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 )
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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