设α∈(0,π/3),β(π/6,π/2),且α,β满足5√3sinα+5cosα=8√2sinβ+√6cosβ =2
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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 )
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,(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;
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