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If $\mathrm{f}\{\mathrm{x}\}=\left[\begin{array}{lcl}\cos \mathrm{x} & -\sin x & 0 \\ \sin \mathrm{x} & \cos \mathrm{x} & 0 \\ 0 & 0 & 1\end{array}\right]$, then which of the following are
correct ?
$1.$ $\quad \mathrm{f}(\theta) \times \mathrm{f}(\phi)=\mathrm{f}(\theta+\phi)$
$2.$ The value of the determinant of the matrix $\mathrm{f}(\theta) \times \mathrm{f}(\phi)$ is 1 .
$3.$ The determinant of $f(x)$ is an even function.
Select the correct answer using the code given below:
Options:
correct ?
$1.$ $\quad \mathrm{f}(\theta) \times \mathrm{f}(\phi)=\mathrm{f}(\theta+\phi)$
$2.$ The value of the determinant of the matrix $\mathrm{f}(\theta) \times \mathrm{f}(\phi)$ is 1 .
$3.$ The determinant of $f(x)$ is an even function.
Select the correct answer using the code given below:
Solution:
2471 Upvotes
Verified Answer
The correct answer is:
1,2 and 3
$\quad f(\theta)=\left[\begin{array}{ccc}\cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{array}\right] \& f(\phi)\left[\begin{array}{ccc}\cos \phi & -\sin \phi & 0 \\ \sin \phi & \cos \phi & 0 \\ 0 & 0 & 1\end{array}\right]$
$\mathbf{f}(\theta) \times \mathrm{f}(\phi)=\left[\begin{array}{ccc}\cos \theta \cos \phi-\sin \theta \sin \phi & -\cos \theta \sin \phi-\sin \theta \cos \phi & 0 \\ \sin \theta \cos \phi+\cos \theta \sin \phi & -\sin \theta \sin \phi+\cos \theta \cos \phi & 0 \\ 0 & 0 & 1\end{array}\right]$
(using Trigonometric Identities)
$\Rightarrow \quad \mathrm{f}(\theta) \times \mathrm{f}(\phi)=\mathrm{f}(\theta+\phi)$
Also, det, $[\mathrm{f}(\theta) \times \mathrm{f}(\phi)]=1\left[\cos ^{2}(\theta+\phi)-\left(-\sin ^{2}(\theta+\phi)\right)\right]$
$=\cos ^{2}(\theta+\phi)+\sin ^{2}(\theta+\phi)=1 .$
$\& \operatorname{det}(\mathrm{f}(x))=\left(\cos ^{2} x-\left(-\sin ^{2} x\right)=\cos ^{2} x+\sin ^{2} x=1 .\right.$
For $x=-x$
det. $(\mathrm{f}(-x))=\cos ^{2}(-x)+\operatorname{Sin}^{2}(-x)=1$
Hence, det. $(\mathrm{f}(-x))=\operatorname{det}(\mathrm{f}(x))$
Hence, $\operatorname{det} .(\mathrm{f}(\mathrm{x}))$ is even function.
$\mathbf{f}(\theta) \times \mathrm{f}(\phi)=\left[\begin{array}{ccc}\cos \theta \cos \phi-\sin \theta \sin \phi & -\cos \theta \sin \phi-\sin \theta \cos \phi & 0 \\ \sin \theta \cos \phi+\cos \theta \sin \phi & -\sin \theta \sin \phi+\cos \theta \cos \phi & 0 \\ 0 & 0 & 1\end{array}\right]$
(using Trigonometric Identities)
$\Rightarrow \quad \mathrm{f}(\theta) \times \mathrm{f}(\phi)=\mathrm{f}(\theta+\phi)$
Also, det, $[\mathrm{f}(\theta) \times \mathrm{f}(\phi)]=1\left[\cos ^{2}(\theta+\phi)-\left(-\sin ^{2}(\theta+\phi)\right)\right]$
$=\cos ^{2}(\theta+\phi)+\sin ^{2}(\theta+\phi)=1 .$
$\& \operatorname{det}(\mathrm{f}(x))=\left(\cos ^{2} x-\left(-\sin ^{2} x\right)=\cos ^{2} x+\sin ^{2} x=1 .\right.$
For $x=-x$
det. $(\mathrm{f}(-x))=\cos ^{2}(-x)+\operatorname{Sin}^{2}(-x)=1$
Hence, det. $(\mathrm{f}(-x))=\operatorname{det}(\mathrm{f}(x))$
Hence, $\operatorname{det} .(\mathrm{f}(\mathrm{x}))$ is even function.
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