Prove, using induction, that if n is a positive integer then (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ), where i² = -1 - Leaving Cert Mathematics - Question 4 - 2018

Assume the hypothesis holds for n = k:

$(cos θ + i sin θ)^k = cos(kθ) + i sin(kθ)$

Now, we must show that it holds for n = k + 1:

$(cos θ + i sin θ)^{k+1} = (cos θ + i sin θ)^k (cos θ + i sin θ)$

Using the assumption for k:

$= (cos(kθ) + i sin(kθ))(cos θ + i sin θ)$

Expanding using the distributive property gives:

$= cos(kθ)cos(θ) - sin(kθ)sin(θ) + i (sin(kθ)cos(θ) + cos(kθ)sin(θ))$

This simplifies to:

$= cos((k + 1)θ) + i sin((k + 1)θ)$

Thus, the equation holds for k + 1.

By the principle of mathematical induction, the proposition is true for all positive integers n.

First, convert to polar form:

1. Modulus:

$r = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = 1$

2. Argument:

$\theta = \tan^{-1}\left(\frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}}\right) = \frac{2\pi}{3}$

Thus,

$-\frac{1}{2} + \frac{\sqrt{3}}{2} i = 1(cos(\frac{2\pi}{3}) + i sin(\frac{2\pi}{3}))$

Using De Moivre's theorem:

$(1(cos(\frac{2\pi}{3}) + i sin(\frac{2\pi}{3})))^3 = 1(cos(2\pi) + i sin(2\pi)) = 1$

Therefore, in its simplest form:

$= 1$