Below is the question:
If $x=\sin t$ and $y=\cos 2t$, find $\frac{dy}{dx}$ in terms of $x$ and prove that $\frac{d^2y}{dx^2} + 4 =0$.
I found $\frac{dy}{dx}$ first (using the chain rule, since what they have given us is parametric equations) and got $$\frac{dy}{dx} = -\frac{2\sin 2t}{\cos t}$$ and then I derived that again to get the second derivative, but was not able to prove it.
For the second derivative I got $$\frac{d^2y}{dx^2} = -\frac{4cost}{\sin t} - \frac{4\cos 2t}{\cos t}.$$ After this I tried to bring the two together getting this: $$\frac{-4 \cos t^{2} - 4\sin t\ cos 2t}{\sin t\cos t}$$
But then did not know how to get it to $-4$ (considering $-4+4=0$).
Can someone help me on how to solve this? As well as point out what I have done wrong?
Thank you.
Hint: use the expansion of $\sin 2t = 2\sin t\cos t$
You have: $$\frac{dy}{dx} = -\frac{2\sin{2t}}{\cos t} = -\frac{4\sin t\cos t}{\cos t} = -4\sin t$$ $$\Rightarrow \frac{dy}{dx} = -4x$$
The rest should be straightforward.
Note: I think the reason you did not get to the same result is because you have not used the quotient rule for the second-order derivative.