如题!已知cos(π-α)=3/5,α∈[0,π),则sin(2α-π/4)=?要有具体步骤啊,

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如题!已知cos(π-α)=3/5,α∈[0,π),则sin(2α-π/4)=?要有具体步骤啊,如题!已知cos(π-α)=3/5,α∈[0,π),则sin(2α-π/4)=?要有具体步骤啊,如题!已

如题!已知cos(π-α)=3/5,α∈[0,π),则sin(2α-π/4)=?要有具体步骤啊,
如题!已知cos(π-α)=3/5,α∈[0,π),则sin(2α-π/4)=?要有具体步骤啊,

如题!已知cos(π-α)=3/5,α∈[0,π),则sin(2α-π/4)=?要有具体步骤啊,
∵cos(π-α)=3/5 ==>-cosα=3/5
∴cosα=-3/5
∵α∈[0,π)
∴sinα=√(1-cos²α)=4/5
故sin(2α-π/4)=sin(2α)cos(π/4)-cos(2α)sin(π/4)
=2sinαcosα*(√2/2)-(2cos²α-1)(√2/2)
=√2(4/5)(-3/5)-√2(-3/5)²+√2/2
=√2(1/2-12/25-9/25)
=-17√2/50