Question

高一数学暑假作业

请高手解决问题。

Answer 1

(1)
因为π<2θ<2π
有π/2<θ<π
tanθ≠±1
于是1-tan²θ≠0
tan2θ=(2tanθ)/(1-tan²θ)=-2√2
2√2tan²θ-2√2-2tanθ=0
√2tan²θ-tanθ-√2=0
(tanθ-√2)(√2tanθ+1)=0
tanθ=√2或者tanθ=-√2 /2
由于θ∈(π/2,π)
所以tanθ<0 故正根舍去
所以tanθ=-√2 /2
(2)
[2cos²(θ/2）-sinθ-1]/√2sin(θ+π/4)
=(cosθ-sinθ)/√2sin(θ+π/4)
=√2 /2(cosθ-sinθ)/sin(θ+π/4)
=[cos(π/4)cosθ-sin（π/4)sinθ]/sin(θ+π/4)
=cos(θ+π/4)/sin(θ+π/4)
=tan(θ+π/4)
=(tanθ+tanπ/4)/(1-tanθ·tanπ/4)
=(-√2/2+1)/ [1-1(-√2/2)]
=(1-√2/2)/(1+√2/2)
=(1-1/2)/(1+√2/2)²
=1/[2(1+√2/2)²]

Answer 2

1.tanθ=-√2 /2
2.1/[2(1+√2/2)²]

Answer 3

万能公式解题，谢谢