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

We already have the equation from part (a):

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

Using the identity for the sine function, we know that:

$\sin(2\theta - \frac{\pi}{2}) = -\cos(2\theta).$

Then substituting this in gives us:

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

We also know that:

$\cos(2\theta) = 2\cos^2 \theta - 1,$

which leads us to:

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

To find the greatest possible value of x in the equation:

$x = \frac{9d}{8} - d(\cos \theta - \frac{1}{4})^2,$

we need to recognize that the expression $(\cos \theta - \frac{1}{4})^2$ reaches its minimum when $\cos \theta = \frac{1}{4},$ giving:

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

Hence, the greatest possible value of x is rac{9d}{8}, and the corresponding value of \cos \theta is rac{1}{4}.

From the geometry of the arm in Figure 3, we use the law of cosines to calculate the distance OQ. We know:

$OQ^2 = d^2 + d^2 - 2d \cdot d \cdot \cos(\theta).$

This gives:

$OQ^2 = 2d^2(1 - \cos(\theta)).$

Substituting in the value of \cos(\theta) = \frac{1}{4}, we find:

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

Therefore, the distance OQ is:

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