Question

(cosx+isinx) ^n=(cosnx+isinnx)
Prove it using the principle of mathematical induction.

Answer 1

Answer:

Step-by-step explanation:

I'm assuming that your 'n' is meant to represent an integer.

DeMoivre's Theorem is true even if n is a complex number (has a real

part and possibly an imaginary part), but when n is an integer we

can prove the formula easily by using some basic trigonometry.

We will prove DeMoivre's Theorem for any positive integer n by

induction.

First, the theorem is clearly true for n = 1, since then both sides

of the equation are the same quantity. So now we will assume that the

theorem has been shown true for n = k. We want to show that it is

true for n = k + 1.

We calculate:

[cos(x) + i sin(x)]^(k+1) =

[cos(x) + i sin(x)]^k * [cos(x) + i sin(x)] (split up factors)

= [cos(kx) + i sin(kx)] * [cos(x) + i sin(x)]

(induction hypothesis)

= cos(kx)cos(x) - sin(kx)sin(x) + i[sin(kx)cos(x) +

cos(kx)sin(x)] (multiply out)

= cos(kx + x) + i sin(kx + x) (use trig formulas)

= cos((k+1)x) + i sin((k+1)x).

(add the angles)

This proves the inductive step, so DeMoivre's Theorem has been shown

to be true for all positive integers n.

To show the theorem is also true for negative integers, we just

compute (still assuming that n itself is positive)

[cos(x) + i sin(x)]^(-n) = [{cos(x) + i sin(x)}^(-1)]^n

= [cos(x) - i sin(x)]^n

= [cos(-x) + i sin(-x)]^n

= cos(-nx) + i sin(-nx).

So DeMoivre's Theorem is true for all integers, positive and negative.

I don't know if you know this already, but both the sine and cosine

functions can be expressed in terms of the exponential function.

Namely,

cos(x) = [e^(ix) + e^(-ix)]/2

sin(x) = [e^(ix) - e^(-ix)]/2i,

from which we get

e^(ix) = cos(x) + i sin(x).

This is true for any complex number x. This is the usual way we state

DeMoivre's formula. We see that

[cos(x) + i sin(x)]^n = [e^(ix)]^n and

cos(nx) + i sin(nx) = e^(inx),

and the two righthand sides above are clearly equal. So this an easier

way of proving the theorem that you stated, although first you have to

know about the exponential function.

A fundamental and truly amazing fact that comes out of this is

e^(i*pi) = -1,

which you get by substituting x = pi in DeMoivre's formula.

Hope it helps u mark as brainliest please

Answer 2

( cos x + i sin x )^n = ( cos n x+ i sin n x )

Let P(n) = ( cos x + i sin x )^n = ( cos n x+ i sin n x )

P(1) = ( cos x + i sin x ) = cos x + i sin x

∴ P (1) = true .

Let P(n) be true for all natural numbers .

P( n + 1 ) = ( cos x + i sin x )^(n + 1 )

Use a^( n + 1 ) = a^n × a

⇒ ( cos x + i sin x )^n ( cos x + i sin x )

Use ( cos x + i sin x )^n = ( cos n x+ i sin n x )

⇒ ( cos x + i sin x )( cos n x + i sin n x )

⇒ cos x cos ( n x ) + i cos x sin x + i sin x cos ( n x ) + i² sin x sin ( n x )

⇒ cos x cos ( n x ) + i cos x sin x + i sin x cos ( n x ) - sin x sin ( n x )

⇒ cos( x )cos( n x ) + i sin ( x + n x ) - sin x sin( n x )  

⇒ cos( x ) cos( n x ) - sin ( x ) sin ( n x) + i sin ( x ( n + 1) )  

⇒ cos( x n + x ) + i sin ( x ( n + 1 ) )

⇒ cos( ( n + 1)x ) + i sin (n + 1 )x

Hence P ( n + 1 ) = true .

Proved !