If Δ1 = |x& sinθ& cosθ - sinθ&-x&1 cosθ&1&x| and Δ2 = |x& sin2θ& cos2 θ - sin2θ&-x&1 cos2θ& 1&x| , x ≠0 ; then for all θ ∈ (0, (π/2)) :

Question

If Δ1 = |x& sinθ& cosθ - sinθ&-x&1 cosθ&1&x| and Δ2 = |x& sin2θ& cos2 θ - sin2θ&-x&1 cos2θ& 1&x| , x ≠0 ; then for all θ ∈ (0, (π/2)) : (A) Δ1 - Δ2 = x ( cos 2 θ - cos 4 θ) (B) Δ1 + Δ 2 = - 2x3 (C) Δ1 - Δ 2 = - 2x3 (D) Δ1 + Δ 2 = - 2 (x3 + x - 1)

Tardigrade Admin · Accepted Answer

Δ1 =f(θ) = |x& sinθ& cosθ - sinθ&-x&1 cosθ&1&x| =-x3 and Δ2 =f(2θ) = |x sin2θ& cos2θ - sin2θ&-x&1 cos2θ&1&x| = -x3 So Δ1 + Δ2 = - 2x3

Q. If $Δ_{1} = ∣ ∣ x - sin θ cos θ sin θ - x 1 cos θ 1 x ∣ ∣$ and $Δ_{2} = ∣ ∣ x - sin 2 θ cos 2 θ sin 2 θ - x 1 cos 2 θ 1 x ∣ ∣, x \neq = 0;$ then for all $θ \in (0, \frac{π}{2}) :$

$Δ_{1} - Δ_{2} = x (cos 2 θ - cos 4 θ)$

$Δ_{1} + Δ_{2} = - 2 x^{3}$

$Δ_{1} - Δ_{2} = - 2 x^{3}$

$Δ_{1} + Δ_{2} = - 2 (x^{3} + x - 1)$

$Δ_{1} = f (θ) = ∣ ∣ x - sin θ cos θ sin θ - x 1 cos θ 1 x ∣ ∣ = - x^{3}$
and $Δ_{2} = f (2 θ) = ∣ ∣ x - sin 2 θ cos 2 θ sin 2 θ - x 1 cos 2 θ 1 x ∣ ∣ = - x^{3}$
So $Δ_{1} + Δ_{2} = - 2 x^{3}$

Q. If Δ1​=∣∣​x−sinθcosθ​sinθ−x1​cosθ1x​∣∣​ and Δ2​=∣∣​x−sin2θcos2θ​sin2θ−x1​cos2θ1x​∣∣​,x=0; then for all θ∈(0,2π​):

Δ1​−Δ2​=x(cos2θ−cos4θ)

Δ1​+Δ2​=−2x3

Δ1​−Δ2​=−2x3

Δ1​+Δ2​=−2(x3+x−1)