Prove (Cos²A)³ - (sin²A)³ = cos2A (1- 1/4 Sin²2A)
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(cos²A)³-(sin²A)³=cos2A[1-(1/4)sin²2A]
Step-by-step explanation:
LHS = (cos²A)³-(sin²A)³
= (cos²A-sin²A)[(cos²A)²+cos²Asin²A+(sin²A)²]
/* By algebraic identity:
a³-b³ =(a-b)(a²+ab+b²) */
= cos2A[(cos²A)²+2cos²Asin²A+(sin²A)²-sin²Acos²A]
/* cos²A-sin²A = cos2A */
= cos2A[(cos²A+sin²A)²-sin²Acos²A]
= cos2A[1-(4sin²Acos²A)/4]
= cos2A[1-(2sinAcosA)²/4]
=cos2A[1-(1/4)sin²2A]
= RHS
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