(a) (i) Prove that cos 2A = cos^2 A - sin^2 A - Leaving Cert Mathematics - Question 4 - 2021

Given sin^2 θ = rac{1}{3}, we can find $cos θ$ using the Pythagorean identity:

cos^2 θ = 1 - sin^2 θ = 1 - rac{1}{3} = rac{2}{3}.

Taking the square root, we find:

cos θ = ±rac{ oot{2}}{1}.

Since $0 ≤ θ ≤ π$, $cos θ$ must be positive in this range. Therefore, we have:

cos θ = rac{ oot{6}}{3}.

We know that $tan(60°) = √3$, so we can set up the equation:

$$tan(θ + 150°) = - √3.$$

This implies:

$$θ + 150° = 180° + 60° ext{ or } θ + 150° = 360° + 60°.$$

Solving the first case:

$$θ + 150° = 240° \implies θ = 90°.$$

For the second case:

$$θ + 150° = 420° \implies θ = 270°.$$

Thus, the solutions are:

$$θ = 90°, 270°.$$