sin^6x + cos^6x - giúp mình với

cos^6x + sin^6x = (sin²x)³ + (cos²x )³

= ( sin²x + cos²x ) * [ ( sinx)^4 - sin²xcos²x + (cosx)^4 ]

= 1 * [ ( sinx)^4 + 2sin²xcos²x + (cosx)^4 - 3sin²xcos²x ]

= [ ( sin²x + cos²x )² - 3sin²xcos²x ]

= 1 - 3sin²xcos²x

= 1 - 3( sin²2x / 4 )

= 1 - (3/4) * (1 - cos4x) / 2

= 1 - 3(1 - cos4x) / 8

cos^6x + sin^6x = (sin²x)³ + (cos²x )³

= ( sin²x + cos²x ) * [ ( sinx)^4 - sin²xcos²x + (cosx)^4 ] 

cos^6x + sin^6x = (sin²x)³ + (cos²x )³

= ( sin²x + cos²x ) * [ ( sinx)^4 - sin²xcos²x + (cosx)^4 ]

= 1 * [ ( sinx)^4 + 2sin²xcos²x + (cosx)^4 - 3sin²xcos²x ]

= [ ( sin²x + cos²x )² - 3sin²xcos²x ]

= 1 - 3sin²xcos²x

= 1 - 3( sin²2x / 4 )

= 1 - (3/4) * (1 - cos4x) / 2

= 1 - 3(1 - cos4x) / 8

= 

Câu này dễ thế mà cũng hỏi ahiihi 😛