1.设cos(α+π/4)=2/3,求sin2α的值；2.已知sinθ=3/5,θ∈(π/2,π),求cos(2θ-2π/3)的值。
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1
sin2α=cos(α-π/2)=-cos(α+π/2)
=-[2cos^2(α+π/4)-1]
=1-2×(2/3)^2
=1/9
2
cosθ=-√(1-sin^2θ)=-4/5;
cos(2θ-2π/3)=2cos^2(θ-π/3)-1
=2[cosθ·cos(π/3)+sinθ·sin(π/3)]-1
=2[(-4/5)(1/2)+(3/5)(√3/2)]-1
=(3√3-9)/5
sin2α=cos(α-π/2)=-cos(α+π/2)
=-[2cos^2(α+π/4)-1]
=1-2×(2/3)^2
=1/9
2
cosθ=-√(1-sin^2θ)=-4/5;
cos(2θ-2π/3)=2cos^2(θ-π/3)-1
=2[cosθ·cos(π/3)+sinθ·sin(π/3)]-1
=2[(-4/5)(1/2)+(3/5)(√3/2)]-1
=(3√3-9)/5
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