sin(3θ)=sin(2θ+θ)=sin(2θ)cos(θ)+cos(2θ)sin(θ)=
2sin(θ)cos2(θ)+(1-2sin2(θ))sin(θ)=
2sin(θ)(1-sin2(θ))+sin(θ)-2sin3(θ)=
3sin(θ)-4sin3(θ) QED
When θ=π/4, 3θ=3π/4 and sin(3π/4)=sin(π/4)=√2/2.
3sin(π/4)-4sin3(π/4)=3√2/2-4(2√2/8)=3√2/2-√2=√2/2 confirms the identity.