Consider the matrix f(x) = begin{bmatrix} cos | vedclass

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(D) To check Statement $I$: We find $f(-x)$ by replacing $x$ with $-x$ in $f(x)$.$f(-x) = \begin{bmatrix} \cos(-x) & -\sin(-x) & 0 \ \sin(-x) & \cos(

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Both Statement $I$ and Statement $II$ are true.

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(D) To check Statement $I$: We find $f(-x)$ by replacing $x$ with $-x$ in $f(x)$.$f(-x) = \begin{bmatrix} \cos(-x) & -\sin(-x) & 0 \ \sin(-x) & \cos(-x) & 0 \ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos x & \sin x & 0 \ -\sin x & \cos x & 0 \ 0 & 0 & 1 \end{bmatrix}$.Now,calculate $f(x) \cdot f(-x)$:$f(x) \cdot f(-x) = \begin{bmatrix} \cos x & -\sin x & 0 \ \sin x & \cos x & 0 \ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos x & \sin x & 0 \ -\sin x & \cos x & 0 \ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos^2 x + \sin^2 x & 0 & 0 \ 0 & \sin^2 x + \cos^2 x & 0 \ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} = I$.Since $f(x) \cdot f(-x) = I$,$f(-x)$ is the inverse of $f(x)$. Thus,Statement $I$ is true.To check Statement $II$: Calculate $f(x) \cdot f(y)$:$f(x) \cdot f(y) = \begin{bmatrix} \cos x & -\sin x & 0 \ \sin x & \cos x & 0 \ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos y & -\sin y & 0 \ \sin y & \cos y & 0 \ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos x \cos y - \sin x \sin y & -(\cos x \sin y + \sin x \cos y) & 0 \ \sin x \cos y + \cos x \sin y & \cos x \cos y - \sin x \sin y & 0 \ 0 & 0 & 1 \end{bmatrix}$.Using trigonometric identities $\cos(x+y) = \cos x \cos y - \sin x \sin y$ and $\sin(x+y) = \sin x \cos y + \cos x \sin y$,we get:$f(x) \cdot f(y) = \begin{bmatrix} \cos(x+y) & -\sin(x+y) & 0 \ \sin(x+y) & \cos(x+y) & 0 \ 0 & 0 & 1 \end{bmatrix} = f(x+y)$.Thus,Statement $II$ is true.

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