Question

Show that :
$2\left(sin^6x+cos^6x\right)-3\left(sin^4x+cos^4x\right)+1=0$​

Answer 1

1)sin^2A+cos^2A=1

2)1-sin^2A=cos^2A

3)1-cos^2A=sin^2A

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

=2sin^6A+2cos^6A-3sin^4A-3cos^4A

+sin^2A+cos^2A

=2sin^6A+2cos^6A-2sin^4A-2cos^4A-sin^4A-cos^4A+sin^2A+cos^2A

=2sin^6A-2sin^4A+2cos^6A-2cos^4A

+sin^2A-sin^4A+cos^2A-cos^4A

=-2sin^4A(1-sin^2A)-2cos^4A(1-cos^2A)

sin^2A(1-sin^2A)+cos^2A(1-cos^2A)

=-2sin^4Acos^2A-2cos^4Asin^2A

+sin^2Acos^2A+cos^2Asin^2A

=-2sin^2Acos^2A(sin^2A+cos^2A)

+2sin^2Acos^2A

=-2sin^2Acos^2A+2sin^2Acos^2A

=0

Hence,

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

Answer 2

SOLUTION

$= > 2(sin {}^{6} \theta + cos {}^{6} \theta) - 3(sin {}^{4} \theta + cos {}^{4} \theta) + 1 \\ \\ = >2(sin {}^{2} \theta)3 + (cos {}^{2} \theta)3 - 3(sin {}^{2} \theta)2 + (cos {}^{2} \theta)2 + 1 \\ \\ = > 2(sin {}^{2} \theta + cos {}^{2} \theta)3 - 3sin {}^{2} \theta \: cos {}^{2} \theta(sin {}^{2} \theta + cos {}^{2} \theta) - 3(sin {}^{2} \theta + cos {}^{2} \theta)2 - (2sin {}^{2} \theta \: cos {}^{2} \theta) + 1 \\ \\ = > the \: algebraic \: identity \\ = > a {}^{3} + b {}^{3} = (a + b) {}^{3} - 3ab(a + b) \: and \: \\ a {}^{2} + b {}^{2} = (a + b) {}^{2} - 2ab \\ are \: used \: in \: the \: above \: step \: where \\ \\ = > a = sin {}^{2} \theta \: and \: b = cos {}^{2} \theta \\ writting \: sin {}^{2} \theta + cos {}^{2} \theta = 1 \: we \: have \\ \\ = > 2(1 - 3sin {}^{2} \theta \: cos {}^{2} \theta) - 3( - 2sin {}^{2} \theta \: cos {}^{2} \theta) + 1 \\ \\ = > 2 - 6sin {}^{2} \theta \: cos {}^{2} \theta - 3 + 6sin {}^{2} \theta \: cos {}^{2} \theta + 1 \\ \\ = > - 3 + 3 = 0 \: proved$

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