If P- cos- 4 -sin- 4- sin- 4 cos- 4 - and X- 1 - 2 - 1 - 2 - Then- P3X is equal to

The correct option is B [ 1; 0 ] Given #160; P=[ cosπ #160; 4 #160; #160; amp; sinπ #160; 4 #160; ...

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Inverse of a Function

If P=[ cosπ...

Question

If $P = ⎛ ⎜ ⎜ ⎝ \begin{matrix} cos \frac{π}{4} & - sin \frac{π}{4} sin \frac{π}{4} & cos \frac{π}{4} \end{matrix} ⎞ ⎟ ⎟ ⎠$ and $X = ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \begin{matrix} \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} \end{matrix} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠$ . Then, $P^{3} X$ is equal to

(\begin{matrix} 01 \end{matrix})

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⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \begin{matrix} - \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} \end{matrix} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠

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(\begin{matrix} - 1 0 \end{matrix})

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⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \begin{matrix} - \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} \end{matrix} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠

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Q = ⎛ ⎜ ⎜ ⎝ \begin{matrix} cos \frac{π}{4} & - sin \frac{π}{4} sin \frac{π}{4} & cos \frac{π}{4} \end{matrix} ⎞ ⎟ ⎟ ⎠

x = ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \begin{matrix} \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} \end{matrix} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠

Q^{3} x

sin \frac{π}{6} cos 0 + sin \frac{π}{4} cos \frac{π}{4} + sin \frac{π}{3} cos \frac{π}{6} = \frac{7}{4}

{sin}^{2} \frac{π}{6} + {cos}^{2} \frac{π}{3} - {tan}^{2} \frac{π}{4} = - \frac{1}{2}

{⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{sin \frac{π}{6} + i (1 - cos \frac{π}{6})}{sin \frac{π}{6} - i (1 - cos \frac{π}{6})} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦}^{2}

i = \sqrt{- 1}

n

{⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} cos \frac{π}{4} & sin \frac{π}{4} - sin \frac{π}{4} & cos \frac{π}{4} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦}^{n}

2

A + B + C = π

s i n \frac{A}{2} + s i n \frac{B}{2} + s i n \frac{C}{2} - 1 = 4 s i n \frac{π - 4}{4} . s i n \frac{π - B}{4} . s i n \frac{π - C}{4}

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