Question

If $f\left(x\right)=\left[\begin{array}{ccc}\cos x& -\sin x& 0\\ \sin x& \cos x& 0\\ 0& 0& 1\end{array}\right]$ then f(x+y) is equal to

• A. f(x) + f(y)
• B. f(x) - f(y)
• C. f(x) . f(y)
• D. None of these

Answer 1

The correct option is C f(x) . f(y)

$f\left(x\right)=\left[\begin{array}{ccc}cosx& -sinx& 0\\ sinx& cosx& 0\\ 0& 0& 1\end{array}\right]$
$f(x+y)=\left[\begin{array}{ccc}cos(x+y)& -sin(x+y)& 0\\ sin(x+y)& cos(x+y)& 0\\ 0& 0& 1\end{array}\right]$
$=\left[\begin{array}{ccc}cosxcosy-sinxsiny& -sinxcosy-cosxsiny& 0\\ sinxcosy+cosxsiny& cosycosy-sinxsiny& 0\\ 0& 0& 1\end{array}\right]$
$\left[\begin{array}{ccc}cosx& -sinx& 0\\ sinx& cosx& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}cosy& -siny& 0\\ siny& cosy& 0\\ 0& 0& 1\end{array}\right]$