已知f(x)=2sinx/2sin(θ -x/2)-1,若f(x)是偶函数,则cosθ/2
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已知f(x)=2sinx/2sin(θ -x/2)-1,若f(x)是偶函数,则cosθ/2
已知f(x)=2sinx/2sin(θ -x/2)-1,若f(x)是偶函数,则cosθ/2
已知f(x)=2sinx/2sin(θ -x/2)-1,若f(x)是偶函数,则cosθ/2
因为f(x)是偶函数,所以
f(-π/2)=f(π/2)
-2/[2sin(θ+π/4)-1]= 2/[2sin(θ-π/4)-1]
1-2sin(θ+π/4)=2sin(θ-π/4)-1
2=2sin(θ-π/4)+2sin(θ+π/4)=4*(√2/2)sinθ
sinθ=√2/2
令cosθ/2=t
cosθ=2t^2-1=±√2/2
2t^2=1±√2/2
t^2=(2±√2)/4
t=±√(2±√2)/2
cosθ/2=±√(2±√2)/2
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