Question

Prove that : sin^6 A + cos^6 A = 1- 3 sin^2 A .cos^2 A

Answer 1

Step-by-step explanation:

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Answer 2

Proof 1:sin^6 x + cos^6 x

= (sin^2 x + cos^2 x) ( sin^4 x - sin^2 cos^2 x + cos^4 x)

= (sin^4 x + cos^4 x) - sin^2 x cos^2 x

= (sin^2 x + cos^2 x)^2 - 2 sin^2 cos^2 x - sin^2 x cos^2 x

= 1 - 3 sin^2 x cos^2 x

Proof 2:

sin^6 x + cos^6 x

= (1 - cos^2 x)^3 + cos^6 x

= 1 - 3cos^2 x + 3cos^4 x - cos^6 x + cos^6 x

= 1 - 3cos^2 x(1-cos^2 x)

= 1 - 3 cos^2 x sin^2 x

Proof 3:

sin^6 x + cos^6 x

= sin^6 x + cos^6 x + 3 sin^2 x cos^2 x - 3 sin^2 x cos^2 x

= sin^6 x + cos^6 s + 3 sin^2 x cos^2 x (sin^2 x + cos^2 x) - 3 sin^2 x cos^2 x

= 1 - 3sin^2 x cos^2 x

Proof 4:

sin^6 x + cos^6 x

= sin^6 x + cos^6 x - 1 + 1

= sin^6 x + cos^6 x - (sin^2 x + cos^2 x)^3 + 1

= sin^6 x + cos^6 x - sin^6 x - 3sin^4 x cos^2 x - 3 sin^2 x cos^4 x - sin^6 x + 1

= 1 - 3 sin^2 x cos^