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