cos(x/4 - α)=12/13,α∈(0,π/4),求(cos2α)/[sin(π/4 + α)]
cos(x/4-α)=12/13,α∈(0,π/4),求(cos2α)/[sin(π/4+α)]...
cos(x/4 - α)=12/13,α∈(0,π/4),求(cos2α)/[sin(π/4 + α)]
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sin(π/4 + α)
=cos[π/2-(π/4 + α)]
=cos(π/4-α)=12/13
sin²(π/4-α)+cos²(π/4-α)=1
0<π/4-α<π/4
所以sin(π/4-α)=5/13
所以cos2=sin(π/2-2α)
=2sin(π/4-α)cos(π/4-α)
=120/169
所以原式=10/13
=cos[π/2-(π/4 + α)]
=cos(π/4-α)=12/13
sin²(π/4-α)+cos²(π/4-α)=1
0<π/4-α<π/4
所以sin(π/4-α)=5/13
所以cos2=sin(π/2-2α)
=2sin(π/4-α)cos(π/4-α)
=120/169
所以原式=10/13
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