If f(theta ) = left| begin{array}{ccc} 1 & co | vedclass

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(C) Given $f(	heta ) = \left| \begin{array}{ccc} 1 & \cos 	heta & 1 \ - \sin 	heta & 1 & - \cos 	heta \ - 1 & \sin 	heta & 1 \end{array} ight

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$(2 + \sqrt{2}, 2 - \sqrt{2})$

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(C) Given $f(	heta ) = \left| \begin{array}{ccc} 1 & \cos 	heta & 1 \ - \sin 	heta & 1 & - \cos 	heta \ - 1 & \sin 	heta & 1 \end{array} ight|$.Expanding the determinant along the first row:$f(	heta ) = 1(1 - (-\sin 	heta \cos 	heta )) - \cos 	heta (-\sin 	heta - \cos 	heta ) + 1(-\sin^2 	heta + 1)$$f(	heta ) = 1 + \sin 	heta \cos 	heta + \sin 	heta \cos 	heta + \cos^2 	heta - \sin^2 	heta + 1$$f(	heta ) = 2 + 2 \sin 	heta \cos 	heta + \cos 2	heta$Using the identity $\sin 2	heta = 2 \sin 	heta \cos 	heta$:$f(	heta ) = 2 + \sin 2	heta + \cos 2	heta$.The expression $a \sin x + b \cos x$ has a maximum value of $\sqrt{a^2 + b^2}$ and a minimum value of $-\sqrt{a^2 + b^2}$.Here,$a = 1$ and $b = 1$,so the range of $\sin 2	heta + \cos 2	heta$ is $[-\sqrt{2}, \sqrt{2}]$.Therefore,the maximum value $A = 2 + \sqrt{2}$ and the minimum value $B = 2 - \sqrt{2}$.Thus,$(A, B) = (2 + \sqrt{2}, 2 - \sqrt{2})$.

If $f(\theta ) = \left| \begin{array}{ccc} 1 & \cos \theta & 1 \\ - \sin \theta & 1 & - \cos \theta \\ - 1 & \sin \theta & 1 \end{array} \right|$ and $A$ and $B$ are respectively the maximum and the minimum values of $f(\theta )$,then $(A, B)$ is equal to

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