Question

Prove that 2(sin^6+cos^6)-3(sin4+cos^4)+1=0​

Answer 1

Answer:

0

Step-by-step explanation:

we know that

cos^2n+sin^2n =1

2(1)-3(1)+1=0

Answer 2

Answer:

Proved

Step-by-step explanation:

2(sin∧6x+cos^6x)-3(sin4x+cos^4x)+1

= 2(sin²x+cos²x)( sin∧4x + cos∧4x - sin²xcos²x) -3(sin∧4x+cos^4x)+1

= 2 ( sin∧4x + cos∧4x - sin²xcos²x) -3(sin∧4x+cos^4x)+1

= 2( sin∧4x + cos∧4x + 2sin²xcos²x-3 sin²xcos²x) -3(sin∧4x+cos^4x)+1

= 2{(sin²x+cos²x)² - 3 sin²xcos²x} - 3(sin∧4x+cos^4x)+1

= 2 { 1 - 3 sin²xcos²x} - 3(sin∧4x+cos^4x)+1

= 2 - 3 ( sin∧4x+cos^4x + 2sin²xcos²x) + 1

= 3 - 3 ( sin²x+cos²x)²

= 3 - 3

= 0