Let A= cosα -sinα sinα cosα , a∈R such that A^32= 0 -1 1 0 . Then, a value of α is:

Correct answer: (C) (π)/64. ∵ A= cosα -sinα sinα cosα ⇒A^2= cos2α -sin2α sin2α cos2α ⇒A^3 = cos3α -sin3α sin3α cos3α Similarly, A^32 = cos32α -sin32α sin32α cos32α Given, A^32 = 0 -1 1 0 ⇒ cos32α -sin32α sin32α cos32α = 0 -1 1 0 On comparing cos32α=0 & sin32α=1, we get, α=(π)/64

Q41 / 90 · JEE Main 2019 (08 Apr Shift 1) · Mathematics

Let $A = [\begin{matrix} \cos α & - \sin α \\ \sin α & \cos α \end{matrix}], (a \in R)$ such that $A^{32} = [\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}] .$ Then, a value of $α$ is:

A
$0$
B
$\frac{π}{16}$
Correct answer
C
$\frac{π}{64}$
D
$\frac{π}{32}$

Solution

$∵ A = [\begin{matrix} \cos α & - \sin α \\ \sin α & \cos α \end{matrix}]$

${\Rightarrow A}^{2} = [\begin{matrix} \cos 2 α & - \sin 2 α \\ \sin 2 α & \cos 2 α \end{matrix}]$

$\Rightarrow A^{3} = [\begin{matrix} \cos 3 α & - \sin 3 α \\ \sin 3 α & \cos 3 α \end{matrix}]$

Similarly, $A^{32} = [\begin{matrix} \cos 32 α & - \sin 32 α \\ \sin 32 α & \cos 32 α \end{matrix}]$
Given,

$A^{32} = [\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}]$

$\Rightarrow [\begin{matrix} \cos 32 α & - \sin 32 α \\ \sin 32 α & \cos 32 α \end{matrix}] = [\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}]$

On comparing $\cos 32 α = 0$ & $\sin 32 α = 1$ , we get,
$α = \frac{π}{64}$

Hence, the correct answer is option: 'C'.

Let A=cosα-sinαsinαcosα, a∈R such that A32=0-110. Then, a value of α is:

Solution

Relevant Questions

Let $A = [\begin{matrix} \cos α & - \sin α \\ \sin α & \cos α \end{matrix}], (a \in R)$ such that $A^{32} = [\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}] .$ Then, a value of $α$ is: