Prove that the equation of the normal at a point of the curve \[x=n\cos t-\cos nt,\qquad y=n\sin t-\sin nt,\] where \(n\) is an integer greater than unity, is \[x\cos \frac{n+1}{2}t + y\sin \frac{n+1}{2}t=(n-1)\cos \frac{n-1}{2}t.\]

How do we calculate the gradient when \(x\) and \(y\) are given parametrically? How do we then calculate the gradient of the normal?

The equation we are aiming for is written in terms of \(\dfrac{n\pm 1}{2}t\) which suggests that the following identities might be useful: \[\begin{align*} \sin a-\sin b&= 2\sin\left(\frac{a-b}{2}\right)\cos\left(\frac{a+b}{2}\right)\\ \cos a-\cos b&= 2\sin\left(\frac{a+b}{2}\right)\sin\left(\frac{b-a}{2}\right) \\ \cos a \cos b&= \frac{1}{2}\cos(a-b)+\frac{1}{2}\cos(a+b) \\ \sin a \sin b&= \frac{1}{2}\cos (a-b) -\frac{1}{2}\cos(a+b) \end{align*}\]

Show that, if \(n\) is an even integer, the normals at the points \(t\) and \(t+\pi\) are perpendicular and intersect on the circle \[x^2+y^2=(n-1)^2.\]

This applet illustrates what is described here. Use the sliders to vary \(t\) and try different values of \(n\).

How would we show that two lines are perpendicular?

You might find it helpful to define a shorthand such as \(A=\dfrac{n+1}{2}t\) and \(P=\dfrac{n+1}{2}\pi\).

We could write down the equations of the two normals from the first part of the question. What might we need to do to them to get them in the form of the required circle equation?