a属于(0,π/4).sin(π/4-a)=3/5 求tan(2a+π/4)

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a属于(0,π/4).sin(π/4-a)=3/5 求tan(2a+π/4)

a属于(0,π/4).sin(π/4-a)=3/5 求tan(2a+π/4)
a属于(0,π/4).sin(π/4-a)=3/5 求tan(2a+π/4)

a属于(0,π/4).sin(π/4-a)=3/5 求tan(2a+π/4)

a∈(0,π/4)
∴ π/4-a∈(0,π/4)
∵ sin(π/4-a)=3/5
∴ cos(π/4-a)=4/5
∴ tan(π/4-a)=3/4
∴ tan[2(π/4-a)]=2tan(π/4-a)/[1-tan²(π/4-a)]=(3/2)/(1-9/16)=24/7
即 tan(π/2-2a)=24/7
∴ tan(2a+π/4)
=tan[3π/4-(π/2-2a)]
=[tan(3π/4)-tan(π/2-2a)]/[1+tan(3π/4)*tan(π/2-2a)]
=(-1-24/7)/(1-24/7)
=-31/(-17)
=31/17