a/2)-tan(a/2)] - cos^(pai/4-b)
= 2cos^(a)/[2cos(a)/sin(a)] - [sin(b)cos(b)+1/2]
= cos(a)sin(a) - sin(b)cos(b) - (1/2)
= cos(a+b) -(1/2)
= cos(2π/3) - (1/2)
= -1/2 -(1/2)
= -1
你设b=0，再算