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 Mechanics - Question 9 - 2021 - Paper 2

We start from the equation derived previously:

$x = d(\cos \theta + \sin(2\theta - \frac{\pi}{2}))$

Using the sine double angle identity, we substitute:
$\sin(2\theta - \frac{\pi}{2}) = -\cos(2\theta)$

Thus, we rewrite:

$x = d(\cos \theta - \cos(2\theta))$

Now, using the identity for cosine of double angle:
$\cos(2\theta) = 2\cos^2 \theta - 1$

Substituting this into the equation, we have:

$x = d(\cos \theta - (2\cos^2 \theta - 1)) = d(1 + \cos \theta - 2\cos^2 \theta)$

From the equation:
$x = \frac{9d}{8} - d(\cos \theta - \frac{1}{4})^2$

To find the maximum value, we need to minimize the term $d(\cos \theta - \frac{1}{4})^2$ which achieves its minimum when $\, \cos \theta = \frac{1}{4}$.
Substituting this back, the greatest possible value of x becomes:

$x_{max} = \frac{9d}{8} - d(0) = \frac{9d}{8}$.

Using the relation derived from cosine law:

$OQ^2 = d^2 + d^2 - 2d^2 \cos \theta$

This can be rewritten considering $\cos \theta = \frac{1}{4}$:

$OQ^2 = d^2 + d^2 - 2d^2\left(\frac{1}{4}\right) = 2d^2(1 - \frac{1}{4}) = \frac{3d^2}{2}$

Thus, taking the square root finally gives:

$OQ = d\sqrt{\frac{3}{2}} = d\frac{\sqrt{6}}{2}$