Math, asked by surya1098p43n9y, 1 year ago

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

Answers

Answered by nitthesh7
8
= 2(sin⁶ x + cos⁶ x) - 3(sin⁴ x + cos⁴ x) + 1

= 2((sin² x)³ + (cos² x)³ - 3(sin⁴ x + cos⁴ x) + 1

= 2((sin² x + cos² x)(sin⁴ x + cos⁴ x - sin² x cos² x) - 3(sin⁴ x + cos⁴ x) + 1

= 2(1)(sin⁴ x + cos⁴ x - sin² x cos² x) - 3(sin⁴ x + cos⁴ x) + 1

= 2 sin⁴ x + 2 sin⁴ x - 2 sin² x cos² x - 3 sin⁴ x - 3 cos⁴ x + 1

= - sin⁴ x - cos⁴ x - 2 sin² x cos² x + 1

= -(sin² x + cos² x)² + 1

= -(1)² + 1

= -1 + 1

= 0
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