For any θ ∈ ((π/4) , (π/2)), the expression 3( sinθ - cosθ)4 + 6 ( sinθ + cosθ)2 + 4 sin6 θ equals :

Question

For any θ ∈ ((π/4) , (π/2)), the expression 3( sinθ - cosθ)4 + 6 ( sinθ + cosθ)2 + 4 sin6 θ equals : (A) 13 - 4 cos6 θ (B) 13 - 4 cos4 θ + 2 sin2θ cos2 θ (C) 13 - 4 cos2 θ + 6 cos4 θ (D) 13 - 4 cos2 θ + 6 sin2 θ cos2 θ

Tardigrade Admin · Accepted Answer

We have , 3( sin θ - cos θ)4 + 6( sin θ + cos θ)2 + 4 sin6 θ = 3(1 - sin2 θ)2 + 6(1 + sin2 θ) + 4 sin6 θ = 3(1 - 2 sin2 θ + sin2 2θ) + 6 + 6 sin 2 θ + 4 sin6 θ = 9 + 12 sin2 θ · cos2 θ + 4(1 - cos2 θ)3 = 13 - 4 cos6 θ

Q. For any $θ \in (\frac{π}{4}, \frac{π}{2}),$ the expression $3 (sin θ - cos θ)^{4} + 6 (sin θ + cos θ)^{2} + 4 sin^{6} θ$ equals :

$13 - 4 cos^{6} θ$

$13 - 4 cos^{4} θ + 2 s i n^{2} θ co s^{2} θ$

$13 - 4 cos^{2} θ + 6 cos^{4} θ$

$13 - 4 cos^{2} θ + 6 sin^{2} θ cos^{2} θ$

We have ,
$3 (sin θ - cos θ)^{4} + 6 (sin θ + cos θ)^{2} + 4 sin^{6} θ$
$= 3 (1 - sin^{2} θ)^{2} + 6 (1 + sin^{2} θ) + 4 sin^{6} θ$
$= 3 (1 - 2 sin^{2} θ + sin^{2} 2 θ) + 6 + 6 sin 2 θ + 4 sin^{6} θ$
$= 9 + 12 sin^{2} θ \cdot cos^{2} θ + 4 (1 - cos^{2} θ)^{3}$
$= 13 - 4 cos^{6} θ$

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

13−4cos6θ

13−4cos4θ+2sin2θcos2θ

13−4cos2θ+6cos4θ

13−4cos2θ+6sin2θcos2θ