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 \[x\cos \frac{n+1}{2}t + y\sin \frac{n+1}{2}t=(n-1)\cos \frac{n-1}{2}t.\]

First we need the gradient of the curve and from that the gradient of the normal. We use the chain rule to differentiate the parametric equation.

We have that \[ x=n\cos t-\cos nt, \] so \[ \frac{dx}{dt}=-n\sin t+n\sin nt \] and \[ y=n\sin t-\sin nt, \] so \[ \frac{dy}{dt}=n\cos t-n\cos nt. \] Now we could find the gradient of the curve since \[ \frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}. \]

The form for the answer we are trying to prove is quite different, so it’s worth having a think before ploughing on.

We notice that the form of the answer contains arguments of \(\frac{n+1}{2}t\) or \(\frac{nt+t}{2}\) which points us toward the following identities

\[\begin{align} \sin a-\sin b&= 2\sin\left(\frac{a-b}{2}\right)\cos\left(\frac{a+b}{2}\right)\label{trig1} \\ \cos a-\cos b&= 2\sin\left(\frac{a+b}{2}\right)\sin\left(\frac{b-a}{2}\right) \label{trig2} \\ \cos a \cos b&= \frac{1}{2}\cos(a-b)+\frac{1}{2}\cos(a+b) \label{trig3} \\ \sin a \sin b&= \frac{1}{2}\cos (a-b) -\frac{1}{2}\cos(a+b) \label{trig4} \end{align}\]

Alternatively, we could use the more familiar compound angle identities. Then some slightly different algebra would lead to the same result.

Applying identity \(\eqref{trig1}\), we have that \[ \frac{dx}{dt}=2n\sin\left(\frac{n-1}{2}t\right)\cos\left(\frac{n+1}{2}t\right). \]

Applying identity \(\eqref{trig2}\) yields \[ \frac{dy}{dt}=2n\sin\left(\frac{n+1}{2}t\right)\sin\left(\frac{n-1}{2}t\right). \]

and so \[ \frac{dy}{dx}=\frac{2n\sin\left(\frac{n+1}{2}t\right)\sin\left(\frac{n-1}{2}t\right)}{2n\sin\left(\frac{n-1}{2}t\right)\cos\left(\frac{n+1}{2}t\right)}=\frac{\sin \left(\frac{n+1}{2}t\right)}{\cos \left(\frac{n+1}{2}t\right)}. \] Therefore the gradient of the normal to the curve is \[\begin{equation} -\frac{\cos \left(\frac{n+1}{2}t\right)}{\sin \left(\frac{n+1}{2}t\right)} \label{eq:grad-of-norm} \end{equation}\] and the normal to the curve has equation \[ y-\left( n\sin t -\sin nt \right)=-\frac{\cos \left(\frac{n+1}{2}t\right)}{\sin \left(\frac{n+1}{2}t\right)}\bigl(x-\left(n\cos t-\cos nt\right)\bigr). \] Multiplying by \(\sin\left(\frac{n+1}{2}t\right)\) gives \[ \sin\left(\frac{n+1}{2}t\right)y-\sin\left(\frac{n+1}{2}t\right)\left( n\sin t -\sin nt \right)=\quad \] \[ \quad-\cos\left(\frac{n+1}{2}t\right)x+\cos\left(\frac{n+1}{2}t\right)\left(n\cos t-\cos nt\right) \] and rearranging gives us \[\begin{align*} \sin\left(\frac{n+1}{2}t\right)y+\cos\left(\frac{n+1}{2}t\right)x &= \sin\left(\frac{n+1}{2}t\right)\left( n\sin t -\sin nt \right) \\ &+\cos\left(\frac{n+1}{2}t\right)\left(n\cos t-\cos nt\right). \end{align*}\] Multiplying out the brackets and applying identities \(\eqref{trig3}\) and \(\eqref{trig4}\) we find that \[\begin{align*} \sin\left(\frac{n+1}{2}t\right)y+\cos\left(\frac{n+1}{2}t\right)x&=n\cos t\cos\left(\frac{n+1}{2}t\right)-\cos nt\cos\left(\frac{n+1}{2}t\right)\\ &\quad+n\sin t\sin\left(\frac{n+1}{2}t\right)-\sin nt\sin\left(\frac{n+1}{2}t\right) \\ &=\frac{n}{2}\left[\cos\left(\frac{n-1}{2}t\right)+\cos\left(\frac{n+3}{2}t\right)\right] \\ &\quad -\frac{1}{2}\left[\cos\left(\frac{n-1}{2}t\right)+\cos\left(\frac{3n+1}{2}t\right)\right] \\ &\quad +\frac{n}{2}\left[\cos\left(\frac{n-1}{2}t\right)-\cos\left(\frac{n+3}{2}t\right)\right] \\ &\quad -\frac{1}{2}\left[\cos\left(\frac{n-1}{2}t\right)-\cos\left(\frac{3n+1}{2}t\right)\right] \\ &=(n-1)\cos\left(\frac{n-1}{2}t\right), \end{align*}\]

which is exactly what we wanted.

Show that, if \(n\) is an even integer, the normals at the points \(t\) and \(t+\pi\) are perpendicular…

From \(\eqref{eq:grad-of-norm}\) we have that the gradient of the normal to the curve at \(t\) is \[ m_N(t) = -\frac{\cos \left(\frac{n+1}{2}t\right)}{\sin \left(\frac{n+1}{2}t\right)}. \]

Let’s use a shorthand \(A=\dfrac{n+1}{2}t\) and \(P=\dfrac{n+1}{2}\pi\). Then we can write the gradients of the normals at \(t\) and \(t+\pi\) as \[ m_N(t) = -\frac{\cos A}{\sin A} \quad\text{and}\quad m_N(t+\pi) = -\frac{\cos (A+P)}{\sin (A+P)}. \]

Multiplying these two gradients together (we hope to get \(-1\)) and then using identities \(\eqref{trig3}\) and \(\eqref{trig4}\), we have \[\begin{align*} m_N(t) \times m_N(t+\pi) &= \frac{\cos A \cos (A+P)}{\sin A \sin (A+P)} \\ &= \frac{\cos(-P)+\cos(2A+P)}{\cos(-P)-\cos(2A+P)} \end{align*}\]

Since \(n\) is an even integer, \(P=\frac{\pi}{2}, \frac{3\pi}{2}, \cdots\) so \(\cos(-P) = 0\). The fraction therefore cancels down, the product of the two gradients is indeed \(-1\) and the normals are perpendicular, as required.

…and intersect on the circle \[x^2+y^2=(n-1)^2.\]

Using our shorthand \(A\) and \(P\) from above and also defining \(B=\dfrac{n-1}{2}t\), we can write the equations of the two normals as \[\begin{align} x\cos A + y\sin A &= (n-1)\cos B \label{eq:n_t}\\ \text{and}\quad x\cos (A+P) + y\sin (A+P) &= (n-1)\cos (B+P-\pi) \label{eq:n_t+pi} \end{align}\]

The form of the equation of the circle we are aiming for implies that we want to square these equations, but first let’s expand the \((A+P)\) expressions.

Using the compound angle identities we have \[\begin{align*} \cos(A+P) &= \cos A \cos P - \sin A \sin P = - \sin A \sin P \\ \sin(A+P) &= \sin A \cos P + \cos A \sin P = \cos A \sin P \\ \cos(B+P-\pi) &= \cos B \cos (P-\pi) - \sin B \sin(P-\pi) = \sin B \sin P \end{align*}\]

We have used the facts that \(\cos P = \cos(P-\pi)=0\) and \(\sin(P-\pi) = -\sin P\). Notice also that \(\sin P\) is either \(1\) or \(-1\) depending on the value of \(n\).

Equation \(\eqref{eq:n_t+pi}\) then becomes \[\begin{align*} -x\sin A \sin P + y\cos A \sin P &= (n-1)\sin B \sin P \\ \implies\quad -x\sin A + y\cos A &= (n-1)\sin B \end{align*}\] Squaring this equation and equation \(\eqref{eq:n_t}\) then adding the results we get \[\begin{align*} x^2\sin^2 A + y^2\cos^2A-2xy\sin A\cos A &= (n-1)^2\sin^2 B \\ \text{and}\quad x^2\cos^2 A + y^2\sin^2A+2xy\sin A\cos A &= (n-1)^2\cos^2 B \\ \implies\quad x^2\left(\sin^2A+\cos^2A\right) + y^2\left(\cos^2A+\sin^2A\right) &= (n-1)^2\left(\sin^2B+\cos^2B\right) \\ \implies\quad x^2 + y^2 &= (n-1)^2 \end{align*}\]

as required.