Question

Evaluate 2(sin^6x+cos^6x)-3(sin^4x+cos^4x)+4(sin^2 + cos^2)=​

Answer 1

""" ❤️ Answer ❤️ """

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

Answer 2

Answer:

We know that

We know that=> sin2A+cos2A=1

We know that=> sin2A+cos2A=1=> 1-sin2A=cos2A

We know that=> sin2A+cos2A=1=> 1-sin2A=cos2A=> 1-cos2A=sin2A

We know that=> sin2A+cos2A=1=> 1-sin2A=cos2A=> 1-cos2A=sin2AApplying these and solving we have:

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

=2(sin6A+cos6A)-3(sin4A+cos4A)+1=2sin6A+2cos6A-3sin4A-3cos4A+sin2A+cos2A

=2(sin6A+cos6A)-3(sin4A+cos4A)+1=2sin6A+2cos6A-3sin4A-3cos4A+sin2A+cos2A=2sin6A+2cos6A-2sin4A-2cos4A-sin4A cos4A+sin2A+cos2A

+sin2A+cos2A=2sin6A-2sin4A+2cos6A-2cos4A+sin2A-sin4A+cos2A-co)+2sin2Acos2)+2sin2Acos2A=-2sin2Acos

2A+2sin2Acos2A=0Hence,

"2(sin6A+cos6A)-3(sin4A+cos4A)+1=0."

"Hope this Helpful."