If f(theta) = begin{bmatrix} cos theta & -sin | vedclass

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Question

(A) Given the function $f(	heta) = \begin{bmatrix} \cos 	heta & -\sin 	heta \ \sin 	heta & -\cos 	heta \end{bmatrix}$.To find $f\left(\frac{\pi}

Vedclass · Accepted Answer

$\begin{bmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \ \frac{1}{2} & -\frac{\sqrt{3}}{2} \end{bmatrix}$

Vedclass · Answer

(A) Given the function $f(	heta) = \begin{bmatrix} \cos 	heta & -\sin 	heta \ \sin 	heta & -\cos 	heta \end{bmatrix}$.To find $f\left(\frac{\pi}{6}ight)$,we substitute $	heta = \frac{\pi}{6}$ into the matrix.$f\left(\frac{\pi}{6}ight) = \begin{bmatrix} \cos(\frac{\pi}{6}) & -\sin(\frac{\pi}{6}) \ \sin(\frac{\pi}{6}) & -\cos(\frac{\pi}{6}) \end{bmatrix}$.We know that $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$ and $\sin(\frac{\pi}{6}) = \frac{1}{2}$.Substituting these values,we get:$f\left(\frac{\pi}{6}ight) = \begin{bmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \ \frac{1}{2} & -\frac{\sqrt{3}}{2} \end{bmatrix}$.Thus,the correct option is $A$.

If $f(\theta) = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & -\cos \theta \end{bmatrix}$,then $f\left(\frac{\pi}{6}\right) = $ . . . . . . .

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