Prove that
sin^6A+cos^6A = 1- 3sin^2Acos^2A
Answers
Answered by
112
sin^6A+cos^6A
(sin²A)³+(cos²A)³ ............[a^6=(a²)³]
(sin²A+cos²A)³-3sin²Acos²A.........[(a+b)³=a³+b³+3ab]
1³-3sin²Acos²A
1-3sin²Acos²A
!!hence proved!!
(sin²A)³+(cos²A)³ ............[a^6=(a²)³]
(sin²A+cos²A)³-3sin²Acos²A.........[(a+b)³=a³+b³+3ab]
1³-3sin²Acos²A
1-3sin²Acos²A
!!hence proved!!
ansss:
hope it helps
u made some mistakes in the answer.
check the formula.
its absolutely correct
(a+b)^3=a^3+b^3+3ab(a+b) (a+b)^3-3ab(a+b)=a^3+b^3
Answered by
89
sin⁶A+cos⁶A = (sin²A)³ + (cos²A)³
= ( sin²A+cos²A)³ - 3(sin²A.cos²A)(sin²A+cos²A)
= 1 - 3(sin²A.cos²A)(1)
= 1 - 3.sin²A.cos²A
= ( sin²A+cos²A)³ - 3(sin²A.cos²A)(sin²A+cos²A)
= 1 - 3(sin²A.cos²A)(1)
= 1 - 3.sin²A.cos²A
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