已知3π/2<θ<2π,化简√(1/2+1/2√(1/2+1/2cosθ))

Question

分步写吧：化简√a
a=(1/2)+(1/2)√b
b=(1/2)+(1/2)cosθ

Answer 1

Answer 2

3π/2<θ<2π

3π/4<θ/2<π

3π/8<θ/4<π/2

所以
原式=√(1/2+1/2√(1/2(1+cosθ))
=√(1/2+1/2√(cos²θ/2))
=√(1/2+1/2|cosθ/2|)
=√(1/2-1/2cosθ/2)
=√[1/2(1-cosθ/2)]
=√sin²θ/4
=|sinθ/4|
=sinθ/4

Answer 3

已知3π/2<α<2π,化简：{[√(1-cosα)]+[√(1+cosα)]}/{[√：{[√(1-cosα)]+[√(1+cosα)]}/{[√(1-cosα)-√(1+cosα

Answer 4

sin（π/4）