if sin^8 θ + cos^8 θ - 1 = 0, then what is the value of cos^2 θ sin^2 θ (if θ ≠ 0 or π/2)?

Given: sin8θ+cos8θ=1 Dividing by 2–√on both sides,12√sin8θ+12√cos8θ=12√—(1) Since, sin45=cos45=12√ LHSof (1)=>cos45sin8θ+sin45cos8θ—(2) RHSof (1)=12√=sin45 Since sin(A+B)=sinAcosB+cosAsinB, (2)=>sin(45+8θ)=12√=sin45=sin135 Considering RHS=sin45 =>45+8θ=45=>8θ=0=>θ=0, but θ is not 0 (given) Considering RHS=sin135 =>45+8θ=135=>8θ=90=>θ=908=454—(3) Requested answer is cos2θ∗sin2θ=(cosθ∗sinθ)2=([2∗cosθ∗sinθ]2)2—(4) Since 2sinθcosθ=sin2θ,(4) can be written as (sin2θ)24=(sin(2∗454))24 =(sin45/2)^2/4------->(5) sin(45/2)=0.3827,(5) can be evaluated to 0.036612