91Ó°ÊÓ

Using mathematical induction, let's prove that de Moivre's formula holds true for positive integer exponents. We first prove it for \(n = 1\): \((\cos x + i\sin x)^1 = \cos x + i\sin x = \cos(1x) + i\sin(1x)\). Now let's assume that the formula holds for some positive integer \(n\): \((\cos x + i\sin x)^n = \cos(nx) + i\sin(nx)\).

We need to show that the formula is true for \(n+1\). To do this, we can multiply both sides of our induction assumption by \((\cos x + i\sin x)\): \((\cos x + i\sin x)^{n+1} = (\cos x + i\sin x)(\cos(nx) + i\sin(nx))\). Now, we can use the trigonometric product-sum identities: \(\cos(x+y) = \cos x\cos y - \sin x\sin y\) and \(\sin(x+y) = \sin x\cos y + \cos x\sin y\). We are left with: \((\cos x + i\sin x)^{n+1} = (\cos x\cos(nx) - \sin x\sin(nx)) + i(\sin x\cos(nx) + \cos x\sin(nx)) = \cos((n+1)x) + i\sin((n+1)x)\). By induction, de Moivre's formula holds for all positive integers \(n\).

Now we need to prove that de Moivre's formula holds for negative integer exponents. If \(n\) is a negative integer, we can write it as \(n = -m\), where \(m\) is a positive integer. Hence, we have: \((\cos x + i\sin x)^{-m} = \frac{1}{(\cos x + i\sin x)^m}\). But we know that de Moivre's formula holds for positive integer exponents from our induction proof in steps 1 and 2, so we can use that: \(\frac{1}{(\cos x + i\sin x)^m} = \frac{1}{(\cos(mx) + i\sin(mx))}\). Next, we can multiply both of these by its complex conjugate to rationalize the denominator: \(\frac{\cos(-mx) - i\sin(-mx)}{\cos^2(mx) + \sin^2(mx)}\). Using the Pythagorean identity, \(\cos^2(mx) + \sin^2(mx) = 1\), we get: \(\frac{\cos(-mx) - i\sin(-mx)}{1} = \cos(-mx) + i\sin(-mx)\). Thus, de Moivre's formula holds for negative integer exponents as well, completing the proof.