1 + 2sin2θ cos2θ - sin4θ - cos4θ चे कमाल मूल्य किती आहे, जेथे 0° < θ < 90° आहे?

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दिलेल्याप्रमाणे:

त्रिकोणमितीय पादवली  1 + 2sin2θ cos2θ - sin4θ - cos4θ जेथे 0° < θ < 90° आहे
 
वापरलेले सूत्र:

2sinθ.cosθ = sin2θ 

sin2θ + cos2θ = 1

a+ b2 = (a + b)2 - 2ab

गणना :

आपल्याकडे आहे 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

⇒ sin2

x = 90° वर sin2x चे कमाल मूल्य= 1
 
2θ = 90° वर sin2 चे कमाल मूल्य =1
 
म्हणजे θ = 45° जे दिलेली स्थिती पूर्ण करते
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