Show that $$\frac{\cos 7A + \cos A}{\sin 7A - \sin A} = \cot 3A.$$ Given that $\cos 2\theta = \frac{1}{9}$ find $\cos \theta$ in the form $\pm \frac{\sqrt{a}}{b}$ where $a, b \in \mathbb{N}$. - Leaving Cert Mathematics - Question 3 - 2016

To prove the equation, we can use the sum-to-product formula for cosines and sines. First, apply the sum-to-product formula:

$\cos x + \cos y = 2 \cos \left( \frac{x+y}{2} \right) \cos \left( \frac{x-y}{2} \right)$

and

$\sin x - \sin y = 2 \sin \left( \frac{x+y}{2} \right) \cos \left( \frac{x-y}{2} \right)$

in our case:

• Set $x = 7A$ and $y = A$, so we have:

$\cos 7A + \cos A = 2 \cos \left( \frac{8A}{2} \right) \cos \left( \frac{6A}{2} \right) = 2 \cos 4A \cos 3A$

and

$\sin 7A - \sin A = 2 \sin \left( \frac{8A}{2} \right) \cos \left( \frac{6A}{2} \right) = 2 \sin 4A \cos 3A$

Substituting these into the original equation:

$\frac{2 \cos 4A \cos 3A}{2 \sin 4A \cos 3A} = \frac{\cos 4A}{\sin 4A} = \cot 4A.$

Now, we can apply the identity $\cot x = \frac{\cos x}{\sin x}$ to rewrite:

$\cot 4A \cdot \frac{1}{\cot 3A} = \cot 3A$

Which establishes the equation.