Using induction, prove that

Question

(cos x + i snin x)^n = cosnx + i sinnx?

Answer 1

Mathematical induction :

It is a method of mathematical proof typically used to establish a given statement for all natural numbers.

The simplest and most common form of mathematical induction infers that a statement involving a natural number n holds for all values of n. The proof consists of two steps:

1. The basis (base case): prove that the statement holds for the first natural number n. Usually, n = 0 or n = 1, rarely, n = –1 (although not a natural number, the extension of the natural numbers to –1 is still a well-ordered set).
2. The inductive step: prove that, if the statement holds for some natural number n, then the statement holds for n + 1.

Consider  S (n) : (cosx+isinx)ⁿ= cos(nx) +isin(nx)

It can be established by mathematical induction for natural numbers, and extended to all integers from there.

Basis (base case):

n = 1

S (1) :(cosx+i sinx)¹= cos(1x)+isin(1x) = cosx+isinx

S (1) is clearly true

Inductive step : prove that S (n) holds for some natural number n, then S (n) holds for n + 1.

Now, considering S (n+1):

S (n+1) = (cosx+isinx)ⁿ⁺¹ = (cosx+isinx)ⁿ(cosx+isinx)

= [cos(nx) +isin(nx)](cosx+isinx)

= cos(nx)cosx+icos(nx)sinx+isin(nx)cosx+i2sin(nx)sinx

= [cos(nx)cosx−sin(nx)sinx] +i[cos(nx)sinx+ sin(nx)cosx]

= cos(nx+x) +isin(nx+x)

= cos[(n+ 1)x] +isin[(n+ 1)x].

We deduce that S(n) implies S(n+1).

By the principle of mathematical induction it follows that the result is true for all natural numbers n.

(cosx+isinx)ⁿ= cos(nx) +isin(nx) is proved by mathematical induction