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Basic Inverse Trigonometric Functions

Let Q = [ c...

Question

Let $Q = ⎡ ⎢ ⎢ ⎣ \begin{matrix} cos \frac{π}{4} & - sin \frac{π}{4} sin \frac{π}{4} & cos \frac{π}{4} \end{matrix} ⎤ ⎥ ⎥ ⎦$ and $x = ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦$ then $Q^{3} x$ is equal to

A

[\begin{matrix} 01 \end{matrix}]

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B

⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} - \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦

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C

[\begin{matrix} - 1 0 \end{matrix}]

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D

⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦

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Solution

The correct option is B $[\begin{matrix} - 1 0 \end{matrix}]$
Given $Q = ⎡ ⎢ ⎢ ⎣ \begin{matrix} cos \frac{π}{4} & - sin \frac{π}{4} sin \frac{π}{4} & cos \frac{π}{4} \end{matrix} ⎤ ⎥ ⎥ ⎦ = \frac{1}{\sqrt{2}} [\begin{matrix} 1 & - 1 1 & 1 \end{matrix}]$ and $x = ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ = \frac{1}{\sqrt{2}} [\begin{matrix} 11 \end{matrix}]$

$\Rightarrow Q^{3} = \frac{1}{2 \sqrt{2}} [\begin{matrix} 1 & - 1 1 & 1 \end{matrix}] [\begin{matrix} 1 & - 1 1 & 1 \end{matrix}] [\begin{matrix} 1 & - 1 1 & 1 \end{matrix}]$

$= \frac{1}{2 \sqrt{2}} [\begin{matrix} 0 & - 2 2 & 0 \end{matrix}] [\begin{matrix} 1 & - 1 1 & 1 \end{matrix}] = \frac{1}{\sqrt{2}} [\begin{matrix} - 1 & - 1 1 & - 1 \end{matrix}]$

$\Rightarrow Q^{3} x = \frac{1}{2} [\begin{matrix} - 1 & - 1 1 & - 1 \end{matrix}] [\begin{matrix} 11 \end{matrix}] = \frac{1}{2} [\begin{matrix} - 1 - 1 1 - 1 \end{matrix}] = [\begin{matrix} - 1 0 \end{matrix}]$

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Similar questions

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}

f (x) = 2 x + 1; x \leq 1

= x^{2} + 2; 1 < x \leq 2

= 4 x^{2} + 2; x > 2

then number of points where

f (x)

is not differentiable is

$A = ⎡ ⎢ ⎣ \begin{matrix} \frac{1}{\sqrt{2}} & \frac{- 1}{\sqrt{2}} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{matrix} ⎤ ⎥ ⎦$ is an orthogonal matrix

Q. The least positive integer

n

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

is an identity matrix of order

2

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