Q. 11

# Write the value o

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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