设α∈(0,π/3),β(π/6,π/2),且α,β满足5√3sinα+5cosα=8√2sinβ+√6cosβ =2 求cos(α+β )设α∈(0,π/3),β(π/6,π/2),且α,β满足53sinα+5cosα=8 /2sinβ+/6cosβ =2 求cos(α+β )
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![设α∈(0,π/3),β(π/6,π/2),且α,β满足5√3sinα+5cosα=8√2sinβ+√6cosβ =2 求cos(α+β )设α∈(0,π/3),β(π/6,π/2),且α,β满足53sinα+5cosα=8 /2sinβ+/6cosβ =2 求cos(α+β )](/uploads/image/z/9301763-11-3.jpg?t=%E8%AE%BE%CE%B1%E2%88%88%280%2C%CF%80%2F3%29%2C%CE%B2%28%CF%80%2F6%2C%CF%80%2F2%29%2C%E4%B8%94%CE%B1%2C%CE%B2%E6%BB%A1%E8%B6%B35%E2%88%9A3sin%CE%B1%2B5cos%CE%B1%3D8%E2%88%9A2sin%CE%B2%2B%E2%88%9A6cos%CE%B2+%3D2+%E6%B1%82cos%28%CE%B1%2B%CE%B2+%29%E8%AE%BE%CE%B1%E2%88%88%280%2C%CF%80%2F3%29%2C%CE%B2%28%CF%80%2F6%2C%CF%80%2F2%29%2C%E4%B8%94%CE%B1%2C%CE%B2%E6%BB%A1%E8%B6%B353sin%CE%B1%2B5cos%CE%B1%3D8+%2F2sin%CE%B2%2B%2F6cos%CE%B2+%3D2+%E6%B1%82cos%28%CE%B1%2B%CE%B2+%29)
设α∈(0,π/3),β(π/6,π/2),且α,β满足5√3sinα+5cosα=8√2sinβ+√6cosβ =2 求cos(α+β )设α∈(0,π/3),β(π/6,π/2),且α,β满足53sinα+5cosα=8 /2sinβ+/6cosβ =2 求cos(α+β )
设α∈(0,π/3),β(π/6,π/2),且α,β满足5√3sinα+5cosα=8√2sinβ+√6cosβ =2 求cos(α+β )
设α∈(0,π/3),β(π/6,π/2),且α,β满足53sinα+5cosα=8 /2sinβ+/6cosβ =2 求cos(α+β )
设α∈(0,π/3),β(π/6,π/2),且α,β满足5√3sinα+5cosα=8√2sinβ+√6cosβ =2 求cos(α+β )设α∈(0,π/3),β(π/6,π/2),且α,β满足53sinα+5cosα=8 /2sinβ+/6cosβ =2 求cos(α+β )
由5√3sinα+5cosα=8,(√3/2)*sinα+(1/2)*cosα=4/5,sin(α+π/6)=4/5,cos(α+π/6)=3/5;
由√2sinβ+√6cosβ=2,(1/2)*sinβ+(√3/2)cosβ=√2/2,sin(β-π/6)=√2/2,cos(β-π/6)=√2/2;
所以 cos(α+β)=cos[(α+π/6)+(β-π/6)]=cos(α+π/6)cos(β-π/6)-sin(α+π/6)sin(β-π/6)
=(3/5)*(√2/2)-(4/5)*(√2/2)=(1/5)*(√2/2)=√2/10;
由√2sinβ+√6cosβ=2得
2√2(1/2*sinβ+√3/2*cosβ)=2
sin(β+π/3)=√2/2
β∈(π/6,π/2)则
β∈(π/2,5π/6)
所以cos(β+π/3)=-√2/2
所以cos(α+β)=sin[π/2+(α+β)]
=sin[(α+π/6)+(β+π/3)]
=sin(α+π/6)coc(...
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由√2sinβ+√6cosβ=2得
2√2(1/2*sinβ+√3/2*cosβ)=2
sin(β+π/3)=√2/2
β∈(π/6,π/2)则
β∈(π/2,5π/6)
所以cos(β+π/3)=-√2/2
所以cos(α+β)=sin[π/2+(α+β)]
=sin[(α+π/6)+(β+π/3)]
=sin(α+π/6)coc(β+π/3)+Cos(α+π/6 )sin(β+π/3)
=4/5*(-√2/2)+3/5*√2/2
=-√2/10
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