1 + 2sin 2 θ cos 2 θ - sin 4 θ - cos 4 θ যেখানে 0° < θ < 90° এর সর্বোচ্চ মান কত?

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Option 1 : 1
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প্রদত্ত:

ত্রিকোণমিতিক রাশি হল 1 + 2sin 2 θ cos 2 θ - sin 4 θ - cos 4 θ যেখানে 0° < θ < 90°

সূত্র ব্যবহৃত:

2sinθ.cosθ = sin2θ

sin 2 θ + cos 2 θ = 1

a 2 + b 2 = (a + b) 2 - 2ab

গণনা:

আমাদের আছে 1 + 2sin 2 θ cos 2 θ - sin 4 θ - cos 4 θ

⇒ 1 + 2sin 2 θ cos 2 θ - (sin 4 θ + cos 4 θ)

⇒ 1 + 2sin 2 θ cos 2 θ - [(sin 2 θ) 2 + (cos 2 θ) 2 ]

⇒ 1 + 2sin 2 θ cos 2 θ - [(sin 2 θ + cos 2 θ) 2 - 2sin 2 θcos 2 θ]

⇒ 1 + 2sin 2 θ cos 2 θ - 1 + 2sin 2 θcos2 θ

⇒ 4sin 2 θ cos 2 θ

⇒ (2sinθ.cosθ) 2

⇒ (sin2θ) 2

⇒ পাপ 2

x = 90° এ সিন 2 x = 1 এর সর্বোচ্চ মান

∴ পাপের সর্বোচ্চ মান 2 2θ = 1 এ 2θ = 90°

অর্থাৎ θ = 45° যা প্রদত্ত শর্ত পূরণ করে

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