Question

Evaluate sin^6A-cos^6A

Answer 1

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

From basic algebraic rules:

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

Hence,

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