Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Find all the points, if any, such that the curve $y=\sin(x-\sin x)$ has horizontal tangents at the $x$- axis.

My Solution

$$\frac{\mathrm{d} y}{\mathrm{d} x}\sin(x-\sin x)=(\cos (x-\sin x))(1-\cos x)$$

Let $\frac{\mathrm{d} y}{\mathrm{d} x}=0$ to find all the horizontal points.

We have, $(\cos (x-\sin x))(1-\cos x)=0$ which implies $(\cos (x-\sin x))=0$ or $(1-\cos x)=0$

Solving $(1-\cos x)=0$ first, we have $x= 2n\pi$ for all integers $n$.

Solving $(\cos (x-\sin x))=0$, I'm plain stuck. Any help thanks!

share|improve this question

1 Answer 1

up vote 1 down vote accepted

The question asks for the $x$-axis to be a (horizontal) tangent line of our curve.

So in addition to having $\frac{dy}{dx}=0$, we need to have $y=\sin(x-\sin x)=0$. We cannot have simultaneously the cosine and sine of $x-\sin x$ equal to $0$. This is a familiar fact about sine and cosine. They are never simultaneously $0$. For a proof, note that $\cos^2 t+\sin^2 t=1$.

So solutions of $\cos(x-\sin x)=0$ are no good for us. Good thing, too, since as you observed, finding the solutions looks like a hard problem!

So we need $1-\cos x=0$, that is, $x=2n\pi$ for some integer $n$. And we still need $y=\sin(x-\sin x)=0$. It should now not be hard to reach the right conclusion.

share|improve this answer
    
Thanks! Didn't read the question carefully! –  Yellow Skies Oct 3 '12 at 2:18

Your Answer

 
discard

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.