Question

已知sin(π/4-α)=a,0<α<π/2,求sin(5π/4+α)

Answer 1

已知sin(π/4-α)=a得
√2/2（cosα-sinα）=a
左右平方得：2cosαsinα=1-2a^2
（cosα+sinα）^2=1+2cosαsinα=2-2a^2
因为0<α<π/2，所以cosα+sinα=√（2-2a^2）

sin(5π/4+α)=-√2/2（cosα+sinα）=-√(1-a^2)

Answer 2

∵0<α<π/2
∴-π/4<π/4-α<π/4
∴cos(π/4-α)>0
而sin(π/4-α)=a
∴cos(π/4-α)=√[1-(sin(π/4-α))^2]=√(1-a^2)
且有sin(α-π/4)=-a
∴sin(5π/4+α)=sin[(α-π/4)+3π/2]
=sin(α-π/4)*cos(3π/2)+cos(α-π/4)*sin(3π/2)
=0-cos(α-π/4)
=-√(1-a^2)