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Given that $x=1+\sin(t)$ , $y=\sin(t) -\frac{1}{2} \cos(2t)$ show that $\frac{\text{d}^2y}{\text{d}x^2}=2$. I am having trouble proving this. Here is my working so far:

\begin{align}\frac{dx}{dt}&= cos(t)\\ \frac{dy}{dt}&= cos(t) + sin(2t)\end{align}

\begin{align}\frac{dy}{dx}&=\frac{\cos(t) + sin(2t)}{cos(t)}\\ \frac{d^2y}{dx^2}&=\frac{2cos(2t)cos(t) - sin(2t)sin(t)}{cos^2(t)}\frac{1}{cos(t)} \end{align}

I think its just a matter of simplifying my expression for $\frac{\text{d}^2y}{\text{d}x^2}$ using trigonometric identities but I can't see the right ones to use.

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@RGB: read the OP's work below; the edit should have been clear from that. – Ron Gordon Aug 5 '13 at 13:48
up vote 2 down vote accepted

$$\frac{dx}{dt}=\cos t, \frac{dy}{dt}=\cos t+\sin2t=\cos t(1+2\sin t) $$

Using Chain Rule (1,2),

$$\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=1+2\sin t (\text{ assuming }\cos t\ne0) $$

Again using Chain Rule,

$$\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dt}\left(\frac{dy}{dx}\right)/\frac{dx}{dt}=\frac{d(1+2\sin t)}{dt}/ \cos t=\frac{2\cos t}{\cos t}=2$$

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Thanks for this! I am pretty weak at spotting when certain trigonometric relationships are useful, I need more practice I guess! – ScottMillsFan Aug 5 '13 at 13:58
@ScottMillsFan, My pleasure. Trigonometric identities are indispensable as well as useful calculus. – lab bhattacharjee Aug 5 '13 at 13:59

Use $\cos{2 t} = 1 - 2 \sin^2{t}$ and $\sin{t}=x-1$. Then

$$y = (x-1)^2+x-1-\frac12 = x^2-x-\frac12$$

It should be clear what to do from here.

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