If ∆₁ = ∆₂ = x ≠ 0; then for all θ ∈ (0, π/2) :
(A) ∆₁ – ∆₂ = –2x³
(B) ∆₁ + ∆₂ = –2 (x³ + x –1)
(C) ∆₁ – ∆₂ = x(cos2θ – cos4θ) (D) ∆₁ + ∆₂ = –2x³
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If ∆₁ = ∆₂ = x ≠ 0; then for all θ ∈ (0, π/2) :
(A) ∆₁ – ∆₂ = –2x³
(B) ∆₁ + ∆₂ = –2 (x³ + x –1)✓✓
(C) ∆₁ – ∆₂ = x(cos2θ – cos4θ) (D) ∆₁ + ∆₂ = –2x³
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Answer:
If ∆₁ = ∆₂ = x ≠ 0; then for all θ ∈ (0, π/2) :
(A) ∆₁ – ∆₂ = –2x³
(B) ∆₁ + ∆₂ = –2 (x³ + x –1)✔✔
(C) ∆₁ – ∆₂ = x(cos2θ – cos4θ) (D) ∆₁ + ∆₂ = –2x³
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