Question

已知sin(π/4-α)=5/13,α∈(0,π/4) 求cos2α/cos(π/4+α)

Answer 1

sin(Π/4 -x)=5/13 √2/2(cosx-sinx)=5/13
x∈(0,Π/4),
cos(Π/4 -x)=12/13 √2/2(cosx+sinx)=12/13 相乘，得
1/2(cos^2x-sin^2x)=60/169
cos2x=120/169

cos(Π/4 +x)=sin(Π/4 -x)=5/13

(cos2x)/[cos(Π/4 +x)]=(120/169)/(5/13)=24/13

追问

cos(Π/4 +x)=sin(Π/4 -x)=5/13 怎么得到的？

追答

诱导公式 cos(π/2-x)=sinx 即两个角互余 (π/2-x)+x=π/2 则两个角的余函数值相等
(π/4 +x)+(π/4 -x)=π/2
所以
cos(Π/4 +x)=sin(Π/4 -x)

Answer 2

∵α∈(0,π/4)
∴π/4-α∈(0,π/4)
∴cos(π/4-α)=√[1-sin²(π/4-α)]=√[1-﹙5/13﹚²]=12/13
cos2α=sin﹙π/2-2α﹚=sin[2﹙π/4-α﹚]=2sin(π/4-α)cos(π/4-α)
=2×﹙5/13﹚×﹙12/13﹚=120/169
cos(π/4+α)=sin[π/2-﹙π/4 ＋α)]=sin(π/4-α)=5/13
∴cos2α/cos(π/4+α)=﹙120/169﹚/﹙5/13﹚
=24/13

Answer 3

两端平方得 [sin(π/4-a)]^2=25/169 ，即 [1-cos(π/2-2a)]/2为锐角，因此 cos(2a)=√[1-(sin2a)^2]=120/169 。