For any θ∈(π/4, π/2), the expression 3(sinθ−cosθ)⁴+6(sinθ+cosθ)²+4sin⁶θ equals.

The correct option is A 13−cos6θ
We have,
S=3(sinθ−cosθ)⁴+6(sinθ+cosθ)²+4sin⁶θ
(sinθ−cosθ)⁴=((sinθ−cosθ)²)²=(sin²θ+cos²θ−2sinθcosθ)²=(1−2sinθcosθ)²
(sinθ+cosθ)²=(sin²θ+cos²θ+2sinθcosθ)²=1+2sinθcosθ
sinθcosθ=sin2θ/2
S=3(1−sin2θ)²+6(1+sin2θ)+4sin⁶θ=3(1−2sin2θ+sin⁴θ)+6+6sin2θ+4sin⁶θ=3−6sin2θ+3sin⁴θ+6+6sin2θ+4sin⁶θ=9+3sin⁴θ+4sin⁶θ
Let sin²θ=x. Then cos²θ=1−x and sinθcosθ=√x(1−x)
S=3(1−2√x(1−x))²+6(1+2√x(1−x))+4x³=3(1−4√x(1−x)+4x(1−x))+6+12√x(1−x)+4x³=3−12√x(1−x)+12x(1−x)+6+12√x(1−x)+4x³=9+12x−12x²+4x³=9+12sin²θ−12sin⁴θ+4sin⁶θ
Using the identity cos3θ = 4cos³θ − 3cosθ, we can write cos³θ = (cos3θ + 3cosθ)/4
cos6θ=cos²3θ−sin²3θ=1−2sin²3θ=1−2(3sinθ−4sin³θ)²
S=3(1−sin2θ)²+6(1+sin2θ)+4sin⁶θ=3(1−2sin2θ+sin⁴θ)+6+6sin2θ+4sin⁶θ=9+3sin⁴θ+4sin⁶θ
Let's try another approach:
S = 3(sinθ−cosθ)⁴+6(sinθ+cosθ)²+4sin⁶θ
= 3(1−sin2θ)²+6(1+sin2θ)+4sin⁶θ
= 3(1−2sin2θ+sin⁴θ)+6+6sin2θ+4sin⁶θ
= 9+3sin⁴θ−6sin2θ+6sin2θ+4sin⁶θ
= 9+3sin⁴θ+4sin⁶θ
This doesn't seem to lead to the given options.
Let's use the identity:
cos2θ = 1−2sin²θ, so sin²θ = (1−cos2θ)/2
cos4θ = 1−2sin²2θ = 1−2(2sinθcosθ)² = 1−8sin²θcos²θ
cos6θ = 4cos³2θ−3cos2θ = 4(1−2sin²θ)³−3(1−2sin²θ)
= 4(1−6sin²θ+12sin⁴θ−8sin⁶θ)−3+6sin²θ
= 1+6sin²θ−48sin⁴θ+32sin⁶θ
This also doesn't lead to the given options. Let's re-examine the problem statement and the options.