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

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asked Jan 25, 2018 in Mathematics by Shiv (20 points)

Solve:  2(sin6x + cos6x) - 3(sin4x + cos4x) + 1

commented Jan 25, 2018 by Shiv (20 points)
Find the value​ of it

2 Answers

+2 votes
answered Jan 25, 2018 by Ankit Agarwal (28,847 points) ● 7 ● 32 ● 68

Solution:

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+2 votes
answered Jan 25, 2018 by Rohit Singh (61,782 points) ● 35 ● 133 ● 357

We know that

=> sin2A+cos2A=1
=> 1-sin2A=cos2A
=> 1-cos2A=sin2A
Applying these and solving we have:
=2(sin6A+cos6A)-3(sin4A+cos4A)+1

=2sin6A+2cos6A-3sin4A-3cos4A+sin2A+cos2A

=2sin6A+2cos6A-2sin4A-2cos4A-sin4A-cos4A+sin2A+cos2A

=2sin6A-2sin4A+2cos6A-2cos4A+sin2A-sin4A+cos2A-cos4A

=-2sin4A(1-sin2A)-2cos4A(1-cos2A)+sin2A(1-sin2A)+cos2A(1-cos2A)

=-2sin4Acos2A-2cos4Asin2A+sin2Acos2A+cos2Asin2A

=-2sin2Acos2A(sin2A+cos2A)+2sin2Acos2A

=-2sin2Acos2A+2sin2Acos2A

=0

Hence,
2(sin6A+cos6A)-3(sin4A+cos4A)+1=0

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