If sin theta + cos theta = x, then {sin ^6}th | vedclass

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(B) Given $\sin 	heta + \cos 	heta = x.$ Squaring both sides,we get $\sin^2 	heta + \cos^2 	heta + 2 \sin 	heta \cos 	heta = x^2.$Since $\sin^2

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${x^2} \le 2$

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(B) Given $\sin 	heta + \cos 	heta = x.$ Squaring both sides,we get $\sin^2 	heta + \cos^2 	heta + 2 \sin 	heta \cos 	heta = x^2.$Since $\sin^2 	heta + \cos^2 	heta = 1,$ we have $1 + \sin 2	heta = x^2,$ which implies $\sin 2	heta = x^2 - 1.$Since $-1 \le \sin 2	heta \le 1,$ we must have $-1 \le x^2 - 1 \le 1,$ which simplifies to $0 \le x^2 \le 2.$Now,consider the expression ${\sin ^6}	heta + {\cos ^6}	heta = (\sin^2 	heta)^3 + (\cos^2 	heta)^3.$Using the identity $a^3 + b^3 = (a+b)^3 - 3ab(a+b),$ we get:${\sin ^6}	heta + {\cos ^6}	heta = (\sin^2 	heta + \cos^2 	heta)^3 - 3 \sin^2 	heta \cos^2 	heta (\sin^2 	heta + \cos^2 	heta) = 1 - 3 \sin^2 	heta \cos^2 	heta.$Substituting $\sin 	heta \cos 	heta = \frac{1}{2} \sin 2	heta,$ we get $1 - 3(\frac{1}{2} \sin 2	heta)^2 = 1 - \frac{3}{4} \sin^2 2	heta.$Substituting $\sin 2	heta = x^2 - 1,$ we get $1 - \frac{3}{4}(x^2 - 1)^2 = \frac{1}{4}[4 - 3(x^2 - 1)^2].$This identity holds true only when $\sin 2	heta$ is defined,which requires $x^2 \le 2.$

If $\sin \theta + \cos \theta = x,$ then ${\sin ^6}\theta + {\cos ^6}\theta = \frac{1}{4}[4 - 3{({x^2} - 1)^2}]$ for

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