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∵sinα=3/5,α∈(π/2,π)∴cosα=-√(1-sin²;α)=-4/5故sin(2α)=2sinαcosα=2(3/5)(-4/5)=-24/25 cos(2α)=cos²;α-sin²;α=(-4/5)²;-(3/5)²;=7/25 tan(2α)=sin(2α)/cos(2α)=(-24/25)/(7/25)=-24/7
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