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 that the proposition holds for n = k:

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

We will now show it holds for k + 1.

For n = k + 1:

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

Substituting the assumption:

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

Using the trigonometric identities:

Thus, the proposition is true for n = k + 1 if it is true for n = k. Therefore, by the principle of mathematical induction, the formula holds true for all positive integers n.

To solve:

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

Where r is the modulus and ( \theta ) is the argument.

Calculate r:

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

Next, find the argument:

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

Thus,

$\left(-\frac{1}{2} + \frac{\sqrt{3}}{2} i \right)^{3} = 1 (cos(2\pi) + i sin(2\pi)) = 1 + 0i = 1$