Question

$(cosθ+i sinθ)^{n}$=cos(nθ)+i sin(nθ),i=$\sqrt-1}$

Answer 1

Question:-

Prove by Principle of Mathematical Induction that $\displaystyle\sf {(\cos\theta+i\sin\theta)^n=\cos (n\theta)+i\sin (n\theta)\ \forall n\in\mathbb{N}}$ where $\displaystyle\sf {i=\sqrt{-1}.}$

Solution:-

Let,

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

Consider P(1).

$\displaystyle\longrightarrow\sf {LHS}$

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

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

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

$\displaystyle\longrightarrow\sf {RHS}$

So P(1) is true. Hence assume P(k) is true.

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

Consider P(k + 1).

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

Let's check whether it's true.

$\displaystyle\longrightarrow\sf {LHS}$

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

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

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

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

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

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

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

$\displaystyle\longrightarrow\sf {RHS}$

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

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