A robotic arm which is attached to a flat surface at the origin O, is used to draw a graphic design - AQA - A-Level Maths Pure - Question 9 - 2021 - Paper 2

Using the double angle identity, we can express \cos(2\theta) in terms of \cos(\theta): $\cos(2\theta) = 2\cos^{2}(\theta) - 1$

Substituting this back into the equation gives us: $x = d \cos \theta - d(2\cos^{2}(\theta) - 1)$

Simplifying further results in: $x = d(1 + \cos \theta - 2\cos^{2}(\theta))$

This shows the relationship as required.

The expression for x is given as: $x = \frac{9d}{8} - d \left(\cos \theta - \frac{1}{4}\right)^{2}$

To find the maximum, we set the squared term to zero, which occurs when: $\cos \theta = \frac{1}{4}$

Substituting this back, we find: $x = \frac{9d}{8}$

Thus, the greatest possible value of x is \frac{9d}{8} when \cos \theta = \frac{1}{4}.

The distance OQ can be calculated using the triangle formed by the points O, P, and Q. Since we know that OP = PQ = d, we may find OQ using the Pythagorean theorem:

Let\s angle \theta at point O: $OQ^{2} = d^{2} + d^{2} - 2d^{2}\cos \theta$

distances yields: $OQ^{2} = 2d^{2}(1 - \cos \theta) = 2d^{2} \sin^{2}\left(\frac{\theta}{2}\right)$

When \cos \theta = \frac{1}{4}, substituting we find: $OQ = d \sqrt{2(1 - \frac{1}{4})} = d \sqrt{\frac{3}{2}} = \frac{d\sqrt{6}}{2}$

Thus, the exact distance OQ is computed.