已知sinα=3/5,α∈(π/2,π),求sin2α,cos2α的值
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已知sinα=3/5,α∈(π/2,π),求sin2α,cos2α的值
已知sinα=3/5,α∈(π/2,π),求sin2α,cos2α的值
已知sinα=3/5,α∈(π/2,π),求sin2α,cos2α的值
∵sinα=3/5,α∈(π/2,π)
∴cosα=-√(1-sin²α)=-4/5
故sin(2α)=2sinαcosα
=2(3/5)(-4/5)
=-24/25
cos(2α)=cos²α-sin²α
=(-4/5)²-(3/5)²
=7/25
cosα=-√(1-9/25)=-4/5;
sin2α=2sinαcosα=-24/25;
cos2α=cos²α-sin²α=16/25-9/25=7/25;
sinα=3/5,α∈(π/2,π),
cosα=-√(1-sin²α)=-4/5
sin2α=2sinαcosα=-24/25
cos2α=1-2sin²α=7/25
-24\25 7\25