Question

求数学大神，求详细解答步骤

Answer 1

追问

简直屌爆！

追答

╭(╯ε╰)╮

Answer 2

f（x）=[cos(x/2)]^2-sin(x/2)cos(x/2)-1/2{[cos(x/2)]^2+[sin(x/2)]^2}=1/2[cos(x/2)]^2-sin(x/2)cos(x/2)- 1/2[sin(x/2)]^2=1/2cosx-1/2sinx=2^0.5/2[(2^0.5/2)*cosx-(2^0.5/2)*sinx]=2^0.5/2*[sin(π/4-x)]
所以f（x）最小正周期为T=2π/|-1|=2π,
值域为[-2^0.2/2, 2^0.2/2]
f(a)=1/2cosa-1/2sina=3*2^0.5/10
则（1/2cosa-1/2sina）^2=9/50
1/4-1/2cosasina=1/4-(sin2a)/4=9/50
则sin2a=7/25

Answer 3

f(x)=1/2-0.5cosx-0.5sinx-1/2
f(x)=-0.5sin(x+π/4)
f(x)的最小正周期为2π，值域为[-0.5,0.5]