Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
@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,

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

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.