cos2α/[sin(α-π/4)]=-根号2 /2
cos2α/[sin(π/4-α)]=根号2 /2
sin(π/2-2α)/[sin(π/4-α)]=根号2 /2
利用二倍角公式,
即 2cos(π/4-α)=√2/2
∴ 2[cos(π/4)cosα+sin(π/4)sinα]=√2/2
即√2(cosα+sinα)=√2/2
∴ cosα+sinα=1/2
cos2α/[sin(α-π/4)]=-根号2 /2,cosα+sinα=?
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