Question

Sin^6a+cos^6a=1-3sin^2a+3sin^4a

Answer 1

Answer:

sin^6A+cos^6A

= (sin²A)³+(cos²A)³

[(a+b)³=a³+b³+3ab]

= (sin²A+cos²A)³-3sin²Acos²A

= 1³-3sin²Acos²A

= 1-3sin²Acos²A

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