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sin6A+cos6A=1-3sin2Acos2A

priyanshi joshi , 10 Years ago
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anser 2 Answers
Y RAJYALAKSHMI

Last Activity: 10 Years ago

sin6A + cos6A = (sin2A)3 + (cos2A)3 = (sin2A + cos2A)(sin4A – sin2Acos2A + cos4A)
= 1 (sin4A – sin2Acos2A + cos4A + 2sin2Acos2A – 2sin2Acos2A)
= (sin2A + cos2A) – 3sin2Acos2A
= 1 – 3sin2Acos2A
 

BShankar

Last Activity: 6 Years ago

sin^2A + cos^2A =1
Cubing on both sides
(sin^2A+cos^2A)^3=1^3
(sin^2A)^³+(cos^2A)^3 + 3sin^2Acos^2A(sin^2A+cos^2A)=1
sin^6A+cos^6A+3sin^2Acos^2A(1)=1
sin^6A+cos^6A=1-3sin^2Acos^2A
Hence, proved.

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