Q $13$(a) for the Pure Maths Paper is on parametric differentiation. I got $2\sqrt{3}\cos (t)$ for the top of the fraction, as $\sqrt{3}\sin 2(t)$ is the same as $\sqrt{3}(2\sin (t)\cos (t))$ and $\sin (t)$ is cancelled by $\sin(t)$ on the bottom.
However, for the denominator I got $2\sin (t)$, rather than $\sin (t)$. Am I missing something here, or have the exam board got it wrong?
I hope this is the question you are asking about!
The curve $C$ has parametric equations $x = 2 \cos t$, $y = \sqrt{3} \cos 2t$, $0 \leq t \leq \pi$. Find an expression for $\dfrac{dy}{dx}$ in terms of $t$.
The solution to the question is as follows:-
$$\dfrac{dy}{dx} = \dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}$$ $$\therefore \dfrac{dy}{dx}=\dfrac{\frac{d}{dt} \left( \sqrt{3} \cos 2t \right)}{\frac{d}{dt} \left( 2 \cos t \right)}$$ $$\therefore \dfrac{dy}{dx}= \dfrac{- 2 \sqrt{3} \sin 2t}{- 2 \sin t}$$ $$\therefore \dfrac{dy}{dx}= \dfrac{4 \sqrt{3} \sin t \cos t}{2 \sin t}$$ $$\therefore \dfrac{dy}{dx} = 2 \sqrt{3} \cos t$$