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

Starting from the previous result, we need to manipulate the expression:

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

Substituting the sine identity: $x = d(\cos \theta + \cos(2\theta))$ Using the angle addition formula, this can be further manipulated:

Using the identity $\cos(2\theta) = 2\cos^2 \theta - 1$, we get: $x = d(\cos \theta + 2\cos^2 \theta - 1)$

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

To find the maximum value of $x = \frac{9d}{8} - d(\cos \theta - \frac{1}{4})^2$, note that the term $d(\cos \theta - \frac{1}{4})^2 \geq 0$. Thus, the maximum occurs when $\cos \theta = \frac{1}{4}$:

Substituting this back into the equation provides: $x_{max} = \frac{9d}{8}$

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

Given that the maximum x-coordinate has been established, we can use the Pythagorean theorem to find the distance $OQ$. We use: $OQ^2 = d^2 + d^2 - 2d^2\cos \theta$

oSubstituting $\\cos \theta = \frac{1}{4}$ into the formula gives: $OQ^2 = d^2 + d^2 - 2d^2\left(\frac{1}{4}\right) = 2d^2 - \frac{1}{2}d^2 = \frac{3}{2}d^2$

Therefore, we have: $OQ = d\sqrt{\frac{3}{2}} = d\frac{\sqrt{6}}{2}$