If left| asin^2 theta + bsin theta cos theta | vedclass

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(A) Let $E = a\sin^2 	heta + b\sin 	heta \cos 	heta + c\cos^2 	heta - \frac{1}{2}(a + c)$.Using the identities $\sin^2 	heta = \frac{1 - \cos 2

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$b^2 + (a - c)^2$

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(A) Let $E = a\sin^2 	heta + b\sin 	heta \cos 	heta + c\cos^2 	heta - \frac{1}{2}(a + c)$.Using the identities $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$ and $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$:$E = a\left( \frac{1 - \cos 2	heta}{2} ight) + b\left( \frac{\sin 2	heta}{2} ight) + c\left( \frac{1 + \cos 2	heta}{2} ight) - \frac{1}{2}(a + c)$$E = \frac{1}{2} [a - a\cos 2	heta + b\sin 2	heta + c + c\cos 2	heta - a - c]$$E = \frac{1}{2} [b\sin 2	heta - (a - c)\cos 2	heta]$We know that for any expression of the form $A\sin x + B\cos x$,the range is $[-\sqrt{A^2 + B^2}, \sqrt{A^2 + B^2}]$.Thus,$|b\sin 2	heta - (a - c)\cos 2	heta| \le \sqrt{b^2 + (a - c)^2}$.Therefore,$|E| = \left| \frac{1}{2} [b\sin 2	heta - (a - c)\cos 2	heta] ight| \le \frac{1}{2} \sqrt{b^2 + (a - c)^2}$.Comparing this with the given inequality $|E| \le \frac{1}{2}k$,we get $k = \sqrt{b^2 + (a - c)^2}$.Squaring both sides,$k^2 = b^2 + (a - c)^2$.

If $\left| a\sin^2 \theta + b\sin \theta \cos \theta + c\cos^2 \theta - \frac{1}{2}(a + c) \right| \le \frac{1}{2}k,$ then $k^2$ is equal to

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