Question

Evaluate sin^6 A - cos^6 A​

Answer 1

Hey mate,

sin^6A-cos^6A\\=(sin^2A)^3-(cos^2A)^3\\

From basic algebraic rules:

a^3-b^3=(a-b)(a^2+b^2+ab)

Hence,

(sin^2A)^3-(cos^2A)^3=(sin^2A-cos^2A)(sin^4A+cos^4A+sin^2Acos^2A)\\=-cos2A[(sin^2A)^2+(cos^2A)^2+sin^2Acos^2A]\\=-cos2A[(sin^2A+cos^2A)^2-2sin^2Acos^2A+sin^2Acos^2A]\\=-cos2A(1-sin^2Acos^2A)\\=-cos2x(1-\frac{sin^22A}{4} )

Hope it will help you.

✨It's. M.S.V.

Answer 2

HIII.....

GOOD MORNING:-)

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