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