Q. 11

Write the value of 2 (sin6 x + cos6 x) 3 (sin4 x + cos4 x) + 1.

Answer :

sin6x + cos6x = (sin2x)3 + (cos2x)3


=(sin2x + cos2x)(sin4x+cos4x-sin2xcos2x)


= 1 (sin4x + cos4x – sin2xcos2x)


Substituting above value in given equation


2(sin4x + cos4x – sin2xcos2x)-3(sin4 x + cos4 x)+1


2sin4x + 2cos4x – 2sin2xcos2x – 3sin4x-3cos4x+1.


-sin4x-cos4x-2sin2xcos2x+1


-[(sin2x)2+(cos2x)2-2sin2xcos2x]+1


-[( sin2x + cos2x)2]+1


-1+1


0.


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