Question

一道数学题，急需，在线等

已知cos(x-π/4)=√2/10,x∈(π/2,3π/4),sinx=？

Answer 1

cos（x－π／4）=cosxcosπ/4+sinxsinπ/4=(√2/2)(cosx+sinx)=√2/10
cosx+sinx=1/5
(cosx+sinx)^2=(cosx)^2+2cosxsinx+(sinx)^2
=1+2sinxcosx
=1/25
sinxcosx=-12/25
∵x∈（π／2，3π／4）
∴sinx>0
则cosx<0
(sinx-cosx)^2=(cosx)^2+2cosxsinx+(sinx)^2-4cosxsinx
=(cosx+sinx)^2-4×(-12/25)
=49/25
sinx-cosx>0
sinx-cosx=7/5
sinx=(1/5+7/5)÷2=4/5

Answer 2

x∈(π/2,3π/4),
x-π/4∈(π/4,π/2),
cos(x-π/4)=√2/10 sin(x-π/4)=7√2/10
sinx=sin[(x-π/4)+π/4]
=sin(x-π/4)cosπ/4+cos(x-π/4)sinπ/4
=7√2/10*√2/2+√2/10*√2/2
=4/5