∵,锐角θ满足tanθ=2√2
∴sin2θ=2sinθcosθ
=(2sinθcosθ)/(sin²θ+cos²θ)
=2tanθ/(1+tan²θ)
=4√2/9
cos2θ
=(cos²θ-sin²θ)/(cos²θ+sin²θ)
=(1-tan²θ)/(1+tan²θ)
=(1-8)/(1+8)
=-7/9
f(X)=6sin (2x-π¼),
f(θ)=6sin(2θ-π/4)
=6(sinθcosπ/4-cosθsinπ/4)
=3√2(sin2θ-cos2θ)
=3√2(4√2/9+7/9)
=(7√2+8)/3
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