已知 cos (π /4 -α)=12/13,α∈(0,π/4)则cos2α/sin(π/4+α)=

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已知cos(π/4-α)=12/13,α∈(0,π/4)则cos2α/sin(π/4+α)=已知cos(π/4-α)=12/13,α∈(0,π/4)则cos2α/sin(π/4+α)=已知cos(π/

已知 cos (π /4 -α)=12/13,α∈(0,π/4)则cos2α/sin(π/4+α)=
已知 cos (π /4 -α)=12/13,α∈(0,π/4)则cos2α/sin(π/4+α)=

已知 cos (π /4 -α)=12/13,α∈(0,π/4)则cos2α/sin(π/4+α)=
∵cos(π/4-α)=12/13
==>cos(π/4)cosα+sin(π/4)sinα=12/13 (应用余弦差角公式)
==>(cosα+sinα)/√2=12/13
==>[(cosα+sinα)/√2]^2=(12/13)^2
==>(1+sin(2α))/2=144/169
∴sin(2α)=119/169
∵α∈(0,π/4)
==>2α∈(0,π/2)
==>cos(2α)>0
∴cos(2α)=√[1-(sin(2α))^2]=120/169
∵cos(π/4-α)=12/13
==>sin[π/2-(π/4-α)]=12/13
∴sin(π/4+α)=12/13
故cos(2α)/sin(π/4+α)=(120/169)/(12/13)=10/13.