Question

solve the 4th one using mathematical induction ​

Answer 1

We have,

$\displaystyle\longrightarrow\sf{P(n):(\cos\theta+i\sin\theta)^n=\cos(n\theta)+i\sin(n\theta)\quad \forall n\in\mathbb{N}}$

Consider P(1).

$\displaystyle\longrightarrow\sf{(\cos\theta+i\sin\theta)^1=\cos(1\cdot\theta)+i\sin(1\cdot\theta)}$

$\displaystyle\longrightarrow\sf{\cos\theta+i\sin\theta=\cos\theta+i\sin\theta}$

We see P(1) is true. Hence assume that P(k) is true.

$\displaystyle\longrightarrow\sf{P(k):(\cos\theta+i\sin\theta)^k=\cos(k\theta)+i\sin(k\theta)\quad \forall k\in\mathbb{N}}$

Consider P(k + 1).

$\displaystyle\longrightarrow\sf{P(k+1):(\cos\theta+i\sin\theta)^{k+1}=\cos((k+1)\theta)+i\sin((k+1)\theta)\quad \forall k\in\mathbb{N}}$

Let's check whether P(k + 1) is true.

$\displaystyle\longrightarrow\sf{LHS}$

$\displaystyle\longrightarrow\sf{(\cos\theta+i\sin\theta)^{k+1}}$

As $\sf{a^{m+n}=a^m\cdot a^n,}$

$\displaystyle\longrightarrow\sf{(\cos\theta+i\sin\theta)^k\cdot(\cos\theta+i\sin\theta)}$

By P(k),

$\displaystyle\longrightarrow\sf{(\cos(k\theta)+i\sin(k\theta))\,(\cos\theta+i\sin\theta)}$

$\displaystyle\longrightarrow\sf{\cos(k\theta)\cos\theta+i\sin\theta\cos(k\theta)+i\sin(k\theta)\cos\theta-\sin(k\theta)\sin\theta}$

$\displaystyle\longrightarrow\sf{[\cos(k\theta)\cos\theta-\sin(k\theta)\sin\theta]+i[\,\sin\theta\cos(k\theta)+\cos\theta\sin(k\theta)\,]}$

As $\sf{\cos A\cos B-\sin A\sin B=\cos(A+B)}$ and $\sf{\sin A\cos B+\cos A\sin B=\sin(A+B),}$

$\displaystyle\longrightarrow\sf{\cos(k\theta+\theta)+i\sin(k\theta+\theta)}$

$\displaystyle\longrightarrow\sf{\cos((k+1)\theta)+i\sin((k+1)\theta)}$

$\displaystyle\longrightarrow\sf{RHS}$

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

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

Hence Proved!