Question

Prove by mathematical induction (cosθ + i sinθ)n = (cosnθ +isinnθ) , where i = √-1​

Answer 1

Answer:

Step-by-step explanation:

$\tt{Let\,\,P_{n}:(cos(\theta)+i\,sin(\theta))^n=cos(n\theta)+i\,sin(n\theta)}$

For n=1, $P_{1}$ is true.

For n=2,

$\sf{(cos(\theta)+i\,sin(\theta))^2=cos^2(\theta)-sin^2(\theta)+i\cdot\,2sin(\theta)cos(\theta)=cos(2\theta)+i\,sin(2\theta)}$

So, $P_{2}$, is true

Let $P_{k}$, is true

Then, $P_{k+1}$, must be true

Now,

$\tt{(cos(\theta)+i\,sin(\theta))^{k+1}}$

$\tt{=(cos(\theta)+i\,sin(\theta))^{k}\cdot\,(cos(\theta)+i\,sin(\theta))}$

$\tt{=(cos(k\theta)+i\,sin(k\theta))\cdot\,(cos(\theta)+i\,sin(\theta))}$

$\tt{=cos(k\theta)cos(\theta)+i\,sin(k\theta)(cos(\theta)+i\,cos(k\theta)sin(\theta)-sin(k\theta)sin(\theta)}$

$\tt{=\{cos(k\theta)cos(\theta)-sin(k\theta)sin(\theta)\}+i\,\{sin(k\theta)(cos(\theta)+i\,cos(k\theta)sin(\theta)\}}$

$\tt{=cos(k\theta+\theta)+i\,sin(k\theta+\theta)}$

$\tt{=cos(k+1)\theta+i\,sin(k+1)\theta}$

So, $P_{k+1}$, is true, whenever $P_{k}$ is true.

Answer 2

$\large\underline{\sf{Solution-}}$

Let assume that,

$\red{\rm :\longmapsto\:P(n): {(cos\theta + isin\theta)}^{n} = cos \: n\theta + isin \: n\theta}$

Step :- 1 For n = 1

$\rm :\longmapsto\: {(cos \: \theta + i \: sin \: \theta)}^{1} = cos\theta + i \: sin\theta$

$\rm :\longmapsto\: {cos \: \theta + i \: sin \: \theta} = cos\theta + i \: sin\theta$

$\bf\implies \:\boxed{ \tt{ \: P(n) \: is \: true \: for \: n = 1 \: }}$

Step :- 2 Assume that P (n) is true for n = k, where k is some natural number

So,

$\red{\rm :\longmapsto\:{(cos\theta + isin\theta)}^{k} = cos \: k\theta + isin \: k\theta}$

Step :- 3 Now, we have to prove that P(n) is true for n = k + 1

$\rm :\longmapsto\:P(k + 1)$

$\rm \: = \: {(cos\theta + i \: sin\theta)}^{k + 1}$

can be rewritten as

$\rm \: = \: {(cos\theta + i \: sin\theta)}^{k} \times (cos\theta + i \: sin\theta)$

$\rm \: = \:(cosk\theta + i \: sink\theta)(cos\theta + i \: sin\theta)$

$\rm \: = \:cosk\theta \: cos\theta + i \: cosk\theta \: sin\theta + i \: sink\theta \: cos\theta + {i}^{2} sink\theta \: sin\theta$

$\rm \: = \:cosk\theta \: cos\theta + i (\: cosk\theta \: sin\theta + sink\theta \: cos\theta) - sink\theta \: sin\theta$

can be rewritten as

$\rm \: = \:cosk\theta \: cos\theta - sink\theta \: sin\theta + i (\: cosk\theta \: sin\theta + sink\theta \: cos\theta)$

We know

$\boxed{ \tt{ \: cos(x + y) = cosxcosy - sinxsiny \: \: }}$

and

$\boxed{ \tt{ \: sin(x + y) = sinxcosy + sinycosx \: \: }}$

So, using this

$\rm \: = \:cos(k\theta + \theta) + i \: sin(k\theta + \theta)$

$\rm \: = \:cos(k + 1)\theta + i \: sin(k + 1)\theta$

$\bf\implies \:\boxed{ \tt{ \: P(n) \: is \: true \: for \: n = k + 1 \: }}$

Hence, By the process of Principal of Mathematical Induction

$\red{\boxed{ \tt{ \: \:{(cos\theta + isin\theta)}^{n} = cos \: n\theta + isin \: n\theta}}}$