Question

Prove:
2(sin^6+cos^6)-3(sin^4+cos^4)+1=0

Answer 1

Answer:

proved

Step-by-step explanation:

Given  

Prove: 2(sin^6+cos^6)-3(sin^4+cos^4)+1=0

We can write this as

a ^3 + b^3 = (a + b)(a^2 – ab + b^2)

2(sin^2 a + cos^2 a)(sin^4 a – sin^2 a cos^2 b + cos^2 a) – 3(sin^4 a + cos^4 a) + 1 = 0

We know that sin^2 a + cos^2 a = 1

2(1)(sin^4 a – sin^2 a cos^2 b + cos^2 a) - 3(sin^4 a + cos^4 a) + 1 = 0

2sin^4 a – 2 sin^2 a cos^2b + 2 cos^4 a – 3 sin^4 a – 3 cos^4 a + 1 = 0

-sin^4a – 2sin^2acos^2b – cos^4 a + 1 = 0

-( sin^4a + 2sin^2acos^2b + cos^4 a) + 1 = 0

-(sin^2 a + cos^2 a)^2 + 1 = 0

-1 + 1 = 0