已知cos(π/4-α)=12/13,α∈(0,π/4),则[cos2α]/[sin(π/4+α)]【答案:10/13】请写出详细过程

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已知cos(π/4-α)=12/13,α∈(0,π/4),则[cos2α]/[sin(π/4+α)]【答案:10/13】请写出详细过程已知cos(π/4-α)=12/13,α∈(0,π/4),则[co

已知cos(π/4-α)=12/13,α∈(0,π/4),则[cos2α]/[sin(π/4+α)]【答案:10/13】请写出详细过程
已知cos(π/4-α)=12/13,α∈(0,π/4),
则[cos2α]/[sin(π/4+α)]
【答案:10/13】
请写出详细过程

已知cos(π/4-α)=12/13,α∈(0,π/4),则[cos2α]/[sin(π/4+α)]【答案:10/13】请写出详细过程
∵cos(π/4-α)=12/13 ==>cos²(π/4-α)=12²/13²
==>{1+cos[2(π/4-α)]}/2=144/169
==>1+cos(π/2-2α)=288/169
==>sin(2α)=119/169
又α∈(0,π/4) ==>2α∈(0,π/2)
==>cos(2α)>0
∴cos(2α)=√[1-sin²(2α)]
=√[1-(119/169)²]
=120/169
∵cos(π/4-α)=12/13 ==>sin(π/2-π/4+α)=12/13
==>sin(π/4+α)=12/13
∴[cos(2α)]/[sin(π/4+α)]=(120/169)/(12/13)
=10/13