Question

solve this please please solve

Answer 1

Answer:

Please refer to attachment .

Step for following the questions as :

1. Let P ( n ) : A

⇒ Aⁿ

2.

For n = 1

3.

For  k + 1 .

We shown all three above condition to prove by Mathematical Induction.

Answer 2

Let,

$\quad$

$\displaystyle\longrightarrow\sf{P(n):A^n=\left [\begin {array}{cc}\sf{\cos (n\theta)}&\sf{\sin (n\theta)}\\\sf{-\sin (n\theta)}&\sf{\cos (n\theta)}\end {array}\right]\ if\ A=\left [\begin {array}{cc}\sf{\cos\theta }&\sf{\sin\theta}\\\sf{-\sin\theta}&\sf{\cos\theta}\end {array}\right]\ \forall n\in\mathbb{N}}$

$\quad$

Consider P(1).

$\quad$

$\displaystyle\longrightarrow\sf{A^1=\left [\begin {array}{cc}\sf{\cos\theta}&\sf{\sin\theta}\\\sf{-\sin\theta}&\sf{\cos\theta}\end {array}\right]}$

$\quad$

P(1) is true. But I consider P(2) too.

$\quad$

$\displaystyle\longrightarrow\sf{A^2=A\cdot A}$

$\quad$

$\displaystyle\longrightarrow\sf{A^2=\left [\begin {array}{cc}\sf{\cos\theta }&\sf{\sin\theta}\\\sf{-\sin\theta}&\sf{\cos\theta}\end {array}\right]\cdot\left [\begin {array}{cc}\sf{\cos\theta }&\sf{\sin\theta}\\\sf{-\sin\theta}&\sf{\cos\theta}\end {array}\right]}$

$\quad$

$\displaystyle\longrightarrow\sf{A^2=\left [\begin {array}{cc}\sf{\cos^2\theta-\sin^2\theta}&\sf{\cos\theta\sin\theta+\sin\theta\cos\theta}\\\sf{-\sin\theta\cos\theta-\cos\theta\sin\theta}&\sf{-\sin^2\theta+\cos^2\theta}\end {array}\right]}$

$\quad$

$\displaystyle\longrightarrow\sf{A^2=\left [\begin {array}{cc}\sf{\cos(2\theta)}&\sf{\sin (2\theta)}\\\sf{-\sin (2\theta)}&\sf{\cos (2\theta)}\end {array}\right]}$

$\quad$

Okay, P(2) is true. Let me assume P(k) is true.

$\quad$

$\displaystyle\longrightarrow\sf{P(k):A^k=\left [\begin {array}{cc}\sf{\cos (k\theta)}&\sf{\sin (k\theta)}\\\sf{-\sin (k\theta)}&\sf{\cos (k\theta)}\end {array}\right]\ if\ A=\left [\begin {array}{cc}\sf{\cos\theta }&\sf{\sin\theta}\\\sf{-\sin\theta}&\sf{\cos\theta}\end {array}\right],\ k\in\mathbb{N}}$

$\quad$

Consider P(k + 1).

$\quad$

$\displaystyle\longrightarrow\sf{P(k+1):A^{k+1}=\left [\begin {array}{cc}\sf{\cos ((k+1)\theta)}&\sf{\sin ((k+1)\theta)}\\\sf{-\sin ((k+1)\theta)}&\sf{\cos ((k+1)\theta)}\end {array}\right]\ if\ A=\left [\begin {array}{cc}\sf{\cos\theta }&\sf{\sin\theta}\\\sf{-\sin\theta}&\sf{\cos\theta}\end {array}\right],\ k\in\mathbb{N}}$

$\quad$

Let's check whether it's true.

$\quad$

$\displaystyle\longrightarrow\sf{A^{k+1}=A^k\cdot A}$

$\quad$

$\displaystyle\longrightarrow\sf{A^{k+1}=\left [\begin {array}{cc}\sf{\cos (k\theta)}&\sf{\sin (k\theta)}\\\sf{-\sin (k\theta)}&\sf{\cos (k\theta)}\end {array}\right]\cdot\left [\begin {array}{cc}\sf{\cos\theta }&\sf{\sin\theta}\\\sf{-\sin\theta}&\sf{\cos\theta}\end {array}\right]}$

$\quad$

$\displaystyle\longrightarrow\sf{A^{k+1}=\left [\begin {array}{cc}\sf{\cos (k\theta)\cos\theta-\sin (k\theta)\sin\theta}&\sf{\cos (k\theta)\sin\theta+\sin (k\theta)\cos\theta}\\\sf{-\sin (k\theta)\cos\theta-\cos (k\theta)\sin\theta}&\sf{-\sin (k\theta)\sin\theta+\cos (k\theta)\cos\theta}\end {array}\right]}$

$\quad$

$\displaystyle\longrightarrow\sf{A^{k+1}=\left [\begin {array}{cc}\sf{\cos (k\theta+\theta)}&\sf{\sin (k\theta+\theta)}\\\sf{-\sin (k\theta+\theta)}&\sf{\cos (k\theta+\theta)}\end {array}\right]}$

$\quad$

$\displaystyle\longrightarrow\sf{A^{k+1}=\left [\begin {array}{cc}\sf{\cos ((k+1)\theta)}&\sf{\sin ((k+1)\theta)}\\\sf{-\sin ((k+1)\theta)}&\sf{\cos ((k+1)\theta)}\end {array}\right]}$

$\quad$

Therefore P(k + 1) is true whenever P(k) is true.

$\quad$

Hence P(n) is true $\displaystyle\sf{\forall n\in\mathbb {N}.}$