Two particles, P and Q, are projected at the same time from a fixed point X, on the ground, so that they travel in the same vertical plane - AQA - A-Level Maths Pure - Question 18 - 2021 - Paper 2

To show that ( \cos 2\theta = \frac{1}{8} ), we start by analyzing the time of flight for both particles.

For particle P, the time of flight ( t_P ) can be determined using:

$$t_P = \frac{u \sin \theta}{g}$$

For particle Q, the time of flight ( t_Q ) is given by:

$$t_Q = \frac{2u \sin 2\theta}{g}$$

Since both particles land at the same point after the same time, we can set ( t_P = t_Q ):

$$\frac{u \sin \theta}{g} = \frac{2u \sin 2\theta}{g}$$

Simplifying gives:

$$\sin \theta = 2 \sin 2\theta$$

Utilizing the identity ( \sin 2\theta = 2 \sin \theta \cos \theta ), the equation becomes:

$$\sin \theta = 2(2 \sin \theta \cos \theta) \Rightarrow 1 = 4 \cos \theta$$

Rearranging the above leads to:

$$\cos \theta = \frac{1}{4}$$

Then, substituting into the cosine double angle formula, we have:

$$\cos 2\theta = 2\cos^2 \theta - 1 = 2\left( \frac{1}{4} \right)^2 - 1 = \frac{1}{8}.$$

Thus, we have shown that ( \cos 2\theta = \frac{1}{8} ).