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Answer:
We know that,
sin2θ + cos2θ = 1
2 sinθ cosθ = sin2θ
(a + b)2 = a2 + b2 + 2ab
Given, X = sin4θ + cos4θ
⇒ X = (sin2θ + cos2θ)2 – 2sin2θcos2θ
⇒ X = 1 – (sin22θ)/2
We know that, 0 ≤ sin2θ ≤ 1
∴ ½ ≤ X ≤ 1
Thank you :)
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