If sin4θ + 8cos2θ = 4, then what is the value of sin2θ – 2sinθ?
If sin4θ + 8cos2θ = 4, then what is the value of sin2θ – 2sinθ? Correct Answer 2
sin4θ + 8cos2θ = 4
sin4θ + 8cos2θ – 8 = 4 – 8
sin4θ + 8(cos2θ – 1) = –4
Since sin2θ + cos2θ = 1
sin4θ – 8sin2θ = –4
8sin2θ – sin4θ = 4
sin2θ(8 – sin2θ) = 4
8 – sin2θ = 4/sin2θ
sin2θ + 4/sin2θ = 8
(sinθ)2 + (2/sinθ)2 – 2 × sinθ × (2/sinθ) = 8 – 2 × sinθ × (2/sinθ)
(sinθ – 2/sinθ)2 = 8 – 4 = 4
sinθ – 2/sinθ = 2
sin2θ – 2 = 2sinθ
sin2θ – 2sinθ = 2
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Feb 20, 2025