The value of 2(sin^6 theta + cos^6 theta) - 3 | vedclass

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(B) We know that $\sin^2 	heta + \cos^2 	heta = 1$.First,consider $(\sin^2 	heta + \cos^2 	heta)^3 = 1^3$.Expanding this,we get $\sin^6 	heta + \

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(B) We know that $\sin^2 	heta + \cos^2 	heta = 1$.First,consider $(\sin^2 	heta + \cos^2 	heta)^3 = 1^3$.Expanding this,we get $\sin^6 	heta + \cos^6 	heta + 3 \sin^2 	heta \cos^2 	heta (\sin^2 	heta + \cos^2 	heta) = 1$.Since $\sin^2 	heta + \cos^2 	heta = 1$,this simplifies to $\sin^6 	heta + \cos^6 	heta = 1 - 3 \sin^2 	heta \cos^2 	heta$.Next,consider $(\sin^2 	heta + \cos^2 	heta)^2 = 1^2$.Expanding this,we get $\sin^4 	heta + \cos^4 	heta + 2 \sin^2 	heta \cos^2 	heta = 1$.This simplifies to $\sin^4 	heta + \cos^4 	heta = 1 - 2 \sin^2 	heta \cos^2 	heta$.Now,substitute these into the given expression:$2(1 - 3 \sin^2 	heta \cos^2 	heta) - 3(1 - 2 \sin^2 	heta \cos^2 	heta) + 1$$= 2 - 6 \sin^2 	heta \cos^2 	heta - 3 + 6 \sin^2 	heta \cos^2 	heta + 1$$= (2 - 3 + 1) + (-6 \sin^2 	heta \cos^2 	heta + 6 \sin^2 	heta \cos^2 	heta)$$= 0 + 0 = 0$.

The value of $2(\sin^6 \theta + \cos^6 \theta) - 3(\sin^4 \theta + \cos^4 \theta) + 1$ is

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