Question
Download Solution PDF1 + 2sin2θ cos2θ - sin4θ - cos4θ का अधिकतम मान क्या है, जहाँ 0° < θ < 90° है?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFदिया गया है:
त्रिकोणमितीय व्यंजक 1 + 2sin2θ cos2θ - sin4θ - cos4θ है जहाँ 0° < θ < 90° है।
प्रयुक्त सूत्र:
2sinθ.cosθ = sin2θ
sin2θ + cos2θ = 1
a2 + b2 = (a + b)2 - 2ab
गणना:
हमारे पास है 1 + 2sin2θ cos2θ - sin4θ - cos4θ
⇒ 1 + 2sin2θ cos2θ - (sin4θ + cos4θ)
⇒ 1 + 2sin2θ cos2θ - [(sin2θ)2 + (cos2θ)2]
⇒ 1 + 2sin2θ cos2θ - [(sin2θ + cos2θ)2 - 2sin2θcos2θ]
⇒ 1 + 2sin2θ cos2θ - 1 + 2sin2θcos2θ
⇒ 4sin2θ cos2θ
⇒ (2sinθ.cosθ)2
⇒ (sin2θ)2
⇒ sin22θ
x = 90° पर sin2x का अधिकतम मान = 1
∴ 2θ = 90° पर sin22θ का अधिकतम मान = 1
अर्थात् θ = 45° जो दिये गए शर्त को संतुष्ट करता है।
Last updated on Jun 18, 2025
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