作业帮f(θ)=[sin(3π-θ)cos(π/2+θ)]/tan(π-θ) 求 ①f（θ）的定义域 ②化简f（θ）的表达式③若f（θ）=1/2,求tanθ sinθ cosθ
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f(θ)=[sin(3π-θ)cos(π/2+θ)]/tan(π-θ) 求 ①f（θ）的定义域 ②化简f（θ）的表达式③若f（θ）=1/2,求tanθ sinθ cosθ
f(θ)=[sin(3π-θ)cos(π/2+θ)]/tan(π-θ) 求 ①f（θ）的定义域 ②化简f（θ）的表达式
③若f（θ）=1/2,求tanθ sinθ cosθ
f(θ)=[sin(3π-θ)cos(π/2+θ)]/tan(π-θ) 求 ①f（θ）的定义域 ②化简f（θ）的表达式③若f（θ）=1/2,求tanθ sinθ cosθ
f(θ)=[sin(3π-θ)cos(π/2+θ)]/tan(π-θ)＝sinθcos(π/2－θ)/tanθ＝sin²θ/tanθ
(1)tanθ≠0 ∴θ≠kπ且θ≠kπ＋π/2
(2)f(θ)＝sin²θ/tanθ＝sinθcosθ＝1/2 sin2θ
(3))f(θ)＝1/2 ∴sin2θ＝1 ∴2θ＝2kπ＋π/2 ∴θ＝kπ＋π/4
∴tanθ＝tan(kπ＋π/4)＝tan(π/4)＝1
sinθ＝sin(kπ＋π/4)＝±√2/2
cosθ＝cos(kπ＋π/4)＝±√2/2