Problem Statement
Let $$z = \cos\theta + i\sin\theta$$.\n\n(a) Show that $$z^n + z^{-n} = 2\cos(n\theta)$$ and $$z^n - z^{-n} = 2i\sin(n\theta)$$. [2 marks]\n\n(b) By expanding $$\left(z + \frac{1}{z}\right)^4$$, prove that $$\cos^4\theta = \frac{1}{8}(\cos 4\theta + 4\cos 2\theta + 3)$$. [5 marks]
Verified Solution & Marking Scheme
Part (a): Apply De Moivre
$$z^n = \cos(n\theta) + i\sin(n\theta)$$ and $$z^{-n} = \cos(-n\theta) + i\sin(-n\theta) = \cos(n\theta) - i\sin(n\theta)$$. Adding gives $$2\cos(n\theta)$$; subtracting gives $$2i\sin(n\theta)$$.
Part (b): Binomial Expansion
$$\left(z + \frac{1}{z}\right)^4 = z^4 + 4z^2 + 6 + \frac{4}{z^2} + \frac{1}{z^4} = \left(z^4 + \frac{1}{z^4}\right) + 4\left(z^2 + \frac{1}{z^2}\right) + 6$$
Substitute trig identities
LHS: $$(2\cos\theta)^4 = 16\cos^4\theta$$.\nRHS: $$2\cos(4\theta) + 4(2\cos(2\theta)) + 6 = 2\cos(4\theta) + 8\cos(2\theta) + 6$$.
Divide by 16
$$\cos^4\theta = \frac{2\cos(4\theta) + 8\cos(2\theta) + 6}{16} = \frac{1}{8}(\cos 4\theta + 4\cos 2\theta + 3)$$ (AG)